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Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion

Very Short Answer Questions

Question 1.
Is it necessary that a mass should be present at the centre of mass of any system?
Answer:
No. Any mass need not be present at the centre of mass of a system.
Ex : a hollow sphere, centre of mass lies at its centre.

Question 2.
What is the difference in the positions of a girl carrying a bag in one of her hands and another girt carrying a bag in each of her two hands?
Answer:
When the girl carries a bag in one hand (left) her centre of mass shifts towards the other hands (right). In order to bring it in the middle, the girl has to lean towards her other side. When the girl carries a bag in each of her two hands (left and right), the centre of mass does not shift. The girl does not bend any side because of the same hags’ .w in her two hands.

Question 3.
Two rigid bodies have same moment of inertia about their axes of symmetry. Or the two, which body will have greater kinetic energy?
Answer:
E = \(\frac{1}{2}\) I ω² = \(\frac{1}{2} \frac{\mathrm{L}^2}{1}\) , E oc \(\frac{1}{l}\) (∵ L = constant)
The rigid body having less moment of inertia will have greater kinetic energy.

Question 4.
Why are spokes provided in a bicycle wheel ?
Answer:
By connecting to the rim of wheel to the axle through the spokes the mass of the wheel gets concentrated at its rim. This increases its moment of inertia. This ensures its uniform speed.

Question 5.
We cannot open or close the door by applying force at the hinges, why ?
Answer:
When the force is applied at the hinges, the line of action of the force passes through the axis of rotation i.e, r = 0, so we can not open or close the door by pushing or pulling it at the hinges.

Question 6.
Why do we prefer a spanner of longer arm as compared to the spanner of shorter arm ?
Answer:
The torque applied on the nut by the spanner is equal to the force multiplied by the perpendicular distance from the axis of rotaion.
A spanner with longer arm provides more torque compared to a spanner with shorter arm. Hence longer arm spanner is preferred.

Question 7.
By spinning eggs on a table top, how will you distinguish a hard boiled egg from a raw egg ? [Mar. 13]
Answer:
A raw egg has some fluid in it and a hard boiled egg is solid form inside. Both eggs are spinning on a table top, the fluid is thrown outwards. Therefore (I_r > I_b) That means M. I of raw egg is greater than boiled egg. As I × ω = constant; ∴ ω_r < ω_b. That means Angular Velocity of raw egg is less than angular velocity of boiled egg.

Question 8.
Why should a helicopter necessarily have two propellers ?
Answer:
If there were only one propeller in the hellicopter then, due tio conservation of angular momentum, the helicopter itself would have turned in the opposite direction. Hence, if should have two propellers.

Question 9.
It the polar ice caps of the earth were to melt, what would the effect of the length of the day be ?
Answer:
Earth rotates about its polar axis. When ice of polar caps of earth melts, mass concentrated near the axis or rotation spreads out. Therefore, moment of Inertia I increases.

As no external torgue acts, L = I × ω = \(I\left(\frac{2 \pi}{T}\right)\)= constant. With increase of I, T will increase i.e. length of the day will increase.

Question 10.
Why is it easier to balance a bicycle in motion?
Answer:
When a bicycle is in motion, it is easy to balance because the principle of conservation of angular momentum is involved.

Short Answer Questions

Question 1.
Distinguish between centre of mass and centre of gravity. [A.P. – Mar. ‘18, ‘16, ‘15, ‘14, ‘13; TS – Mar. ‘16, ‘15, ‘14, ‘13]
Answer:
Centre of mass

1. Point at which entire mass of the body is supposed to be concentrated, and the motion of the point represents motion of the body.
2. It refers mass of to body.
3. In a uniform gravitational field centre of mass and centre of gravity coincide
4. Centre of mass of the body is defined to describe the nature of motion of a body as a whole.

Centre of gravity

1. Fixed point through which the weight of the body act.
2. It refers to the weight acting on all particles of the body
3. In a non-uniform gravitational field, centre of gravity and centre of mass do not coincide.
4. Centre of gravity of body is defined to know the amount of stability of the body when supported.

Question 2.
Show that a system of particles moving under the influence of an external force, moves as if the force is applied at its centre of mass.
Answer:
Consider \(\overrightarrow{r_1}, \overrightarrow{r_2}, \overrightarrow{r_3} \ldots \ldots \overrightarrow{r_n}\) be the position vectors of masses m₁, m₂, m₃ …………. m_n of n particle system.
According to Defination of centre of mass.

Differentiating the above equation w.r.t. time, we obtain

Where F_ext represents the sum of all external forces acting on the particles of the system. This equation states that the centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point.

Question 3.
Explain about the centre of mass of earth-moon system and its rotation around the sun.
Answer:
In the solar system the planets have different velocities and have complex two dimensional motion. But the motion of the centre of mass of the planet is simple and translational. consider the earth and moon system. We consider that the earth is moving around the sun in an elliptical path. But actually the centre of mass of earth and moon moves in an elliptical path round the sun. But the motion of either earth or moon is complicated when considered separately, more over we say that moon goes round the earth.

But actually earth and moon are revolving round their centre of mass such that they are always on opposite sides of the centre of mass. Here the forces of attraction between earth and moon are internal forces.

Question 4.
Define vector product. Explain the properties of a vector product with two examples. [T.S. AP – Mar. 15]
Answer:
The cross product of two vectors is given by \(\vec{C}=\vec{A} \times \vec{B}\). The magnitude of the vector defined from cross product of two vectors is equal to product of magnitudes of the vectors and sine of angle between the vectors.

Direction of the vectors is given by right hand corkscrew rule and is perpendicular to the plane containing the vectors.
∴ \(|\vec{C}|\) = AB sin θ. and \(\vec{C}\) = AB sin θ \(\hat{n}\). Where, \(\hat{n}\) is the unit vector perpendicular to the plane containing the vectors \(\vec{A}\) and \(\vec{B}\)
Example: 1) Torque is cross product of position vector and Force. i.e., \(\overrightarrow{\mathrm{T}}=\vec{r} \times \vec{F}\)
2) Angular momentum is cross product of position vector and momentum
i.e., \(\vec{L}=\vec{r} \times \vec{p}\)

Properties:

1. Cross product does not obey commutative law. But its magnitude obeys commutative low.
   \(\vec{A} \times \vec{B} \neq \vec{B} \times \vec{A} \Rightarrow(\vec{A} \times \vec{B})=-(\vec{B} \times \vec{A}),|\vec{A} \times \vec{B}|=|\vec{B} \times \vec{A}|\)
2. It obeys distributive law \(\vec{A} \times(\vec{B} \times \vec{C})=\vec{A} \times \vec{B}+\vec{A} \times \vec{C}\)
3. The magnitude of cross product of two vectors which are parallel is zero.
   Since θ = 0; \(|\vec{A} \times \vec{B}|\) = AB sin 0° = 0
4. For perpendicular vectors, θ = 90°, \(|\vec{A} \times \vec{B}|\) = AB sin 90° \(|\hat{n}|\)= AB

Question 5.
Define angular velocity(u). Derive v = r ω. [T.S. Mar. 16]
Answer:
Angular velocity (ω):
The rate of change of angular displacement of a body is called angular velocity, i.e., ω = \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\)
Derivation of v = rω
consider a rigid body be moving with uniform speed (v)along the circumference of a circle of radius r. Let the body be displaced from A to B in a small interval of time At making an angle ∆θ at the cantre. Let the linear displacement be ∆x from A to B.
From the property of circle, length of arc = radius × angle
∆x = r ∆θ
This equation is divided by ∆t, and taking

Question 6.
Define angular acceleration and torque. Establish the relation between angular acceleration and torque. [T.S. Mar. 18]
Answer:
Angular acceleration :
The rate of change of angular velocity is called angular acceleration
i.e., α = \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\)
Torque : The rate of change of angular momentum is called torque or The moment of Force is called Torque.

Relation between angular acceleration and Torque: Consider a rigid body of mass ‘M’ rotating in a circular path of radius ‘R’ with angular velocity ‘ωω’ about fixed axis.

By definition, τ = \(\frac{\mathrm{dL}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{I} \omega)}{\mathrm{dt}}\)
Where I = MR² = moment of inertia of a body
τ = I \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\) [∵ I = constant]
But \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\) α
∴ τ = Iα

Question 7.
Write the equations of motion for a particle rotating about a fixed axis?
Answer:
Equations of motion for a particle rotating about a fixed axis:
1. ω_f = ω_f + αt [∵ like v = u + at]
2. θ = \(\left(\frac{\omega_f+\omega_t}{2}\right) t\) [∵ like v = \(\left(\frac{v_1+v_2}{2}\right) t\)]
3. θ = ω_it + \(\frac{1}{2}\) α t² [∵ like s = ut + \(\frac{1}{2}\) at²]
4. ω_f² – ω_i² = 2 α θ [∵ like v² – u² = 2as]
Where ω_i = Initial angular velocity
ω_f = Final angular velocity
α = Angular acceleration
θ = Angujar displacement
t = time

Question 8.
Derive expressions for the final velocity and total energy of a body rolling without slipping.
Answer:
Expression of velocity of a body Rolling down an inclined plane:
Consider a rigid body of mass M and radius R rolling down an inclined plane from a height h . Let v the linear speed acquired by the body when it reaches the bottom of the plane and k is its radius of gyration.

According to law of conservation of Energy, we have P.E of body on top of inclined plane = K.E of a body at the bottom of inclimed plane
When a body rolls down an incline of height h, we apply the principle of conservation of energy.
P.E at the top – K.E of translation + K.E of rotation

Expression of Total energy of a body Rolling down on an inclined plane:
Suppose a body (Sphere) is rolling on a surface. Its motion can be treated as a combination of the translation of the centre of mass and rotation about an axis passing through the centre of mass. The total kinetic energy E can written as
E = E_T + E_R
Where E_T = Translational kinetic energy
and E_R = Rotational kinetic energy
E = \(\frac{1}{2}\) Mv² + \(\frac{1}{2}\) Iω²
= \(\frac{1}{2}\) MR² ω² + \(\frac{1}{2}\) Mk²ω²
E = \(\frac{1}{2}\) M ω² (R² + K²)
where k is radius of gyration
E = \(\frac{1}{2}\) Mv² (1 + \(\frac{K^2}{R^2}\)) [∵ ω = \(\frac{V}{R}\)]

Long Answer Questions

Question 1.
a) State and prove parallel axis theorem.
Answer:
Statement: The moment of inertia of a rigid body about any axis is equals to the sum of moment of inertia about a parallel axis passing through centre of mass (lg) and product of mass of body and square of perpendicular distance between two parallel axis.
∴ I = I_G + mr²
Proof : Consider a rigid body of mass M. Let I and I_G are the moment of inertia of a body about parallel axes X and Y respectively.

Let ‘r’ be the perpendicular distance between two axes.
Moment of inertia about ‘X’ axes is
I = Σm(OP)² ………………. (1)
Moment of inertia about Y axes is
I_G = Σm(GP)² …………….. (2)
from ∆⁴ OQP, OP² = (OQ)² + (PQ)²
= (OG + GQ)² + (PQ)²
= OG² + (GQ)² + 2.OG.GQ + (PQ)²
= OG² + (GP)² + 2(OG) (GQ) (∴ (GP)² = (GQ)² + (PQ)²)
Multiplying both sides by Σm,
⇒ Σm(OP)² = Σm(OG)² + Σm(GP)² + 2Σm(OG)(GQ)
from (1) and (2), I = mr² + IG + 2Σm (OG) (GQ)
But sum of moments of particles about centre of mass is zero.
i.e., Σm (OG) (GQ) = 0
∴ I = I_G + mr²
Hence proves.

b) For a thin flat circular disk, the radius of gyration about a diameter as axis is k. If the disk is cut along a diameter AB as shown in to two equal pieces, then find the radius of gyration of each piece about AB.

Answer:
For thin circular disk, the radius of gyration about a diameter
AB is, K = \(\sqrt{\frac{I}{M}}\)
Where M = mass of disk
I = M.I of disk
The disk is cut into two halves about AB, when each,
Mass M’ = \(\frac{\mathrm{M}}{\mathrm{2}}\) and each M.I, I = \(\frac{1}{2}\)
Therefore Radius of gyration of each piece is
K’ = \(\sqrt{\frac{I^{\prime}}{M^{\prime}}}=\sqrt{\frac{\left(\frac{I}{2}\right)}{\left(\frac{M}{2}\right)}}=\sqrt{\frac{I}{M}}\) = k

Question 2.
a) State and prove perpendicular axes theorem.
Answer:
Statement: The moment of inertia of a plane lamina about an axes perpendicular to its plane is equal to sum of moment of inertia of lamina about the perpendicular axes in its plane intersecting each other at a point, where the perpendicular axes passes.
i.e., I_z = I_x + I_y
Proof’ Consider a particle of mass’m’ at p. Let it be at a distance Y from Z-axis. Here ‘X’ and Y axes are in plane Ramina and Z-axes perpendicular to plane lamina.

Now,
Moment of inertia about X-axix, I_x = Σmx²
Moment of inertia about Y-axis, I_y = Σmy²
Moment of inertia about Z-axis, I_z = Σmr²
From ∆⁴ OQP, r² = x² + y²
multiplying both side with Σm,
⇒ Σmr² = Σmx² + Σmy²
⇒ I_z = I_x + I_y
Hence proved.

b) If a thin circular ring and a thin flat circular disk of same mass have same moment of inertia about their respective diameters as axes. Then find the ratio of their radii.
Answer:
For a thin flat circular dist, M.I., I_r = \(\frac{\mathrm{MR}_{\mathrm{r}}^2}{2}\)
For a thin flat circular dist, M.I., I_d = \(\frac{\mathrm{MR}_{\mathrm{d}}^2}{2}\)
Given that I_r = I_d
\(\frac{M R_r^2}{2}=\frac{M R_d^2}{4} \Rightarrow \frac{R_r^2}{R_d^2}=\frac{2}{4}=\frac{1}{2}\)
∴ \(\frac{R_r}{R_d}=\frac{1}{\sqrt{2}} \text { or } R_r: R_d=1: \sqrt{2}\)

Question 3.
State and prove the principle of conservation of angular momentum. Explain the principle of conservation of angular momentum with examples. [A.P. Mar 16]
Answer:
Statement: Angular momentum of a body remains constant when the external torque is zero.
L = I ω = constant K.
or I₁ ω₁ = I₂ ω₂.
If the moment of Inertia of a body is lowered, the angular velocity of the body co increases.
Proof:
By derfination, the rate of change of angular momentum is called Torque.
i.e., τ = \(\frac{\mathrm{dL}}{\mathrm{dt}}\)
If τ = 0 ⇒ \(\frac{\mathrm{dL}}{\mathrm{dt}}\) = 0
or L = constant k
⇒ L₁ = L₂
∴ I₁ ω₁ = I₂ ω₂
Example 1): When a man with stretched out arms stands on a turn table which is revolving then his moment of inertia is high. If he folded his hands, the moment of inertia decreases and hence the angular velocity, linear velocity increase, but the period decreases. In both cases angular momentum remains constant.

Example 2): An acrobat from a swing in a circus, leaves the swing with certain angular momentum, with his arms and legs stretched. As soon as he leaves the swing he pulls his hands and legs together thus lowering his M.I. and increasing his angular velocity. He then quickly makes somersaults in air and finally lands on a net or ground.

Problems

Question 1.
Show that a(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.
Solution:
Let a parallele piped be formed on three vectors.

Which is equal in magnitude to the volume of the parallel piped.

Question 2.
A rope of negligible mass is wound 4. round a hollow cylinder of mass 3kg and redius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N ? What is the linear acceleration of the rope ? Assume that there is no slipping.
Solution:
Here M = 3kg, R = 40 cm = 0.4 m
Moment of intertia of the hollow cylinder about its axis
I = MR² = 3(0.4)² = 0.48 kgm²
Force applied F = 30 N
∴ Torque, τ = F × R= 30 × 0.4 = 12 N – m.
If α is the angular acceleration, produced, then from τ = Iα
α = \(\frac{\tau}{\mathrm{l}}=\frac{12}{0.48}\) = 25 rads^-2
Linear acceleration, a = Rα = 0.4 × 25 = 10 m/s²

Question 3.
A coin is kept a distance of 10 cm from the centre of a circular turntable. If the coefficient of static friction between the table and the coin is 0.8 find the frequency of rotation of the disc at which the coin will just begin to slip.
Solution:
Here, radius, r = 10cm = 0.1 m; μ_s = 0.8
F = μmg
mrω² = μmg
rω² = μg
ω = \(\frac{\mu \mathrm{g}}{\mathrm{r}}=\sqrt{\frac{0.8 \times 9.8}{0.1}}=\sqrt{8 \times 9.8}\)
= 8.854 rad/s
w = 2πn
∴ frequency n = \(\frac{\omega}{2 \pi}=\frac{8.854}{2 \times 3.14}\) = 1 .409rps
n = 1.409 × 60 = 84.54 rpm

Question 4.
Particles of masses 1g, 2g, 3g…. 100g are kept at the marks 1 cm. 2cm, 3cm ….100cm respectively on 2 meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale. –
Solution:
By adding all the masses, we get M = 5050g
= 5.050 kg
= 5.1 kg
and L = 1 m
M.I. of the meter scale = \(\frac{\mathrm{mL}^2}{12}=\frac{5.1 \times 1^2}{12}\)
= 0.425kg m²
= 0.43kg -m²

Question 5.
Three particles each of mass 100g are placed at the vertices of an equilateral triangle of side length 10 cm. Find the moment of inertia of the system about an axis passing through the centroid Of the triangle and perpendicular to its plane.
Solution:
m = 100 g = 100 × 10^-3kg
Side a = 10 cm

Question 6.
Four particles each of mass 100g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system.
Solution:
mass m = 100g
= 100 × 10^-3kg
M = 4m = 400 × 10^-3kg
radius r = 10cm
= 10 × 10^-2m
Moment of Inertia I = Mr²

Question 7.
Two uniform circular discs, each of mass 1 kg and radius 20 cm, are kept in contact about the tangent passing through the point of contact. Find the moment of inertia of the system about the tangent passing through the point of contact.
Solution:
Mass m = 1kg; r = 20cm = 20 × 10^-2m
I = I₁ + I<sub2

I₁ = \(\frac{\mathrm{MR}^2}{4}\) + MR²
I₁ = \(\frac{5\mathrm{MR}^2}{4}\)
Similarly I₂ = \(\frac{5\mathrm{MR}^2}{4}\)
∴ I = \(\frac{10 \mathrm{MR}^2}{4}=\frac{10 \times 1 \times\left(20 \times 10^{-2}\right)^2}{4}\) = 0.1 kg-m²

Question 8.
Four spheres, each diameter 2a and mass ‘m’ are placed with their centres on the four corners of a spuare of the side b. Calculate the moment of inertia of the system about any side of the square.
Solution:
I₁ = mb²
I₂ = \(\frac{2}{5}\) ma²
I₃ = \(\frac{2}{5}\) ma²
I₄ = mb²
∴ Moment of inertia of the system
I = I₁ + I₂ + I₃ + I₄
= mb² + \(\frac{2}{5}\) ma² + \(\frac{2}{5}\) ma² + mb²
I = \(\frac{4}{5}\) ma² + 2mb²

Question 9.
To maintain a rotor at a uniform angular speed or 200 rads^-1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine ? (Note : uniform angular velocity in the absence of friction implies zero torque. In practice applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Solution:
Here, ω = 200 rad/s, torque τ = 180 N – m, power p = ?
As p = τ ω
∴ p = 180 × 200 = 3600.watt = 36 kw.

Question 10.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick ?
Solution:
Let M be the mass of stick concentrated at L, the 50 cm, mark.
For equilibrium about G’ the 45 cm mark,
10g (45 – 12) = mg(50 – 45)
10 g × 33 = mg × 5
M = \(\frac{10 \times 33}{5}\) = 66 gram.

Question 11.
Determine the kinetic energy of a circular disc rotating with a speed of 60 rpm about an axis passing through a point on its circumference and perpendicular to its plane. The circular disc has a mass of 5kg and radius 1m.
Solution:
Here M = 5kg; R = 1 m;
ω = 2π × \(\frac{\mathrm{N}}{\mathrm{t}}\) = 2n × \(\frac{60}{60}\) rad /s = 2π rad/s
The M.I of disc about parallel axis passing through a point on its circumferance
I = \(\frac{\mathrm{MR}^2}{2}\) + MR² = \(\frac{3}{2}\) MR²
∴ Kinetic energy = \(\frac{1}{2}\) I ω²
= \(\frac{1}{2}\) × \(\frac{3}{2}\) MR² ω² = \(\frac{3}{4}\) × 5 × (1)² × (2π)²
= \(\frac{3}{4}\) × 5 × 4π² = 15 × (\(\frac{22}{7}\)²
∴ K.E = 148.16 J.

Question 12.
Two particles each of mass m and speed v Travel in opposite directions along parallel lines separated by a distance. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.
Solution:
Given that m₁ = m₂ = m;
\(\vec{V}_1=\vec{V}_2=\vec{V}\)
momentum of 1^st particle p₁ = \(\mathrm{m} \vec{\mathrm{V}}\)
∴ p₁ = \(\mathrm{m} \vec{\mathrm{V}}\)
momentum of 2^nd particle \(\vec{\mathrm{P}}_2=\mathrm{mV}\)

If two particles moves oppositely on
Circumference of circle, distance d = 2r
Angular momentum of 1^st particle w.r.t
Centre ‘O’ is \(\vec{L_1}=\vec{r} \times \vec{P}_1=\vec{r} \times m \vec{V}\)
Angular momentum of 2^nd particle w.r.t
centre ‘O’ is \(\vec{\mathrm{L}_2}=\vec{\mathrm{r}} \times \vec{\mathrm{P}_2}=\vec{\mathrm{r}} \times \mathrm{m} \vec{\mathrm{V}}\)
∴ \(\vec{L_1}=\vec{L_2}\)

Question 13.
The moment of inertia of a fly wheel making 300 revolutions per minute is 0.3 kgm ² Find the torque required to bring it to rest in 20s.
Solution:

Question 14.
When 100 J of work is done on a fly wheel, its angular velocity is increased from 60 rpm to 180 rpm. What is the moment of inertia of the wheel ?
Solution:
Here, Initial frequency
n₁ = \(\frac{60}{60}\) = 1Hz
Initial angular velocity
ω₁ = 2π n₁ = 2π rad /sec
Final frequency n₂ = \(\frac{180}{60}\) = 3Hz
Final angular velocity
ω₂ = 2π n₂ = 2π × 3 = 6π rad/sec
Work done = 100 J from work – energy therorem,
Workdone = change in K.E.
W = \(\frac{1}{2}\) I ω₂² – \(\frac{1}{2}\) I ω₁²
100 = \(\frac{1}{2}\) I [(6π)² – (2π)²]
100 = \(\frac{1}{2}\) I (32π)²
I = \(\frac{200}{32 \pi^2}\) = 0.634 kg – m²
∴ I = 0.634 kg- m²

Additional Problems

Question 1.
Give the location of the centre of mass of a (i) sphere (ii) cylinder (iii) ring and (iv) cube. each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
Solution:
In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres.
No, it is not necessary that the center of mass of a body should lie on the body for example in caSe of a circular ring. Center of mass is at the centre of the ring, where there is no mass.

Question 2.
In the HCl molecule the separation between the nuclei of the two atoms Is about 1.27 Å(1 Å = 10^-10m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom Is concentrated in its nucleus.
Solution:
Let the mass of the H atom = m unit, mass of the Cl atom = 35.5 m units
Let cm be at the distance xÅ from Htom
∴Distance of cm from CL atom = (1.27 – x) Å
It cm is taken at the origin,
then mx + (1.27 – x) 55.5m = 0
mx = (1.27 – x) 35.5 m.
Negative sign indicates that if chlorine atom is on the right side of cm (+), the hydrogen atom is on the left side of cm, so leavning negative sign, we get
x + 35.5 x = 1.27 × 35.5
36.5 x = 45.085
x = \(\frac{45.085}{36.5}\)
= 1.235
x = 1.235 Å
Hence cm is located on the line joining centres of H and d atoms at a distance 1.235 Å from H.

Question 3.
A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?
Solution:
The speed of the centre of mass of the system , (trolley + child) shall remain unchanged, when the child gets up and runs about on the trolley in any manner. This is because forces involved in the exercise are purely internal i.e., from within the system. No external force acts on the system and hence there is no change in velocity of the system.

Question 4.
Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b
Solution:
Let \(\vec{\mathrm{a}}\) represented by \(\vec{\mathrm{OP}}\) and \(\vec{\mathrm{b}}\) be represented by \(\vec{\mathrm{OQ}}\).
Let ∠POQ = θ

∴ area of ∆ OPQ = \(\frac{1}{2}|\vec{a} \times \vec{b}|\), which was to be proved.

Question 5.
Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.
Solution:
Let a parallel piped be formed on three vectors.

Where \(\hat{n}\) is unit vector along \(\overrightarrow{\mathrm{OA}}\) perpendicular to the plane containing \(\) and \(\vec{\mathrm{b}}\). Now \(\vec{\mathrm{b}}\) = (a) (be) cos 0°
= abc
Which is equal in magnitude to the volume of the parallelepiped.

Question 6.
Find the components along the x. y. z axes of the angular momentum 1 of a particle whose position vector is r with components x, y, z and momentum is p with components p_x, p_y, and p_z,. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Solution:
For motion in 3D, the position vector \(\vec{\mathrm{r}}\) and linear momentum vector \(\vec{\mathrm{p}}\) can be written in terms of their rectangular components as follows.

Comparing the coefficient of k on both sides, we have L_z = xp_y – yp_x.
Therefore the particle moves only in x-y plane. The angular momentum has only z component.

Question 7.
Two particles, each of mass m and speed v. travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same what ever be the point about which the angular momentum is taken.
Solution:
From fig, vector angular momentum of the two particle system any point A on x₁ y₁ is.

Similary, vector angular momentum of the tw.o particle system, a bout any pt. B on x₂y₂ is
\(\vec{L_B}=m \vec{v} \times d+m \vec{v} \times 0=m \vec{v} d\)
Let us consider any other point (on AB, where AC = x)
∴ Vector angular momentum of the two particle system about c is
\(\vec{L_c}=m \vec{v}(x)+m \vec{v}(d-x)=m \vec{v} d\)
Clearly, \(\vec{\mathrm{L}_{\mathrm{A}}}=\vec{\mathrm{L}_{\mathrm{B}}}=\vec{\mathrm{L}_C}\)
Which has to be proved.

Question 8.
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in figure. The angles made by the strings with the vertical are 36.9 and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
Solution:

From fig (b)
θ₁ = 36.9°, θ₂ = 53.1°
If T₁, T₂ are the tensions in two strings, then for equilibrium along the horizontal,
T₁ sin θ₁ = T₂ sin θ₂
(or) \(\frac{T_1}{T_2}=\frac{\sin \theta_2}{\sin \theta_1}=\frac{\sin 53.1^{\circ}}{\sin 36.9^{\circ}}\)
= \(\frac{0.7407}{0.5477}\) = 1.3523
Let d be the distance of center of gravity c of the bar from the left end.
For rotational equilibrium about c,
T₁ cos θ₁ Xd = T₂ cos θ₂ (2 – d)
T₁ cos 36.9° × d = T₂ cos 53.1° (2 – d)
T₁ × 0.8366 d = T₂ × 0.6718 (2 – d)
Put T₁ = 1.3523 T₂ and solve to get
d = 0.745 m.

Question 9.
A car weight 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Solution:
Here, m = 1800 kg
Distance between front and .back axles = 1.8 m
Distance of center of gravity (c) behind the front axle = 1.05 m
Let R₁ and R₂ be the forces exerted by the level ground on each front wheel and each back wheel. As it is dear from fig.
R₁ + R₂ = mg = 1800 × 9.8

For rotational equilibrium about C,
R₁ × 1.05 = R₂ (1.8 – 1.05) = R₂ × 0.75 ………….. (i)
\(\frac{R_1}{R_2}=\frac{0.75}{1.05}=\frac{5}{7}\)
Putting in (i)
\(\frac{5}{7}\) R₂ + R₂ = 1800 × 9.8
R₂ = \(\frac{7 \times 1800 \times 9.8}{12}\)
= 10290 N
R₁ = \(\frac{5}{7}\) R₂
= \(\frac{5}{7}\) × 10290
= 7350 N.

Question 10.
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 MR²/5. where M is the mass of the sphere and R is the radius of the sphere.
Solution:
a) Moment of inertia of sphere about any diameter = \(\frac{2}{5}\)MR²
Applying theorem of parallel axes,
Moment of inertia of sphere about a tangent to the sphere = \(\frac{2}{5}\)MR² + M(R)²
= \(\frac{7}{5}\)MR².

(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR²/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Solution:
We are given, moment of inertia of the disc about any of its diameters = \(\frac{1}{4}\)MR².
i) Using theorem of perpindicular axes, moment of inertia of the disc about an axis passing through its center and normal to the disc = 2 × \(\frac{1}{4}\)MR² = \(\frac{1}{2}\)MR²
ii) Using theorem of parallel axes, moment of inertia of the disc passing through a point on its edge and normal to the disc
= \(\frac{1}{2}\)R² + MR² = \(\frac{3}{2}\) MR².

Question 11.
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.
Solution:
If M is mass and R is radius of the hollow cylinder and the solid sphere, then
M.I of hollow cylinder about its axis of symmetry I₁ = MR² and
M.I of solid sphere about an axis through its 2 , centre, I₁ = \(\frac{2}{5}\) MR²
Torque applied, I = I₁ α₁ = I₂ α₂
\(\frac{\alpha_2}{\alpha_1}=\frac{l_1}{l_2}=\frac{M^2}{\frac{2}{5} M R^2}=\frac{5}{2}\)
α₂ > α₁
From ω = ω₀ + αt, we find that for given ω₀ and t, ω₂ > ω₁ to, i.e. angular speed of solid sphere will be greater than angular speed of hallow sphere.

Question 12.
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s^-1. The radius of the cylinder is 0.25m. What is the kinetic energy associated with the rotation of the cylinder ? What is the magnitude of angular momentum of the cylinder about its axis ?
Solution:
Here, M = 20 kg, R = 0.25 m, w – 100 g^-1
Moment of inertia of solid cylinder
= \(\frac{\mathrm{MR}^2}{2}=\frac{20 \times(0.25)^2}{2}\) = 0.625 kg/m²
K.E of rotation = \(\frac{1}{2}\) Iω² = \(\frac{1}{2}\) × 0.625 × (100)² = 3125 J
Angular momentum, L = Iω
= 0.625 × 100
= 62.5 Js.

Question 13.
a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/miri. How much is the angular spped of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction.
Solution:
Here, intial angular speed ω₁ = 40 rev/min, ω₂ = ?
Final moment of inertia, I₂ = \(\frac{2}{5}\)I, Intial moment of Inertia
As no external torque acts in the process, therefore
Iω = constant
i.e. I₂ω₂ = I₁ω₁
ω₂ = \(\frac{I_1}{I_2}\) ω₁ = \(\frac{5}{2}\) × 40
= 100 rpm

b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy ?
Solution:
Final K.E of rotation, E₂ = \(\frac{1}{2}\)I₂ω₂²
Intial K.E of rotation, E₁ = \(\frac{1}{2}\)I₂ω₁²
\(\frac{E_2}{E_1}=\frac{\frac{1}{2} l_2 \omega_2^2}{\frac{1}{2} l_1 \omega_1^2}=\left(\frac{l_2}{l_1}\right)\left(\frac{\omega_2}{\omega_1}\right)^2\)
= \(\frac{2}{5} \times\left(\frac{100}{40}\right)^2=\frac{5}{2}\) = 2.5
∴ K.E of rotation increase. This is because child spends internal energy folding back his hands.

Question 14.
A rope of negligible mass is wound roung a hollow cylinder of mass 3kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? what is the linear acceleration of the rope ? Assume that there is no slipping.
Solution:
Here M = 3kg, R = 40 cm = 0.4 m
Moment of Inertia of the hollow cylinder about its axis
I = MR² = 3(0.4)² = 0.48 kgm²
Force applied F = 30 N
∴ Torque, τ = F × R = 30 × 0.4 = 12 N – m.
If α is the angular acceleration, produced, then from τ = Iα
α = \(\frac{\tau}{\mathrm{I}}=\frac{12}{0.48}\) = 25 rads^-2
Linear acceleration, a = Rα = 0.4 × 25 = 10 m/s².

Question 15.
To maintain a rotor at a uniform angular speed of 200 rad s^-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note : uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Solution:
Here, ω = 200 rad/s, torque τ = 180 N – m, power p = ?
As p = τω
∴ p = 180 × 200 = 3600 watt = 36 kw.

Question 16.
From a uniform disc of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is a R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Solution:
Suppose mass per unit area of the disc = M
∴ Mass of original disc

M = πR² × m
Mass of portion removed from the disc
M’ = π\(\left(\frac{R}{2}\right)^2 \times m=\frac{\pi R^2}{4} m=\frac{M}{4}\)
In fig, mass M is connected at O and mass
M’ is concentrated at O’, where OO’ = \(\frac{\mathrm{R}}{\mathrm{2}}\).
After the circular disc of mass M’ is removed, the remaining portion can be considered as a system of two masses M at 0 and – M’ = \(\frac{\mathrm{-M}}{\mathrm{4}}\) at O’. If x is the distance of centre of 4 mass (p) of the remaining part, then
x = \(\frac{M \times O-M^{\prime} \times \frac{R}{2}}{M-M^{\prime}}\)
= \(\frac{\frac{-M}{4} \times \frac{R}{2}}{M-\frac{M}{4}}=\frac{-M R}{8} \times \frac{4}{3 M}=\frac{-R}{6}\)
Negative sign shows that p is the left O.

Question 17.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick ?

Solution:
Let m be the mass of stick concentrated at c, the 50 cm, mark,
For equilibrium about c’ the 45 cm mark,
10g (45 – 12) = mg(50 – 45)
10 g × 33 = mg × 5
m = \(\frac{10 \times 33}{5}\)
= 66 gram.

Question 18.
A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination, (a) Will it reach the bottom with the same speed in each case ? (b) Will jt take longer to roll down one plane than the other ? (c) If so, which one and why ?
Solution:
Let v be the speed of the solid sphere at the bottom of the incine. Applying principle of conservation of energy, we get
\(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω² = mgh
As I = \(\frac{2}{5}\) mr²; \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (\(\frac{2}{5}\) mr²)ω² = mgh
as rω = v, \(\frac{1}{2}\) mv² + \(\frac{1}{5}\) mv² = mgh
v = \(\sqrt{\frac{10}{7} \mathrm{gh}} .\)
As h is same in two cases, v must be same i.e., It will reach the bottom with the same speed. Time taken to roll down the two planes will also be the same, as their height is the same.

Question 19.
A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it ?
Solution:
Here, R = 2m, M = 100 kg v = 20 cm/s = 0.2 m/s.
Total energy of the hoop = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω²
= \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (MR)²ω² = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) mv²
= mv².
Work required to stop the hoop = total energy of the hoop
w = mv² = 100(0.2)² = 4 joule.

Question 20.
The oxygen molecule has a mass of 5.30 × 10^-26 kg and a moment of inertia of 1.94 × 10^-46 kg m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of trans-lation. Find the average angular velocity of the molecule.
Solution:
Here, m = 5.30 × 10^-26 kg
I = 1.94 × 10-^-46 . kgm²
v = 500 m/s
If \(\frac{\mathrm{m}}{\mathrm{2}}\) is mass of each atom of oxygen and 2r is the distance between the two atoms as shown in fig. then

= 6.7 × 10-¹² rod/s.

Question 21.
A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
a) How far will the cylinder go up the plane ?
b) How long will is take to return to the bottom ?
Solution:
Here θ = 30°, v = θm/s
Let the cylinder go up the plane upto a height h
From \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω² = mgh
\(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (\(\frac{1}{2}\) mr²) ω² = mgh
\(\frac{3}{4}\) mv² = mgh
h = \(\frac{3 v^2}{4 g}=\frac{3 \times 5^2}{4 \times 9.8}\) = 1.913 m
If s is the distance up the inclined plane, then as sin θ = \(\frac{\mathrm{h}}{\mathrm{s}^2}\)
s = \(\frac{\mathrm{h}}{\sin \theta}=\frac{1.913}{\sin 30^{\circ}}\) = 3.826 m
Time taken to return to the bottom
t = \(\sqrt{\frac{2 s\left(1+\frac{k^2}{r^2}\right)}{g \sin \theta}}=t \sqrt{\frac{2 \times 3.826\left(1+\frac{1}{2}\right)}{9.8 \sin 30^{\circ}}}\) = 1.53 s.

Question 22.
As shown in Fig. the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F.1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder.
(Take g = 9.8 m/s²)

(Hint: Consider the equilibrium of each side of the ladder separately.)
Solution:
Data seems to be insufficient

Question 23.
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand . The angular speed of the platform is 30 revolutions per minute. The man then brings his arms dose to his body with the distance of each weight from the axis changing from 90cm to 20 cm. Thd moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m²
(a) What is his new angular speed ? (Neglect friction.)
b) Is kinetic energy conserved in the process ? If not, form where does the change come about ?
Solution:
Here l₁ = 7.6 × 2 × 5(0.9)² = 15.7 kgm²
ω₁ = 30 rpm
l₂ = 7.6 + 2 × 5(0.2)² = 8.0 kgm²
ω₂ = ?
According to the principle of conservation of angular momentum
l₂ω₂ = l₁ω₁
ω₂ = \(\frac{l_1}{l_2} \omega_1=\frac{15.7 \times 30}{8.0}\) = 58.88 rpm

Question 24.
A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.
(Hint : The moment of inertia of the door about the vertical axis at one end is ML²/3.)
Solution:
Angular momentum imparted by the bullet
I_ = mv × r = (10 × 10^-3) × 500 × \(\frac{1}{2}\) = 2.5
Also, L = \(\frac{M L^2}{3}=\frac{12 \times 1.0^2}{3}\) = 4kgm
As L = Lω
∴ ω = \(\frac{\mathrm{L}}{1}=\frac{2.5}{4}\) = 0.625 rad/sec.

Question 25.
Two discs of moments of inertia I, and I2 about their respective axes (normal to the disc and passing through the centre), and rotawing with angular speeds ω₁, and ω₂ are brought into contact face to face with their axes of rotation coincident,
(a) .What is the angular speed of the two-disc system ?
(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy ? Take ω₁ ≠ ω₁.
Solution:
Here, total intial angular momentum of the two discs L₁ = I₁ω₁ + I₂ω₂
Under, the given conditions, moment of intertia of the two disc system = (I₁ + I₂)
If ω is angular speed of the combined system, the final angular momentum of the system
L₂ = (I₁ + I₂)ω
As no external torque is involved in this excercise, therefore, L₂ = L₁
(I₁ + I₂)ω = I₁ω₁ + I₂ω₂
ω = \(\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}\)

b) Initial K.E of two disc E₁ = \(\frac{1}{2}\) I₁ω₁² + \(\frac{1}{2}\) I₂ω₂²
Final K.E of the system E₂ = \(\frac{1}{2}\) (I₁ + I₂)ω²

∴ E₁ – E₂ > 0 or E₁ > E₂ or E₂ < E₁
Hence there occurs a loss of K.E in the process. Loss of energy = E₁ – E₂. This loss must be due to friction in the contact of the two discs.

Question 26.
a) Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point (x, y) in the x – y plane from an axis through the origin and perpendicular to the plane is x² + y²).
Answer:
Statement: The sum of moments of inertia of a plane lamina about any two perpendicular axes in its plane is equal to its moment of inertia about an axis perpendicular to the plane and passing through the point of intersection of the first two axes.

Proof : Consider a plane lamina revolving about the Z axis. Let ‘O’ be the origin of the axis. Imagine a particle of mass ‘m’ lying at a distance ‘r’ from point ‘o’ on the plane. Let x, y be the coordinates of the point P.
Thus r² = x² + y²
Then the moment of the body about x-aixs
I_x = Σm y²
The moment of inertia of the body about y-axis.
I_y = Σm x²
Then the moment of inertia of the body about Z-axis

I_z = Σm r²
I_z = Σm(x² + y²)
I_z = Σm x² + Σm y² = I_y + I_x
∴ I_z = I_x + I_y
Hence perpendicular axes theorem is proved.

b) Prove the theorem of parallel axes.
(Hint: If the centre of mass is chosen to be the origin Σm_ir_i = 0).
Answer:
Statement: The moment of inertia of a plane lamina about an-axis is equal to the sum of the moment of inertia about a parallel axis passing through the centre of mass and the product of its mass and square of the distance between the two axes i.e., I₀ = I_G + Mr²
Let I_G is the moment of inertia of the plane lamina about the axis Z₂ passing through the centre of mass.
I₀ is the moment of inertia of the plane lamina about an axis Z₁
Let M be the mass of the lamina and r be the distance between the two axes . Then
I₀ = I_G + mr².
Proof : Let a particle of mass m is situated at P. Moment of inertia about the axis passing through 0 is
dl = m op² or I = Σm op².
Join the lines PO and PG and draw the line PQ and Join with the line extending from OG.
From the trainagle POQ, OP² = OQ² + PQ²
OP² – (OG + GQ)² + PQ² + OQ = OG + GQ
OP² = OG² + 2OG . GQ + (GQ² + PQ²)
OP² – OG² + 2OG. GQ + GP²
OP² = OG² + GP² + 2OG.GQ
[∵ From the ∆le PGQ, GP² = PQ² + GQ²]

Multiplying with Σm on both side
Σm OP² = Σm OG² + Σm GP² + Σm OG. GQ
But Σm OG² = Mr²
(∵ OG is constant and Σm = M, total mass of the body)
Σm GP² = I_G
Σm OP² = I₀
∴ I₀ = Mr² – I_G + 2r ΣmGQ
Σm.GQ = 0
[∵ The moment of all the particles about the centre of mass is always zero]
I₀ = I_G + Mr²
Thus the theorem is proved.

Question 27.
Prove the result that the velocity u of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by υ² = \(\frac{2 g h}{\left(1+k^2 / R^2\right.}\) using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
Solution:
When a body rolls down an incline of height h, we apply the principle of conservation of energy.

K.E of translation + K.E of rotation = P.E at the top
i.e . \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) Iω² = mgh
\(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (mk²) ω² = mgh
As w = \(\frac{\mathrm{V}}{\mathrm{R}}\)
∴ \(\frac{1}{2}\)mv² + \(\frac{1}{2}\)m \(\frac{k^2}{R^2}\)v² = mgh
or mv² (1 + \(\frac{k^2}{R^2}\)) = mgh
v² = \(\frac{2 g h}{\left(1+\frac{K^2}{R^2}\right)}\)

Question 28.
A disc rotating about its axis with angular speed ω₀ is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. ? Will the disc roll in the direction indicated ?

Solution:
Using the reaction v = rω,
we get
for point A, V_A = Rω₀, along AX
for point B, V_B = Rω₀, along BX
for point C, V_c = (\(\frac{\mathrm{R}}{\mathrm{2}}\) ω₀ parallel to AX,
The disc will not rotate, because it is placed on a perfectly frictionless table, without, rolling is not possible.

Question 29.
Explain why friction is necessary to make the disc in Figure roll in the direction indicated.

(a) Give the direction of frictional force at B and the sense of frictional torque before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins ?
Solution:
To roll a disc, we require a torque, which can be proved only by a tangential force. As force of friction is the only tangential force in this case, it is necessary.
(a) As frictional force at B opposes the velocity of point B, which is to the left, the frictional force must be to the right. The sense of frictional torque will be perpendicular to the plane of the disc and outwards.

(b) As frictional force at B decreases the velocity of the point of contact B with the surface, the perfect rolling begins only when velocity of point B becomes zero. Also, force of friction would become zero at this stage.

Question 30.
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s^-1. Which of the two will start to roll earlier ? The co-efficient of kinetic friction is μ_k = 0.2.
Solution:
Here, intial velocity of centre of mass is zero i.e, u = 0.
Frictional force causes the GM to accelerate
μ_k mg = ma ∴ a = μ_k g
As v = u + at ∴ v = 0 + μ_k gt
Torque due to friction causes retardation in the intial angular speed ω₀.
i.e. μ_k mg × R = – Iα
α = \(\frac{\mu_k \mathrm{mgR}}{\mathrm{I}}\)
ω = ω₀ + αt
∴ ω = ω₀ – \(\frac{\mu_k \mathrm{mgRt}}{\mathrm{I}}\)
Rolling begins, when v = Rω from (ii) and (iv)

Comparing (vi) and (vii) we find that the disc would begin to roll earlier than the ring We can calculate the values of t from (vi) and (vii) using known values of μ_k, g, R and ω₀.

Question 31.
A cylinder of mass 10kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction μ_s = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?
Solution:
Here, m = 10 kg, r = 15 cm = 0.15 m
θ = 30°, μ_s = 0.25
Accelaration of the cylinder down the incline,
α = \(\frac{2}{3}\)g sin θ = \(\frac{2}{3}\) × 9.8 sin 30° = \(\frac{9.8}{3}\) m/s²
a) Force of friction, f = mg sin θ – ma = m(g sin θ – α) = 10(9.8 sin 30° – \(\frac{9.8}{3}\)) = 16.4 N
b) During rolling the point of contact is at rest. Therefore work done against friction is zero
c) For rolling without slipping/skidding μ = \(\frac{1}{3}\) tan θ
tan θ = 3μ
= 3 × 0.25 = 0.75
θ = 37°.

Question 32.
Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly italic inclined plane will undergo slipping (not roiling) motion.
Solution:
a) The statement is false.

b) True. This is because rolling body can be imagined to be rotating about an axis passing through the point of contact of the body with the ground.
Hence its instantaneous speed is zero.

c) This is not true. This is because when the body is rotating its instantaneous acc is not zero.

d) It is true. This is because once the perfect rolling begins, force of friction becomes zero. Hence work done against friction is zero.

e) The statement is true. This is because rolling occurs only on account of friction which is tangential force capable of providing torque. When the inclined plane is perfectly smooth it will simply slip under the effect of its own weights.

Question 33.
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass :
(a) Show P = \(\mathrm{P}_{\mathrm{i}}^{-}+\mathrm{m}_{\mathrm{i}} \mathrm{V}\)
Where P_i is the momentum of the i_i the particle (of mass m_i) and p’_i + m_iv’_i. Note relative velocity of the ith particle relative to the centre of mass. Also, prove using the definition of the centre of mass Σp’_i = 0
(b) Show K = K’ + 1/2MV²
Where K is the total kinetic energy of the system of particles. K’ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV²/2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.
(c) Show L = L’ + R × MV
Where L’ = Σr_i P_i is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember r’_i = r_i – R rest of the notation is the standard notation used in the chapter. Note L1 and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
(d) Show \(\frac{d l^{\prime}}{d t}=\sum r_i^{\prime} \frac{d p^{\prime}}{d t}\)
Further, show that \(\frac{\mathrm{dL}^{\mathrm{l}}}{\mathrm{dt}}\) = τ’_ext
Whereτ’_ext is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)
Solution:
a) Let m₁, m₂, … m_i mass points have the position vectors \(\overrightarrow{r_1}, \overrightarrow{r_2}, \overrightarrow{r_i}\) w.r.t origin ‘O’. The position vector of C.M. say \(\overrightarrow{\mathrm{OP}}\)
i.e., \(\overrightarrow{\mathrm{OP}}=\frac{m_1 \overrightarrow{r_1}+m_2 \overrightarrow{r_2}+\ldots}{m_1+m_2+\ldots}=\frac{m_i r_i}{M}\)
where i = 1, 2, 3 .
Now, let us change the origin to O’ and assume that the C.M is now at p’ with

Multiplying eqn. (1) by m_i and differentiating w.r.t to time, we get

even if we change the origin, the position of
centre of mass will not change i.e., Σp’_i = 0

b) In rotational kinematics, K.E_T of the system of particles = K.E_T of the system when the particle velocities are taken w.r.t to CM + K.E of the translation of the system as a whole (i.e. of the C.M motion of the system)
i.e. \(\frac{1}{2}\) m₁v₁² + \(\frac{1}{2}\) m₂v₂² + …..
= [\(\frac{1}{2}\) m₁v₁² + \(\frac{1}{2}\)m₂v₂ +……] + \(\frac{1}{2}\)Mv²
\(\frac{1}{2}\) m_iv_i² = \(\frac{1}{2}\)m_iv_i² + \(\frac{1}{2}\) mv²
∴ k = k’ + \(\frac{1}{2}\)
where M = total mass of particles
and y = velocity of C.M motion of the system.

c) From eq. (1), we have

Textual Examples

Question 1.
Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150 g, and 200 g respectively. Each side of the equilateral triangle is 0.5 long. [A.P. & T.S. Mar.18]
Solution:

With the x – and y-axes chosen as shown in Fig. 7.9, the coordinates of points O, A and B forming the equilateral triangle are
respectively (0, 0), (0, 5, 0), (0.25. \(\sqrt{3}\) ). Let the masses 100 g, 150g and 200g be located at O, A and B be respectively. Then.

Question 2.
Find the centre of mass of a triangular lamina.
Solution:
The lamina (∆LMN) may be subdivided into narrow strips each parallel to the base (MN) as shown in Fig.

By symmetry each strip has its centre of mass at its midpoint. If we joint the midpoint of all the strips we get the median LP. The centre of mass of the triangle as a whole therefore, has to lie on the median LP. Similarly, we can argue that it lies on the median MQ and NR. This means the centre of mass lies on the point of concurrence of the medians, i.e. on the centroid G of the triangle.

Question 3.
The density of the atmosphere at sea level is 1.29 kg/m³. Assume that it does not change with altitude. Then how high would the atmosphere extend ?
Solution:
The centres of mass C₁, C₂ and C₃ of the squares are, by symmetry, their geometric centres and have coordinates (1/2,1/2), (3/ 2,1/2), (1/2,3/2) respectively.

Question 4.
A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet reston the floor 1 m from the wall as shown in Fig. Find the reaction forcesof the wall and the floor.
Solution:

Let AB is 3 m long, A is at distance AC = 1 m from the wall. Pythagoras theorem,
BC = 2\(\sqrt{2}\) m.
For translational equilibrium, taking the forces in the vertical direction, N – W = 0 ……….. (i)
Taking the forces in the horizontal direction,
F – F₁ = 0 ……………. (ii)
taking the moments of the forces about A,
2\(\sqrt{2}\) F1 – (1/2) W = 0 ………….. (iii)
Now W = 20 g = 20 × 9.8 N = 1960.0 N
From (i) N = 196.0
From (iii) = F₁ = W/4 \(\sqrt{2}\) = 196.0/4 \(\sqrt{2}\)
= 34.6 N
From (ii) F = F₁ = 34.6 N
F₂ = \(\sqrt{F^2+N^2}\) = 199.0 N

Question 5.
Find the scalar and vector products of two vectors
a = \((3 \vec{i}-4 \vec{j}+5 \vec{k})\) and b = \((-2 \vec{i}+\vec{j}-3 \vec{k})\)
Solution:
a.b = (\((3 \vec{i}-4 \vec{j}+5 \vec{k})\)) . (\((-2 \vec{i}+\vec{j}-3 \vec{k})\))
= -6 – 4 – 15 = -25
a × b = \(\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
3 & -4 & 5 \\
-2 & 1 & -3
\end{array}\right|=7 \vec{i}-\vec{j}-5 \vec{k}\)
Note b × a = \(7 \vec{i}-\vec{j}-5 \vec{k}\)

Question 6.
Obtain Equation from first principles.
Solution:
The angular acceleration is uniform,
ω = αt + c (as α is comstant)
At t = 0, ω = ω₀ (given)
From (i) we get at t = 0, ω = c = ω₀
Thus, ω = αt + ω₀ as required

Question 7.
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm is 16 seconds, (i) What is its angular acceleration, assuming the acceleration to be uniform ? (ii) How many revolutions does the engine make during this time ?
Solution:
(i) We shall use ω = ω₀ + αt
ω₀ = initial angular speed in rad/s
= 2π × angular speed in rev/s
= \(\frac{2 \pi \times \text { angular speed in rev } / \mathrm{min}}{60 \mathrm{~s} / \mathrm{min}}\)
= \(\frac{2 \pi \times 1200}{60} \mathrm{rad} / \mathrm{s}\)
= 40π rad/ s
Similarly ω = final angular speed in rad /s
= \(\frac{2 \pi \times 3120}{60} \mathrm{rad} / \mathrm{s}\)
= 2π × 52 rad/s = 104 π rad/s
∴ Angular acceleration
α = \(\frac{\omega-\omega_0}{t}\) = 4π rad/s²
The angular acceleration of the engine = 4π rad/s²

(ii) The angular displacement in time t is given by
θ = ω₀t + \(\frac{1}{2}\) αt²
= (40π × 16 + \(\frac{1}{2}\) × 4π × 16²) rad
= (640 π+ 512 π )rad
= 1150π rad
Number of revolutions = \(\frac{1152 \pi}{2 \pi}\) = 576

Question 8.
Find the torque of a force \((\bar{i}-\bar{j}+\bar{k})\) about the origin. The force acts on a particle whose position vector is \((\bar{i}-\bar{j}+\bar{k})\) (Mar.’14, 13)
Answer:
Here r = \(\vec{i}-\vec{j}+\vec{k}\)
and F = \(7 \vec{i}-3 \vec{j}+5 \vec{k}\)
We shall use the determinant rule to find the torque τ = r × F
τ = \(\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & -1 & 1 \\
7 & 3 & -5
\end{array}\right|=(5-3) \vec{i}-(-5-7) \vec{j}\) + \((3-(-7)) \vec{k}\)
or τ = \(2 \vec{i}+12 \vec{j}+10 \vec{k}\)

Question 9.
Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.
Solution:
Let the particle with velocity v be at point P at some instant t. We want to calculate the angular momentum of the particle about an arbitrary point O.

The angular momentum is 1 = r × mv. Its magnitude is mvr. sine, where θ is the angle between r and v as shown in Fig. Although the particle changes position with time, the line of direction of v remains the same and hence OM = r sin θ ; a constant.

Question 10.
Show that moment of a couple does not depend on the point about which you take the moments.
Solution:

The moment of the couple = sum of the moments of the two forces making the couple
= r₁ × (-F) + r₂ × F
= r₂ × F + r₁ × F
= (r₂ – r₁) × F
But r₂ + AB = r₂, and hence AB = r₂ – r₁.
The moment of the couple, therefore, is AB × F.

Question 11.
What is the moment of intertia of a disc about one of its diameters ?
Solution:

x and y-axes lie in the plane of the disc and z is perpendicular I_z = I_x = I_y
Now, x and y axes are along two diameters of the disc, and by symmetry the moment of inertia of the disc is the same about any diameter. Hence
I_x = I_y
and I_z = 2I_y
But I_x = I_y
So finally, I_x = I_z/2 = MR²/4
Thus the moment of inertia of a disc about any of its diameter is MR²/4.
Find similarly the moment of inertia of a ring about any of its diameter. Will the theorem be applicable to a solid cylinder?

Question 12.
What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end ?
Solution:
For the rod of mass M and length l, I = Ml²/ 12. Using the parallel axes theorem, I’ = I + Ma² with a = l/2 we get.
I’ = \(\mathrm{M} \frac{l^2}{12}+\mathrm{M}\left(\frac{l}{2}\right)^2=\frac{\mathrm{M} l^2}{3}\)
We can check this independently since I is half the moment of inertia of a rod of mass 2M and length 21 about its midpoint,
I’ = \(2 \mathrm{M} \frac{4 l^2}{12} \times \frac{1}{2}=\frac{\mathrm{M} l^2}{3}\)

Question 13.
What is the moment of inertia of a ring about a tangent to the circle of the ring ?
Solution:
The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring. The distance between these two parallel axes is R, the radius of the ring. Using the parallel axes theorem.

= 0.4kg m²
α = angular acceleration
= 5.0 Nm/0.4 kg m² = 12.5 S^-2

b)Work done by the pull unwinding 2m of the cord
= 25 N × 2 m = 50 J

(c) Let ω be the final angular velocity. The kinetic energy gained = \(\frac{1}{2}\) Iω²
since the wheel starts from rest. Now,
ω₂ = ω²₀ +2αθ, ω₀ = 0
The angular displacement θ = length of unwound string / radius of wheel = 2m / 0.2 m = 10 rad
ω² = 2 × \(\frac{1}{2}\) × 12.5 × 10.0 = 250 (rad/s)²
∴ K.E. gained = \(\frac{1}{2}\) × 0.4 × 250 = 50 J

(d) The answers are the same, i.e. the kinetic energy gained by the wheel work done by the force. There is no loss of energy due to friction.

Question 14.
A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. The flywheel is mounted on a horizontal axle -with frictionless bearings.
(a) Compute the angular acceleration of the wheel.
(b) Find the work done by the pull, when 2m of the cord is unwound
(c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest.
(d) Compare answers to parts (b) and (c).
Solution:

a) We use Iα = τ
the toque τ = FR
= 25 × 0.20 Nm (as R = 0.20m)
= 5.0 Nm

Question 15.
Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity ?
Solution:
We assume conservation of energy of the rolling body, i.e. there is no loss of energy due to friction etc. The protential energy lost by the body in rolling down the inclined plane (= mgh) must, therefore, be equal to kinetic energy gained. (See fig.) Since the bodies start from rest the kinetic energy gained is equal tot he final kinetic energy of the odies. From K = \(\frac{1}{2} m v_{\mathrm{cm}}^2\left[1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right]\)
K = \(\frac{1}{2} m v^2 1+\left(\frac{K^2}{R^2}\right)\) Where v is the final velocity of (the centre of mass of) the body.
Equating K and mgh

Note v² is independent of the mass of the rolling body:
For a ring, k² = R²
v_ring = \(\sqrt{\frac{2 g h}{1+1}}=\sqrt{g h}\)
For a solid cylinder K² = R²/0.2
v_disc = \(\sqrt{\frac{2 g h}{1+1 / 2}}=\sqrt{\frac{4 g h}{3}}\)
For a solid sphere K² = 2R²/5
v_sphere = \(\sqrt{\frac{2 g h}{1+1 / 52}}=\sqrt{\frac{10 g h}{7}}\)

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 5th Lesson Laws of Motion Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 5th Lesson Laws of Motion

Very Short Answer Questions

Question 1.
What is inertia ? What gives the measure of inertia ? [T.S. Mar. 17]
Answer:
The resistance of the body to change its state of rest of state of uniform motion in a straight line is called inertia of the body.
Acceleration = \(\frac{\text { Force }}{\text { Mass }}\)
∴ The more is the mass, less is the acceleration and more is the inertia. The mass of a body is a quantitative measure of its inertia.

Question 2.
According to Newton’s third law, every force is accompanied by an equal and opposite force. How can a movement ever take place ?
Answer:
Because both action and reaction are taking place on different bodies.

Question 3.
When a bullet is fired from a gun, the gun gives a kick in the backward direction. Explain. [A.P. Mar. 15]
Answer:
Due to law of conservation of momentum, Recoil of the gun V = \(\frac{\mathrm{mu}}{\mathrm{M}}\)
Where M – mass of the gun; m = mass of the bullet; u = velocity of the bullet

Question 4.
Why does a heavy rifle not recoil as strongly as a light rifle using the same cartridges ?
Answer:
Recoil of the Gun V = \(\frac{\mathrm{mu}}{\mathrm{M}}\)
Due to heavy mass of rifle the recoil is less.

Question 5.
If a bomb at . _st explodes into two pieces, the pieces must travel in opposite directions. Explain. [T.S. Mar. 16, 15]
Answer:
According to law of conservation of momentum,
Mu = m₁v₁ + m₂v₂
Initially the bomb is at rest u = 0
m₁v₁ + m₂v₂ = 0
or m₁v₁ – m₂v₂
(Negative sign indicates that the pieces must travel in opposite direction)

Question 6.
Define force. What are the basic forces in nature ?
Answer:
The force is on which changes or tends to change the state of rest or motion of a body. Basic forces :

1. Gravitational force
2. Electromagnetic force
3. Nuclear force
4. Weak interaction force

Question 7.
Can the coefficient of friction be greater than one ?
Answer:
Yes, coefficient of friction may be greater than one. In some particular cases it is possible. They are

1. Due to increase the inner molecular attractive forces between surfaces when the contact surfaces are highly polished.
2. When the contact surfaces of the bodies are inter locking the coefficient friction may be greater than one.

Question 8.
Why does the car with a flattened tyre stop sooner than the one with inflated tyres ?
Answer:
Flattened deforms more than the inflated tyre. Due to greater deformation of the type rolling friction is large nence it stops soon.

Question 9.
A horse has to pull harder during the start of the motion than later. Explain. [Mar. 13]
Answer:
We know the limiting frictional force is greater than kinetic frictional force. For starting motion of the cart, the limiting friction is to be overcome. Once motion is set, frictional force reduces. Therefore, the horse has to pull harder during starting of the cart.

Question 10.
What happens to the coefficient of friction if the weight of the body is doubled? [A.P. – Mar. 16]
Answer:
If weight of a body is doubled, coefficient of friction does not change. Coefficient of friction is independent of normal reaction. If weight is doubled, normal reaction doubled and correspondingly frictional force doubled. So, coefficient of friction does not change.

Short Answer Questions

Question 1.
A stone of mass 0.1 kg is thrown vertically upwards. Give the magnitude and direction of the net force on the stone
(a) during its upward motion.
(b) during the downward motion.
(c) at the highest point, where it momentarily comes to rest.
Answer:
Given that, mass of stone, m = 0.1 kg, g = 9.8 ms^-2.
a) During upward motion: Magnitude of net force on the stone,
F = |-mg|; F = 0.1 × 9.8 = 0.98N.
Direction of net force is in upward direction.

b) During downward motion: Magnitude of net force on the stone,
F = ma = 0.1 × 9.8 = 0.98N.
Direction of net force is in downward direction.

c) At the heighest point : Magnitude of net force, F = mg = 0.1 × 9.8 = 0.98N..
At highest point of stone, direction is indeterminate.

Question 2.
Define the terms momentum and impulse. State and explain the law of conservation of linear momentum. Give examples.
Answer:
Momentum : The product of mass and velocity of a body is called momentum momentum p = mv .
Impulse : The product of force and time that produces finite change in momentum of the body is called impulse.
Impulse (I) = Force × time duration = mat = \(m \frac{(v-u)}{t} t\)
= (mv – mu)
Law of conservation of linear momentum : The total momentum of an isolated system of interacting particles remains constant it there is no resultant external force acting on it.

Explanation : Consider two smooth, non-rotating spheres of masses m₁ and m₂ (m₁ > m₂). Let u₁ and u₂ be their initial velocities. Let v₁ and v₂ be final velocities after head on collision. According to law of conservation of linear momentum, we have
Momentum of the system before collision = Momentum of the system after collision.
t.e., m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Examples :

1. Motion of a Rocket
2. Bullet-Gun motion

Question 3.
Why are shock absorbers used in motor cycles and cars ?
Answer:
When a scooter or a car moves on a rough road it receives an impulse due to the Jerkey motion. In case the shockers are used in the vehicle, the time of impact increases, and decreases the impulsive force, due to increased value of the time of impact the force of impact is reduced. So it saves the vehicle and its occupants from experiencing reverse jerks.

Question 4.
Explain the terms limiting friction, dynamic friction and rolling friction.
Answer:
Limiting friction : The maximum value of static friction is called “Limiting friction”.
It is denoted by F_L = F_s(max) [∵ F_s ≤ μ_sN]
Dynamic friction (Kinetic friction) : The resistance encountered by a sliding body on a surface is called kinetic or dynamic friction F_k.
If the applied force overcomes the limiting friction and sets the body into motion. Then motion of the body is resisted by another friction called “Dynamic friction” or “Kinetic friction”.

Rolling friction : ‘The resistance encountered by a rolling body on a surface is called “Rolling friction”.
If a wheel or a cylinder or a spherical body like a marble rolls on horizontal surface, the speed of rolling gradually decreases and it finally stops.

Question 5.
Explain advantages and disadvantages of friction. [T.S., A.P. Mar. 15]
Answer:
Advantages of friction :

1. Safe walking on the floor, motion of vehicles etc., are possible only due to friction.
2. Nails, screws are driven into walls (or) wooden surfaces due to friction.
3. Friction helps the fingures hold the things (or) objects like pen, pencil and water tumbler etc.
4. Speed running vehicles etc. can be stopped suddenly when friction is present, otherwise accidents become large. Due to friction vehicles move on the roads without slipping and they can be stopped.
5. The mechanical power transmission of belt drive is possible due to friction.

Disadvantages of friction:

1. Due to friction there is large amount of power loss in machines and engines.
2. Due to friction wear and tear of the machines increases and reducing their life.
3. Due to friction some energy gets converted into heat which goes as waste.

Question 6.
Mention the methods used in decrease friction [A.P. Mar. 18; T.S. Mar. 16, Mar. 14]
Answer:

1. Polishing : By polishing the surfaces of contact, friction can be reduced.
2. Bearings : The rolling friction.is less than the sliding friction hence free wheels of a cycle, motor
   car, dynamos etc., are provided with ball bearings to reduce friction. Bearings convert sliding motion into rolling motion.
3. Lubricants: The lubricant forms a thin layer between surfaces of contact. It reduces the friction. In light vehicles or machines, oils like “three in one” are used as lubricants. In heavy machines greasure is used. In addition to this they guard the mechanical parts from over heating.
4. Streamlining : Automobiles and aeroplanes are streamlined to reduce the friction due to air.

Question 7.
State the laws of rolling friction.
Answer:
Laws of friction – rolling friction :

1. The smaller the area of contact, the lesser will be the rolling friction.
2. The larger the radius of the rolling body, the lesser will be the rolling friction.
3. The rolling friction is directly proportional to the normal reaction.
   If F_R is the rolling friction and ‘N’ is the normal reaction at the contact, then F_R ∝ N
   F_R = μ_RN; where μ_R is the coefficient of rolling friction.

Question 8.
Why is pulling the lawn roller preferred in pushing it ?
Answer:
Pulling of lawn roller : Let a lawn roller be pulled on a horizontal road by a force ‘F, which makes an angle θ with the horizontal, to the right as shown in the figure. The weight of the body “mg” acts vertically downwards.

Let the force ‘F’ be resolved into two mutually perpendicular components F sin θ, vertically upwards and F cos θ horizontally along the road.
∴ The normal reaction N = mg – F sin θ
Then the frictional force acting towards left is F_R = μ_RN, where μ_R is the coefficient of rolling friction between he roller and the road, or F_R= μ_R (mg – F sin θ)
∴ The net pulling force on roller is P = F cos θ – F_R = F cos θ – (μ_R mg – F sin θ)
or P = F(cos θ + μ_R sin θ) – μ_R mg ………………(1)

Pushing of lawn roller : When a lawn roller is pushed by a force ‘F1, which makes an angle ‘θ’ with the horizontal, the component of force acting vertically downwards is F sin θ. The horizontal component F cos θ pushes the roller to the right as shown in figure.
The weight ‘mg’ of the lawn roller acts vertically downwards. Therefore the normal reaction N of the surface on the roller
N = mg + F sin θ
Then the frictional force acting towards left.
F_R = μ_RN = μ_R (mg + F sin θ)
The net pushing force on roller is
P’ = F cos θ – F_R = F cos θ – μ_R (mg + F sin θ)
or P’ = F(cos θ – μ_R sin θ) – μ_R mg ……………… (2)
From equations (1) and (2) that it is easier to pull than push a lawn roller.

Long Answer Questions

Question 1.
a) State Newton’s second law of motion. Hence derive the equation of motion F = ma from it. [T.S. Mar. 18; A.P. Mar. 16, Mar. 13]
b) A body is moving along a circular path such that its speed always remains constant. Should there be a force acting on the body ?
Answer:
a) Newton’s Second of motion : “The rate of change of momentum of a body is directly proportional to the external force applied and takes place in the same direction”.
To show F = ma : Let a body of mass ‘m’ moving with a velocity ‘v’ under the action of an external force F in the direction of velocity.
Momentum ‘P’ of a body is the product of the mass and velocity V.
∴ P = mv ……………… (1)
According to Newton’s second law of motion, we have
\(\frac{\mathrm{dp}}{\mathrm{dx}}\) ∝ F, where F = external force
(or)F = K\(\frac{\mathrm{dp}}{\mathrm{dx}}\) ………………. (2)
From equations (1) and (2) we have
F = \(K \frac{d(m v)}{d t}=K \cdot m \frac{d v}{d t}\)
= Kma …………………. (3)
Since the rate of change of velocity \(\frac{\mathrm{dv}}{\mathrm{dx}}\) is the acceleration ‘a’ of the body.
In SI system the unit of force is Newton and is defined as that force which when acting on a body of mass 1 kg produces in it an acceleration of 1 ms^-2.
i.e., from equation (3),
If F = 1, m = 1 and a = 1
we get K = 1
Hence F = \(\frac{\mathrm{dp}}{\mathrm{dx}}\) = ma .
∴ F = ma

b) Suppose a body is moving along a circular part though its speed always remains constant its velocity changes at every point and resultant force acts on the body.

Question 2.
a) Define angle of friction and angle of repose. Show that angle of friction is equal to angle of repose for a rough inclined place.
Answer:
Angle of friction : The angle of friction is defined as the angle made by the resultant of the normal reaction and the limiting friction with the normal reaction is called angle of friction.

Angle of repose : The angle of repose is defined as the angle of inclination of a plane with respect to horizontal for which the body will be in equilibrium on the inclined plane is called angle of repose.

Angle of friction is equal to angle of repose for a rough inclined plane : Let us consider a body of mass’m1 on a rough inclined plane. The angle of inclination of the rough surface is ‘θ’. By increasing the angle of inclination at one end, the body tends to slide on the surface. Then the angle of inclination ’0′ is called angle of repose.

The weight (mg) of the body resolved into two components, the component mg cos 0 acts perpendicular to the inclined surface, which is equal to normal reaction ‘N‘.
i.e., N = mg cos θ ……………. (1)
The other component mg sin θ acts parallel to inclined plane and opposite to the frictional force ‘f’.
f_s = my sin θ ……………… (2)
From \(\frac{(2)}{(1)} \Rightarrow \frac{f_s}{N}=\frac{m g \sin \theta}{m g{con} \theta}\)
= \(\frac{f_s}{N}\) = tan θ = µ_s
∴ µ_s = tan θ …………………. (3)
when ‘α’ is angle of friction then from the definition of co-efficient of static friction,
µ_s = tan α ………………… (4)
from (3) and (4)
⇒ tan θ = tan α
θ = α
Hence angle of friction is equal to angle of repose.

b) A block of mass 4 kg is resting on a rough horizontal plane and is about to move when a horizontal force of 30 N is applied on it. If g = 10 m/s². Find the total contact force exerted by the plane on the block.
Solution:
Given m = 4kg
F = 30 N
g = 10 ms^-2
a = \(\frac{F}{m}=\frac{30}{4}\) = 7.5 ms^-2
µ = \(\frac{\mathrm{a}}{\mathrm{g}}=\frac{7.5}{10}=\frac{3}{4}\)
∴ Contact force = Frictional force
= µ mg
= \(\frac{3}{4}\) × 4 × 10 = 30 N.

Problems

Question 1.
The linear momentum of a particle as a function of time t is given by p = a + bt, where a and b are positive constants. What is the force acting on the particle ?
Answer:
Linear momentum of a particle p = a + bt
Force F = \(\frac{\mathrm{dp}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(a + bt) = 0 + b
∴ F = b

Question 2.
Calculate the time needed for a net force of 5 N to change the velocity of a 10 kg mass by 2 m/s.
Answer:
F = 5N, m = 10kg; (v-u) = 2m s^-1, t = ?
F = \(m \frac{(v-u)}{t} \Rightarrow 5=\frac{10 \times 2}{t}\)
∴ t = 4s.

Question 3.
A ball of mass m is thrown vertically upward from the ground and reaches a height h before momentarily coming to rest. If g is acceleration due to gravity. What is the impulse received by the ball due to gravity force during its flight ? (neglect air resistance).
Answer:
mass = m
Initial velocity of a ball to reach h is,
u = \(\sqrt{2 g h}\)

On return journey, velocity of a ball to reach the ground, v = –\(\sqrt{2 g h}\)
Impluse I = m (u – v) = m [\(\sqrt{2 g h}\) – (-\(\sqrt{2 g h}\))]
= m 2 \(\sqrt{2 g h}\) = 2m \(\sqrt{2 g h}\)
∴ I = \(\sqrt{8 m^2 g h}\)

Question 4.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 in m s^-1 to 3.5 m s^-1 in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force ?
Answer:
m = 3.0 kg; u = 2.0 ms^-1,
v = 3.5ms^-1, t = 25 s;
F = m \(\left(\frac{v-u}{t}\right)\) = 3\(\left(\frac{3.5-2}{25}\right)\)
= \(\frac{3 \times 15}{25}=\frac{0.9}{5}\) = 0.18 N
This force acts in the direction of change in velocity.

Question 5.
A man in a lift feels an apparent weight W when the lift is moving up with a uniform acceleration of 1/3^rd of the acceleration due to gravity. If the same man were in the same lift now moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity, then what is his apparent weight ?
Answer:
When the lift is moving up, a = \(\frac{g}{3}\)
Apparent weight W = m(g + a)
= m(g + \(\frac{g}{3}\)) = \(\frac{4mg}{3}\)
⇒ mg = \(\frac{3 W}{4}\) …………….. (1)
When the lift is moving down, a = \(\frac{g}{2}\)
Apparent weight W¹ = m(g – a)
= m(g – \(\frac{g}{2}\)) = \(\frac{mg}{2}\)
∴ |W¹| = \(\frac{\left[\frac{3 W}{4}\right]}{2}=\frac{3 W}{8}\)

Question 6.
A container of mass 200 kg rests on the back of an open truck, if the truck accelerates at 1.5 m/s², what is the minimum coefficient of static friction between the container and the bed of the truck required to prevent the container from sliding off the back of the truck ?
Answer:
m = 200kg, a = 1.5 ms^-2, g = 9.8 ms^-2
ma = μ_smg
μ_s = \(\frac{a}{g}=\frac{1.5}{9.8}\) = 0.153

Question 7.
A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments. One of them starts moving vertically down- wards with an initial speed of 10 m/s. If acceleration due to gravity is 10 m/s². What is the separa-tion between the fragments 2s after the explosion ?
Answer:
Bomb explodes into two fragments say 1 and 2.
For 1^st fragment, u₁ = 10m/s, t = 2 sec; g = 10 m/s^-2, s₁ = ?
Now displacement of 1^st fragment,
s₁ = u₁t + \(\frac{1}{2}\) gt²
= 10 × 2 × \(\frac{1}{2}\) × 10 × 2² = 40m

2^nd fragment is move opposite to 1^st fragment like an object from a tower.
For 2^nd fragment u₂ = – u₁ = 10m/s
t = 2 sec; g = 10 m/s²
Now displacement of 2^nd fragment
s₂ = u₂t + \(\frac{1}{2}\) gt²
= 10 × 2 × \(\frac{1}{2}\) × 10 × 2² = 0
∴ The seperation of two fragments
= s₁ + s₂ = 40 + 0 = 40 m

Question 8.
A fixed pulley with a smooth grove has a light string passing over it with a 4 kg attached on one side and a 3 kg on the other side. Another 3 kg is hung from the other 3 kg as shown with another light string. If the system is released from rest, find the common acceleration ? (g = 10 m/s²).

Answer:
From fig,
m₁ = 3 + 3
= 6 kg
m₂ = 4 kg
g = 10ms^-2
Acceleration of the system,
a = \(\left(\frac{\dot{m}_1-m_2}{m_1+m_2}\right) g=\left(\frac{6-4}{6+4}\right) \times 10\) = 2 ms^-2

Question 9.
A block of mass of 2 kg slides on an inclined plane that makes an angle of 30% with the horizontal. The coefficient of friction between the block and the surface is \(\frac{\sqrt{3}}{2}\).
(a) What force should be applied to the block so that it moves up without any acceleration ?
(b) What force should be applied to the block so that it moves up without any acceleration ?
Answer:
m = 2kg; θ = 30°; µ = \(\frac{\sqrt{3}}{2}\)
i) F_down = mg (sinθ – µ cosθ)
= 2 × 9.8 (sin30° – \(\frac{\sqrt{3}}{2}\)cos30°)
= 2 × 98 [\(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) × \(\frac{\sqrt{3}}{2}\)] = 49 N

ii) F_up = mg (sinθ + µ cosθ)
= 2 × 9.8 [sin30° + \(\frac{\sqrt{3}}{2}\) × cos30°]
= 2 × 9.8 [\(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) × \(\frac{\sqrt{3}}{2}\)] = 24.5 N

Question 10.
A block is placed on a ramp of parabolic shape given by the equation g = \(\frac{x^2}{20}\), see If i_s = 0.5, what is the maximum height above the ground at which the block can be placed without slipping ? (tan θ = µ_s = \(\frac{\mathrm{d} y}{\mathrm{~d} x}\))

Answer:

Question 11.
A block of metal of mass 2 kg on a horizontal table is attached to a mass of 0.45 kg by a light string passing over a frictionless pulley at the edge of the table. The block is subjected to a horizontal force by allowing the 0.45 kg mass to fall. The coefficient of sliding friction between the block and table is 0.2.

Calculate (a) the initial acceleration, (b) the tension in the string, (c) the distance the block would continue to move if. after 2 s of motion, the string should break.
Answer:
Here, m₁ = 0.45kg
m₂ = 2kg
µ = 0.2
a) Initial acceleration.
a = \(\left(\frac{m_1-\mu m_2}{m_1+m_2}\right) g=\left[\frac{0.45-0.2 \times 2}{0.45+2}\right] \times 9.8\)
a = 0.2ms^-2

b) From fig, we have
T – f = m₂ a
T – 3.92 = 2 × 0.2
[∵ f = µm₂g = 0.2 × 2 × 9.8] = 3.92 N
⇒ N = 0.4 + 3.92 = 4.32 N

c) Velocity of string after 2 sec
= u in this case: µ’ = 0
Stoping distance s = \(\frac{\mu^2}{2 \mu g}=\frac{0.4 \times 0.4}{2 \times 0.2 \times 9.8}\) = 0.0408 m

Question 12.
On a smooth horizontal . surface a block A of mass 10 kg is kept. On this block a second block B of mass 5 kg is kept. The coefficient of friction between the two blocks is 0.4. A horizontal force of 30 N is applied on the lower block as shown.

The force of friction between the blocks is (take g = 10 m/s²).
Answer:
Here m_A = 10kg; m_B = 5kg;
F = 30N; µ = 0.4
F = (m_A + m_B)a
⇒ a = \(\frac{F}{\left(m_A+m_B\right)}\)
= \(\frac{30}{10+5}\) = 2ms^-2
f = m_Ba = 5 × 2 = 10 N

Additional Problems

(For simplicity in numerical calculations, take g = 10 ms^-2)

Question 1.
Give the magnitude and direction of the net force acting on
a) a drop of rain falling down with a constant speed,
b) a cork of mass 10 g floating on water,
c) a kite skillfully held stationary in the sky,
d) a car moving with a constant velocity of 30 km/h on a rough road,
e) a high-speed electron in space far from all material objects and free of electric and magnetic fields.
Answer:
a) As the rain drop is falling with a constant speed, its acceleration a = 0. Hence net force F = ma = 0.
b) As the cork is floating on water, its weight is being balanced by the upthrust (equal-to weight of water displaced). Hence net force on the cork is zero.
c) As the kite is held stationary, net force on the kite is zero, in accordance with Newton’s first law.
d) Force is being applied to overcome the force of friction. But as velocity of the car is constant, its acceleration, a = 0. Hence net force on the car F = ma = 0.
e) As no field (gravitational / electric / magnetic) is acting on the electron, net force on it is zero.

Question 2.
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble.
a) during its upward motion,
b) during its downward motion
c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at ah angle of 45° with the horizontal direction ?
Ignore are resistance.
Answer:
When a body is thrown vertically upwards (or) it moves vertically downwards, gravitational pull of earth gives it a uniform acceleration a = + g = + 9.8 ms^-2 in the downward direction. Therefore, the net force on the pebble in all the three cases is vertically downwards.
As m = 0.05 kg and a = + 9.8 m/s²
In all the three cases,
∴ F = ma
= 0.05 × 9.8 = 0.49 N, vertically downwards.
If the pebble were thrown at an angle of 45° with the horizontal direction, it will have horizontal and vertical components of velocity. These components do not affect the force on the pebble. Hence our answers do not alter in any case. However in each case (C), the pebble will not be at rest. It will have horizontal component of velocity at highest point.

Question 3.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg.
a) just after it is dropped from the window of stationary train,
b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
c) just after it is dropped from the window of a train accelerating with 1 ms^-2,
d) lying on the floor of a train which is accelerating with 1 ms-2, the stone being at rest relative to the train.
Answer:
a) Here, m – 0.1 kg, a = +g = 9.8 m/s²
Net force, F = ma = 0.1 × 9.8 = 0.98 N This force acts vertically downwards.

b) When the train is running at a constant velocity its acc. = 0. No force acts on the stone due to this motion.
Therefore, force on the stone F = weight of stone mg = 0.1 × 9.8 = 0.98 N
This force also acts vertically downwards.

c) When the train is accelerating with 1m/s², an additional force F¹ = ma = 0.1 × 1 = 0.1 N acts on the stone in the horizontal direction. But once the stone is dropped from the train, F¹ becomes zero and the net force on the stone is F = mg = 0.1 × 9.8 = 0.98 N, acting vertically downwards.

d) As the stone is lying on the horizontal direction of motion of the train. Note the weight of the stone in this case is being balanced by the normal reaction.

Question 4.
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is ;
i) T,
ii) T – \(\frac{\mathrm{m} v^2}{l}\),
iii) T + \(\frac{\mathrm{m} v^2}{l}\),
iv) 0
T is the tension in the string. (Choose the correct Answer)
Answer:
The net force on the particle directed towards the center is T. This provides the necessary centripetal force to the particle moving in the circle.

Question 5.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop ?
Answer:
Here, F = -50 N, m = 20 kg
μ = 15 m/s, v = 0, t = ?
From F = ma,
a = \(\frac{F}{m}=\frac{-50}{20}\) = -2.5 m/s²
From v = u + at
0 = 15 – 2.5t
t = \(\frac{15}{2.5}\) = 6s.

Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms^-1 to 3.5 ms^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ?
Answer:
Here m = 3.0 kg
μ = 2.0 m/s
v = 3.5 m/s,
t = 25s, F = ?
F = ma = \(\frac{m(v-u)}{t}=\frac{3.0(3.5-2.0)}{25}\)
= 0.18 N.
The force is along the direction of motion.

Question 7.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body ?
Answer:

θ = 36° 52
This the direction of resultant force and hence the direction of acceleration of the body, fig.
Also a = \(\frac{F}{m}=\frac{10}{5}\) = 2ms^-2

Question 8.
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three wheeler is 400 kg and the mass of the driver is 65 kg.
Answer:
Here, u = 36km/h = 10 m/s, v = 0, t = s
m = 400 + 65 = 465 kg
Retarding force
F = ma = \(\frac{m(v-u)}{t}=\frac{465(0-10)}{4}\) = -1162 N.

Question 9.
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Answer:
Here m = 20,000 kg = 2 × 10⁴ kg
Initial ace = 5 m/s²
Thrust, F = ?
Clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 m/s². Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 m/s² As thrust = force = mass × acceleration
∴ F = 2 × 10⁴ × 14.8 = 2.96 × 10⁵N

Question 10.
A body of mass 0.40 kg moving initially with a constant speed of 10 ms-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at the time to be x = 0 and predict its position at t = -5 s, 25 s, 100 s.
Answer:
Here, m = 0.40 kg, µ = 10m/s due N
F = – 8.0 N
a = \(\frac{F}{m}=\frac{-8.0}{0.40}\) = -20 m/s²
for 0 ≤ t ≤ 30s.
i) At t = -5s, x = Ut = 10 × (-5) = -50 m

ii) At t = 25s, x = Ut + \(\frac{1}{2}\) at²
= 10 × 25 + \(\frac{1}{2}\) (-20) (25)² = – 6000m

iii) At t = 100s, The problem is divided into two parts, upto 30s, there is force/acc.
∴ from x₁ = Ut + \(\frac{1}{2}\) at²
= 10 × 30 + \(\frac{1}{2}\) (-20) (30)²
= -8700
At t = 30s, v = U + at = 10 – 20 × 30 = – 590 m/s,
∴ for motion from 30s to 100s
x₂ = vt = – 590 × 70 = – 41300 m
x = x₁ + x₂ = -8700 – 41300
= -50,000 m = – 50km.

Question 11.
A truck starts from rest and accelerates uniformly at 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity and (b) acceleration of the stone at t = 11 s ? (Neglect air resistance).
Answer:
Here, u = 0, a = 2 m/s², t = 10s

Let v be the velocity of the truck when the stone is dropped.
From v = u + at
v = 0 + 2 × 10 = 20m/s
a) Horizontal velocity of stone, when it is dropped, v_x = v = 20 m/s.
As air Resistance is neglected, v_x remains constant.
In the vertical direction, initial velocity of stone, µ = 0, a = g = 9.8 m/s²,
time t = 11 – 10 = 1s
From v = u + at
v_y = 0 + 9.8 × 1 = 9.8 ms^-1
Resultant velocity of stone, OC is given by
v = \(\sqrt{v_x^2+v_y^2}=\sqrt{20^2+(9.8)^2}\)
v = 22.3 m/s.
Let θ is the angle with the resultant velocity OC of stone makes with the horizontal direction OA, then from fig.
tan θ = \(\frac{v_y}{v_x}=\frac{9.8}{20}\) = 0.49
∴ θ = 29°

b) The moment the stone is dropped from the car, horizontal force on the stone = 0. The only acceleration the path followed by the stone is, however parabolic.

Question 12.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms^-1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
Answer:
a) We shall study in unit x that at each extreme position, velocity of the bob is zero. If the string is cut at the extreme position, it is only under the action of ‘g’. Hence the bob will fall vertically downwards.

b) At the mean position, velocity of the bob is 1m/s. along the tangent to the arc, which is in the horizontal direction. If the string is let at mean position, the bob will be have as a horizontal projectile. Hence it will follow a parabolic path.

Question 13.
A man of mass 70 kg stands on a weighing scale in a lift which is moving
a) upwards with a uniform speed of 10 ms^-1
b) downwards with a uniform acceleration of 5 ms^-2
c) upwards with a uniform acceleration of 5 ms^-2
What would be the readings on the scale in each case ?
d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ?
Answer:
Here, m = 70 kg, g = 9.8 m/s²
The weighing machine in each case measures the reaction R i.e. the apparent weight.
a) When lift moves upwards with a uniform speed, its accelerations is zero.
R = mg = 70 × 9.8 = 686N

b) When the lift moves downwards with a = 5 m/s²
R = m(g – a) = 70 (9.8 – 5) = 336 N

c) When the lift moves upwards with a = 5 m/s²
R = m(g + a) = 70 (9.8 + 5) = 1036 N
If the lift were to come down freely under gravity, downward acc. a = g
∴ R = m (g – a) = m(g – g) = zero

Question 14.
Figure shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t < 4 s, 0 < t < 4s ? (b) impulse at t = 0 and t = 4 s ?
(Consider one-dimensional motion only)

Answer:
i) For t < 0, the position time graph is OA which means displacement of the particle is zero.
i.e. particle is at rest at the origin. Hence force on the particle must be zero.

ii) For 0 < t < 4s, s, the position time graph OB has a constant slope. Therefore, velocity of the particle is constant in this interval i.e. particle has zero acceleration. Hence force on the particle must be zero. iii) For t > 4s, the position time graph BC is parallel to time axis. Therefore, the particle remains at a distance 3m from the origin, i.e. it is at rest. Hence force on the particle is zero.

iv) Impulse at t = 0 :
We know. Impulse = change in linear momentum, Before t = 0 particle is at rest i.e. u = 0. After t = 0, particle has a constant velocity v = \(\frac{3}{4}\)
= 0.75 m/s

∴ Impulse = m(v – u)
= u (0.75-0)
= 3kg m/s
∴ Impulse at t = 4s
Before t = 4s, particle has a constant velocity u = 0.75 m/s
After t = 4s, particle is at rest i.e. v = 0
Impulse = m(v – u) = 4 (0 – 0.75) = 3kg m/s.

Question 15.
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string, a horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of String. What is the tension in the string in each case ?
Answer:
Here, F = 500 N
m₁ = 10kg, m₂ = 20kg
Let T be the tension in the string and a be the acceleration of the system, in the direction of the force applied.
a = \(\frac{F}{m_1+m_2}=\frac{500}{10+20}=\frac{50}{3}\) m/s²
a) When force is applied on heavier block,
T = m₁ a = 10 × \(\frac{50}{3}\) N
T = 166.66 N

b) When force is applied on lighter block,
T = m₂a = 20 × \(\frac{50}{3}\) N
= 333.33 N
Which is different from value T in case (a) Hence our answer depends on which mass end, the force is applied.

Question 16.
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Answer:
Here, m₂ = 8 kg, m₁ = 12kg
as a = \(\frac{\left(m_1-m_2\right) g}{m_1+m_2}\)
a = \(\frac{(12-8) 9.8}{12+8}=\frac{39.2}{20}\)
= 1.96 m/s²
Again
T = \(\frac{2 m_1 m_2 g}{m_1+m_2}=\frac{2 \times 12 \times 8 \times 9.8}{(12+8)}\) = 94.1 N

Question 17.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Answer:
Let m₁m₂ be the masses of products and \(\vec{v}_1, \vec{v}_2\) be their respective velocities. Therefore total linear momentum after disintegration = \(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}\). Before disintegration, the nucleus is at rest. There, its linear momentum before dis-integration is zero. According to the principle of conservation of linear momentum.
\(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}\) (Or) \(\vec{v}_2=\frac{-m_1 \vec{v}_1}{m_2}\)
Negative sign shows that \(\overrightarrow{v_1}\) and \(\overrightarrow{v_2}\) are in opposite direction.

Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other ?
Answer:
Here, initial momentum of the ball,
A = 0.05 (6) = 0.3 kg m/s
As the speed is reversed on collision, final momentum of the ball A = 0.05 (-6)
= – 0.3 kg ms^-1
Impulse imparted to ball A = Change in momentum of ball A = Final momentum – Intial momentum = – 0.3 – 0.3 = – 0.6 kg m/s

Question 19.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun ?
Answer:
Here, mass of shell m = 0.02 kg
Mass of gun M = 100 kg
Muzzle speed of shell v = 80 m/s
Recoil speed of gun V = ?
According to the principle of conservation of linear momentum mv + MV = 0
(Or) V = \(\frac{-\mathrm{mv}}{\mathrm{M}}=\frac{-0.02 \times 80}{100}\) = 0.016 m/s

Question 20.
A batsman deflects a ball by the angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball ? (Mass of the balls is 0.15 kg).
Answer:
In fig. the ball hits the bat KL along AO and is deflected by the bat along OB. Where LAOB = 45°. ON is normal to the portion of the bat KL deflecting the ball.

∴ θ = LNOA
= 45°/2 = 22.5°
Intial vel along AO = u = 54 km/h
= 15 m/S₁ and mass of ball m = 0.15 kg Intial velocity along AO has the two rectangular components : u cos θ along NO produced and u sin 6 along the horizontal OL.
Final velocity along OB has the same magnitude = u
It is resolved into two rectangular components u cos 6 along ON and u sin θ along OL. We observe that there is no change in velocity along the horizontal, but velocity along vertical is just reserved.
∴ Impulse imparted to the ball
= Change in linear momentum of the ball
= m u cos θ – (- m u cos θ)
= 2 m u cos θ
= 2 × 0.15 × 15 cos 22.5°
= 4.5 × 0.9239
= 4.16 kg m/s

Question 21.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./ min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Answer:
Here, m = 0.25 kg, r = 15m
n = 40 rpm = \(\frac{40}{60}\), rps = \(\frac{2}{3}\), T = ?
T = mrw² = mr(2 π)² = 4 π²m²
T = 4 × \(\left(\frac{22}{7}\right)^2\) × 0.25 × 1.5 × \(\left(\frac{2}{3}\right)^2\) = 6.6N
If T_max = 200 N. then from
T_max = \(\frac{\mathrm{mv}_{\max }^2}{r}\)
\(v_{\max }^2=\frac{T_{\max } \times r}{m}=\frac{200 \times 1.5}{0.25}\) = 1200
v_max = \(\sqrt{1200}\) = 34.6 m/s

Question 22.
If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :
a) the stone moves radially outwards,
b) the stone flies off tangentially from the instant the string breaks,
c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?
Answer:
The instant the string, breaks, the stone flies off tangentialy, as per Newton’s first law of motion.

Question 23.
Explain why
a) a horse cannot pull a cart and run in empty space,
b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
c) it is easier to pull a lawn mower than to push it.
d) a cricketer moves his hands back-wards while holding a catch.
Answer:
a) While trying to pull a cart, a horses pushes the ground backwards with a certain force at an angle. The ground offers an equal reaction in the opposite direction on the feet of the horse. The forward component of this reaction is responsible for the motion of the cart. In empty space, there is no reaction and hence a horse cannot pull the cart and run.

b) This is due to inertia of motion. When the speeding bus stops suddenly, lower part of the body in contact with the seats stops. The upper part of the body of the passengers tend to maintain its uniform motion. Hence the passengers are thrown forward.

c) While pulling a lawn roller, force is applied upwards along the handle. The vertical component of this force is upwards and reduces the effective weight of the roller, fig. (a) while pushing a lawn roller. Face is applied downwards along the handle. The vertical component of this force is downwards and increases the effective weight of the roller, fig. (b). As the effective weight is lesser in case of pulling than in a case of pushing, therefore pulling is easier than pushing.

d) While holding a catch, the impulse receives by the hands F × t = Change in linear momentum of the ball is constant. By moving his hands, backwards, the cricketer increases the time of impact (t) to complete the catch. As t increases, F decreases and as a reaction, his hands are not hurt severly.

Question 24.
Figure shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body ? What is the magnitude of. each impulse ?

Answer:
Here, m = 0.04 kg. position time graph shows that the particle moves from x = 0 at 0 to x = 2 cm at A in 2 sec.
As x – t graph is a st. line, the motion is with a constant velocity.
μ = \(\frac{(2-0) \mathrm{cm}}{(2-0) \mathrm{s}}\) = 1 cm s^-1
= 10^-2 ms^-1

From x = 2 cm at A, particle goes to x = 0 at B in 2 sec.
As AB is a stline, motion is with constant velocity v = 1 cm/s = 10^-2 m/s
Negative sign indicates the reversal of direction of motion. This is being repeated. We can visualise a ball moving between two walls located at x = 0 and x = 2 cm, getting rebounded repeatedly on striking against each wall on every collision with a wall, linear momentum of the ball changes. Therefore, the ball receives impulse after every two seconds. Magnitude of impulse = Total change in linear momentum.
= mu -(mv) = mu – mv = m(u – v)
= 0.04(10^-2 + 10^-2)
= 0.08 × 10^-2
= 8 × 10^-4 kg m/s

Question 25.
Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms-2. What is the net force on the man ? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt ? (Mass of the man = 65 kg).

Answer:
Here, acceleration of conveyer belt, a = 1 m/s²
As the man is standing stationary w.r.t the belt, acceleration of the man = Acceleration of belt = a = 1 m/s²
As m = 65 kg
∴ Net force on the man, F = ma = 65 × 1 = 65 N
Now, µ = 0.2
Force of limiting friction- F = µR = µmg
It the man remains stationary upto max.acc. a of the belt, then F = ma¹ = g mg
a¹ = mg = 0.2 × 9.8 = 1.96 ms^-2

Question 26.
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : (Choose the correct alternative)
Lowest Point – Highest Point
a) mg – T₁ – mg + T₂
b) mg + T₁ – mg – T₂
c) mg + T₁ – (mυ₁²)/R – mg – T₂ + (mυ₁²)/R
d) mg – T₁ – (mυ₁²/R – mg + T₂ + (mυ₁²)/R
T₁ and υ₁ denote the tension and speed at the lowest point T₂ and υ₂ denote corresponding values at the highest point.
Answer:
The net force at the lowest point α is
F_L = (mg – T₁) and the net force at the highest point H is F_H = mg + T₂. Therefore, alternative (a) is correct.

Question 27.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms^-2. The crew and the passengers weight 300 kg. Give the magnitude and direction of the
a) force on the floor by the crew and passengers,
b) action of the rotor of the helicopter on the surrounding air,
c) force on the helicopter due to the surrounding air.
Answer:
Here, mass of the helicopter, m₁ = 100 kg
Mass of the crew and passengers m₂ = 300 kg
Upward acceleration a = 15 ms^-2 and g = 10 ms^-2
a) Force on the floor of helicopter by the crew and passengers = appeard weight of crew and passengers = m₂(g + a)
= 300(10 + 15) = 7500 N

b) Action of rotor of helicopter on surrounding air is obviously vertically downwards because helicopter rises on account reaction to this force. Thus, force of action F = (m₁ + m₂) (g + a)
= (1000 + 300) (10 + 15)
= 1300 × 25 = 32500 N

c) Force on the helicopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore, force of reaction F¹ = 32500 N, vertically upwards.

Question 28.
A stream of water flowing horizontally with a speed of 15 ms^-1 gushes out of a tube of cross-sectional area 10^-2m² and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?
Answer:
Here v = 15 ms^-1
Area of cross section a = 10² m^-2, F = ?
Volume of water pushing out/sec
= a × v = 10^-12 × m³ s^-1
As density of water is 10³ kg/m², therefore, mass of water striking the wall per sec.
m = (15 × 10^-2) × 10³ = 150 kg/s
Change in linear momentum
As F = \(\frac{\text { Change in linear momentum }}{\text { Time }}\)
∴ F = \(\frac{m \times v}{t}=\frac{150 \times 15}{1}\) = 2250 N

Question 29.
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of
a) the force on the 7^th coin (counted from the bottom) due to all the coins on its top
b) the force on the 7^th coin by the eighth coin,
c) the reaction of the 6^th coin on the 7^th coin.
Answer:
a) The force on 7^th coin is due to weight of the three coins lying above it. Therefore,
F (3m) kgf = (3mg)N
Where g is acceleration due to gravity. This force acts vertically downwards.

b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e.,
F = 2m + m = 3(m) kgf = (3mg)N
The force acts vertically downwards.

c) The sixth coin is under the weight of four coins above it.
Reaction, R = – F = -4m(kgf) = -(4 mg)N
Minor sign indicates that the reaction acts vertically upwards, opposite to the weight.

Question 30.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop ?
Answer:
Here θ = 15°
v = 720 km/h =\(\frac{720 \times 1000}{60 \times 60}\) = 200 ms^-1
g = 9.8 ms^-2
From tan θ = \(\frac{\mathrm{v}^2}{\mathrm{rg}}\)
v² = rg tan θ
r = \(\frac{v^2}{g \tan \theta}=\frac{(200)^2}{9.8 \times \tan 15^{\circ}}\)
= 15232 m = 15.232 km

Question 31.
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What provides the centripetal force required for this purpose – The engine or the rails ? What is the angle of banking required to prevent wearing out of the rail ?
Answer:
The centripetal force is provided by the lateral thrust exerted by the rails on the wheels. By Newton’s 3^rd law, the train exerts an equal and opposite thrust on the rails causing its wear and tear.
Obviously, the outer rail will wear out faster due to the larger force exerted by the train on it.
Here v = 54 km/h =\(\frac{54 \times 1000}{60 \times 60}\) = 15 m/s
g = 9.8 ms^-2
As tan θ = \(\frac{v^2}{r g}=\frac{15 \times 15}{30 \times 9.8}\) = 0.76
∴ θ = tan^-1 0.76 = 37.4°

Question 32.
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ?

Answer:
Here, mass of block m = 25 kg
Mass of man M = 50 kg
Force applied to lift the block
F = mg = 25 × 9.8 = 245 N
Weight of Man W = mg = 50 × 9.8 = 490 N
a) When block is raised by man as shown in Fig. force is applied by the man in the upward direction. This increases the apparent weight of the man. Hence action on the floor.
W¹ = W + F
= 490 + 245
= 735 N

b) When block is raised by man as shown in Fig. force is app ed by the man in the downward direction. This decreases the apparent weight of the man. Hence action on the floor in this case would be
W¹ = W – F = 490 – 245 = 245 N
As the floor yeilds to a normal force of 700 N, the mode (b) has to be adopted by the man to lift the block

Question 33.
A monkey of mass 40 kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break : the money
a) climbs up with an acceleration of 6 m s^-2
b) climbs down with an acceleration of 4 ms^-2
c) climbs up with a uniform speed of 5 ms^-1
d) falls down the rope nearly freely under gravity ?
(ignore the mass of the rope)

Answer:
Here, mass of monkey m = 40 kg
Maximum tension the rope can stand T = 600 N
In each case, actual tension in the rope will be equal to apparent weight of money (R). The rope will break when R exceeds T.
a) When monkey climbs up with a = 6 ms^-2
R = m(g + a)
= 40(10 + 6)
= 640 N (Which is greater than T)
Hence the rope will break.

b) When monkey climbs down with a = 4 ms^-2,
R = m(g – a) = 40(10 – 4) = 240 N
Which is less than T.
∴ The rope will not break.

c) When monkey climbs up with a uniform speed v = 5 ms^-1.
Its acceleration, a = 0
∴ R = mg = 40 × 10 = 400 N,
which is less than T.
∴ The role will not break

d) When monkey falls down the rope nearly freely under gravity
a = g
∴ R = m(g – a) = m(g – g) = zero
Hence the rope will not break.

Question 34.
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig.). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the wall is removed ? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μ_s and μ_k.

Answer:
Here, mass of body A, m₁ = 5 kg
Mass of body B, m₂ = 10 kg

Coefficient of friction between the bodies and the table, μ = 0.15
Horizontal force applied on A,
F = 200 N

a) Force of limiting friction acting to the left
f = μ (m₁ + m₂)g
= 0.15(5 + 10) × 9.8 = 22.05 N
∴ Net force to the right exerted on the partition
F’ = 200 – 22.05 = 177.95 N
Reaction of partition = 177.95 N to the left.

b) Force of limiting friction acting on body A
f₁ = μm₁g = 0.15 × 5 × 9.8
= 7.35 N
∴ Net force exerted by body A on body B.
F” = F -f₁ = 200 – 7.35
= 19265 N
This is to the right
Reaction of body B on body A = 192.65
N to the left when the portion is removed, the system of two bodies will move under the action of net force.
F₁ = 177.95 IM
Acceralation produced in the system
a = \(\frac{F^1}{m_1+m_2}=\frac{177.95}{5+10}\)
= 11.86 ms^-2
Force producing motion in body A
F₁ = m₁ a = 5 × 11.86
= 59.3 N
∴ Net force exerted by A on body B, when partition is removed
= F” – F₁ = 192.65 – 59.3
= 133.35 N.
Hence the reaction of body B on A, when partition is removed = 133.35 N to the left.
Thus answers to (b) do change.

Question 35.
A block of mass 15 kg is placed on a long trolley. The cofficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Answer:
a) Flere, m = 15 kg;
μ = 0.18,
a = 0.5 ms^-2
t = 20 s

Force on the block due to motion of the trolly F’ = ma = 15 × 0.5 = 7.5 N
Force of limiting friction on the block
= F = μR = μmg
= 0.18 × 15 × 9.8 = 26.46 N
This opposes the motion of the block. The block shall not move. The force of static friction F will adjust itself equal and opposite to F’, the applied force.

Flence to a stationary observer on the ground, the block will appear to be at rest relative to the trolly. When trolly moves with uniform velocity, the block will continue to be stationary. Because in that case, forward force is zero. Force of friction alone is acting on the block.

b) An observer moving with the trolly has accelerated motion. The observer is therefore non-inertial.. The law of inertia is no longer valid.

Question 36.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms^-2. At what distance from the starting point does the box fall of the truck?

Answer:
Here mass of the box m = 40 kg
Acceleration of truck a = 2 ms^-2
Distance of box from open end S = 5m
Coeff. of friction μ = 0.15
Force on the box due to accelarated motion of the truck, F = ma = 40 × 2 = 80 N
This force F is in the forward direction Reaction F’ on the box is equal to F = 80 N in the backward direction. This is opposed by force of limiting friction
f = μ R = μ mg
= 0.15 × 40 × 9.8
= 58.8 N in the forward direction
∴ Net force on the box in the backward direction is p = F’ – F = 80 – 58.8 = 21.2 N
Backward acceleration produced in the box
a = \(\frac{p}{m}=\frac{21.2}{40}\) = 0.53 ms^-2
It t is time taken by the box to travel S = 5 metre and fall off the truck, then from
S = ut + \(\frac{1}{2}\) at²
5 = 0 × t + \(\frac{1}{2}\) × 0.53 t²
t = \(\frac{\sqrt{5 \times 2}}{0.53}\) = 4.345
If the truck travels a distance x during this time, then again from
S = ut + \(\frac{1}{2}\) at²
x = 0 × 4.34 + \(\frac{1}{2}\) × 2(4.34)² = 18.84 m

Question 37.
A disc revolves with a speed of 33\(\frac{1}{3}\) rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ?
Answer:
The coin revolves with the record in the case when the force of friction is enough to provide the necessary centripetal force. If this force is not sufficient to provide centripetal force, the coin slips on the record.
Now the frictional force is µR where R is the
normal reaction and R = mg
Hence force of friction = µ mg and centripetal force required is \(\frac{m v^2}{r}\) or mrw²
µ₁w are same for both the coins and we have different values of r for the two coins.
So to prevent slipping i.e., causing coins to rotate µ mg > mrω² or µg > rω² …………….. (1)
For 1^st coin
r = 4 cm = \(\frac{4}{100}\) m
n = 33 \(\frac{1}{2}\) rev/min = \(\frac{100}{3 \times 60}\) rev/sec
w = 2πn = 2π × \(\frac{100}{180}\) = 3.49 S^-1
∴ rw² = \(\frac{4}{100}\) × (3.49)² = 0.49 ms^-2 and
µg = 0.15 × 9.8 = 1.47 ms^-2
As µg > rw², there fore this coin Will revolve with the record.
Note: We have nothing to do with the radius of the record = 15 cm

Question 38.
You may have seen in a circus a motor-cyclist driving in verticle loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a verticle loop if the radius of the chamber is 25 m ?
Answer:
At the uppermost point of the death well, with no support from below, the motorcyclist does not drop down. This is because his weight is being balance by the centrifugal force. Infact, the weight of the motorcyclist is spent up in providing the necessary centripetal force to the motorcyclist and hence he does not drop drown.

At the uppermost point, R + mg = \(\frac{m v^2}{r}\),
where R is the normal reaction (downwards) on the motor cyclist by the ceiling of the chamber.
Speed will be minimum, when N = 0
∴ mg = \(\frac{m v^2}{r}\) or
v = \(\sqrt{\mathrm{rg}}=\sqrt{25 \times 10}\)
= 15.8 m/s

Question 39.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed ?
Answer:
Here m = 70 kg, r = 3
n = 200 rpm = \(\frac{200}{60}\) rps
µ = 0.15
w = ?
The horizontal force N by the wall on the man provides the necessary centripetal force = mrω². The frictional force (f) in this case is vertically upwards opposing the weight (mg) of the man.
After the floor is removed, the man will remain stuck to the wall, when
mg = f < µ N i.e., mg < µ mr ω² or g < µ r ω²
∴ Minimum angular speed of rotation of the cylinder is ω = \(\sqrt{\frac{g}{\mu \mathrm{r}}}=\sqrt{\frac{10}{0.15 \times 3}}\)
= 4.7 rad/s

Question 40.
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency no. Show that a small bead on the wire loop remains at its
lowermost point for ω ≤ \(\sqrt{g / R}\) . What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = 2g/R ?
Answer:
In fig. we have show that radius vector joining the bead to the centre of the wire makes as angle θ with the vertical downward direction. It N is normal reaction, then as it is clear from the figure,

mg = N cos θ ………… (1)
m rω² = N sin θ …………….. (2)
Or m(R sin θ) ω² = N sin θ
Or mRω² = N
From (i) mg = mRω² cos θ
Or cos θ = \(\frac{\mathrm{g}}{\mathrm{R} \omega^2}\) ……………….. (3)
As |cos θ| ≤ 1, therefore, bead will remain at its lower most point for
\(\frac{\mathrm{g}}{\mathrm{R} \omega^2}\) ≤ 1 or w ≤ \(\sqrt{\frac{g}{R}}\)
When ω = \(\sqrt{\frac{2 g}{R}}\), from (iii),
cos θ = \(\frac{g}{R}\left(\frac{R}{2 g}\right)=\frac{1}{2}\)
∴ θ = 60°

Textual Examples

Question 1.
An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a constant rate of 100 ms^-2. What is the acceleration of the astronaut the instant after he is outside the spaceship ? (Assume that there are no nearby stars to exert gravitational force on him).
Answer:
Since there are no nearby starts to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero. By the first law of motion the acceleration of the astronaut is zero.

Question 2.
A bullet of mass 0.04 kg moving with a speed of 90 m s^-1 enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet ?
Answer:
The retardation ‘a’ of the bullet (assumed constant) is given by
a = \(\frac{-u^2}{2 s}=\frac{-90 \times 90}{2 \times 0.6}\) m s^-2 = -6750 m s^-2
The retarding force, by the second law of motion, is = 0.04 kg × 6750 ms^-2 = 270 N
The actual resistive force, and therefore, retardation of the bullet may not be uniform. The answer therefore, only indicates the average resistive force.

Question 3.
The motion of a particle of mass m is described by y = ut + \(\frac{1}{2}\)at². Find the force acting on the particle.
Answer:
We know, y = ut + \(\frac{1}{2}\) gt²
Now, υ = \(\frac{\mathrm{dy}}{\mathrm{dt}}\)u + gt
acceleration, a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = g
Then the force is given by F = \(\frac{\mathrm{dp}}{\mathrm{dt}}\) = ma
F = ma = mg

Question 4.
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s^-1. If the mass of the ball is 0.15 kg determine the impulse imparted to the ball. (Assume linerar motion of the ball) [A.P. Mar. 17]
Answer:
Change in momentum
= 0.15 × 12-(-0.15 × 12) = 3.6 N s
Impulse = 3.6 N s, in the direction from the batsman to the bowler.
This is an example where the force on the ball by the batsman and the time of contact of the ball and the bat are difficult to know, but the impulse is readily calculated.

Question 5.
Two identical billiard balls strike a rigid wall with the same speed but at different angles and get reflected without any change in speed, as shown in Fig. What is (i) the direction of the force on the wall due to each ball ? (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall ?

Answer:
An instinctive answer to (i) might be that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at 30° to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases. How to find the force on the wall ? The trick . is to consider the force on the wall ? The trick is to consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer (i). Let u be the speed of each ball before and after collision with the wall and m the mass of each bah. Choose the x and y axes as shown in the figure, and consider the change in momentum of the ball in each case :
Case (a) :
(P_x)_initial = mu
(P_x)_initial = 0
(P_x)_final = – mu
(P_y)_final = 0
Impulse is the change in momentum vector.
Therefore,
x-component of impulse = -2mu
y-component of impulse – 0
Impulse and force are in the same direction. Clearly, from above, the force on the ball due to the wall is normal to the wall, along the negative x-direction. Using Newton’s third law of motion, the force on the wall due to the ball is normal to the wall along the positive x-direction. The magnitude of force cannot be ascertained since the small time taken for the collision has not been specified in the problem. .
Case (b) :
(P_x)_initial = mu cos 30°
(P_y)_initial = – m u sin 30°
(P_x)_final = – mu cos 30°
(P_y)_final = -m u sin 30°
Note, while p_x changes sign after collision, p_y does not. Therefore,
x-component of impulse = -2 m u cos 30°
y-component of impulse 0
The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative x direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive x direction.
The ratio of the magnitudes of the impulses imparted to the balls in(a) and (b) is
2 mu / (2 m u cos 30°) = \(\frac{2}{\sqrt{3}}\) ≈ 1.2

Question 6.
See Fig. A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the mid point P of the rope, as show. What is the angle the rope makes with the vertical in equilibrium ? (Take g = 10 ms^-2). Neglect the mass of the rope.

Answer:
Figures (b) and (c) are known as free-body diagrams. Figure (b) is the free-body diagram of W and Fig. (c) is the free-body diagram of point P.
Consider the equilibrium of the weight W.
Clearly, T₂ = 6 × 10 = 60 N.
Consider the equilibrium of the point P under the action of three forces – the tensions T₁ and T₂, and the horizontal force 50 N. The horizontal and vertical components of the resultant force must vanish separately :
T₁ cos θ = T₂ = 60N
T₂ sin θ = T₂ = 50N
which gives that tan θ = \(\frac{5}{6}\) or
θ = tan^-1 (\(\frac{5}{6}\)) = 40°
Note the answer does not depend on the length of the rope (assumed massless) nor on the point at which the horizontal force is applied.

Question 7.
Determine the maximum acceleration of the train in which a box lying on the floor will remain stationary, given that the coefficient of static friction between the box and the train’s floor is 0.15.
Answer:
Since the acceleration of the box is due to the static friction,
ma = f_s ≤ μ_s N = μ_s m g
i.e. a ≤ μ_s g
∴ a_maximum = μ_s g = 0.15 × 10m s^-2
= 1.5 ms^-2

Question 8.
See Fig. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined unti at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?

Answer:
The forces acting on a block of mass …. at rest on an inclined plane are (i) the weight mg acting vertically downwards (ii) the normal force N of the plane on the block, and (iii) the static frictional force fs opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight mg along the two directions shown, we have
m g sin θ = f_s, m g cos θ = N
As θ increases, the self-adjusting frictional force f_s increases until at θ = θ_max, f_s achieves its maximum value, (f_s)max = μ_s N.
Therefore,
tan θ_max = μ_s, or θ_max = tan^-1 μ_s
When θ becomes just a little more than there is a small net force on the block and θ_nm begins to slide. Note that θ_max depends only on μ_s and is independent of the mass of the block
θ_max = 15°, μ_s = tan 15° = t = 0.27

Question 9.
What is the acceleration of the book and trolley system shown in the Fig., if the coefficient of kinetic friction between the trolley and the suriace is 0.04 what is the tension in the string. Take g 10 ms^-2). Neglect the mass of the string.

Answer:
As the string is inextensible, and the pully is smooth, the 3 kg block and the 20 kg trolley both have same magnitude of acceleration. Applying second law to motion of the block (Fig. (b).
30 – T = 3a
Apply the second law to motion of the trolley (Fig. (c))
T – f_k = 20a
Now f_k = μ_k N,
Here μ_k = 0.04
N = 20 × 10 = 200N
Thus the equation for the motion of the trolley is
T – 0.04 × 200 = 20 a or T – 8 = 20a
These equations give a = \(\frac{22}{23}\) m s^-2 = 0.96 m s^-2 and T = 27.1 N.

Question 10.
A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The co-efficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn ?
Answer:
On an unbanked road, frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular turn without slipping. If the speed is too large, or if the turn is too sharp (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by
v_max = \(\sqrt{\mu_s R_g}\) v₂ ≤ μ_s Rg
Now, R = 3m, g = 9.8 m s^-2, μ_s = 0.1.
That is, μ_s R g = 2.94 m² s^-2. v = 18 km/h = 5 m s^-1; i.e., = 25 v². The condition is not obeyed. The cyclist will slip while taking the circular turn.

Question 11.
A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the(a) optimum speed of the race-car to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping?
Answer:
On a banked road, the horizontal component of the normal force and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping. At the optimum speed, the normal reaction’s component is enough to provide the needed centripetal force, and the frictional force is not needed. The optimum speed v₀ is given by
v₀ = (R g tan θ)^1/2-
Here R = 300 m, θ = 15°, g = 9.8 m s^-2; we have
v₀ = 28.1 ms^-1
The maximum permissible speed v_max is given by
v_max = R g\(\left(\frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)^{1 / 2}\) = 38.1 m s^-1

Question 12.
See (Fig.) A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of 0.1 ms^-2. What is the action of the block on the floor (a) before and (b) after the floor yields ? Take g = 10 m s^-2. Identify the action- reaction pairs in the problem.

Answer:
(a) The block is at rest on the floor. Its free- body diagram shows two forces on the block, the force of gravitational attraction by the earth equal to 2 × 10 = 20 N; and the normal force R of the floor on the block. By the First Law, the net force on the block must be zero i.e., R = 20 N. Using third law the action of the block (i.e. the force exerted on the floor by the block) is equal to 20 N and directed vertically downwards.

(b)The system (block cylinder) accelerates downwards with 0.1 ms^-2. The free – body diagram of the system shows two forces on the system : the force of gravity due to the earth (270 N); and the normal force R’ by the floor. Note, the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system,
270 – R’ = 27 × 0.1 N
i.e, R’ = 267.3 N
By the third law, the action of the system on the floor is equal to 267.3 N vertically downward.
Action-reaction pairs
For (a) :
(i) the force of gravity (20N) on the block by the earth (say, action); the force of gravity on the earth by the block (reaction) equal to 20 N directed upwards (not shown in the figure).
(ii) the force on the floor by the block (action); the force on the block by the floor (reaction).

For (b):
(i) the force of gravity (270 N) on the system by the earth (say, action), the force of gravity on the earth by the system (reaction), equal to 270 N, directed upwards (not shown in the figure).

(ii) the force on the floor by the system (action); the force on the system by the floor (reaction). In addition, for (b), the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair.

The important thing to remember is that an action-reaction pair consists of mutual fortes which are always equal and opposite between two bodies. Two forces on the same body which happen to be. equal and opposite can never constitute an action-reaction pair. The force of gravity on the mass in (a) or (b) and the normal force on the mass by the floor are not action-reaction pairs. These forces happen to be equal and opposite for (a) since the mass is at rest. They are not so for case (b), as seen already. The weight of the system is 270 N, while the normal force R’ is 267.3 N.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 3rd Lesson Motion in a Straight Line Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 3rd Lesson Motion in a Straight Line

Very Short Answer Questions

Question 1.
The states of motion and rest are relative. Explain.
Answer:
Rest and motion are relative. They are not absolute. A body can be in the rest or in motion w.r.t. reference frame. A man in a moving train is a rest w.r. to a co-passenger, but he is in motion w.r.t a man on the ground.

Question 2.
How is average velocity different from instantaneous velocity? [Mar. 13]
Answer:
The average velocity does not give any details of the motion of the particle. It gives only the result of the motion. The instantaneous velocity defines how fast the particle moves at a particular instant of time.
In uniform motion, the instantaneous velocity is equal to the average velocity.

Question 3.
Give an example where the velocity of an object is zero but its acceleration is not zero.
Answer:
When the body is projected vertically upwards, at the highest point, its velocity is zero. But its acceleration (a = g) is not equal to zero.

Question 4.
A vehicle travels half the distance L with speed v₁ and the other half with speed v₂. What is the average speed ?
Answer:
Average speed = \(\frac{\text { Total length of the path }}{\text { Total time taken }}=\frac{\mathrm{L}}{\frac{\left(\frac{\mathrm{L}}{2}\right)}{\mathrm{V}_1}+\frac{\left(\frac{\mathrm{L}}{2}\right)}{\mathrm{V}_2}}=\frac{2 \mathrm{~V}_1 \mathrm{~V}_2}{\mathrm{~V}_1+\mathrm{V}_2}\)

Question 5.
A lift coming down is just about to reach the ground floor. Taking the ground floor as origin and positive direction upwards for all quantities, which one of the following is correct
a) x < 0, v < 0, a > 0, b) x > 0, v < 0, a < 0, c) x > 0, v < 0, a > 0, d)x > 0, V > 0, a > 0
Answer:
While lift is moving towards ground floor (origin), position x decreases, velocity decreases. Hence x < 0, v < 0. But a > 0. So(a) is correct option.

Question 6.
A uniformly moving cricket ball is hit with a bat for a very short time and is turned back. Show the variation of its acceleration with time taking the acceleration in the backward direction as positive.
Answer:

Question 7.
Give an example of one-dimensional motion where a particle moving along the positive x-direction comes to rest periodically and moves forward.
Answer:
When simple harmonic oscillator starts from left extreme position and comes to rest at that point periodically and moves forward.

Question 8.
An object falling through a fluid is observed to have an acceleration given by a = g – bv where g is the gravitational accelaration and b is a constant. After a long time it is observed to fall with a constant velocity. What would be the value of this constant velocity ?
Answer:
Given acceleration a = g – bv
\(\frac{\mathrm{dv}}{\mathrm{dt}}\) = g – bv [∵ a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\)]
If an object is moving through fluid with constant velocity, dv = 0
0 = g – bv ∴ v = \(\frac{\mathrm{g}}{\mathrm{b}}\)

Question 9.
If the trajectory of a body is parabolic in one frame, can it be parabolic in another frame that moves with a constant velocity with respect to the first frame ? If not, what can it be ?
Answer:
No. the trajectory is a vertical straight line.

Question 10.
A spring with one end attached to a mass and the other to a rigid support is stretched and released. When is the magnitude of acceleration a maximum ?
Answer:
The magnitude of acceleration is maximum at Extreme Positions.

Short Answer Questions

Question 1.
Can the equations of kinematics be used when the acceleration varies with time ? If not, what form would these equations take ?
Answer:
No, the equations of kinematics be used when the acceleration varies with time.
If an object moves along a straight line with uniform acceleration (a), equations of kinematics are

1. v = v₀ + at;
2. x = v₀t + \(\frac{1}{2}\) at²;
3. v² = v₀² + 2ax

Where x is displacement, v₀ is velocity at t = 0, v is velocity at time t, a is acceleration.
These are kinematic equations of rectilinear motion for constant acceleration.
If an object moves with non-uniform acceleration, the equations of motion are,

1. v = v₀ + at
2. x = x₀ + v₀t + \(\frac{1}{2}\)at²
3. v² = v₀² + 2a (x – x₀)

Question 2.
A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is v₁ and at time t = t is v₂. The average velocity of the particle in this time interval is (v₁ + v₂)/2. Is this correct ? Substantiate your answer.
Answer:
Consider a particle moving with uniform acceleration a.
At t = 0, the (initial) velocity = v₁
At t = t, the (final) velocity = v₂
time = t

∴ The given statement is true.

Question 3.
Can the velocity of an object be in a direction other than the direction of acceleration of the object ? If so, give an example.
Answer:
Yes, the velocity of an object be in a direction other than the direction of acceleration of the object.

Ex. : In the case of the upward motion of a projectile, the angle between velocity and acceleration is 180°. During its journey, the direction of velocity is in upwards and the direction of acceleration is in downwards.

Question 4.
A parachutist flying in an aeroplane jumps when it is at a height of 3 km above ground. He opens his parachute when he is about 1 km above ground. Describe his motion.
Answer:

1. A parachutist is jumping from an air-plane at a height of 3 km from the ground. Upto 1 km above the ground, his motion is like a freely falling body. He falls with constant acceleration 9.8 ms^-2.
2. At a height of 1 km above the ground, when he opened the parachute, the air drag opposes the force of gravity resultant on it. The acceleration of parachutist gradually decreases since velocity increases (a = g – bv) and becomes zero.
3. Further, the parachutist attains terminal speed (constant speed), where air drag (in upward direction) is equal to force of gravity (in downward).
4. Hopefully this terminal speed is slow enough, so he can touch the ground without much difficulty.

Question 5.
A bird holds a fruit in its beak and flies parallel to the ground. It lets go of the fruit at some height. Describe the trajectory of the fruit as it falls to the ground as seen by
(a) the bird
(b) a person on the ground.
Answer:
A bird holds a fruit in its beak and flies parallel to the ground.

It lets go of the fruit at some height, the trajectory of the fruit as it falls to the ground as seen by (a) the bird is a straight line (b) a person on the ground is a parabola.

Question 6.
A man runs across the roof of a tall building and jumps horizontally on to the (lower) roof of an adjacent building. If his speed is 9 ms^-1 and the horizontal distance between the buildings is 10 m and the height difference between the roofs is 9 m, will he be able to land on the next building ? (take g = 10 ms^-2).
Answer:
Difference of heights between two buildings h = 9 m; g = 10 ms^-2

Time of flight of a man t = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 9}{10}}\)
= 1.341 sec
Horizontal speed of a man u = 9 ms^-1
Horizontal distance travelled by a man d_m
= Horizontal speed × Time of flight
= u × t = 9 × 1.341 = 12.07 m
Given, the horizontal distance between the buildings d_b = 10 m
Since d_m > d_b, the man can able to land on the next building.

Question 7.
A ball is dropped from the roof of a tall building and simultaneously another ball is thrown horizontally with some velocity from the same roof. Which ball lands first? Explain your answer.
Answer:
Let height of the building = Displacement of ball = h

For first ball u = 0; S = h. a = g; t = t₁
Substituting these values in S = ut + \(\frac{1}{2}\) at²
⇒ h = 0 + \(\frac{1}{2}\) gt₁²
∴ t₁ = \(\sqrt{\frac{2 h}{g}}\) …………… (1)
For second ball, u_x = u (say) u_y = 0, a_Y = g, S_Y = h; t = t₂
Substituting these values in
S_Y = u_Yt + \(\frac{1}{2}\) a_Yt²
h = 0 + \(\frac{1}{2}\) gt₂²
∴ t₂ = \(\sqrt{\frac{2 h}{g}}\)
From equation (1) and equation (2), t₁ = t₂
∴ Two balls will reach the ground in same time.

Question 8.
A ball is dropped from a building and simultaneously another ball is projected upward with some velocity. Describe the change in relative velocities of the balls as a function of time.
Answer:
For first ball u = u₁; v = v₁; a = g; t = t

Substituting these values in equation
v = u + at, we get
v₁ = u₁ + gt₁
For second ball u = u<sub2; v = v₂; a = – g; t = t₂
Substituting these values in equation, v = u + at,
We get, v₂ = u₂ + gt₂ ………………. (2)
(1) – (2) ⇒ (v₁ – v₂) = (u₁ – u₂) + g(t₁ + t₂)
∴ (v₁ – v₂) – (u₁ – u₂) = g(t₁ + t₂)
[∵ u₁ = 0]
∴ (v₁ – v₂) – (0 – u₂) = g(t₁ + t₂)
∴ The change in final relative velocity and initial relative velocity of two balls = Function of time.

Question 9.
A typical raindrop is about 4 mm in diameter. If a raindrop falls from a cloud which is km above the ground, estimate its momentum when it hits the ground.
Answer:
Diameter of rain drop, D = 4 mm
Radius of ram drop r = 2 mm = 2 × 10^-3 m
Volume of the rain atop, V = \(\frac{4}{3}\) πr³ = \(\frac{4}{3}\) × \(\frac{22}{7}\) × (2 × 10^-3)³
Density of water drop = 10³ kg/m³
Mass of water drop M = Vd = \(\frac{4}{3}\) × \(\frac{22}{7}\) × 8 × 10^-9 × 10³
= 33.5 × 2 × 10^-6 kg
The height of raindrop falls from a cloud, h = 1 km = 1000 m
Velocity of raindrop just before touching the ground V = \(\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 1000}\) = 140 ms^-1
Momentum of raindrop when it hits the ground P = mV = 33.52 × 10^-6 × 140 = 469.28 × 10^-5
= 0.004692 kg ms^-1

Question 10.
Show that the maximum height reached by a projectile launched at an angle of 45° is one quarter of its range.
Answer:

Problems

Question 1.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the (a) magnitude of average velocity and (b) average speed of the man over the time interval 0 to 50 min.
Answer:
From home to market:
X₁ = 2.5 km; v₁ = 5 km h^-1;
t₁ = \(\frac{\mathrm{X_1}}{\mathrm{v_1}}=\frac{2.5}{5}=\frac{1}{2}\) h = 30 min.
From market to home:
X₂ = 2.5 km; v₂ = 7.5 km h^-1
t₂ = \(\frac{\mathrm{X_2}}{\mathrm{v_2}}=\frac{2.5}{7.5}=\frac{1}{3}\) h = 20 min.
a) Magnitude of average velocity
= \(\frac{\Delta \mathrm{X}}{\Delta \mathrm{t}}=\frac{\mathrm{X}_1-\mathrm{X}_2}{\mathrm{t}_1+\mathrm{t}_2}=\frac{2.5-2.5}{50}\) = 0
b) Average speed
= \(\frac{X_1+X_2}{t_1+t_2}=\frac{2.5-2.5}{(30+20) \min }=\frac{5 \mathrm{~km}}{50 \times \frac{h}{60}}\)
= 6 km h^-1

Question 2.
A car travels the first third of a distance with a speed of 10 kmph, the second third at 20 kmph and the last third at 60 kmph. What is its mean speed over the entire distance ? [T.S. Mar 16]
Answer:
v₁ = 10 km ph; v₂ = 20 km ph; v₁ = 60 km ph; v = ?
\(\frac{3}{v}=\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}\)
\(\frac{3}{v}=\frac{1}{10}+\frac{1}{20}+\frac{1}{60}, \frac{3}{v}=\frac{6+3+1}{60}=\frac{3}{v}=\frac{10}{60}\)
∴ v = 18 km ph

Question 3.
A bullet moving with a speed of 150 ms-1 strikes a tree and penetrates 3.5 cm before stopping. What is the magnitude of its retardation in the tree and the time taken for it to stop after striking the tree ?
Answer:
u = 150 m/s, s = 3.5 cm = 0.035 m, v = 0
v² – u² = 2as
0 – 150² = 2 × a × 0.035
a = \(\left|\frac{-150 \times 150}{2 \times 0.035}\right|\)
= -3.214 × 10⁵ m/s² = -3.214 × 10⁵ m/s²
Time = \(\frac{v-u}{a}=\frac{0-150}{3.214 \times 10^5}=\frac{-150}{3.214 \times 10^5}\)
= 4.67 × 10^-4 sec.

Question 4.
A motorist drives north for 30 min at 85 km/h and then stop for 15 min. He continues travelling north and covers 130 km in 2 hours. What is his total displacement and average velocity ?
Answer:
v₁ = 85 kmph, t = 35.0 min, S₂ = 130 km
S₁ = Displacement = \(\frac{85}{60}\) × 30 = 42.5 km
S₂ = 130 km
a) S = S₁ + S₂ = 42.5 + 130 = 172.50 km
b) Avg.velocity = \(\frac{S_1+S_2+S_3}{t_1+t_2+t_3}\)
= \(\frac{42.5+0+130}{\frac{30}{60}+\frac{15}{60}+2}\)
= \(\frac{172.5 \times 60}{165}\)
= 62.7 km/hr

Question 5.
A ball A is dropped from the top of a .building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what fraction of the height of the building did the collison occur ?
Answer:
Let height of the building =H
Let two balls collide at a height = h
For ball A, u = 0; V = V_A s = H – h; t = t; a = g
Substituting these values in s = ut + \(\frac{1}{2}\) at²
H – h = 0 + \(\frac{1}{2}\) gt²
H – h = \(\frac{1}{2}\) gt² …………. (1)
and V_A = gt ……………… (2)
For ball B, u = u; v = v_B; s = h; a = -g
Substituting these values in s = ut + \(\frac{1}{2}\) at²
⇒ h = ut – \(\frac{1}{2}\) gt² ………. (3)
and V_B = u – gt ……….. (4)
given V_A = 2V_B
gt = 2(u – gt)
u = \(\frac{3}{2}\) gt …………… (5)

Question 6.
Drops of water fall at regular intervals from the roof of abuilding of height 16 m. The first drop strikes the ground at the same moment as the fifth drop leaves the roof. Find the distances between successive drops.
Answer:
H = 16 m
Time taken by the first drop to touch the ground t = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 16}{9.8}=\sqrt{3.26}}\)
sec = 1.8 sec
Time interval between two successive drops
= \(\frac{t}{n-1}\) Where n = no. of drops
= \(\frac{1.8}{5-1}=\frac{1.8}{4}\) = 0.45 sec

For second drop h₂ = \(\frac{1}{2}\) gt²
= \(\frac{1}{2}\) × 9.8 × 1.35 × 1.35 = 8.93 m
d₁₂ = 16 – 8.93 = 7.06 = 7 m
For 3^rd drop h₃ = \(\frac{1}{2}\) × 9.8 × 0.90 × 0.90
= 3.97
d₂₃ = 8.93 – 3.97 = 4.961 = 5 m
For 4^th drop h₄ = \(\frac{1}{2}\) × 9.8 × 0.45 × 0.45
= 0.9922
d₃₄ = 3.97 – 0.9922
d₃₄ = 2.9778 = 3 m
Similarly for d₄₅ = 0.9922 – 0 = 0.9922 = 1 m

Question 7.
A hunter aims a gun at a monkey hanging from a tree some distance away. The monkey drops from the branch at the moment he fires the gun hoping to avoid the bullet. Explain why the monkey made a wrong move.
Answer:
Suppose a hunter aims the gun at a monkey hanging from a high tree branch some distance away d. The instant the monkey observes the flash of the gun it drops from the tree.
Here the time taken by the monkey from the tree to the ground is t₁ = \(\sqrt{\frac{2 h}{g}}\) ……………. (1)
The path of bullet from the gun is like a horizontal Projectile, vertical velocity u_y = 0, Let S = h. Let t₂ be the time taken by the bullet to reach the ground.
∴ S = ut + \(\frac{1}{2}\)a t₂²
S = 0 × t + \(\frac{1}{2}\)a t₂²
∴ h = \(\frac{1}{2}\)g t₂²
t₂ = \(\sqrt{\frac{2 h}{g}}\) …………….. (2)
It is observed from equation (1) and (2),
t₁ = t₂
Both bullet and monkey reach the ground simultaneously.
Hence the bullet hits the monkey. Therefore Monkey made a wrong move.

Question 8.
A food packet is dropped from an aeroplane, moving with a speed of 360 kmph in a horizontal direction, from a height of 500 m. Find (i) its time of descent (ii) the horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped.
Answer:
Velocity of aeroplane v = 360 kmph

v = 360 × \(\frac{5}{18}\) = 100 m/s
h = 500 m
i) Time of descent =
t = \(\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 500}{10}}\) = 10 sec
ii) Horizontal range R = u × \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\)
= 100 × 10 sec = 1000 sec

Question 9.
A ball is tossed from the window of a buliding with an initial velocity of 8 ms-1 at an angle of 20° below the horizontal. It strikes the ground 3 s later. From what height was the ball thrown ? How far from the base of the building does the ball strike the ground ?
Answer:
u = 8 m/s, θ = 20°, t = 35
a) Horizontal distance = (u cos θ) t = 8 cos 20° × 3 = 8 × 0.9397 × 3 = 22.6 m

b) Height h = (u sin θ)t + \(\frac{1}{2}\) gt²
= 8 sin 20° × 3 + \(\frac{1}{2}\) × 9.8 × 9
= 8.208 + 44.1 = 52.31 m

c) The ball is thrown from a height of 44.1 m 1
h₁ = (u sin θ)t₁ + \(\frac{1}{2}\) gt₁²
10 = (8 sin 20°)t₁ + \(\frac{1}{2}\) 9.8% t₁²
= 2.736 t₁ + 4.9 t₁²
⇒ 4.9 t₁² + 2.736 t₁ – 10 = 0
t₁ = \(\frac{-2.736 \pm \sqrt{(2.736)^2+4 \times 4.9 \times 10}}{2 \times 4.9}\)
⇒ t₁ = \(\frac{-2.736 \pm \sqrt{203.48}}{9.8}\)
t₁ = \(\frac{-2.736+14.265}{9.8}=\frac{11.5288}{9.8}\)
= 1.176 = 1.18 sec

Question 10.
Two balls are projected from the same point in directions 30° and 60° with respect to the horizontal. What is the ratio of their initial velocities if they
(a) attain the same height ?
(b) have the same range 1
Answer:
θ₁ = 30°, θ₂ = 60°
a) First ball maximum height H₁

Additional Problems

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:
a) a railway carriage moving without jerks between two stations.
b) a monkey sitting on top of a man cycling smoothly on a circular track.
c) a spinning cricket ball that turns sharply on hitting the ground.
d) a tumbling beaker that has slipped off the edge of a table.
Solution:
a) The carriage can be considered a point object because the distance between two stations is very large as compared to the size of the railway carriage.

b) The monkey can be considered as a point object if the cyclist describes a circular track of very large radius because in that case the distance covered by the cyclist is quite large as compared to the size of monkey. The monkey can not be considered as a point object. If the cyclist describes a circular track of small radius, because in that case the distance covered by the cyclist is not very large as compared to the size of the monkey.

c) The cricket ball can not be considered as a point object because the size of the spinning cricket ball is quite appreciable as compared to the distance through which the ball may turn on hitting the ground.

d) A beaker sleeping off the edge of a table cannot be considered as a point, object because the size of the beaker is not negligable as compared to the height of the table.

Question 2.
The position-time (x – t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in fig. Choose the correct entries in the brackets below :

a) (A/B) lives.closer to the school than (B/A).
b) (A/B) starts from the school earlier than (B/A).
c) (A/B) walks faster than (B/A).
d) A and B reach home at the (same/ different) time.
e) (A/B) overtakes (B/A) on the road (once/twice).
Solution:
a) As OP < OQ, A lives closer to the school than B.

b) For x = 0, t = 0 for A; while t has some finite value for B.
Therefore, A starts from the school earlier than B.

c) Since the velocity is equal to slope of x -1 graph in case of uniform motion and slope of x -1 graph for B is greater than for A, hence B walks faster than A.

d) Corresponding to points P and Q, the value of t from x – t graph for A and B is same, which can be checked by drawing lines through P and Q parallel to x axis. Thus both A and B reach home at the same time.

e) The x – t graph for A and B intersect each other only once. Since B starts from the school after- wards, therefore B overtakes A on the road once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5 km h^-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm and returns home by an auto with a speed of 25 km h^-1. Choose suitable scales and plot the x-t graph of her motion.
Answer:
Time taken in reaching office
= \(\frac{\text { Distance }}{\text { Speed }}=\frac{2.5}{5}\) = 0.5 hr.
Time taken in returning from office = \(\frac{2.5}{25}\)
= 0.1 hr = 6 min

It means the woman reaches the office at 9.30 am and returns home at 5.06 p.m. The x – t graph of this motion will be as shown in fig.

Question 4.
A man walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:
The effective distance travelled by drunkard in 8 steps = 5 – 3 = 2 m.

Therefore he take 32 steps to walk 8 meters. Now he will have to cover 5 mt more to reach the pit, for which he has to take only 5 forward steps.

Therefore he will have to take = 32 + 5 = 37 steps to move 13 mts. Thus he will fall into pit after taking 37 steps i.e., after 37 seconds from the start.

Question 5.
A jet airplane travelling at the speed of 500 km h^-1 ejects its products of combustion at the speed of 1500 km h^-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?
Answer:
Lt v_p be the velocity of the products w.r.to ground
Lt us consider the direction of motion of airplane to be positive direction of x-axis.
Here, speed of jet plane v_A = 500 km h^-1
Relative speed of products of combustion w.r.to jet plane v_ρA = -1500 km h^-1
Relative velocity of the products w.r.to air plane is
v_PA – v_P – v_A = -1500
(Or)
v_P = v_A – 1500 = 500 – 1500
= – 1000 km h^-1
Here -ve sign shows the direction of products of combustion is opposite to that of airplane. Thus the magnitude of relative velocity is 1000 km h^-1.

Question 6.
A car moving along a straight highway with speed of 126 km h^-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop ?
Answer:
Here, u = 126 km h^-1 = \(\frac{126 \times 1000}{60 \times 60}\) ms^-1
= 35 ms^-1; v = 0, S = 200 m, a = ?
and t = ?
We know v² = u² + 2as
0 = (35)² + 2 × a × 200 (Or)
a = \(\frac{-(35)^2}{2 \times 200}=\frac{-49}{16}\)
= -3.06 ms^-2
As v = u + at
0 = 35 + \(\left(\frac{-49}{16}\right) \mathrm{t}\) (Or)
t = \(\frac{35 \times 16}{49}=\frac{80}{7}\) = 11.43s

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s^-2. If after 50 s, the guard of B just moves past the driver of A, what was the original distance between them ?
Solution:
For train A; u = 72 km h^-1 = \(\frac{72 \times 1000}{60 \times 60}\)
= 20 m s^-2; t = 50s; a = 0, S = S_A;
As S = ut + \(\frac{1}{2}\) at²
∴ S_A = 20 × 50 + \(\frac{1}{2}\) × 0 × 50²
= 1000 m
For train B; u = 72 kms^-1 = 20 ms^-2;
a = 1 ms^-2, t = 50/S, S = s_-B
As, S = ut + \(\frac{1}{2}\) at²
∴ S_B = 20 × 50 + \(\frac{1}{2}\) × 1 × 50² = 2250 m
Taking the guard of the train B in the last compartment of the train B, it follows that original distance between two trains + length of train A + length of train B = S_B – S_A.
(Or) Original distance between the two trains
400 + 400 = 2250 – 1000 = 1250
(Or) Original distance between the two trains
= 1250 – 800 = 450 m

Question 8.
On a two-lane road, car A is travelling with a speed of 36 km h^-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?
Solution:
Velocity of car A = 36 km h^-1 = 10 ms^-1
Velocity of car B or C = 54 km h^-1 = 15 ms^-1
Relative velocity of B w.r.to
A = 15 – 10 = 5 ms^-1
Relative velocity of C w.r.to
A = 15 + 10 = 25 ms^-1
As, AB = AC = 1 km = 1000 m
Time available to B (Or) C for crossing
A = \(\frac{1000}{25}\) = 40s
If car B accelerates with acceleration a, to cross A before car C does, then
u = 5 ms^-1, t = 40s, s = 1000 m, a = ?
Using s = ut + \(\frac{1}{2}\) at²
1000 = 5 × 40 + \(\frac{1}{2}\) × a × 40² (Or)
1000-200 = 800 a (Or)
a = 1 m/s²

Question 9.
Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h^-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road ?
Solution:
Let v km h^-1 be the constant speed with which the buses ply between the towns A and B.
The relative velocity of the bus (for the motion A to B) w.r.to the cyclist C i.e., in the direction in which the cyclist is going = (v – 20) kmh^-1.
The relative velocity of the bus from B to A W r to the cyclist = (v + 20) kmh^-1.
The distance travelled by the bus in time T(min) = vT
As per question \(\frac{v T}{v-20}\) = 18 (Or) vT
= 18v – 18 × 20 ………………… (i)
and \(\frac{v T}{v+20}\) = 6 (Or) vT = 6v + 20 × 6 …………….. (ii)
Equations (i) and (ii) we get
18v- 18 × 20 = 6v + 20 × 6 (Or)
12v = 20 × 6 + 18 × 20 = 480
(Or) v = 40 kmh^-1
Putting this value of v in (i) we get 40
T = 18 × 40- 18 × 20 = 18 × 20
(Or) T = 18 × 20/40 = 9 min.

Question 10.
A player throws a ball upwards with an initial speed of 29.4 ms^-1.
a) What is the direction of acceleration during the upward motion of the ball ?
b) What are the velocity and accele-ration of the ball at the highest point of its motion ?
c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
d) To what height does the ball rise and after how long does the ball return to the player’s hands ?
(Take g = 9.8 ms^-2 and neglect air resistance).
Solution:
a) Since the ball is moving under the effect of gravity, the direction of acceration due the gravity is always vertically downwards.

b) At the highest point, the velocity of the ball becomes zero and acceleration is equal to the acceleration due to gravity = 9.8 ms^-2 in vertically downward direction.

c) When the highest point is chosen as the location for x = 0 and t = 0 and vertically downward direction to be positive direction of x- axis and upward direction as negative direction of x-axis.
During upward motion, sign of position is negative, sign of velocity is negative and sign of acceleration is positive.
During downward motion, sign of position is positive, sign of acceleration is also positive.

d) Let the t be time taken by the ball to reach the highest point where height from ground to be S.
Taking vertical upward motion of the ball,
we have
u = -29.4 m/s^-1, a = 9.8 m/s^-2,
v = 0, S = 5, t = 2
As v² – u² = 2as
0 – (29.4)² = 2 × 9.8 × s (Or)
S = \(\frac{-(29.4)^2}{2 \times 9.8}\) = -44.1 m
Here -ve sign shows that distance is covered in upward direction.
As v = u + at
∴ 0 = -29.4 + 9.8 × t (or) t = \(\frac{-29.4}{9.8}\) = 3s
It means time of ascent = 3s
When an objective move under the effect of gravity alone, the time of ascent is always equal to time of descent.
Therefore total time after which the ball returns the player’s hand = 3 + 3 = 6s

Question 11.
Read each statement below carefully and state with reasons and examples, if it is true or false :
A particle in one-dimensional motion
a) with zero speed at an instant may have non-zero acceleration at that instant
b) with zero speed may have non-zero velocity.
c) with constant speed must have zero acceleration.
d) with positive value of acceleration must be speeding up.
Solution:
a) True, when a body is thrown vertically upwards in the space. Then at the highest point, the body has zero speed but has downward acceleration equal to acceleration due to gravity.

b) False, because velocity is the speed of body in a given direction. When speed is zero, the magnitude of velocity of body is zero, hence velocity is zero.

c) True, when a particle is moving along a strait line with a constant speed, its velocity remains constant with time. Therefore, acceleration ( = Change in velocity/time) is zero.

d) The statement depends upon the choice of instant of time as origin, when the body is moving along a strait line with positive acceleration. The velocity of the body at on instant of time is v = u + at.

The given statement is not correct. If a is positive and μ is negative, at the instant of time taken as origin. Then for all times before the time for which v vanishes, there is slowing down of the particle i.e., The speed of the particle keeps on decreasing with time. It happens when body is projected vertically upwards. However the given statement is true if u is positive, at the instant of time as origin. When the body is moving along a straight line with positive acceleration. The velocity of body at any instant of time t is v = u + at.

The given statement is not correct if a is positive and u is negative, at the instant of time taken as origin. Then for all times before the time for which v vanishes, there is slowing down of the particle i.e., the speed of particle keeps on decreasing with time. It happens when body is projected vertically upwards. However, the given statement is true if u is positive and a is positive, at the instant of time taken as origin. It is so when the body is falling vertically upwards.

Question 12.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Solution:
Taking vertical downward motion of ball from a height 90 m We have
u = 0, a = 10 m/s², S = 90 m, t = ?, v = ?
t = \(\sqrt{\frac{25}{a}}=\sqrt{\frac{2 \times 90}{10}}=3 \sqrt{25}\) = 4.245
V = \(\sqrt{2 a \mathrm{~s}}=\sqrt{2 \times 10 \times 30}=30 \sqrt{2} \mathrm{~m} / \mathrm{s}\)

Rebound velocity of ball,
u¹ =\(\frac{9}{10} v=\frac{9}{10} \times 30 \sqrt{2}=27 \sqrt{2}\) m/g
Time to reach the highest point is
t¹ = \(\frac{u^1}{a}=\frac{27 \sqrt{2}}{10}=2.7 \sqrt{2}\) = 3.81 S
Total time = t + t¹ = 4.24 + 3.81 = 8.05 S
The ball will take further 3.1 S to fall back to floor, where its velocity before striking the floor = 2.7\(\sqrt{2}\) m/s.
Time to reach the highest point is,
t¹ = \(\frac{u^1}{a}=\frac{27 \sqrt{2}}{10}=2.7 \sqrt{2}\) = 3.81 S
Total time = t + t¹ = 4.24 + 3.81 = 8.05 S
The ball will take further 3.81 S fall back to floor, where its velocity before striking the
floor = 2.7\(\sqrt{2}\) m/s.
Velocity of ball after striking the floor
= \(\frac{9}{10}\) \(\sqrt{2}\) = 24.3 \(\sqrt{2}\) m/s.
Total time elapsed before upward motion of ball.
= 8.05 + 3.81 = 11.86 S
Thus the speed – time graph of this motion will be as shown in fig.

Question 13.
Explain clearly, with examples, the distinction between :
a) magnitude of displacement (sometimes called distance) over an interval of time and the total length of path covered by a particle over the same interval:
b) magnitude of average velocity over an interval of time and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
Answer:
a) Magnitude of displacement of a particle in motion for a given time is shortest distance between the initial and final position of the particle in that time, where as the total length of the path covered by particle is actual path traversed by the particle in the given time. If particle goes from A to B and B to C in time t as shown in fig. Then magnitude of displacement = Distance AC.
Total path length = Distance AB + Distance AC
From the above we note that total path length (AB + AC) is greater than magnitude of displacement (AC).

If there is motion of the particle in one dimension i.e., along a straight line, then magnitude of displacement becomes equal to totalpath length transversed by the particle in the given time.

b) Magnitude of average velocity
= \(\frac{\text { Magnitude of displacement }}{\text { Time interval }}\)
= \(\frac{A C}{t}\) and average speed
= \(\frac{\text { Total path length }}{\text { Time interval }}=\frac{(A B+B C)}{t}\)
As, (AB + BC) AC, So average speed is greates than the magnitude of average velocity. If the particle is moving along a straight line, then in a given time the magnitude of displacement is equal to total path length transvered by particle in that time, so average speed is equal to magnitude of average velocity.

Question 14.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the
a) magnitude of average velocity and
b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 m in, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Solution:
Time taken by man to go from his home to market t₁ = \(\frac{\text { Distance }}{\text { Speed }}=\frac{2.5}{5}=\frac{1}{2}\) h
Time taken by man to go from market to his home t₂ = \(\frac{2.5}{7.5}=\frac{1}{3}\) h
Total time taken = t₁ + t₂ = \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
h = 50 min.
i) 0 to 30 min
a) Average velocity
= \(\frac{\text { Displacement }}{\text { Time }}=\frac{2.5}{1 / 2}\) = 5 km/h
b) Average speed
= \(\frac{\text { Displacement }}{\text { Time }}=\frac{2.5}{1 / 2}\) = 5 km/h

ii) 0 to 50 min
Total distance travelled
= 2.5 + 2.5 = 5 km
Total displacement = zero
a) Average velocity = \(\frac{\text { Displacement }}{\text { Time }}\) = 0
b) Average speed = \(\frac{\text { Distance }}{\text { Time }}=\frac{5}{5 / 6}\) = 6 km/h

iii) 0 to 40 min
Distance moved in 30 min (from home to market) = 2.5 km
Distance moved in 10 min (from market to home) with speed 7.5 km/h = 7.5 × \(\frac{10}{60}\)
= 12.5 km
So displacement = 2.5 – 1.25 = 1.25 km
Distance travelled = 2.5 + 1.25 = 3.75 km
a) Avg velocity = \(\frac{1.25}{(40 / 60)}\) = 1.875 km/h
b) Avg speed = \(\frac{3.75}{(40 / 60)}=\) = 5.625 km/h

Question 15.
In Exercises 3.13and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider in ‘antaneous speed and magnitude of velocity. The instantaneous speed is alway equal to the magnitude of instantaneous velocity. Why ?
Solution:
Instaneous speed (v_ins) of the particle at an instant is the first derivative of the distance w.r.to time at that instant of time
i.e., v_ins = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)
Since in instantaneous speed we take only a small interval of time (dt) during which direction of motion of a body is not supposed to change, here there is no difference between total path length and magnitude of displacement for small interval of time dt. Hence instantaneous speed is always equal to magnitude of instantaneous velocity.

Question 16.
Look at the graphs (a) to (d) (fig.) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Solution:
a) This graph does not represent one dimensional motion because, at the given instant of time, the particle will have two positions, which is not possible in one dimensional motion.

b) This graph does not represent one dimensional motion because, at the given instant of time, particle will have velocity in positive as well as in negative direction which is not possible in one dimensional motion.

c) It also does not represent one dimensional motion, because this graph tells that the particle can have the negative speed but the speed of the particle can never be negative.

d) It also does not represent one dimensional motion, because this graph tells that the total path length decreases after certain time but total path length of a moving particle can never decrease with time.

Question 17.
Figure shows the x – t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves In a straight line for t < 0 and on a parabolic path for t > 0 ? If not, suggest a suitable physical context for this graph.

Solution:
No, because the x – t graph does not represent the trajectory of the path followed by a particle.

From the graph, it is noted that at t = 0, x = 0 context. The above graph can represent the motion of a body falling freely from a tower uncles gravity.

Question 18.
A police van moving on a highway with a speed of 30 km h^-1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 ms^-1 with what speed does the bullet hit the thief’s car ? (Note : Obtain that speed which is relevant for damaging the thief’s car),
Solution:
Muzzle speed of bullet, υ_B = 150 m/s = 540 km h^-1
Speed of police van, υ_p = 30 km/h
Speed of theif car, υ_T = 192 km/h
Since the bullet is sharing the velocity of the police van, its effective velocity is
υ_B = υ_B + υ_P = 540 + 30 = 570 km/h
The speed of bullet w.r.to the theifs car moving in the same direction.
v_BT = v_B – v_T = 570 – 192 = 378 378 km/h
= \(\frac{378 \times 1000}{60 \times 60}\) = 105 m/s

Question 19.
Suggest a suitable physical situation for each of the following graphs (Fig.):

Solution:
In fig(a) : The x – t graph shows that intially x is zero i.e, at rest, then it increases with time, attains a constant value and again reduces to zero with time, then it increases in opposite direction till it again attains a constant value i.e., comes to rest. The similar In fig(b) : The velocity changes sign again and again with passage of time and every time some similar speed is lost. The similar physical situation arises when a ball is thrown up with some velocity, returns back and falls freely on striking the floor, it rebounds with reduced speed each it strikes against the floor.

Question 20.
Figure gives the x – t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, -1.2 s.

Solution:
In the S.H.M. acceleration a = ω²x, where co (i.e., angular frequency) is constant.

i) At time t = 0.35, x is negative, the slope of x -1 graph plot is negative, hence position and velocity are negative, Since a =ω²x, hence, acceleration is positive.
ii) At time t = 1.25, x is positive, the slope of x – t plot is also positive hence position and velocity are positive. Since a = -ω²x, hence acceleration is negative.
iii) At t = 1.2S, x is negative, the slope of x – t plot is also negative. But since both x and t are negative here, hence velocity is positive. Finally acceleration ‘a’ is also positive.

Question 21.
Figure gives the x – t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest and in which is it the least ? Give the sign of average velocity for each interval.

Solution:
We know that average speed in a small interval of time is equal to slope of x -1 graph in that interval of time. The avg speed is the greatest in the interval 3 because slope is greatest and the average speed is least in interval 2 because slope is last three.

Question 22.
Figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?

Solution:
We know that average acceleration in a small interval of time is equal to slope of velocity-time graph in that interval. As the slope of v – t graph is maximum in interval 2 as compared to other intervals 1 and 3, hence the magnitude of average acceleration is greatest in interval 2.
The average speed is greatest in interval 3 for obvious reasons.
In interval 1, the speed of v – t graph is positive. Hence acceleration a is positive. The speed u is positive in this interval due to obvious reasons.

In interval 2, the slope of v – t graph is negative, hence acceleration a is negative. The speed u is positive in this interval due to obvious reasons.

In interval 3, the v – t graph is parallel to time axis, therefore acceleration a is zero in this interval but u is positive due to obvious reasons. At points A, B, C and D the v – t graph is parallel to time axis. Therefore acceleration is zero at ail the four points.

Question 23.
A three-wheeler starts from rest, accelerates uniformly with 1 ms^-2 on a straight road for 10 s and then moves with uniform velocity. Plot the distance covered by the vehicle during the n^th second (n = 1, 2, 3 ) versus n. What do you expect this plot to be during accelerated motion : a straight line or a parabola ?
Solution:
Here u = 0, a = 1 m/s²
Distance covered in n th second is
D_n = u + \(\frac{a}{2}\)(2n-1) = 0 + \(\frac{1}{2}\) (2n – 1) = 0.5 (2n – 1)
Putting n = 1, 2, 3, …………… we can find the value of D_n. The various values of n and corresponding values of D_n are shown below :

On plotting a graph between D_n and n, we get a st. line AB as shown in fig. From (1) D_n∝_n so the graph is a straight line. After 10S the graph is straight line BC parallel to time axis.

Question 24.
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 ms^-1. How much time does the ball take to return to his hands ? If the lift starts moving up with a uniform speed of 5 ms^-1 and the boy again throws the ball up with the maximum speed he can how long does the ball take to return to his hands ?
Solution:
Taking vertical upward direction as the positive direction of x-axis.
When lift is stationary, consider the motion of the ball going vertically upwards and coming down to the hands of the body. We have
u = 49 m/s, a = 9.8 m/s², t = ? x – x₀ = S = 0
As S = ut + \(\frac{1}{2}\) at²
0 = 49 t + \(\frac{1}{2}\) (-9.8)t² (Or) 49t = 4.9 t² (Or)
t = 49/4.9 = 10 sec
When lift starts moving with uniform speed. As the lift starts moving upwards with uniform speed of 5 m/s, there is no change in the relative velocity of the ball w.r.to the boy which remains 49 m/s. Hence even in the case, the ball will return to the boys hand after 10 sec.

Question 25.
On a long horizontally moving belt a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the
a) speed of the child running in the direction of motion of the belt ?
b) speed of the child running opposite to the direction of motion of the belt ?
c) time taken by the child in (a) and (b) ?
Which of the answer alter if motion is viewed by one of the parents ?

Solution:
Let us consider left to right to be the positive direction of x-axis.
a) Here, the velocity of belt υ_B = + 4 km/h; Speed of child w.r. to belt υ_C
= + 9 Km/h = \(\frac{5}{2}\) m/s
Speed of the child w.r.to stationary observer,
υ_C¹ = υ_C + υ_B = 9 + 4 = 13 km/h

b) Here, υ_B = + 4 km/h, υ_C = -9 km/h
Speed of the child w.r.to stationary observer
υ_C¹ = υ_C + υ_B = -9 + 4 = -5 km/h
Here negative sign shows that the child will appears to run in a opposite to the direction f motion of the belt.

c) Distance between the parents S = 540 m Since parents and child are located on the same belt, the speed of the child as observed by stationary observer in either direction (Either from mother to father (or) from father to mother) will be 9 km/h.
Time taken by child in case (a) and (b) is
t = \(\frac{50}{(5 / 2)}\) = 20 S
If motion is observed by one of the parents, answer to case (a) (Or) case (b) will be altered.
It is so because speed of child w.r. to either of mother (or) father is 9 km/h. But answer (c) remains unaltered due to the fact that parents and child are on the same belt and such all are equal affected by the motion of the belt.

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 ms^-1 and 30 ms^-1. Verify that the graph shown in Fig. correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 ms^-2. Give the equations for the linear and curved parts of the plot.

Solution:
Time vertical upward motion of the first tone for time t, we have
x₀ = 200 m, u = 15 m/s, a = -10 m/s², t = t₁, x = x₁
As x = x₀ + ut + \(\frac{1}{2}\) at²
x₁ = 200 + 15 t + \(\frac{1}{2}\) (-10)t² (Or)
x₁ = 200 + 15 t – 5 t² ……………….. (i)
Taking vertical upward motion of the second stone for time t,
We have
x₀ = 200 m, u = 30 m/s^-1, a = -10 m/s^-2, t = t₁, x = x₂
Then x₂ = 200 + 30 t – \(\frac{1}{2}\) × 10 t²
= 200 + 30 t – 5t²
When the first stone hits the ground x₁ = 0,
So t² – 3t – 40 = 0
(Or) (t – 8) (t + 5) = 0 ……………….. (ii)
∴ Either t = 8 S (Or) – 5S
Since t = 0 corresponds to the instant, when the stone was projected. Hence negative time has no meaning in this case. 50 t = 8S. When the second stone hits the ground, x₂ = 0.
0 = 200 + 30 t – 5t² (Or) t² – 6t – 40 = 0 (Or) (t – 10) (t + 4) = 0
Therefore, either t = 10 s (Or) t = -4s
Since t = -4s is meaningless, So t = 10s
Relative position of second stone w.r.to first is
= x₂ – x₁ = 15 t …………………….. (ii)
From (i) and (ii)
Since (x₂ – x₁) and t are linearly related, therefore the graph is a straight line till t = 8s For maximum separation t = 8 S, so maximum separation = 15 × 8 = 120 m After 8 seconds only second tone would be in motion for 2 seconds, so the graph is in accordance with the aquadratic equation x₂ = 200 + 30t – 5t² for the interval of time 8 seconds to 10 seconds.

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown in Fig. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2s to 6s.

What is the average speed of the particle over the intervals in (a) and (b) ?
Solution:
a) Distance travelled by the particle between 0 to 10s will be
= Area of ∆ OAB, whose base is 10s and height is 12 m/s
= \(\frac{1}{2}\) × 10 × 12 = 60 m 60
Average speed = \(\frac{60}{10}\) = 6mS^-1

b) Let s₁ and s₂ be the distances covered by the particle in the time interval t₁ = 2s to 5s and t₂ = 5s to 6s, then total distance covered in time interval t = 2s to 6s will be s = s₁ + s₂ ……………….. (i)

To find s₁: let us consider u₁ is the velocity of particle after 2 seconds and a1 is the acceleration of the particle during the time interval zero to 5 seconds.
Then u₁ = 0, v = 12 m/s,
a = a , a₁ and t = 5s
We have a₁ = \(\frac{v-u}{t}=\frac{12-0}{5}=\frac{12}{5}\)
= 2.4 m/s²
∴ u₁ = υ + a₁t = 0 + 2.4 × 2 = 4.8 m/s^-1 Thus for the distance travelled by particle in 3 seconds (i.e, time interval 2s to 5s), we have
u₁ = 4.8 m/s, t₁ = 3s, a₁ = 2.4 m/s², s₁ = ?
As s₁ = u₁t₁ + \(\frac{1}{2}\) a₁t₁²
S₁ = 4.8 + 3 × \(\frac{1}{2}\) × 2.4 × 32 = 25.2 m

To find s₂ : let a₂ be the acceleration of the particle during the motion t = 5s to t = 10s
We have a₂ = \(\frac{0-12}{10-5}\) = -2.4 m/s²
Taking motion of the particle in time interval t = 5s to t = 6s We have
u₁ = 12 m/s^-1, a₂ = -2.4 m/s², t₂ = 1s, s₂ = ?
As s₂ = u₂t + \(\frac{1}{2}\) a₂t₂²
s₂ = 12 × 1 + \(\frac{1}{2}\) (-2.4) 1² = 10.8 m
∴ Total distance travelled s = 25.2 + 10.8
= 36m
Average velocity = \(\frac{36}{6-2}=\frac{36}{4}\) = 9 m/s

Question 28.
The velocity-time graph of a particle in onedimensional motion is shown in Fig.

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂ :
a) x(t₂) = x(t₁) + v(t₁) (t₂ – t₁) + (\(\frac{1}{2}\)) a(t₂ – t₁)²
b) v(t₂) = v(t₁) + a(t₂ – t₁)
c) v_average = (x(t₂) – x(t₁)) / (t₂ – t₁)
d) v_average = (v(t₂) – v(t₁)) / (t₂ – t₁)
e) x(t₂) = x(t₂) + V_average (t₂ – t₂) + (\(\frac{1}{2}\)) a_average (t₂ – t₁)²
f) x(t₂) = x(t₁) = area under the y – t curve bounded by the t-axis and the dotted line shown.
Solution:
From the graph we note that the slope is not constant and is not uniform, hence the relations (iii), (iv), (v) are correct.

Textual Examples

Question 1.
A car is moving along a straight line. Say OP Fig. 3.1. It moves from O to P in 1 8s and returns from P to Q in 6.0s. What are the average velocity and average speed of the car in going (a) from O to P? arid (b) from O to P and back to Q?
Solution:

Thus, in this case the average speed is equal to the magnitude of the average velocity.

b) In this case,

Question 2.
The position of an object moving along x-axis is given by x = a + bt² where a = 8.5 m, b = 2.5 ms^-2 and t is measured in seconds. What is its velocity at t = 0s and t = 2-.0s. What is the average velocity between t = 2.0s and t = 4.0s ?
Solution:
In notation of differential calculus, the velocity is
υ = \(\frac{d x}{d t}=\frac{d}{d t}\)(a + bt²) = 2b t = 5.0 t ms^-1
At t = 0s, υ = 0 ms^-1 and at t = 2.0s, υ = 10 ms^-1
Average velocity = \(\frac{x(4.0)-x(2.0)}{4.0-2.0}\)
= \(\frac{a+16 b-a-4 b}{2.0}\) = 6.0 × b
= 6.0 × 2.5 = 15 ms^-1

Question 3.
Obtain equations of motion for cons-tant acceleration using method of calculus.
Answer:
By definition a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\)
dυ = a dt
Integrating both sides
\(\int_{v_0}^v d v=\int_0^t a d t\)
= \(a \int_0^t d t\) (a is constant)
υ – υ₀ = at
υ = υ₀ + at
Further υ = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)
dx = υ dt
Integrating both sides
\(\int_{x_0}^x d x=\int_0^t v d t=\int_0^t\left(v_0+a t\right) d t\)
x – x₀ = υ₀ t + \(\frac{1}{2}\) a t²
x = x₀ + υ₀ t + \(\frac{1}{2}\) a t²
We can write
a = \(\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}=v \frac{d v}{d x}\)
or, υ dυ = a dx
Integrating both sides,
\(\int_{v_0}^v v d v=\int_{x_0}^x a d x\)
\(\frac{v^2-v_0^2}{2}\) = a(x – x₀)
υ² = υ₀² + 2a(x – x₀)
The advantage of this method is that it can be used for motion with non-uniform acceleration also.
Now, we shall use these equations to some important cases.

Question 4.
A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. [T.S. A.P. Mar. 15]
(a) How high will the ball rise ? and
(b) how long will it be before the ball hits the ground ? Take g = 10 ms^-2. (Actual valqp is 9.8 ms^-2)
Solution:
a) Let us take the y—axis in the vertically upward direction with zero at the vertically upward direction with zero at the ground, as shown in fig. 3.13.
Now υ₀ = + 20 ms^-1
a = -g = -10 ms^-2,
υ = 0 ms^-1
If the ball rises to height y from the point of launch, then using the equation
υ² + υ₀² + 2a(y – y₀)
we get
0 = (20)² + 2(-10) (y – y₀)
Solving, we get, (y – y₀) = 20 m

b) We can solve this part of the problem in two ways. Note carefully the methods used.

First Method : In the first method, we split the path in two parts : the upward motion (A to B) and the downward motion (B to C) and calculate the corresponding time taken t₁ and t₂. Since the velocity at B is zero, we have :
υ = υ₀ + at
0 = 20 – 10 t₁
t₁ = 2s
Or,
This is the time in going from A to B. From B, or the point of the maximum height, the ball falls freely under the acceleration due to gravity. The ball is moving in negative y direction. We use equation
y = y₀ + υ₀t + \(\frac{1}{2}\) at²
We have, y₀ = 45 m, y = 0, υ₀ = 0,
a = -g = – 10 ms^-2
0 = 45 + (\(\frac{1}{2}\)) (-10) t₂²
Solving, we get t₂ = 3s
Therefore, the total time taken by the ball
before it hits the ground = t₁ + t₂ = 2s + 3s = 5s.

Second method : The total time taken can also be calculated by nothing the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation.
y = y₀ + υ₀t + \(\frac{1}{2}\) at²
Now y₀ = 25 m, y = 0 m
υ₀ = 20 ms^-1, a = -10 ms^-2, t = ?
0 = 25 + 20t + (\(\frac{1}{2}\))(-10)t²
Or, 5t² -20t – 25 = 0
Solving this quadratic equation for t, we get t = 5s

Question 5.
Free-fall : Discuss the motion of an object under fall. Neglect air resistance.
Solution:
An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the earth’s radius, g can be taken to be constant, equal to 9.8 ms^-2. Free fall is thus a case of motion with uniform acceleration.

We assume that the motion is in y-direction, more correctly in -y-direction because we choose upward direction as positive Since that acceleration due to gravity is always downward, it is in the negative direction and we have
a = – g = -9.8 ms^-2
The object is released from rest at y = 0. Therefore, υ₀ = 0 and the equations of motion become :


Motion of an object under free fall.
a) Variation of acceleration with time.
b) Variation of velocity with time.
c) Variation of distance with time.

Question 6.
Galileo’s law of odd numbers : The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity (namely 1:3:5:7 )ⁿ Prove it.
Solution:
We have y = –\(\frac{1}{2}\) gt² Using this equation, we can calculate the position of the object after different time intervals, 0, τ, τ2, τ3 which are given in second column of table 3.2. If we take (-1/2) gτ² as y₀ – the position coordinate after first time interval τ, then third column gives the positions in the unit of y₀. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3 : 5 : 7 : 9 :11 as shown in the last column.

Question 7.
Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (-υ₀) and the braking capacity or deceleration, – a that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of υ₀ and a.
Solution:
Let the distance travelled by the vehicle before it stops be ds. Then, using equation of motion υ² = υ₀² + 2 ax and noting that υ = 0, we have the stopping distance
d_s = \(\frac{-v_0^2}{2 a}\)
Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).

For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15,20 and 25 m/s which are nearly consistent with the above formula.

Question 8.
Reaction time : When a situation demands our immediate action. It takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then, the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual.

You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Fig. 3.15). After you catch it, find the distance d travelled by the ruler. In a particular case, d was found to be 21.0 cm. Estimate reaction time.

Solution:
The ruler drops under free fall. Therefore υ₀ = 0 and a = -g = -9.8 ms^-2. The distance travelled d and the reaction time t, are related by d = –\(\frac{1}{2}\) gt_r² Or t_r = \(\sqrt{\frac{2 \mathrm{~d}}{\mathrm{~g}}}\)s.
Given d = 21.0 cm and g = 0.8 ms^-2 the reaction time is
t_r = \(\sqrt{\frac{2 \times 0.21}{9.8}}\) s ≅ 0.2 s

Question 9.
Two parallel rail tracks run north- south. Train A moves north with a speed of 54 km h^-1 and train B moves south with a speed of 90 kmh^-1. What is the
a) velocity of B with respect to A ?
b) velocity of ground with respect to B ?
c) velocity of a monkey running on the roof of the train A against. Its motion (with a velocity of 18 kmh^-1 with respect to the train A) as observed by a man standing on the ground ?
Solution:
Choose the positive direction of x-axis to be from south to north. Then,
υ_A = + 54 km h^-1 = 15 ms^-1
υ_B = – 90 km h^-1 = -25 ms^-1
Relative velocity of B with respect to A = υ_A – υ_B = – 40 ms^-1, i.e., the train B appears to A to move with a speed of 40 ms^-1 from north to souch.
Relative velocity of ground with respect to B = 0 – υ_B = 25 ms^-1
In.(c), let the velocity of the monkey with respect to ground be υ_M. Relative velocity of the monkey with respect to A,
υ_MA = υ_M – υ_A = – 18 km h^– = – 5 mh^-1
Therefore, υ_M = (1 5 – 5)mh^-1 = 10 mh^-1.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements

Very Short Answer Questions

Question 1.
Distinguish between accuracy and precision. [A.P. Mar. 16, 15; T.S. Mar. 15, 13]
Answer:
Accuracy

1. The accuracy of measurement of any physical quantity made by any measuring instrument is a measure of how close the measured value is to the true value of the quantity.
2. The accuracy depends on errors.

Precision

1. The precision of the measuring instrument denotes upto what limit (or) resolution the quantity can be measured with the given instrument.
2. The precision does not depend on errors.

Question 2.
What are the different types of errors that can occur in a measurement?
Answer:
Mainly there are three types of errors.

1. Systematic errors
2. Random errors
3. Gross errors.

Question 3.
How can systematic errors be minimised or eliminated ? [Mar. 14]
Answer:
Systematic errors can be minimised by improving experimental techniques, selecting better instruments and removing personal bias as far as possible. For a given set up, these errors may be estimated to a certain extent and the necessary corrections may be applied to the readings.

Question 4.
Illustrate how the result of a measurement is to be reported indicating the error involved. [T.S., A.P. – Mar. 17]
Answer:
To measure any physical quantity, we compare it with a standard (unit) of that quantity. No measurement is perfect as the errors involved in the process cannot be removed completely. Hence inspite of our best efforts, the measured value is always some what different from its actual value (or) true value.

Question 5.
What are significant figures and what do they represent when reporting the result of a measurement ? [Mar. 13]
Answer:
The digits of a number that are definitely known plus one more digit that is estimated are called significant digits (or) significant figures.
Example : Time period of a simple pendulum is 1.62, the digits 1 and 6 are reliable while the digit 2 is uncertain. The measured value has three significant figures.

Question 6.
Distinguish between fundamental units and derived units. [T.S. – Mar. 16]
Answer:

1. Units of fundamental quantities are called fundamental units. Fundamental units can neither derived from one another, nor can they be resolved into other units.
2. Units of derived quantities are called derived units.

Question 7.
Why do we have different units for the same physical quantity ?
Answer:
We have different systems like C.G.S system, M.K.S system, F.P.S system and S.l system. Hence we have different physical units for the same physical quantity.

Question 8.
What is dimensional analysis ?
Answer:
Dimensional analysis is the representation of derived physical quantities in terms of units of fundamental quantities.

With the help of dimensional analysis to check the correctness of the equation, convert one system of units into other system and derive certain equations relating physical quantities.

Question 9.
How many orders of magnitude greater is the radius of the atom as compared to that of the nucleus ?
Answer:
Size of atomic nucleus = 10^-14 m
Size of atom = 10^-10 m
Hence size of atom is 1CT4 m greater the size of the nucleus.

Question 10.
Express unified atomic mass unit in kg.
Answer:
1 unified atomic mass unit = \(\frac{1}{12}\) of the mass of carbon – 12 atom.
1 a.m.u = 1.66 × 10^-27 kg

Short Answer Questions

Question 1.
The vernier scale of an instrument has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, then using this instrument what would be the minimum inaccuracy in the measurement of distance ?
Answer:
Value of each main scale division = 0.5 mm
No. of vernier scale divisions = 50
In vernier callipers, L.C = \(\frac{S}{N}=\frac{\text { Value of one main scale division }}{\text { No. of vernier scale divisions }}\)
L.C. = \(\frac{0.5}{50}\) = 0.01 m.m
The minimum inaccuracy in the measurement of distance is 0.01 m.m. (Or)
Value of 1 MSD = 0.5 m.m
NV.S.D = (N – 1) M.S.D
50 V.S.D – 49 M.S.D
1 V.S.D = \(\) M.S.D. = \(\frac{49}{50}\) × 0.5
L.C = 1 M.S.D – 1 V.S.D
(0.5 – \(\frac{49}{50}\) × 0.5) = (1 – \(\frac{49}{50}\))0.5 = \(\frac{1}{50}\) × 0.5
L.C. = 0.01 m.
∴ Minimum inaccuracy in the measurement of distance is 0.01 m.m.

Question 2.
In a system of units, the unit of force is 100 IM. unit of length is 10 m and the unit of time is 100s. What is the unit of mass in this system ?
Answer:
Force (F) = 100 N; Length (L) = 10 m; Time (T) = 100 s F = ma
mass (m) = \(\frac{F}{a}=\frac{F}{L T^{-2}}\); m = \(\frac{\mathrm{FT}^2}{\mathrm{~L}}=\frac{100 \times(100)^2}{10}\) = 10⁵ kg

Question 3.
The distance of a galaxy from Earth is of the order of 10²⁵ m. Calculate the order of magnitude of the time taken by light to reach us from the galaxy.
Answer:
Distance of galaxy from earth = 10²⁵ m
Velocity of light (C) = 3 × 10⁸ m/s
Time taken by the light (t) = \(\frac{\text { Distance of galaxy from earth }}{\text { velocity of light }}\)
t = \(\frac{\mathrm{d}}{\mathrm{c}}\) ⇒ t = \(\frac{10^{25}}{3 \times 10^8}\) = 0.33 × 10¹⁷ s ⇒ t = 3.3 × 10¹⁶ sec.

Question 4.
The Earth-Moon distance is about 60 Earth radius. What will be the approximate diameter of the Earth as seen from the Moon?
Answer:
Distance between Earth-Moon = 60 R
Radius of the earth = R
r = 60R = 60 × 6400 × 10³ (R = 6400km)
θ = 1 sec = \(\frac{1}{60}\) Min = \(\frac{1}{60 \times 60}\) degree ⇒ θ = \(\frac{1}{60 \times 60} \times \frac{\pi}{180}\) radian
Then r = \(\frac{l}{\theta}\) ⇒ l = rθ ⇒ l = 60 × 6400 × 10³ × \(\frac{1}{60 \times 60} \times \frac{\pi}{180}\)
l = 11.16 × 10³ km ⇒ Diamter (l) = 11.16 × 10³ km

Question 5.
Three measurements of the time for 20 oscillations of a pendulum give t₁ = 39.6 s, t₂ = 39.9 s and t₃ = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurements?
Answer:
No. of oscillations = 20
t₁ = 39.6 sec, t₂ = 39.9 sec, t₃ = 39.5 sec
Mean value = \(\frac{t_1+t_2+t_3}{3}=\frac{39.6+39.9+39.5}{3}=\frac{119}{3}\) = 39.66
Mean value = 39.7 sec;
Precision = 0.1 sec.
Accuracy is the closeness of measured value with true value.
Hence 39.6 s is accuracy.

Question 6.
1 calorie = 4.2J where 1J = 1 kg m²s². Suppose we employ a system of units in which the unit of mass is \(\hat{a}\) kg, the unit of length is \(\hat{a}\) m and the unit of time is \(\tilde{a}\) s, show that a calorie has a magnitude 4.2 \(\hat{\mathrm{a}}^{-1} \hat{a}^{-2} \tilde{\mathrm{a}}^{-2}\) in the new system.
Answer:
1 Calorie = 4. 2J ⇒ 1 J = 1 kg m² s^-2
1 calorie = 4.2 kg m² s^-2
In new system, 1 calorie = 4.2 \(\hat{a} \hat{a}^2 \hat{a}^{-2}\)

Question 7.
A new unit of length is chosen so that the speed of light in vacuum is 1 ms-2. If light takes 8 min and 20 $ to cover this distance, what is the distance between the Sun and Earth in terms of the new unit ?
Answer:
V = Speed of light in vacuum = 1 m/s
Time taken (t) = 8 min 20 sec = 500 sec
Distance between the sun and earth (d) = \(\frac{\mathrm{V}}{\mathrm{t}}\) ⇒ d = \(\frac{1}{500}\) = 0.002 m.

Question 8.
A student measures the thickness of a human hair using a microscope of magnification 100. He makes 20 observations and finds that the average thickness (as viewed in the microscope) is 3.5 mm. What is the estimate of the thickness of hair ?
Answer:
Magnification of microscope = M = 100
Observed thickness = 3.5 m.m
Magnification (M) = \(\frac{\text { Observed thickness }}{\text { Real thickness }}\) ⇒ 100 = \(\frac{3.5}{\text { Real thickness }(t)}\) ⇒ t = \(\frac{3.5}{100}\) = 0.035 m.m.

Question 9.
A physical quantity X is related to four measureable quantities a, b, c and d as follows. X = a² b³ c^5/2d^-2
The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. What is the percentage error in X ?
Answer:
X = a² b³ c^5/2d^-2
\(\frac{\Delta \mathrm{a}}{\mathrm{a}}\) × 100 = 1%, \(\frac{\Delta \mathrm{b}}{\mathrm{b}}\) × 100 = 2%, \(\frac{\Delta \mathrm{c}}{\mathrm{c}}\) × 100 = 3%, \(\frac{\Delta \mathrm{d}}{\mathrm{d}}\) × 100 = 4%
Percentage error in X is
(\(\frac{\Delta \mathrm{X}}{\mathrm{X}}\)) × 100 = 2(\(\frac{\Delta \mathrm{a}}{\mathrm{a}}\) × 100) + 3(\(\frac{\Delta \mathrm{b}}{\mathrm{b}}\) × 100) + \(\frac{5}{2}\) (\(\frac{\Delta \mathrm{b}}{\mathrm{b}}\) × 100) + 2 \(\frac{\Delta \mathrm{d}}{\mathrm{d}}\) × 100
= 2 × 1 + 3 × 2 + \(\frac{5}{2}\) × 3 + 2 × 4 = 2 + 6 + \(\frac{15}{2}\) + 8
⇒ (\(\frac{\Delta \mathrm{X}}{\mathrm{X}}\) 100 = 23.5 %

Question 10.
The velocity of a body is given by v = At² + Bt + C. If v and t are expressed in SI what are the units of A, B and C ?
Answer:
Given V = At² + Bt + C
According to principle of homogeneity,

1. V = At² ⇒ A =\(\frac{\mathrm{V}}{\mathrm{t}^2}=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^2}\) = [LT^-3] ⇒ A = ms^-3
2. V = Bt ⇒ B = \(\frac{\mathrm{V}}{\mathrm{t}}=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}}\) = [LT^-1] ⇒ B = ms^-2
3. V = C ⇒ C = LT^-1 ⇒ C = ms^-1

Problems

Question 1.
In the expression P = E l² m^-5 G^-2 the quantities E, l, m and G denote energy, angular momentum, mass and gravitational constant respectively. Show that P is a dimension-less quantity.
Solution:
P = E L² m^-5 G^-2
Energy (E) = [ML² T^-2]
Angular momentum (L) = ML² T^-1
Mass = [M]
Universal gravitational constant
(G) = [M^-1L³T^-2]
P = [ML² T^-2] [ML² T^-1]² [M]^-5 [M^-1 L³ T^-2]^-2
P = M^1+2-5+2 L^2+4-6 T^-2-2+4
P = [M⁰ L⁰ T⁰]
Hence P is dimensional less quantity.

Question 2.
If the velocity of light c. Planck’s constant h and the gravitational ‘ constant G are taken as fundamental quantities : then express mass, length and time in terms of dimensions of these quantities.
Solution:
i) M ∝ G^x C^y h^z
[M¹ L⁰ T⁰] = [M^-1 L³ T^-2]^x [LT^-1]^y [ML² T^-1]^z
[M¹L⁰T⁰] = M^-x+z L^3x+y+2z T^-2x-y-z = 0
– x + z = 1, 3x + y + 2z = 0, – 2x – y – z = 0
Solving these equations, we get
x = \(\frac{-1}{2}\), y = \(\frac{1}{2}\), z = \(\frac{1}{2}\)
M = G^{\(\frac{-1}{2}\)} C^{\(\frac{1}{2}\)} h^{\(\frac{1}{2}\)}
M = \(\sqrt{\frac{\mathrm{hc}}{\mathrm{G}}}\)

ii) Length (l) ∝ G^x C^y h^z
[M⁰ L¹ T⁰] = [M^-1 L³ T^-2]^x [LT^-1]^y [ML² T^-1]^z
[M⁰L¹T⁰] = M^-x+z L^3x+y+2z T^-2x-y-z
Applying principle of homogeneity
-x + z = 0, 3x + y + 2z = 0, -2x – y – z = 0
Solving these equations, we get
x = \(\frac{1}{2}\), y = \(\frac{-3}{2}\), z = \(\frac{1}{2}\)
From equation (1), l = G^{\(\frac{1}{2}\)} C^{\(\frac{-3}{2}\)} h^{\(\frac{1}{2}\)}
Length (l) = \(\sqrt{\frac{\mathrm{Gh}}{\mathrm{C}^3}}\)

iii) Time (T) ∝ G^x C^y h^z
[M⁰ L⁰ T¹] = [M^-1 L³ T^-2]^x [LT^-1]^y [ML² T^-1]^z
[M⁰L⁰T¹] = M^-x+z L^3x+y+2z T^-2x-y-z
Applying principle of homogeneity
-x + z = 0, 3x + y + 2z = 0, -2x – y – z = 1
Solving these equations, we get
x = \(\frac{1}{2}\), y = \(\frac{-5}{2}\), z = \(\frac{1}{2}\)
From equation (1), l = G^{\(\frac{1}{2}\)} C^{\(\frac{-5}{2}\)} h^{\(\frac{1}{2}\)}
Length (l) = \(\sqrt{\frac{\mathrm{Gh}}{\mathrm{C}^5}}\)

Question 3.
An artificial satellite is revolving around a planet of mass M and radius R in a circular orbit of radius r. Using dimensional analysis show that the period of the satellite
T = \(\frac{k}{R} \sqrt{\frac{r^3}{g}}\)
where k is a dimensionless constant and g is acceleration due to gravity.
Solution:
T = \(\frac{k}{R} \sqrt{\frac{r^3}{g}}\)
L.H.S. : Time period = [T]
R.H.S. : \(\frac{k}{R} \sqrt{\frac{r^3}{g}}=\frac{1}{L} \sqrt{\frac{L^3}{L T^{-2}}}\) = [T]
∴ L.H.S. = R.H.S.
Above equation is correct.

Question 4.
State the number of significant figures in the following.
a) 6729
b) 0.024
c) 0.08240
d). 6.032
e) 4.57x
Solution:
a) 6729 – 4     Significant figures
b) 0024 – 2            ”
c) 008240 – 4         ”
d) 6.032 –  4           ”
e) 4.57 × 10⁸ – 3     ”

Question 5.
A stick has a length of 12.132 cm and another has a length of 12.4 cm. If the two sticks are placed end to end what is the total length ? If the two sticks are placed side by side, what is the difference in their lengths?
Solution:
a) Let lengths of the rods are
l₁ = 12.132 cm, l₂ = 12.4 cm
Here has one decimal place and l₁ has to be rounded to have only two decimal places.
l = l₁ + l₂ = 12.13 + 12.4 = 24.53
This is to be rounded off to have one decimal place only.
∴ The result is 24.5 cm

b) l₁ = 12.132 cm, l₂ = 12.4 cm
Here l₁ is rounded only two decimal places,
l₂ – l₁ = 12.4 – 12.13 = 0.27
This should be rounded off to have only one decimal place, l₂ – l₁ = 0.3

Question 6.
Each side of a cube is measured to be 7.203 m. What is
(i) the total surface area and
(ii) the volume of the cube, to appropriate significant figures ?
Solution:
Length of a side = 7.203 m
(i) Total surface area = 6a²
= 6 × (7.203)²
= 311.29
This should be rounded off to four significant figure as 7.203 as 4 significant figures.
∴ The result = 311.3m²

(ii) Volume of the cube = a³
= (7.203)³
= 373.71
This should be rounded off to four significant figures has 7.203 has four significant figures.
∴ Volume = 373.7 m³

Question 7.
The measured mass and volume of a body are 2.42 g and 4.7 cm³ respectively with possible errors 0.01 g and 0.1 cm³. Find the maximum error in density.
Solution:
M = 2.42 g, V = 4.7 cm³
∆M = 0.01 g, ∆V = 0.1 cm³
Density (ρ) = \(\frac{M}{V}\)
\(\frac{\Delta \rho}{\rho}=\frac{\Delta \mathrm{M}}{\mathrm{M}}+\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{0.01}{2.42}+\frac{0.1}{4.7}\)
= 0.004 + 0.02 = 0.024
% error is \(\frac{\Delta \rho}{\rho}\) × 100 = 0.024 × 100
= 2.4% ≈ 2%

Question 8.
The error in measurement of radius of a sphere is 1%. What is the error in the measurement of volume ?
Solution:
Radius of the sphere \(\frac{\Delta \mathrm{r}}{\mathrm{r}}\) × 100 = 1%
Volume (V) = \(\frac{4}{3}\) πr³
\(\frac{\Delta \mathrm{V}}{\mathrm{V}}\) × 100 = 3 × \(\frac{\Delta \mathrm{r}}{\mathrm{r}}\) × 100 = 3 × 1% = 3%

Question 9.
The percentage error in the mass and speed are 2% and 3% respectively. What is the maximum error in kinetic energy calculated using these quantities ?
Solution:
\(\frac{\Delta \mathrm{M}}{\mathrm{M}}\) × 100 = 2%, \(\frac{\Delta \mathrm{V}}{\mathrm{V}}\) × 100 = 3%;
Kinetic energy = \(\frac{1}{2}\) mV²
\(\frac{\Delta \mathrm{K}}{\mathrm{K}}\) × 100 = \(\frac{\Delta \mathrm{M}}{\mathrm{M}}\) × 100 + 2 \(\frac{\Delta \mathrm{V}}{\mathrm{V}}\) × 100
= 2 + 2(3) = 2 + 6 = 8%

Question 10.
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). If the size of the hydrogen molecule is about 1A°, what is the ratio of molar volume to the atomic volume of a mole of hydrogen ?
Solution:
Molar volume = 22.4 lit = 22.4 × 1000 c.c.
= 22.4 × 10^-3 m³
Diameter of the hydrogen molecule
= 1 A° = 10^-10 m
Radius (r) = \(\frac{D}{2}=\frac{10^{-10}}{2}\) = 0.5 × 10^-10 m
Volume (V) = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × 3.14 × (0.5 × 10^-10)³
v = \(\frac{1.57}{3}\) × 10^-30 = 0.5233 × 10^-30 m³
Atomic volume = V × Avagadro’s number (N)
= 0.5233 × 10^-30 × 6.023 × 10²³ = 3.151 × 10^-7
∴ \(\frac{\text { Molar volume }}{\text { Atomic volume }}=\frac{22.4 \times 10^{-3}}{3.151 \times 10^{-7}}\)
= 7.108 × 10^-4

Additional Problems

Note : In stating numerical answers, take care of significant figures.

Question 1.
Fill in the blanks
a) The volume of a cube of side 1 cm is equal to …………….. m³
b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ……………. (mm)²
c) A vehicle moving with a speed of 18 km h^-1 covers ……………. m in 1 s
d) The relative density of lead is 11.3. Its density is …………….. g cm^-3 or …………….. kg m^-3.
Solution:
a) Here, length of slide L = 1 cm = 10^-2 m
Volume of cube = L³ = (10^-2 m)³ = 10^-6 m³

b) Here, r = 2.0 cm = 20 mm
h = 10.0 cm = 100 mm
Surface area of solid cylinder = (2rcr) × h
= 2 × \(\frac{22}{7}\) × 20 × 100 mm²
= 1.26 × 10⁴ mm²

c) Here, speed V = 18 km
h^-1 = \(\frac{18 \times 1000 \mathrm{~m}}{60 \times 60 \mathrm{~s}}\) = 5 ms^-1
∴ Distance covered in 1 second = 5 m

d) Here, relative density = 11.3
density = 11.3 g k.c = \(\frac{11.3 \times 10^{-3}}{\left(10^{-2} \mathrm{~m}\right)^3}\)
= 11.3 × 10³ kg m^-3

Question 2.
Fill in the blanks by suitable conversion of units.
a) 1 kg m² s^-2 = ………….. g cm² s^-2
b) 1 m = ………….. 1y
c) 3.0 ms^-2 = ………….. km h^-2
d) G = 6.67 × 10^-11 N m² (kg)^-2 = …………. (cm)³ s^-2 g^-1.
Solution:
a) 1 kg m² s²
= 1 × 10³ g (10² cm)²s^-2 = 10⁷ g cm² s^-2

b) We know, 1 light year = 9.46 × 10¹⁵ m
∴ 1 m = \(\frac{1}{9.46 \times 10^{15}}\) light year
= 1.053 × 10^-16 light year

c) 3ms^-2 = 3 × 10^-3 km \(\left(\frac{1}{60 \times 60} h\right)^{-2}\)
= 3 × 10^-3 × 3600 × 3600 km h^-2

d) G = 6.61 × 10^-11 Nm² kg^-2
= 6.67 × 10^-11 (kg ms^-2) m² kg^-2
= 6.67 × 10^-11 m³ s^-2 kg^-1
= 6.67 × 10^-11 (100 cm)³ s^-2 (1000 g)^-1
= 6.67 × 10^-8 cm³ s^-2 g^-1

Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m² s^-2. Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals b m, the unit of time is g s. Show that a calorie has a magnitude 4.2 a^-1b^-2 g² in terms of the new units.
Solution:
Here 1 calorie = 4.2 J = 4.2 Kg m^-3S^-1 …………… (1)
As new unit of mass = 1kg
∴ 1 kg = \(\frac{1}{\mathrm{a}}\) new unit of mass = a^-1
similarly, 1m = \(\frac{1}{\mathrm{b}}\) = b^-1
1s = \(\frac{1}{\mathrm{g}}\) = g^-1
Putting these values in eq(1), we obtains.
1 calorie = 4.2 (a^-1 new unit of mass)
(b^-1 new unit of length)²
(g^-1 new unit of time)²
1 calorie = 4.2 a^-1 b^-2 g².

Question 4.
Explain this statement clearly :
‘To call a dimensional quantity ‘large1 or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary :
a) atoms are very small objects
b) a jet plane moves with great speed
c) the mass of Jupiter is very large
d) the air inside this room contains a large number of molecules
e) a proton is much more massive than an electron
f) the speed of sound is much smaller than the speed of light.
Solution:
i) The statement is true. This is because a dimensionless quantity can be large or small only in comparison to some standard. For example, angle is dimensionless ∠θ = 60° is larger than ∠θ = 30°, but smaller than ∠θ = 90°

ii) a) The size of an atom is smaller than the sharp tin of a pin.
b) A jet plane moves faster than a superfast train.
c) The mass of Jupiter is very large compared to the mass of earth.
d) The air inside this room contains more number of molecules than in one mole of air.
e) The statement is already correct.
f) The statement is already correct.

Question 5.
A new unit of length is chosen such that the speed of light in vacuum in unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ?
Solution:
We are given that velocity of light in vacuum, c = 1 new unit of length s^-1
Time taken by light of Sun to reach the Earth,
t = 8 min 20 s = 8 × 60 + 20 = 500 s
∴ Distance between the Sun and the Earth,
x = C × t =1 new unit of length s^-1 × 500 s
= 500 new units of length

Question 6.
Which of the following is the most
precise device for measuring length :
a) a vernier cuHipers with 20 divisions on the sliding scale
b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
c) an optical instrument that can measure length to within a wave-length of light ?
Solution:
The most precise device is that whose least count is minimum. Now
a) Least count of this vernier callipers
= 1 SD – 19 SD = 1 SD = \(\frac{19}{20}\) SD = \(\frac{1}{20}\) SD
= \(\frac{1}{20}\) mm = \(\frac{1}{200}\) cm = 0.005 cm

b) Least count of screw gauge
= \(\frac{\text { Pitch }}{\text { No. of divisions on circular scale }}\)
= \(\frac{1}{100}\) mm = \(\frac{1}{1000}\) cm = 0.01 cm

c) Wavelength of light,
λ = 10^-5 cm = 0.00001 cm
Obviously, the most precise measurement is with optical instrument.

Question 7.
A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?
Soution:
Magnification m = \(\frac{\text { observed with }(\mathrm{y})}{\text { real width }(\mathrm{x})}\)
x = \(\frac{\mathrm{y}}{\mathrm{m}}=\frac{3.5 \mathrm{~mm}}{100}\) = 0.035 mm

Question 8.
Answer the following :
a) You are given a thread and a metre scale. How will you estimate the diameter of the thread ?
b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?
c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?
Solution:
a) The diameter of a thread is so small that it cannot be measured using a metre scale. We wind a number of turns of the thread on the metre scale. So that the turns are closely touching one another. Measure the length (l) of the windings on the scale which contains n number of turns.
∴ Diameter of thread = \(\frac{1}{n}\)

b) As least count
= \(\frac{\text { Pitch }}{\text { Number of divisions on circular scale }}\)
∴ Theoretically speaking, least count decreases on increasing the number of divisions on the circular scale. Hence accuracy would increase. Practically, it may not be possible to take the reading precisely due to low resolution of human eye.

c) A large number of observations will give more reliable result than smaller number of observations. This is because of probability of making a positive random error of certain magnitude is equal to that of making a negative random errors are likely to cancel and the result may be more reliable.

Question 9.
The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on to a screen and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement.
Solution:
Here, area of object = 1.75 cm² and area of image = 1.55 m² = 1.55 × 10⁴ cm²
∴ Areal magnification = \(\frac{\text { area of image }}{\text { area of object }}\)
= \(\frac{1.55 \times 10^4}{1.75}\) = 8857
Linear magnification = \(\sqrt{8857}\) = 94.1

Question 10.
State the number of significant figures in the following :
a) 0.007 m²
b) 2.64 × 10²⁴ kg
c) 0.2370 g cm^-3
d) 6.320 J
e) 6.032 N m^-2
f) 0.0006032 m²
Solution:
The number of significant figures is as given below.
a) one
b) three
c) four
d) four
e) four
f) four

Question 11.
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Solution:
Here, length l = 4.234 m;
Breath, b = 1.005 m
Thickness, t = 2.01 cm = 2.01 × 10^-2 m
Area of the sheet = 2(l × b + b × t + t × l)
= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
2(4.3604739) = 8.7209478 m²
As area can contain a maximum of three significant digits, therefore, rounding off, we get
Area = 8.72 m²
Also, volume = l × b × t
V = 4.234 × 1.005 × 0.0201
V = 0.0855289
V = 0.855 m³

Question 12.
The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ?
Solution:
Here, mass of the box, m = 2.3 kg
Mass of one gold piece,
m₁ = 20.15 g = 0.02015 kg
Mass of another gold piece,
m₂ = 20.17g = 0.02017 kg
a) Total mass = m + m₁ + m₁
= 2.3 + 0.02015 + 0.02017
= 2.34032 kg
As the result is correct only upto one place of decimal, therefore, on rounding off total mass = 2.3 kg

b) Difference in masses
= m₂ – m₁ = 20.17 – 20.15
= 0.02g (correct upto two places of decimal)

Question 13.
A physical quantity P is related to four observables a, b, c and d as follows :
P = a³b²/ (\(\sqrt{c}\)d )
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?
Solution:
Here, P = \(\frac{a^3 b^2}{\sqrt{c} d}\)
Maximum fractional error in P is given by \(\frac{\Delta \mathrm{P}}{\mathrm{P}}\)
= \(3 \frac{\Delta \mathrm{a}}{\mathrm{a}}+2 \frac{\Delta \mathrm{b}}{\mathrm{b}}+\frac{1}{2} \frac{\Delta \mathrm{c}}{\mathrm{c}}+\frac{\Delta \mathrm{d}}{\mathrm{d}}\)
= \(3\left(\frac{1}{100}\right)+2\left(\frac{3}{100}\right)+\frac{1}{2}\left(\frac{4}{100}\right)+\frac{2}{100}\)
= \(\frac{13}{100}\) = 0.13
Percentage error in
P = \(\frac{\Delta \mathrm{P}}{\mathrm{P}}\) × 100 = 0.13 × 100 = 13%
As the result (13% error) has two significant figures, therefore if P turns out to be 3.763, the result would be rounded of to 3.8.

Question 14.
A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion :
a) y = a sin 2 π t/T) b) y = a sin vt,
c) y = (a/T) sin t/a,
d) y = (a\(\sqrt{2}\)) (sin 2πt /T + cos 2 πt/T)
(a = maximum displacement of the particle, v = speed of the particle. T = time- period of motion). Rule out the wrong formulas on dimensional grounds.
Solution:
The argument of a trigonometrical function i.e., angle is dimensionless. Now in
a) \(\frac{2 \pi t}{T}=\frac{T}{T}\) = 1 = (M⁰L⁰T⁰)
…………… dimensionless
b) vt = (LF^-1) (T) = L = (M⁰L¹T⁰)
………….. not dimensionless

c) \(\frac{\mathrm{t}}{\mathrm{a}}=\frac{\mathrm{T}}{\mathrm{L}}\) = (L^-1T^-1)
………….. not dimensionless

d) \(\frac{2 \pi t}{\mathrm{~T}}=\frac{\mathrm{T}}{\mathrm{T}}\) = 1 = [M⁰L⁰T⁰]
…………… dimensionless

Question 15.
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einsterin). A boy recalls the relation almost correctly but for-gets where to put the constant c. He writes :
m = \(\frac{m_0}{\left(1-v^2\right)^{1 / 2}}\)
Guess where to put the missing c.
Solution:
According to the principle of homogenity of dimensions power of M, L, T on either side of the formula must be equal. For this, on RHS, the denominator (1 – v²)^1/2 should be dimensionless. Therefore, instead of (1 – v²)^1/2,
we should write \(\left(\frac{1-v^2}{c^2}\right)^{1 / 2}\)
Hence the correct formula would be
m = \(\frac{m_0}{\left(\frac{1-v^2}{c^2}\right)^{1 / 2}}\)

Question 16.
The unit of length convenient on the atomic scale is known as an angstrom and is denoted byA° : 1A° = 1010m. The size of a hydrogen atom is about 0.5 A°. What is the total atomic volume in m³ of a mole of hydrogen atoms ?
Solution:
Here r = 0.5 A° = 0.5 × 10^-10 m
Volume of each atom of hydrogen = \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\) × 3.14(0.5 × 10^-10) = 5.236 × 10^-31 m³
Number of hydrogen atoms in one gram mole of hydrogen = Avagadro’s number
= 6.023 × 10²³
∴ Atomic volume of one gram mole of hydrogen atom
= 5.236 × 10^-31 × 6.023 × 10²³
= 3.154 × 10^-7 m³

Question 17.
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L of (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to about 1A°). Why is this ratio so large ?
Solution:
Atomic volume = \(\frac{4}{3}\)πR³ × N
= \(\frac{4}{3}\) (0.5 × 10^-10)³ × 6.023 × 10²³
= 3.154 × 10^-7 m³
Molar volume = 22.4 lit = 22.4 × 10^-3 m³
\(\frac{\text { Molar volume }}{\text { Atomic volume }}=\frac{22.4 \times 10^{-3}}{3.154 \times 10^{-7}}\) = 7.1 × 10⁴
The ratio is large due to large intermolecular separations.

Question 18.
Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby tr6es, houses etc., seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving; these distant objects seem to move with you).
Solution:
The line joining the object to the eye is called the line of sight. When a train moves rapidly, the line of sight of a near by tree changes its direction of motion rapidly. Therefore, the trees appear to run in opposite direction. On the contrary, the line of sight of far off objects does not change its direction. So much, due to extremely large distance from the eye. Hence distant hill tops, moon, the stars etc., appear stationary.

Question 19.
The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart 21. in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit 3 × 10¹¹ m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1″ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1″ (second) of arc from opposite ends of a baseline equal to the distance from the Earth to Sol. the Sun. How much is a parsec in terms of metres ?
Solution:
Here, length of baseline
= distance from each to the sun
= 1 A.U = 1.5 × 10¹¹ m
Parallax angle, θ = 1
\(\frac{1^1}{60}=\frac{1^{\circ}}{60 \times 60}=\frac{\pi}{180} \times \frac{1}{60 \times 60}\) radian
r = 1 par sec = ?; From l = rθ = \(\frac{l}{\theta}\)
= \(\frac{1.5 \times 10^{\prime \prime}}{\pi / 180 \times 60 \times 60} \mathrm{~m}\) = 3.1 × 10¹⁶ m
Hence 1 parsec = 3.1 × 10¹⁶ m

Question 20.
The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parasecs ? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?
Solution:
x = 4.29 ly = 4.29 × 9.46 × 10¹⁵ m
= \(\frac{4.29 \times 9.46 \times 10^{15}}{3.08 \times 10^{16}}\) par sec
= 1.323 par sec
θ = \(\frac{l}{r}=\frac{2 A u}{x}\)
= \(\frac{2 \times 1.496 \times 10^4}{4.29 \times 9.46 \times 10^{15}}\) radian = 1.512 sec

Question 21.
Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modem science where precise measurements of length, time, mass etc., are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Solution:
Precise measurements of physical quantities like length, mass and time are the primary requirements for development of quantitative laws of physics or any other science. For example, in the measurement of distance of moon from earth by laser beam, very accurate measurement of time taken is required. Similarly, for measuring distance, elevation and velocity of an aeroplane by radar method, time measurement has to be accurate. For measuring distances of near by stars, accurate measurement of parallax angle is required.

In the field of crystallography, precise measurement of length is needed to determine interatomic distances using a mass spectrometer, the precision measurement of masses of atoms are made.

Question 22.
Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
a) the total mass of rain-bearing clouds over India during the Monsoon
b) the mass of an elephant
c) the wind speed during a storm
d) the number of strands of hair on * your head
e) the number of air molecules in your classroom.
Solution:
a) During the Monsoon, meteorologist record about 100 cm of rainfall, i.e.,
h – 100 cm = 1m
Area of our country,
A = 3.3 million square km
= 3.3 × 10⁶ (10³m²)
= 3.3 × 10¹² m⁶
∴ Volume of rain water,
v = A × h = 3.3 × 10¹² 1m³
As density of water,
P = 103 km/m³
∴ Mass of rain water
= vP = 3.3 × 10¹² × 10³ kg
= 3.3 × 10¹⁵ kg
This must be the total mass of rain bearing clouds over India.

b) To estimate the mass of an elephant, we take a boat of known base area A. Measure the depth of boat in water. Let it be x₁. Therefore, volume of water displaced by the boat , V₁ = Ax₁ move the elephant into this boat. The boat gets deeper into water. Measure the depth of boat now into water, Let it be x₁
∴ Volume of water displaced by boat and elephant V₂ = Ax₂
∴ Volume of water displaced by the elephant V = V₂ – V₁ = A(x₂ – x₁)
If ρ is density of water, then mass of elephant
= mass of water displaced by it
= V_ρ = A(X₂ – X₁) ρ

c) The wind speed during a storm can be estimated using a gas filled balloon. In figure OA is normal position of a gas filled balloon, when there is no wind. As the wind blows to the right, the balloon drifts to position B in one second. The angle of drift ∠AOB = θ is measured, if h is the height of the balloon, then AB = d = hθ This is the distance travelled by the balloon in one second it must be the wind speed.

d) For this, we measure the area of the head . that carrier the hair let it be A. Using a screw guage, we measure thickness of hair, let it bad.
∴ Area of cross section of hair = πd²
Assuming that the distribution of hair over the head is uniform.
The number of strands of hair
= \(\frac{\text { total area }}{\text { area of cross section of each hair }}=\frac{\mathrm{A}}{\pi \mathrm{d}^2}\)
Calculations show that number of strands of hair on human head is of the order of one million.

e) Measure the volume of room. We know that one mole of air at NTP occupies a volume of 22.4 lit i.e., 22.4 × 10^-3 m³
∴ Number of air molecules in 22.4 × 10^-3 m³ = 6.023 × 10²³
Number of air molecules in volume v of room = \(\frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}}\)v

Question 23.
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10⁷ K and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid of liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 10³⁰ kg, radius of the Sun = 7.0 × 10⁸ m.
Solution:
Here, M = 2.0 × 10³⁰ kg; R = 7.0 × 10⁸ m
Density ρ = ?
ρ = \(\frac{\text { mass }}{\text { volume }}=\frac{M}{\frac{4}{3} \pi R^3}=\frac{3 M}{4 \pi R^3}\)
= \(\frac{3 \times 2.0 \times 10^{30}}{4 \times 3.14 \times\left(7 \times 10^8\right)^3}\)
= 1.392 × 10³ kg/m³
This is the order of density of solids and liquids; and not gases.
The high density of Sun is due to inward gravitational attraction on outer layers, due to inner layers of the Sun.

Question 24.
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72, of arc. Calculate the diameter of Jupiter.
Solution:
Here, r = 824.7 × 10⁶ km
θ = 35.72″ radian
= \(\frac{35.72}{60 \times 60} \times \frac{\pi}{180}\) radian
Diameter, l = ?; As l = rθ
∴ l = 824.7 × 10⁶ × \(\frac{35.72 \times \pi}{60 \times 60 \times 180}\) km
= 1.429 × 10⁵km

Question 25.
A man walking briskly in rain with speed u must slant his umbrella forward making an angle q with the vertical. A student derives the following relation between q and v : tan θ = v and checks that the relation has a correct limit: as v → 0, θ →0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.
Solution:
The relation tan θ = v has a correct limit, as v → 0, θ → 0.
However, RHS = tan θ = [M⁰L⁰T⁰] and L.H.S = v = [M⁰L¹T^-1]
Therefore, the relation is not correct dimensionally.
We shall find the correct relation is
tan θ = \(\frac{\mathrm{v^2}}{\mathrm{rg}}\)

Question 26.
It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ?
Solution:
Error in 100 years = 0.02s
Error in 1 sec

Hence the accuracy of the standard cesium clock is measuring a time interval of 1s is 10^-12 s.

Question 27.
Estimate the average mass density of a sodium atom assuming its size to be about 2.5 A°. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m^-3. Are the two densities of the same order of magnitude ? If so why ?
Solution:
Atomic volume = \(\frac{4}{3}\)πR³ × N
= \(\frac{4}{3}\) × \(\frac{22}{7}\) (1.25 × 10^-10)³ 6.023 × 10²³ m
= 4.93 × 10^-6 m³
Average mass density
= \(\frac{\text { mass }}{\text { volume }}=\frac{23 \times 10^{-3}}{4.93 \times 10^{-6}}\)
= 4.67 × 10³ kg/m³
The two densities are not of the same order. This is due to interatomic spacing in the crystalline phase.

Question 28.
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10^-15 m. Nuclear sizes obey roughly the following empirical relation :
r = r₀A^1/3
where r is the radius of the nucleus. A its mass number, and r₀ is a constant equal to about 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different density of a sodium atom obtained in Exercise. 27.
Answer:
Let m be the average mass of nucleon (neutron or proton)
As the nucleus contains A nucleons,
∴ mass of nucleus μ = mA
Radius of nucleus r = r₀A^1/3
Nuclear density ρ = \(\frac{\text { mass }}{\text { volume }}=\frac{\mu}{\frac{4}{3} \pi r^3}\)
= \(\frac{3 m A}{4 \pi\left(r_0 A^{1 / 3}\right)^3}=\frac{3 m}{4 \pi r_0^3}\)
As m and r₀ are constant, therefore, nuclear density is constant for all nuclei.
using m = 1.66 × 10^-27 kg and r₀
= 1.2f = 1.2 × 10^-15 m
We get ρ = \(\frac{3 m}{4 \pi r_0^3}=\frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14\left(1.2 \times 10^{-15}\right)^3}\)
= 2.29 × 10¹⁷ kg m^-3
As ρ is constant for all nuclei, this must be the density of sodium nucleus also.
Density of sodium atom
ρ’ = 4.67 × 10³ kg m^-3
= \(\frac{\rho}{\rho^{\prime}}=\frac{2.29 \times 10^{17}}{4.67 \times 10^3}\) = 4.67 × 10^-3

Question 29.
A LASER is a source or very intense, monochromatic and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth ?
Solution:
Here, t = 2.56s
Velocity of laser light in vacuum,
c = 3 × 10⁸ m/s
The radius of lunar orbit is the distance of moon from the earth. Let it be x.
As x = \(\frac{c \times t}{2}\)
∴ x = \(\frac{3 \times 10^8 \times 2.56}{2}\) = 3 84 × 10⁸ m

Question 30.
A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in waiter = 1450 ms^-1).
Solution:
Here, t = 77.0 s, x = ? V = 1450 ms^-1
As x = \(\frac{\mathrm{V} \times \mathrm{t}}{2}\)
= \(\frac{1450 \times 77.0}{2}\)m; x = 55825 m

Question 31.
The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzlingn features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ?
Solution:
Here, x = ?
Time taken t = 30 billion years = 3 × 10⁹ yr
= 3 × 10⁹ × 365 × 24 × 60 × 60 s
Velocity of light in vacuum, c = 3 × 10⁸ m/s
= 3 × 10⁵ km/s
As distance = velocity × time
x = (3 × 10⁵) × 3 × 10⁹ × 365 × 24 × 60 × 60 km.
= 2.84 × 10²² km

Question 32.
It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 3 and 4. determine the approximate diameter of the moon.
Solution:
Distance of moon from earth,
ME = 3.84 × 10⁸ m
Distance of Sun from earth,

SE = 1.496 × 10¹¹ m
Distance of Sun AB = 1.39 × 10⁹ m
The situation during total solar eclipse is shown in figure.
As ∆^s ABE of CDE are similar, therefore,
\(\frac{A B}{C D}=\frac{S E}{M E}\)
CD = AB × \(\frac{M E}{S E}=\frac{1.39 \times 10^9 \times 3.84 \times 10^8}{1.496 \times 10^{11}}\)
= 3.5679 × 10⁶ m = 3567.9 km

Question 33.
A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational costant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~ 15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants ?
Solution:
Trying out with basic constants of atomic physics (speed of light c, charge on electron e, mass of electron m_c, mass of proton m_p) and universal gravitational constant G, we can arrive at a quantity which has the dimensions of time. One such quantity is
t = \(\left(\frac{c^2}{4 \pi \epsilon_0}\right)^2 \times \frac{1}{m_p m_{e^2} c^3 G}\)
Put e = 1.6 × 10^-19 c
\(\frac{1}{4 \pi \epsilon_0}\) = 9 × 10⁹,
c = 3 × 10<sup8 m/s and
G = 6.67 × 10^-11 Nm² kg^-2
t = (1.6 × 10^-19)⁴ × (9 × 10⁹)² × \(\frac{1}{1.67 \times 10^{-27} \times\left(9 \times 10^{-31}\right) \times\left(3 \times 10^8\right)^3 \times 6.67 \times 10^{-11}}\)
t = 2.18 × 10¹⁶ sec; This time is of the order of age of universe.

Textual Examples

Question 1.
Calculate the angle at (a) 1° (degree) (b) V (minute of arc or arcmin) and (c) 1″ (second of arc or are second) in radians. Use 360° = 2π rad, 1° = 60′ and 1′ = 60″.
Answer:
(a) We have 360° = 2π rad
1° = (π/180) rad = 1.745 × 10^-2 rad

(b) 1′ = 60′ = 1.745 × 10^-2 rad
V = 2.908 × 10^-4 rad, 2.91 × 10^-4 rad

(c) 1′ = 60″ = 2.908 × 10^-4 rad
1″ = 4.847 × 10^-6 rad, 4.85 × 10^-6 rad.

Question 2.
A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 m and looks at O and C again. Since O is very distant, the direction BO is practically the same as AO; but he finds the line of sight of C shifted from the original line of sight by an angle θ = 40° (θ is known as parallax’) estimate the distance of the tower C from his original position A.

Answer:
Parallax angle θ = 40°; AB = AC tan θ
AC = AB/tan θ = 100 m/tan 40°
= 100 m/0.8391 = 119 m.

Question 3.
The moon is observed from two diametrically opposite points A and B on Earth. The angle 6 subtended at the moon by the two directions of observation is 1°54‘. Given the diameter of the Earth to be about 1.276 × 10⁷ m, compute the distance of the moon from the Earth.
Solution:
We have θ = 1°54′ = 114
= (114 × 60)” × (4.85 × 10^-6) rad
= 3.32 × 10^-2 rad.
Since 1″ = 4.85 × 10^-6,; rad
Also b = AB = 1.276 × 10⁷ m
Hence from D = b/θ, we have the earth-moon distance,
D = b/θ
= \(\frac{1.276 \times 10^7}{3.32 \times 10^{-2}}\) = 3.84 × 10⁸ m.

Question 4.
The Sun’s angular diamater is mea-sured to be 1920″. The distance D of the Sun from the Earth is 1.496 × 10″ m. What is the diameter of the Sun ?
Answer:
Sun’s angular diameter α = 1920”
= 1920 × 4.85 × 10^-6 rad
= 9.31 × 10^-3 rad
Sun’s diameter d = α
D = (9.31 × 10^-3) × (1.496 × 10¹¹) m
= 1.39 × 10⁹ m.

Question 5.
If the size of a nucleus in the range of 10^-15 to 10^-14 m) is scaled up to the tip of a sharp pin, what roughly is the size of an atom ? Assume tip of the pin to be in the range 10^-5 m to 10^-4 m.
Answer:
The size of a nucleus is in the range of 10^-15m and 10^-14 m. The tip of a sharp pin is taken to be in the range of 10^-5 m and 10^-4 m. Thus we are scaling up by a factor of 10¹⁰. An atom roughly of size 10^-10 m will be scaled up to a size of 1 m. Thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre long.

Question 6.
Two clocks are being tested against a standard clock located in a national laboratory. At 12 : 00 : 00 noon by the standard dock, the readings of the two clocks are :

If you are doing an experiment that requires ‘precision time interval’ measurements, which of the two clocks will you prefer ?
Answer:
The range of variation over the seven days of observations is 162 s for clock 1 and 31 s for clock 2. The average reading of clock 1 is much closer to the standard time than the average reading of clock 2. The important point is that a clock’s zero error is not as significant for precision work as its variation, because a ‘zero-error’ can always be easily corrected. Hence clock 2 is to be preferred to clock 1.

Question 7.
We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s. 2.56 s, 2.42 s, 2.71 is and 2.80 s. Calculate the absolute errors relative error or percentage error.
Answer:
The mean period of oscillation of the pendulum
T = \(\frac{(2.63+2.56+2.42+2.71+2.80) s}{5}\)
= \(\frac{13.22}{5}\) s = 2.624 s = 2.62 s
As the periods are measured to a resolution of 0.01 s, all times are to the second decimal; it is proper to put this mean period also to the second decimal.
The errors in the measurements are
2.63 s – 2.62 s = 0.01 s
2.56 s – 2.62 s = -0.06 s
2.42 s – 2.62 s = -0.20 s
2.71 s – 2.62 s = 0.09 s
2.80 s – 2.62 s = 0.18 s
Note that the errors have the same units as the quantity to be measured.
The arithmetic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes) is
∆T_mean = [(0.01 + 0.06 + 0.20 + 0.09 + 0.18)s]/5 = 0.54 s/5 = 0.11 s
That means, the period of oscillation of the simple pendulum is (2.62 ± 0.11) s i.e. it lies between (2.62 + 0.11) s and (2.62 – 0.11) s or between 2.73 s and 2.51 s. As the arithmetic mean of all the absolute errors is 0.11 s, there is already an error in the tenth of a second. Hence there is no point in giving the period to a hundredth. A more correct way will be to write T = 2.6 ± 0.1 s

Note that the last numeral 6 is unreliable, since it may be anything between 5 and 7. We indicate this by saying that the measurement has two significant figures. In this case, the two significant figures are 2, which is reliable and 6, which has an error associated with it. You will learn more about the significant figures in section 2.7.
For this example, the relative error or the percentage error is δa = \(\frac{0.1}{2.6}\) × 100 = 4.

Question 8.
The temperatures of two bodies measured by a thermometer are t₁ = 20 °C ± 0.5 °C and t₂ = 50 °C ± 0.5°C. Calculate the temperature difference and the error therein.
Answer:
t’ = t₂ – t₁
= (50 °C ± 0.5 °C) – (20 °C ± 0.5 °C)
t’ = 30 °C ± 1 °C.

Question 9.
The resistance R = V/I where V = (100 ± 5) V and I = (10 ± 0.2) A. Find the percentage error in R.
Answer:
The percentage error in V is 5 and in I it is 2.
The total error in R would therefore be 5 + 2 = 7%.

Question 10.
Two resistors of resistances R₁ = 100 ± 3 ohm and R₂ = 200 ± 4 ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the relation R
= R₁ + R₂ and for (b) \(\frac{1}{R^{\prime}}+\frac{1}{R_1}+\frac{1}{R_2}\) and \(\frac{\Delta R^{\prime}}{R^{\prime 2}}=\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2}\)
Answer:
a) The equivalent resistance of series, combination
R = R₁ + R₂ = (100 ± 3)
ohm + (200 ± 4) ohm = 300 ± 7 ohm.

b) The equivalent resistance of parallel combination

Then, R’ = 66.7 ± 1.8 ohm
(Here, ∆R is expressed as 1.8 instead of 2 to keep in conformity with the rules of significant figures).

Question 11.
Find the relative error in Z, if Z = A⁴B^1/3/CD^3/2.
Answer:
The relative error in Z is ∆Z/Z = 4(∆A/A) + (1/3) (∆B/B) + (∆C/C) + (3/2) (∆D/D).

Question 12.
The period of oscillation of a simple pendulum is T = \(2 \pi \sqrt{L / g}\). Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. What is the accuracy in the determination of g ?
Answer:
g = 4π²L/T²; Here, T = \(\frac{t}{n}\) and ∆T = \(\frac{\Delta t}{n}\)
Therefore, \(\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{\Delta \mathrm{t}}{\mathrm{t}}\)
The errors in both L and t are the least count errors. Therefore, (∆g/g) = (∆L/L) + 2(∆T/T)
= \(\frac{0.1}{20.0}+2\left(\frac{1}{90}\right)\) = 0.027
Thus, the percentage error in g is 100 (∆g/g)
= 100(∆L/L) + 2 × 100 (∆T/T) = 3.

Question 13.
Each side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the cube to appro¬priate significant figures ?
Answer:
The number of significant figures in the measured length is 4. The calculated area and the volume should therefore be rounded off to 4 significant figures.
Surface area of the cube = 6(7.203)² m²
= 311.299254 m²
= 311.3 m²
Volume of the cube = (7.203)³ m³
= 373.714754 m³
= 373.7 m³

Question 14.
5.74 g of a substance occupies 1.2 cm³. Express its density by keeping the significant figures in view.
Answer:
There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume. Hence the density should be expressed to only 2 significant figures.
Density = \(\frac{5.74}{1.2}\) cm^-3 = 4.8 g cm^-3.

Question 15.
Let us consider an equation \(\frac{1}{2}\) mv² = mgh where m is the mass of the body, v its velocity, g is the acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct.
Answer:
The dimensions of LHS are
[M] [LT^-1]² = [M] [L²T^-2] = [M L²T^-2]
The dimensions of RHS are
[M] [LT^-2] [L] = [M] [L²T^-2] = [M L²T^-2]
The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.

Question 16.
The SI unit of energy is J = kg m²s^-2; that of speed v is ms-1 and of acceleration a is ms^-2. Which of the formulae for kinetic energy (K) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body):
(a) K = m²V³
(b) K = (1/2) mv²
(c) K = ma
(d) K = (3/16) mv²
(e) K = (1/2) mv² + ma.
Answer:
Every correct formula or equation must have the same dimensions on both sides of the equation. Also, only quantities with the same physical dimensions can be added or subtracted. The dimensions of the quantity on the right side are [M² L³ T^-3] for (a) : [ML²T^-2] for (b) and (d); [MLT^-2] for (c). The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added. Since the kinetic energy K has the dimensions of [ML²T^-2], formulas (a), (c) and (e) are ruled out. Note that dimensional arguments cannot tell which of the two, (b) or (d), is the correct formula. For this, one must turn to the actual definition of kinetic energy. The correct formula for kinetic energy is given by (b).

Question 17.
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its period using method of dimensions.
Answer:
The dependence of time period T on the quantities l, g and m as a product may be written as :
T = k lx g^y m^z
Where k is dimensionless constant and x, y and z are the exponents.
By considering dimensions on both sides, we have
[L°M°T¹] = [L¹]x [L¹T^-2]y [M¹]² = L^x+yT^-2y M^z
On equating the dimensions on both sides, we have
x + y = 0; -2y = 1 and z = 0
So that x = \(\frac{1}{2}\), y = \(\frac{1}{2}\), z = 0
Then, T = kl^1/2 g^1/2 or T = \(\mathrm{k} \sqrt{\frac{l}{\mathrm{~g}}}\)
Note that value of constant k can hot be obtained by the method of dimensions. Here it does not matter if some number multiplies the right side of this formula because that does not affect its dimensions.
Actually, k = 2π so that T = \(2 \pi \sqrt{\frac{l}{g}}\)

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 13th Lesson Ecological Adaptation, Succession and Ecological Services Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 13th Lesson Ecological Adaptation, Succession and Ecological Services

Very Short Answer Questions

Question 1.
The climax stage is achieved quickly in secondary succession as compared to primary succession. Why?
Answer:
In secondary succession, the species that Invade depend on the condition of the soil, availability of water, the environment and also the seeds or other propagules present. Since soil is already there, the rate of succession is much faster and hence climax stage is also reached more quickly than primary succession.

Question 2.
Among Bryophytes, lichens, and ferns which one is a pioneer species in a xeric succession?
Answer:
Among Bryophytes, lichens, and Ferns, lichens are the pioneer species in a xeric succession.

Question 3.
Give any two examples of xerarch succession.
Answer:
Crustose lichens – Rhizocarpon, Lecanora.
Foliose lichens – Parmelia, Dermetocarpon.
Mossess – Funaria

Question 4.
Name the type of land plants that can tolerate the salinities of the sea.
Answer:
Halophytes.
Ex : Rhizophora.

Question 5.
Define Heliophytes and Sciophytes. Name a plant from your locality that i$ either Heliophyte or Sciophyte.
Answer:
Plants grow in direct sunlight are called heliophytes.
Ex : Tridax, grass.

Plants grow in shady places are called Sciophytes.
Ex : Ferns, Mosses.

Question 6.
Define population and community.
Answer:
A group of similar individuals belonging to the same species found in an area is called population. An assemblage of all the populations belonging to different species occuring in an area is called community.

Question 7.
Define communities? Who classified plant communities into hydrophytes, meso- phytes and xerophytes?
Answer:
An assemblage of all the populations belonging to different species occuring in an area are called communities. “Eugene Warming” classified plant communities into Hydrophytes, mesophytes and xerophytes.

Question 8.
Hydrophytes show reduced xylem. Why?
Answer:
All submerged organs are capable of absorbing water that’s why Hydrophytes show reduced xylem.

Short Answer Type Questions

Question 1.
What are hydrophytes? Briefly discuss the different kinds of hydrophytes with example.
Answer:
Hydrophytes are the plants which grow in water or in very wet places. According to their relation to water and air hydrophytes are classified into five categories. They are

1. Free floating Hydrophytes :
They float freely on the surface of the water and have no contact with soil.
E.g. : Pistia, Eichhornia, Wolffia, Salvinia, Azolla.

2. Rooted hydrophytes with floating leaves :
They are attached to the muddy soil by roots. Their leaves have long petioles which keep them floating on the surface of water.
E.g. : Nymphaea, Nelumbo and Victoria regia.

3. Submerged suspended hydrophytes :
They are completely submerged and suspended in water, but not rooted in the mud and have no contact with air.
E.g. : Hydrilla, Ceratophyllum, Utricularia and Najas.

4. Submerged rooted hydrophytes :
These plants are completely submerged in water. They are attached to the muddy soil by roots.
E.g. : Vallisneria, Potamogeton etc.

5. Amphibious plants :
These live partly in water and partly in air.
E.g. : Sagittaria, Ranunculus, Limnophila.

Some amphibious plants grow around water bodies, with water touching them. They will survive in dry periods also.
E.g. : Typha, Cyperus etc.

Question 2.
Enumerate the morphological adaptations of hydrophytes,
Answer:

1. Roots may be absent or poorly developed. In some plants (Salvinia) submerged leaves compensate for roots.
2. Root caps are usually absent. In some amphibious plants which grow in mud, roots are well developed with root caps. In some plants root caps are replaced by root pockets.
   E.g. : Eichhornia.
3. Roots if present, are generally fibrous, reduced in length, uribranched or poorly branched.
4. Stem is long, slender and flexible.
5. Leaves are thin and either long and ribbon shaped (vallisnaria) or long and linear {potamogeton) or finely dissected (caratophyllum).
6. Floating leaves are large and flat with their upper surfaces coated with wax (Nymphaea, Nelumbium).

Question 3.
List out the anatomical adaptations of hydrophytes.
Answer:

1. Cuticle is totally absent in the submerged parts of the plants. It may be present in the form of a thin film on the surface of parts exposed to atmosphere.
2. The epidermis is composed of thin walled cells. They perform absorption and assimilation as all cells contain chloroplasts.
3. Stomata are totally absent in submerged hydrophytes as the gaseous exchange takes place by diffusion.
4. In Nymphaea, Nelumbium, the leaves are epistomatous.
5. All Hydrophytes contain aerenchyma that helps in gaseous exchange and buoyancy.
6. Mechanical tissues like Collenchyma and Sclerenchyma are poorly developed.
7. Xylem is poorly developed.
8. Secondary growth is absent

Question 4.
Write a brief account on classification of xerophytes.
Answer:
On the basis of their Drought resisting Capacity, Xerophytes are generally classified into the following three categories.

1. Ephemerals :
They are called “drought evaders” or “drought escapers”. They are annuals which complete their life cycle with in a period of 6-8 weeks. They are found in dry zones, e.g. : Tribulus, chenopodium.

2. Succulents :
These are called “drought avoiding plants”. They are fleshy due to storage of water in the form of mucilage. The stored water is sparingly utilised during dry periods.
Examples :
a) Stem succulents : Opuntia, Euphorbia.
b) Leaf succulents : Bryophyllum, Aloe, Agave etc.
c) Root succulents : Asparagus, Ceiba.

Non-succulents :
They are called true xerophytes. These are perennial plants which can withstand prolonged period of drought.
E.g. : Casuarina, Nerium, Ziziphus, Calotropis, Acacia, etc. .

Question 5.
Enumerate the morphological adaptations of xerophytes.
Answer:

1. Roots are long with extensive branching spread over wide areas.
2. Root hairs and root caps are very well developed.
3. Mostly the stem is stunted, woody, hard and covered with thick bark.
4. Stems are usually covered by hairs or waxy coatings.
5. Leaves are very much reduced to small, scale like and sometimes modified into spines to reduce the rate of transpiration.
6. Certain Xerophytes shed their leaves during the dry period.
   Ex : Capparis.

Question 6.
Give in detail the anatomical adaptations shown by xerophytes.
Answer:

1. Epidermis is covered with thick cuticle to reduce the rate of transpiration.
2. Epidermal cells may have silica crystals.
3. Epidermis may be multilayered as in leaves of Nerium.
4. Stomata are generally confined to lower epidermis of leaves called hypostomatous. They are present in pits called sunken stomata.
5. Hypodermis is parenchymatous to check evaporation of water.
   Ex : Calotropis.
6. Mechanical tissues are very well developed.
7. Vascular tissues are very well developed.

Question 7.
Define plant succession. Differentiate primary and secondary successions.
Answer:
The gradual and fairly predictable change in the species composition of a given area is called plant succession.

Differences :

Question 8.
Define ecosystem/ecological services. Explain in brief with regard to pollination.
Answer:
The processes by which the environment produces resources that we often take for granted such as cleen water, timber and habitat for fisheries and pollination of native and agricultural plants is called Ecosystem /Ecological services.

The transfer of pollengrains to fertilize the ovaries of flowers is called pollination. It is an essential part of a healthy ecosystem. Most flowering plants require pollinators to produce fruits and seeds. So pollinators play a significant role in the production of more food crops in the world. Declines of pollinator activity could have serious economic repercussions throughout the world.

The most important pollinator for Agriculture is Honeybee. Over 1,00,000 invertebrate species such as bees, moths, butterflies, beetles and flies serve as pollinators worldwide. At least 1035 species of vertebrates including birds, mammals and reptiles also pollinate many plant species. Continued declines in pollinator activity could mean rising costs for pollinator dependent fruits and vegetables and the disruption of entire ecological systems.

Question 9.
Write about the measure to be taken to sustain ecological functions.
Answer:

1. Choose products produced with methods that conserve resources, minimize waste and reduce or eliminate environmental damage.
2. Prefer products made with methods that reduce the use of pesticides and artificial fertilizers.
3. Reduce consumption and waste production.
4. Support usage of renewable energy alternatives.
5. Use public transit, cycle or walk to conserve natural resources and to reduce pollution and enjoy the health benefits.
6. Participate in developing community garden and tree plantation programmes.
7. Avoid the usage of pesticides and follow methods of natural pest control.
8. Use native plants in the garden and provide habitat for wildlife.

Question 10.
What measure do you suggest to protect the pollinators?
Answer:

1. Creating own pollinator friendly garden using a wide variety of native flowering plants.
2. Reducing the use of pesticides used in and around your home.
3. Encouraging local clubs or school groups to build artificial habiutats like butterfly gardens, bee boards and bee boxes.
4. Supporting agriculture enterprises with pollinato-friendly practices to minimize pesticide use.
5. Encouraging government agencies to take into account the full economic benefits of wild pollinators when formulating policies for agriculture and other land uses.
6. To develop techniques for cultivating native pollinator species for crot pollination.

Long Answer Type Question

Question 1.
Give an account of ecosystem services with reference to carbon fixation and oxygen release.
Answer:
Trees are essential to carbon sequestration keeping excess carbon from entering the atmosphere. The main chemical flow between forest and atmosphere is the exchange of CO₂ and O₂. Forests provide a vast bank for CO₂ and a large amount of CO₂ is deposited in its timber. It plays an essential role in maintaining a dynamic balance between CO₂ & O₂ in atmosphere. According to photosynthesis equation 180 gm of Glucose and 193 gm O₂ are produced by using 264 gm of CO₂ and 108 gm of water and 677.2 K.cal. of solar energy.

180 gm of Glucose can be transformed to 162 gm of polysaccharide inside the plant. So whenever plant produces, 162 gm of dry organic matter 264 gm of CO₂ will be fixed. Then the total amount of the dry organic matter of the reserve forests can be estimated. It provides a foundation for reckoning the total amount of CO₂ fixation by the forests in the reserve.

Natural ecosystems may have helped to stabilize climate and prevent overheating of the earth by removing more of the greenhouse gas, CO₂ from the atmosphere. Many countries have established a carbon tax system to reduce emissions of the greenhouse gases, especially to cut down CO₂ and CO in atmosphere.

Ecosystem services – oxygen release :
Trees and plankton play a big role in release of oxygen; which depends on the species of tree, its age, its health, and also on the trees surroundings. “A mature leafy tree produces as much oxygen in a season as 10 peole in hale in a year”, or A single mature tree can absorb carbon dioxide at a rate of 48 Ibs/year and release enough oxygen back into the atmosphere to support 2 human beings.

One acre of trees annually consumes the amount of CO₂ equivalent to that produced by driving an average can for 26,000 miles. That same acre of trees also produces enough oxygen for 18 people to breathe for a year.

Submerged macrophytes release O₂ and enrich dissolved O₂ in water. The plants and planktons are described as “the Lungs of the World”, taking billion of tonnes of CO₂ and exhaling billions of tones of O₂.

Micro organisms also contribute to the oxygen release in direct and indirect ways.

Ex : Cyanobacteria releases O₂ in a direct way. The other supporting services include Nutrient cycling through decomposition of fallen Logs in forests, soil formation by bacteria and lichens.

Intext Questions

Question 1.
Categorise the following plants into hydrophytes halophytes, mesophytes and xerophytes and give reasons.
a) Salvinia b) Opuntia c) Rhizophora d) Mangifera
Answer:
a) Salvinia is a hydrophyte, It grows on the surface of water.
b) Opuntia is a xerophyte, grows in xeric (dry) areas.
c) Rhizophora is a Halophyte which tolerates the salinities of the sea.
d) Mangifera is a mesophyte, grows in habitats where water is neither scarce nor not abundant.

Question 2.
In a pond, we see plants which are free-floating ; rooted-submerged; rooted emergent; rooted with floating leaves, write the type of plants against each of them.
Answer:

Question 3.
Undertake the following a part of learning process.
a) Identify and assess ecological services found in your area.
b) Think of measures or means to sustain such ecological services.
c) Observe the type of plants or crops grown in your area.
d) Enumerate ecological servies of your area.
e) find out the ecological goods of natural forests commonly used in your area.
f) Observe the biotic agents of pollination for ornamental flowring plants and or agricultural crops in your locality.
Answer:
a) Ecological services :

1. Purification of air and water.
2. Detoxification and decomposition of water.

b) Measures to sustain Ecological services are

1. Reduce consumption and waste production.
2. Avoid the usage of pesticides.

c) Crops grown in our area are
a) Paddy b) Maize c) Black gram d) green gram e) Crotalaria (fodder) g) vegetables.

d) Ecological services :

1. Purification of air and water.
2. Mitigation of floods and droughts
3. Decomposition of wastes

e) Ecological goods :
a) Clean air b) Fresh water c) Food d) fibre e) Timber f) Medicines.

f) Biotic agents of pollination
a) Insects b) Birds c) Animals (Bats, Snails, Snakes).

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 10th Lesson Biomolecules Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 10th Lesson Biomolecules

Very Short Answer Questions

Question 1.
Medicines are either man-made (i.e., synthetic) or obtained from living organisms like plants, bacteria, animals etc. and hence the latter are called natural products. Sometimes natural products are chemically altered by man to reduce toxicity or side effects. Write against each of the following whether they were initially obtained as a natural product or as a synthetic chemical.
a) Pencillin_ b) Sulfonamide_ c) Vitamin C_ d) Growth Hormone_
Answer:
a) Penicillin – Natural
b) Sulfonamide – Synthetic
c) Vitamin C – Natural
d) Growth Hormone – Natural.

Question 2.
Select an appropriate chemical bond among ester bond, glycosidic bond, peptide bond and hydrogen bond and write against each of the following.
a) Polysaccharide b) Protein c) Fat d) Water
Answer:
a) Polysacharide – Glycosidic Bond
b) Protein – Peptide Bond
c) Fat – Ester Bond
d) Water – Hydrogen Bond.

Question 3.
Give one example for each of aminoacids, sugars, nucleotides and fatty acids.
Answer:
Aminoacids – Glycine, Alanine
Sugars – Cellulose [Glucose, Ribose)
Nucleotides – “Adenylic acid”
Fatty acids – Lecithin.

Question 4.
Explain the Zwitterionic form of an amino acid.
Answer:
At a specific PH the amino acid carries both the +ve and -ve charges in equal number and exists as dipolar ion. It is also called as zwitterionic form. At his point, the net charge on it is zero.

Question 5.
What constituents of DNA are linked by glylosidic bond?
Answer:
Individual Monosacharides are linked by glycosidic bond.

Question 6.
Glycine and Alanine are different with respect to one substituent on the a cabron. What are the other common substituent groups?
Answer:
Hydrogen, Carboxyl group, amino group and a variable group designated as R group.

Question 7.
Starch, Cellulose, Glycogen, Chitin are polysaccharides found among the following. Choose the one appropriate and write against each.
a) Cotton fibre_ b) Exoskeleton of cockroach_ c) Liver_ d) Peeled potato_
Answer:
a) Cotton fibre – Cellulose
b) Exoskeleton of cockroach – Chitin
c) Liver – Glycogen
d) Peeled Potato – Starch.

Question 8.
What are primary and secondary metabolites? Give examples.
Answer:
Primary metabolites :
The metabolites have identifiable functions and play known roles in normal physiological processes are called as primary metabolites.
Ex : Hydrogen, Carbon, Oxygen, Nitrogen etc.

Secondary metabolites :
Metabolic products that do not have identifiable functions in the host organism are called as secondary metabolites.
Ex : Alkaloids, Flavonoids, Rubbers, Resins etc.

Short Answer Type Questions

Question 1.
Explain briefly the metabolic basis for ‘living’.
Anwer:
Metabolic pathways can lead to a more complex structure from a simpler structure. (Anabolic pathways). [For example lactic acid becomes cholesterol or sucrose formation from C02 and water in Mesophyll] or lead to a simple structure from a complex structure (catabolic pathways) [glucose becomes lactic acid in our skeletol muscle). Anabolic pathways consume energy whereas catabolic pathways lead to the release of energy.

For example when glucose is degraded to lactic acid in our skeletal muscle, energy is liberated called Glycolysis, Living organisms have learnt to trap this energy and store it in the form of chemical bonds. When ever requires, this energy is utilised for biosynthetic, osmotic and mechanical work that we perform. The most important form of energy currency in living systems is adenorine triphosphate (ATP).

Question 2.
Schematically represent primary, secondary and tertiary structures of a hypothetical polymer using protein as an example.
Answer:
The sequence of amino acids, i.e., the positional information in a protein is called the primary structure of a protein. A protein is imagined as a line, the left end represented by the first amino acid called N- terminal amino acid and the right end represented by the last aminoacid called C-terminal aminoacid In proteins, only, right handed helics are observed. Other regions of the protein thread are folded into other forms in what is called the secondary structure. In addition, the long protein chain is also folded upon itself like a hollow woolen ball giving rise to the tertiary structure.

Question 3.
Nucleic acids exhibits secondary structure. Justify with example.
Answer:
Nucleic acids exhibit a wide variety of seconary structure. For example, one of the secondary structures exhibited by DNA is Watson – Crick model. According to this, DNA exists as a double helix. The two stands of polynucleotides are antiparallel i.e., run in the opposite direction. The back bone is formed by the sugar-phosphate, sugar chain. The nitrogen bases are projected more or less perpendiular to this backbone but face inside. A combines with T by two hydrogen bonds where G combines with C by three hydrogen bonds. One full turn of the helical strand would involve ten base pairs. The length of one coil is 34 A and the distance between base pair is 3.4 A . This type of DNA is called B-DNA.

Question 4.
Comment on the statement living sate is a non-equilibrium steady – state to be able to perform work1.
Answer:
Several chemical compounds are present in a living organism called metabolites or biomolecules are present at concentrations characteristic of each of them. For example the blood concentration of Glucose in a normal healthy individual is 4.5 – 5.0 mp. All living organisms exist in a steady state characterised by concentrations of each of these biomolecules. These are in a metabolic flux. Any chemical or physical process moves spontaneously to equilibrium, The steady state is a non equilibrium state. The systems cannot work at equilibrium. As living organisms work continuously, they cannot afford to reach equilibrium. Hence the living state is a non-equilibrium steady state to be able to perform work.

Question 5.
Is rubber a primary metabolite or a secondary metabolite ? Write four sentences about rubber.
Answer:
Rubber is a secondary metabolite. Rubber is an elastic hydro carbon polymer that was orginally derived from latex, d milky colloid produced-by some plants. The purified form of natural rubber is the chemical polyisoprene. It is used extensively in many products, as a synthetic rubber. It is normally very strechy and flexible and extremely waterproof.

Question 6.
Dynamic state of body constituents is a more realistic concept than the fixed concentrations of body constituents at any point of time. Elaborate.
Answer:
All the living organisms contains biomolecules in certain concentrations, which have a turn over. They are constantly being changed into some other biomolecules and also made from some other biomolecules. This breaking and making occurs through chemical reactions called metabolism. Each of the metabolic reactions results in the transformation of biomolecules.

Example :
Removal of CO₂ from amino acids making into an amine removal of amino group in a nucleotide base. Majority of metabolic reactions do not occur in isolation but are linked to some other reactions. Metabolites are converted into each other by a series of linked reactions called metabolic pathways. Flow of metabolites through metabolic pathways has a definite rate and direction like auto¬mobile traffic. This metabolic flow is called the dynamic state of body constituents.

Long Answer Type Questions

Question 1.
What are secondary metabolites? Enlist them indicating their usefulness to man.
Answer:
Metabolic products that do not have identifiable functions in the host organism are called secondary Metabolites. They are Alkaloids, Flavinoids, Rubber, Essential oils. Antibiotics, Coloured pigments, Scents, Gums and Spices.

Alkaloids :

1. Alkaloids from plant extracts have been used as ingredients in lotions (liquid medicine) and poisons.
2. Ancient people used plant extracts containing alkaloids for treating a large number of ailments including snake bite, fever and insanity.

Flavinoids :
These are a widely distributed group of polyphenolic compounds with health related properties which include anticancer, antiviral, anti inflamatory activities, effects on capillary fregility and can ability to inhibit human platelet aggregation.

Rubber :

1. Uncured rubber is used for cements for adhesive, insulating and friction tapes. The flexibility of rubber is dften used in hose, tires and rollers for a wide variety of devices.
2. Its elasticity makes it suitable for various kinds of shock absorbers.
3. It is impermeable to gases, it is useful in the manufacture of articles such as air hoses, balloons, bulls and cushions.

Essential oils :

1. An essential oil is a concentrated hydrophobic liquid containing volatile aroma compounds from plants.
2. These are also known as volatile oils, etheral oils or aetherolea. These are used in aromatherapy.

Antibiotics :

1. Antibiotics are defined as chemicals of natural organic origin that will till certain harmful pathogens.
2. They should not be toxic or have side effects to the host.
3. An antibiotic is a substance that harms or destroys the bacteria that cause infection and disease. We take antibiotics to fight bacterial infections.

Spices :

1. Asafoetida is a good remedy for whooping cough and stomach ache caused due to gas.
2. Cardamon (elachi) helps to control bad breath and digestive disorder.

Question 2.
What are the processes used to analyse elemental composition, organic constituents and inorganic constituents of living tissue? What are the inferences on the most abundant constituents of living tissue? Support the inferences with appropriate data.
Answer:
Take a living tissue (a vegetable or a piece of liver) and grind it in trichloroacetic acid with the help of Mortar and pestle. The thick slurry obtained was strained through a cheese cloth or cotton, two fractions are formed. The first one is filtrate or acid soluble pool which consists of thousands of organic compounds and the second one is retentate or acid insoluble fraction. All the carbon compounds that we get from living tissues can be called biomolecules.

Weigh a small amount of living tissue and dry it. After the evaporation of water the material is burnt to oxidise all the carbon compounds. The remaining ash contains inorganic elements like sodium, potassium, calcium and magnesium and inorganic compounds like sulphate, phosphate etc., therefore chemical analysis gives elemental composition of living tissues in the form of hydrogen, oxygen, chlorine, carbon etc. From a biological point of view the organic constituents are classified into amino acids, nucleotide bases, fatty acids etc.

Question 3.
Nucleic acids exhibit secondary structure. Describe through Watson – Crick Model.
Answer:
Nucleic acids exhibit a wide variety of secondary structure. For example, one of the secondary structures exhibited by DNA is Watson – Crick model. According to this, DNA exists as a double helix. The two stands of polynucleotides are antiparallel i.e., run in the opposite direction. The back bone is formed by the sugar-phosphate, sugar chain. The nitrogen bases are projected more or less perpendiular to this backbone but face inside. A combines with T by two hydrogen bonds where G combines with C by three hydrogen bonds. One full turn of the helical strand would involve ten base pairs. The length of one coil is 34 A° and the distance between base pair is 3.4 A°. This type of DNA is called B-DNA.

Question 4.
What is the difference between a nucleotide and nucleoside? Give two examples of each with their structure.
Answer:

Question 5.
Describe various forms of lipid using a few examples.
Answer:
Lipids are orgnaic compounds which are insoluble in water. They are fats and fatty acids, oils, triglycerides, phospholipids, steroids, waxes etc.,

Fattyacids :
They have a carboxyl group attahed to an R group. The R group could be a methyl or ehtyl or higher number of CH₂ groups. For example pulmitic acid has 16 carbons including arboxyl carbon. Fatty acids could be saturated or unsaturated.

Glycolipids :
They are composed mainly of mono-di and trisubstituted Glycerols, the most well-known being the fatty acid esters of Glycerol called Triglycerides. In these compounds, the three hydroxyl groups of Glycerol are each esterified usually by different fatty acids. They function as a food store.

Phospholipids :
Some lipids have phophorous and phsophorylated organic compound in them called phospholipids. They are found in cell membrane.
Ex : Lecithin.

Intext Questions

Question 1.
What are macromolecules? Give examples.
Answer:
Acid Insoluble, high molecular weight substances of the living tissue are called macromolecules or Biomacromolecules.
Ex : Polysacharides, Polypeptides, Nucleic acid.

Question 2.
Illustrate a glycosidic, peptide and a phospho-diester bond.
Answer:
Glycosidic Bond :
Bond involving carbons of adjacent sugar molecules.

Peptide bond :
Bond between aminoacids in a protein.

Phosphodiester bond :
The bond between the phosphate and hydroxyl group of sugar is called ester bond.

The ester bond an either side of phosphate is called phosphodjester bond.

Question 3.
What is meant by tertiary structure of proteins?
Answer:
3 Dimensional view of a protein essential for many biological activities is called Tertiary structure of proteins.

Question 4.
Find and write down structures of 10 interesting small molecular weight bio-molecules. Find if there is any industry, which manufactures the compounds by isolation. Find out who are the buyers.
Answer:
Aminoacids, Monosacharide and disacharide sugars, fatty acids, Glycerol, Nucleotides, Nucleosides, nitrogen bases are the biomolecules which have low molecular weight.

Question 5.
Proteins have primary sturcture. If you are given a method to known which aminoacid is at either of the two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?
Answer:
Proteins are the polymers of a-amino acids. The linear sephence of amino acid is called primary structure. Most of proteins starts with an amino acids called Methionine but they are not homologous. Hence, the detection of N termeni (Or) C-termeni amino acid is not give information of homogeneity of a protein.

Question 6.
Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (eg. Cosmotics etc.,)
Answer:
Erythroproteins, Monoclenal Antibodies. Interferons thrombin and fibrinogen. Insulin, Renin, Proteins are also commonly used in the manufacture of cosmotics, toxins and as biological buffers.

Question 7.
Explain the composition of triglyceride.
Answer:
Triglycerides consist entirely of just three elements, carbon, Hydrogen and oxygen. The molecules of life all have carbon ‘back bones’ meaning the basic shape of the molecule comes from of connected carbon atoms. There are many hydrogen atoms connected to the carbon backbone of a molecule of life. Each triglyceride contains a small amount of O₂.

Triglycerides have four basic parts, one glycerol molecule and three fatty acids. A glycerol is a three carbon molecule the fatty acids are long chains of carbon and hydrogen with two oxygen atoms at one end to produce a Triglyceride.

Question 8.
Can you describe what happens when milk is converted into curd or yoghurt based on your understanding of proteins.
Answer:
Proteins are macromolecules formed by the polymerization of amino acids. Structurally proteins are of 4 types.

1. Primary
2. Secondary
3. Tertiary
4. quaternary.

Structure :
More than are polyreptide chains assemble to form the quaternary structure. When milk is converted to curd, these complex proteins get denatured, thus converting globuler proteins into fibrous proteins. Therefore by the process of denaturation, the secondary and tertiary structures of proteins are destroyed.

Question 9.
Draw the sturcture of the amino acid, alanine.
Answer:

Question 10.
What are gums made of? Is Fevicol different?
Answer:
Gums are hetero polysacharides. They are made from two or more different types of mono sacharides. On the other hand, Fevicol is polyvinyl alcohol (PVA) glue. It is not a polysacharide.

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 9th Lesson Cell: The Unit of Life Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 9th Lesson Cell: The Unit of Life

Very Short Answer Questions

Question 1.
What is the significance of vacuole in a plant cell?
Answer:
In plnats, vacuole contains sap mainly composed of water, metabolic byproducts, excretions and other waste materials. They also play an important role in osmoregulation.

Question 2.
What does ‘s’ refer to a 70s and 80s ribosome?
Answer:
In both 70 ‘s’ ribosomes, and 80 ‘S’ ribosomes, ‘s’ stands for the sedimentation coefficient, [expressed in Svedberg unit]. It is indirectly a measure of density and size.

Question 3.
Mention a single membrane bound organelle which is rich in hydrolytic enzymes.
Answer:
Lysosome.

Question 4.
What are gas vacuoles 7 State their functions.
Answer:
Gas vacuoles are the aggregates of a number of small hollow cylindrical vesicles present in the cytoplasm of the floating purple and green photosynthetic bacteria. They are peroneable to atmospheric gases and help the bacteria to help floating on the surface of water.

Question 5.
What is the function of a polysome?
Answer:
Several ribosomes may attach to a single m-RNA and form a chain called polyribosomes or polysome. The ribosomes of a polysome translate the m-RNA into proteins.

Question 6.
What is the feature of a metacentric chromosome?
Answer:
The metacentric chromosome has middle centromere forming two equal arms of the chromosome.

Question 7.
What is refered to as satellite chromosome?
Answer:
A few chromosomes have non-staining secondary constrictions at a constant location, which gives the appearence of a small fragment called satellite. The chromosome with satellite is called satellite chromosome.

Question 8.
What are microbodies? What do they contain?
Answer:
Peroxysomes and glyoxisomes are called Microbodies. Peroxysomes are involved in the convertion of fatty acids into carbohydrates and in photorespiration. Glyoxysomes contain the enzymes of glyoxylatic cycle which convert stored lipids to carbohydrates.

Question 9.
What is middle lamella made of? What is its functional significance?
Answer:
Middle lamella is mainly composed of calcium pectate and holds or glues the different . neighbouring cells together.

Question 10.
What is osmosis?
Answer:
Movement of molecules or ions or water from low concentrated place to high concentrated place through semi permeable membrane is called osmosis.

Question 11.
Which part of the Bacterial cell is targeted in gram staining?
Answer:
Chemical composition of the cell envelope.

Question 12.
Which of the following is not correct? a) Robert Brown discovered the Cell. b) Schleden and Schwann formulated the cell theory c) Virchow explained that cells ae formed from pre existing cells d) Aunicellular organism carries out its lofe activities with in a single cell.
Answer:
a) is not correct. The cell was discovered by “Robert Hooke”.
b) is correct.
c) is correct.
d) is correct.

Question 13.
New cells generate from a) Bacterical Fermentation b) regeneration of old cells c) Pre existing cells d) Abiotic materials.
Answer:
Pre existing cells.

Question 14.
Match the following :

Answer:

Question 15.
Which of the following is correct : a) Cells of all living organisms have a nucleus b) Both animals and plant cells have a well defined cell wall, c) In Prokaryotes, there are no membrane bound organcells. d) Cells are formed de novo from abiotic materials.
Answer:
“C” is correct. In Prokaryotes, there are no membrane bound organelles like chloroplast. Mitochondria, ER of Golgi complex.

Short Answer Type Questions

Question 1.
Describe the cell organelle which contains chlorophyll Pitments.
Answer:
The chlorophyll containing cell organelle is chloroplast. They are found in mesophyll cells of the leaves. They are lens shaped, oval, spherical or discoid or ribbon shaped, having 5 – 10 pm length and 2 – 4 pm width. They varies in number from one (chlamydomonus) to 20 – 40 per cell in the masophyll.

Each chloroplast is a double membrane bound structure with periplastidial space in between them. Inner to Inner membrane fluid filled space is present called the stroma. A number of organised flattened membranous sacs called the thylakoids are present in the stroma, which are arranged like a pile of coins called grana. The grana are inter connected by flat membranous tubules called stroma lamellae.

The stroma of the chlorplast contains enzymes required for the synthe is of carbohydrates proteins, small circular double stranded DNA molecules and 70s Ribosomes, photosynthetic pigments are present in thylakoids, involve in high reaction of photosynthesis.

Question 2.
Describe the structure and function of celi organelle that can be considered as power house of the cell.
Answer:
Mitochondria is considered as power house of the cell.

Each mitochondria is sausage shaped or cylindrical having a diameter of 0.2 to 1.0 µm and length of 1.0 to 4.1 µm. Mitochondria is a double membrane bound cell organelle of which outer membrane is smooth and inner membrane forms a number of infoldings towards inside called the cristae.

The space inner to inner membrane is fluid filled called matrix. The cristae contains several stalked particles called oxysomes or elementary particles. Mitochondria are the sites of aerobic respiration. They produce cellular energy in the form of ATP hence they are called power house of the cell. The matrix also contains single circular DNA molecule, 80s RNA molecules and 70s Ribosomes.

Question 3.
Describe the structure of nucleus.
Answer:
The cell organelle that controls all the metabolic activities of the cell is called nucleus. It was first described by Robert Brown in 1831. Each nucleus is a spherical ball like structure, consists of 4 parts. They are A) Nuclear membrane, B) Nucleoplasm, C) Chromatin material and D) Nucleolus.

A) Nuclear membrane :
Nucleus is covered by double layered lipoprotenous membrane with perinuclear space in between them. At certain places, Nuclear pores are present which acts as passages between nucleoplasm and the cyloplasm in both directions.

B) Nucleoplasm :
Inner to nuclear membrane, nucleoplasm in present. It consists of chromatin material, nucleolus and RNA molecules nucleoplasm is also called karyoplasm.

C) Chromation material:
Inter phase nucleus has a loose and indistinct network of nucleoprotein fibres called chromatin. It contains DNA and some basic proteins called histones some on histone proteins and also RNA. The chromation is furthur classified into Heterochromatin and Euchromatin of these, Euchromatin is delicate, less condensed and active.

D) Nucleolus :
It is a dense, spherical shaped structure present inside the nucleus. It plays an indirect role in protein synthesis by producing rib somes.

Functions :

1. It controls the heriditary characterestics of an organism.
2. It is responsible for the protein synthesis, cell division, growth and differentiation.
3. It controls all the metabolic activities of the cell.

Question 4.
Comment on the cartwheel Structure of centriole.
Answer:
Centrosome is an organelle usually containing two cylindrical structure called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like “cartwheel”. They are made up of nine evenly spaced peripheral fibrils of tubulin. Each of the peripheral fibril is a triplet. The adjacent triplets are also linked.

The central part of the centriole is also proteinaceous and called the ‘Hub’, which is connected with tubules of the peripheral triplets by radial spokes made of protein. The centrioles form the basal body of cilia and flagella and spindle fifres that give rise to spindle apparatus during cell division in animal cells.

Question 5.
Briefly describe the cell theory.
Answer:
Schwann proposed the hypothesis that the bodies of animals and plants are composed of cells and products of cells. Schleiden, examined a large number of plants and observed that all plants are composed of different kinds of cells which form the tissues of the plant. Schleiden and Schwann together formulated cell theory. But this theory did not explain as to how new cells were formed. Rudolf virchow first explained that cells divided and new cells are formed from pre-existing cells. He modified the hypothesis of Schleiden and Schwann and give a definite shape to cell theofy- Which states that,

1. all living organisms are composed of cells and products of cells
2. all cells arise from the pre-existing cells.

Question 6.
Differentiate between Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER).
Answer:

Question 7.
Give the biochemical composition of plasma membrane. How are lipid molecules arranged in the membrane?
Answer:
Plasma membrane is made up of lipids that are arranged in fila. Later biochemical investigations clearly revealed that the cell membrane also posses protein and Carbohydrate. The lipids are arranged with in the membrane, with the polar (hydrophilic) head towards the outer sides and the tail (hydrophobic)towards the inner part. With in the lipids, proteins are classified into integral or peripheral proteins.

An improved model Nicholson (1972) widely accepted as “Fluid Mosaic Model”, According to this, the quasi fluid nature of lipid enables lateral movement of proteins within the overall bilayer.

Question 8.
What are plasmids? Describe their role in Bacteria.
Answer:
Plasmids are small circular DNA molecules outside the genomic DNA. The plasmid DNA conferms certain unique phenotypic characters to Bacteria i.e„ resistance to antibiotics, the plasmid DNA is used to monitor bacterial transformation with foreign DNA.

Question 9.
What are histones? What are their functions?
Answer:
Histones are the proteins closely associated with DNA molecules. They are responsible for structure of chromatin and play an important role in the regulation of gene expression. Five types of histone have been identified H1, H2A, H2B, H3and H4. The other 4 types of histones associate with DNA to form Nucleosomes. They plan an instrumental role in the regulation of many important biological process involving DNA such as transcription, DNA repair & cell cycle.

Question 10.
What is cytoskeleton? What functions is it involved in?
Answer:
An elaborate network of filamentous proteinaceous structures present in the cytoplasm is called cytoskeleton. Eukaryotic cells contain three major components of cytoskeleton, microfilaments, intermediate filaments and microtubules. Cytoskeleton is involved in mechanical support, maintanance of cell shape, cell motility, intracellular transport, signaling across the cell and Karyokinesis.

Question 11.
What is endomembrane system? What cell organelles are not included in it? Why?
Answer:
A group of cell organelles with coordinate functions is called endomembrane system. Mitochondria, chloroplast and peroxisomes are not associated with endomembrane system. Because their functions are not coordinated with the ER, Golgicomplx, lysosomes and vacuoles.

Question 12.
Distinguish between Active transport and Passive transport?
Answer:

Question 13.
What are mesosomes? What do they help in?
Answer:
Extensions of the plasma membrane into the cell are called mesosomes. These extensions are in the form of vesicles, tubules and cisternae. They help in cell wall formation, DNA replication and its distribution to daughter cells, help in respiration, secretion processes to increase the surface area of the plasma membrane to absorb nutrients and enzymetic content.

Question 14.
What are nucleosomes? What are they made of?
Answer:
Under electron Microscope, chromation appears as “beads on string”. These beads are known as Nucleosomes. Atypical nucleosome contains 20,0 bp of DNA double helix wrapped around a core of histone octamer having two copies of each of four types of histone proteins viz H2A, H2B, H3 and H4. HI Histone lies outside the nucleosome core and seals the two turns of DNA by binding at the point where DNA enters and leaves the core. The DNA continues between two nucleosomes is called linker DNA.

Question 15.
How do neutral solutes move across the Plasma membrane? Can the polar molecules also move a cross it in the same way? If not then how are these transported across the membrane.
Answer:
Neutral solutes move a cross the plasma membrane by the process of simple diffusion along the concentration gradient i.e., from higher concentration to the lower concentration. Polar molecules can not pass through the non polar lipid bilayer. Beacuse they require a carriers protein of the membrane to facilitate their transport across the membrane.

Question 16.
Name two cell organelles that are double membrane bound. What are the .characteristics of these two organelles. State their functions and draw labelled diagrams of both.
Answer:
Mitochondria and chloroplasts are double membrane found cell organelles. Mitochondria is a rod shaped cell organelle or sausage shaped or cylindrical having a diameter of 0.2 to 1.0 mµ and length of 1.0 to 4 mµ. They are the sites of aerobic respiration and produce cellular energy in the form of ATP hence they are called “Power houses of the cell”.

Chloroplasts are lens shaped, oval, spherical or ribbon like bodies having 5-10 mp length and 2.4 mµ width. They contain enzymes required for the synthesis of carbohydrates and proteins.

Question 17.
What are the characteristics of a prokaryotic cell?
Answer:

1. Nuclear envelope is absent. Cell shows a single circular, supercoiled naked DNA called nucleoid.
2. Nucleous is absent.
3. Endomembrane system is absent.
4. Mitochondria, plastids, lysosomes, peroxisomes and cytoskeleton are absent.
5. Respiratory enzymes are located in cell membrane.
6. 70 s type of ribosomes are present.
7. Cell is surrounded by a cell wall, made up of polysaccharides, lipids and proteins.
8. Cells divide amitotically.

Question 18.
Cell is the basic unit of life. Discuss in brief.
Answer:
Schleiden explained a large number of plants and observed that all plants are composed Of different kinds of cells which form the tissues of the plant. All living organisms are composed of cells and products of cells. All cells arise from pre-existing cells. A cell is the basic unit of life. It is the smallest unit (or building block) of a living organism that is capable of life. Within cells, there are structures that work togehter to allow the cell to live. Some structures transport materials throughout the cell other structures make food and others release energy for the cell to use.

When you look at a green plant, there is a green chemical called chlorophyll. This chemical enables’ plants to use the sun’s energy to make food for themselves. This is found in the cell part called the chloroplast. The nucleus of plant cells is the control centre of the cells. It directs everything a cell’does. The cytoplasm’ is a fluid inside the cell and the cell parts float in it.

Question 19.
Both lysosomes and vacuoles are endomembrane structures, yet they differ in terms of their functions. Give a critical comment.
Answer:
Lysosomes are the membrane found vesicular structures, filled with hydrolytic enzymes capable of digesting proteins, lipids and nucleic acids. These enzymes are optimally active at the acidic pH. Under starvation, lysosomes digest cellular contents by releasing hydrylyzing enzymes and cause death of cells called acetolysis.

Vacuole is the membrane found space present in the cytoplasm, contains mainly water, metabolic bye products, excretions and other waste materials. In some plant cells, vacuolar sap contains some pigments like anthocyanin which impart colour to the plant part; The vacuole is covered by single membrane called Tonoplast. The tonoplast facilitates the transport of number of ions and other materials against concentration gradients, into the vacuole. Vacuoles play an important role in osmoregulation.

Question 20.
Briefly give the contributions of the following scientists in formulating the cell theory, a) Rudolf Virchow b) Schleiden and Schwann.
Answer:
a) Rudolf Virchow :
He first explained that cells divided and new cells are formed from pre existing cells (Orhnis cellula – e cellula). He modified the hypothesis of Schleiden and Schwann to give the cell theory a final shape cell theory states that all living organisms are composed of cells and porducts of cells. All cells arise from pre-existing cells.

b) Schleiden and Schwann :
In 1838, Mathias Schleiden, examined a large number of plants and observed that cell plants are composed of different kinds of cells which form the tissues of the plant. Theodoe Schwann (1839), a British Zoologist, studied different types of animal cells and reported that cells had a thin outer layer which is called plasma membrane. He also concluded, based on his studies ‘ on plant tissues, the presence of cell wall is a unique character of the plant cells. Based on this, Schwann proposed the hypothesis that the bodies of animals and plants are composed of cells and products of cells.

Question 21.
Is extra genomic DNA present in prokaryotes and eukaryotes? If yes, indicate their location in both the types of organisms.
Answer:
Yes, Extra genome DNA is present in prokaryotes and Eukaryotes. In Prokaryotes, extra genome DNA (plasmid) is present along with Nucleoid and floats freely with in the cytoplasm. In Eukaryotes well defined nucleus is present with specific number of chromosomes. But a extra genome DNA is present in cell organelles like Mitochondria and chloroplast. These two organelles are believed to have originally been independent prokaryotes.

Question 22.
Structure and function are correlatabie in living organisms. Can you justify this by taking plasma membrane as an example?
Answer:
Plasma membrane is made up of bilayered lipids. Later biochemical investigations clearly revealed that the plasma membrane also possess protein and carbohydrate. The lipid component of the membrane is phosphoglycerides. The proteins present in the membrane are integral or peripheral type. Integral proteins are burried in the membrane. Peripheral proteins are on the surface of the membrane. Plasma membrane play an important role in transport of the molecules into and out of cells. This membrane is selectively permeable to some molecules present on either side of it. Many molecules can move freely across the membrane without utilising energy is called passive transport of ions.

As the polar molecules cannot pass through non-polar lipid bilayer, they require carrier protein of the membrane to facilitate their transport across the membrane against concentration gradient i.e, from lower to higher concentration. Such a transport by utilising energy is called active transport of ions.
Ex : Na⁺/K⁺ Pump.

Question 23.
Discuss briefly the role of nucleolus in the cells actively involved in protein synthesis.
Answer:
The nucleoplasm contains nucleous and chromatin material. The nucledi are spherical structures, involved in the synthesis of ribosomal RNA. Longer and more numerous nucledi are present in cells actively carrying out protein synthesis. The nucleous develops from the secondary constriction of a specialized chromosome known as the “nucleolar organiser”. The nucleolus consists of RNA, protein and small amount of DNA. It disappears during the end of prophase and reappears at the end of Telophase. It is also called Plasmosome or ‘Ribosomal factory1. –

Question 24.
Explain the association of Carbohydrate to the plasma membrane and its significance.
Answer:
Plasma membrane is made up of lipids and proteins. Later, biochemicals investigations clearly revealed that the cell membrance also possess protein and carbohydrate. They are present s short, unfranched or branched chains of sugars (oligosaccharides) attached either to exterior ectoproteins or the polar ends of phospholipids at the external surface of the plasma membrane. All types of oligosaccharides of the plasma membrane are formed by various combinations of six principle sugars like D-galactose, D-mannose, L-fructose, N-acetyl neuramic acid (sialic acid), N-acetyl D-glucosamine and N-acetyl – D – gaiactosamine.

Significance :

1. Glycophorins are found to contain certain antigenic determinants for the ABO Blood groups and MN flood groups.
2. Sialic acid contents a high negative change to the cell surface of erythrocyte.

Question 25.
Multicellular organisms have division of labour. Explain.
Answer:
Multicellular organisms are made up of millions and trillions of cells. All these cells perform specific functions. All the cells specialised for performing similar functions are grouped together as tissues in the body. Hence a particular function is carried out by a group of cells at a definite plae in the body. Similarly different functions are carried out by different groups of cells in an organism. By dividing labour, multicelled organisms are able to complete more complex tasks.

Question 26.
What are nuclear pores? State their function.
Answer:
The nuclear membrane is interupted by a minute pores. Which are formed by the fusion of its two membranes. These pores are called Nuclear pores. These regulate the transportation of molecular, between the nucleus and cytoplasm. In eukaryotic cells, the nucleus is separated from the cytoplasm by nuclear envelope. The envelope safeguards the DNA contained in the nucleus. Inspite of this barrier, there is still communication between the nucleus and cytoplasm by nuclear pores.

Each nuclear pore is a large complex of proteins that allows small molecules and ions to freely pass of diffuse into or out of the nucleus. Nuclear pores also allow necessary proteins to enter the nucleus from the cytoplasm ; if the proteins have special sequences that indicate, they belong to the nucleus. These sequences tags are known as nuclear localization signals. Similarly RNA transcribed in the nucleus and proteins are destined to enter the cytoplasm have nuclear export sequences that tag them for release through the nuclear pores.

Long Answer Type Questions

Question 1.
What structural and functional attributes must a cell have to be called a living cell?
Answer:
All living organisms are made up of cells and product of cells. Cell is the basic structural and functional unit of living organism. Each cell have different organelles and perform different functions.

1. Cells obey laws of energetics i.e., they transform energy.
2. Cells highly structured with emergent properties.
3. Cells have an evolutionary origin.
4. Cells metabolize means possess metabolic pathways, process nutrients, self adjust to environment.
5. Cells self replicate – nucleic acids, ribosomes.
6. Cells osmoregulate – vacuoles, vesicles.
7. Cells communicate – Glycoproteins.
8. Cells shows animation Cyclosis.
9. Cell grow, divide and differentiate.
10. Cells die.

Question 2.
Eukaryotic cells have organelles which may
a) Not be bound by a membrane
b) Bound by a single membrane
c) Bound by a double membrane.
Give the various sub-cellular organelles into these three categories.
Answer:
a) Not be bound by a membrane : Nucleolus
b) Bound by a single membrane : Lysosomes, vacuoles.
c) Bound by a double membrane : Mitochondria, chloroplast, nucleus..

Question 3.
The genomic content of the nucleus is constant for a given species where as the extra chromosomal DNA is found to be variable among the members of a population. Explain.
Answer:
In Prokaryotes (Bacteria) in addition to the genomic DNA, small circular DNA molecules are present in the cytoplasm. These small DNA molecules are called plasmids. They confers unique phenotypic character to such bacteria (i.e.,) resistance to antibiotics. It is also used to monitor bacterial transformation with foreign DNA.

In Eukaryotes, extra DNA molecules are present both in chloroplast (stroma) and mitochondria (matrix). Because of the presence of this DNA molecules, they are treated as self – autonomous cell organelles.

Question 4.
Justify the statement. “Mitochondria are power houses of the cell.
Answer:
Mitochondria is sausage – shaped or cylindrical structure having a diameter of 0.2 to 1.0 µm and length 1.0 to 4.1 µm. Each mitochondrion is a double membrane bound structure with outer membrane and inner membrane dividing its lumen distinctly into two aqueous compartments. The inner compartment is called the matrix. The outer membrane forms the continuous limiting boundary of the organelle. The inner membrane forms a number of infoldings called the cristae towards the matrix.

The cristae increase the surface area. The mitochondria are the sites of aerobic respiration. They produce cellular energy in the form of ATP, hence they are called “power houses” of the cell. The matrix also possesses a single circular DNA molecule, a few RNA molecules, ribosomes (70s) and the compartments required for the synthesis of proteins.

Question 5.
Is there a species specific or region specific type of plastids? How does on distinguish one from the other?
Answer:
Plastids are species specific and are found in all plant cells and in euglenoides. They bear some specific pigments thus imparting specific colours to the part of the plant which posseses them. Based on the type of pigments, plastids are classified into three types. They are Leucoplasts, Chromoplasts, Chloroplasts.

Leucoplasts :
They are the colourless plastids which store food materials. Based on the storage product, they are of 3 types namely Amyloplasts (store starch), elaioplasts (store oils) and aleuroplasts (store proteins).

Chromoplasts :
They are coloured pigments which were in yellow, orange or red in colour. In these plastids, fat soluble carotenoids like carotene and xanthophylls are present which imparts orange, red or yellow colour.

Chlorplasts :
These are green coloured plastids which help in synthesis of food materials by photosynthesis. They contain chlorophyll and carotenoid pigments which trap light energy. Each chloroplast is a oval or spherical, double membrane bound cell organelle. The space present inner to inner membrane is called stroma. A number of organised flattened membranas sacs called thyloukoids are present in the stroma.

Thylakoids are arranged in stacks like the piles of coins called grana. The thylakoids of the different grana are connected by membranous tubules called the stroma lamellae. The stroma of the chloroplast contains enzymes required for the syntheis of carbohydrates and proteins.

Question 6.
Write the functions of the following.
a) Centromere b) Cell wall c) Smooth ER d) Golgi complex e) Centrioles
Answer:
a) Centromere :
It is required for proper chromosome seggregation. The centromere keeps the two sister chromatids together. It is also where the chromosomes attaches to the spindle apparatus during Mitosis and Meiosis.

b) Cell wall:
It gives a definite shape to the cell and protects the cell from mechanical damage and infection. It also helps in cell-to-cell interaction and acts as a barrier to undesirable macromolecules.

c) Smooth ER :
It helps in synthesis of lipids, metabolism of carbohydrates and calcium concentration, drug detoxification and attachment of receptors on cell membrane proteins. The smooth ER also contains the enzyme Glucose – 6 – Phosphatase which converts Glucose – 6 – Phosphate to Glucose.

d) Golgi complex :
It is the important site for the formation of glycoproteins and glycolipids. It is also involved in the synthesis of cell wall materials and also plays a main role in the formation of cell plate during cell division.

e) Centrioles :
They form the basal body of cilia and flagella and spindle fibres that give rise to spindle apparatus during cell division in animal cells.

Question 7.
Are the different types of plastids interchangeable? If yes, give examples where they are getting converted from one type to another.
Answer:
Yes. Plastids are inter changeable in form. Generally three types of plastids are present in plant cells namely, leucoplasts (storage), chromoplasts (coloured, attraction) and chloroplasts (synthesis of food). Chromoplasts are coloured plastids (orange, yellow or red), occurs in the cells of petals, fruits etc.

They contain less chlorophylls.and more carotenes (orange) or (red) and xanthophylls (yellow). The red colour of tomato is due to the presence of lycopene in the chromoplasts. The chromoplasts of red algae contain phycocyanin and phycoerythrin. The chromoplasts of Brown algae contain fucoxanthin.

Depending upon circumstances, one type of plastid may be converted into another type.
For Ex :

1. The leucoplasts in stem tubers of potato, on exposure to sunlight transofrm into chloroplasts.
2. In capsicum, the cells of ovary consists of leucoplasts. When ovary changes into fruit, leucoplasts are transformed into chloroplasts. When the fruit ripens chloroplasts are changed into chromoplasts.

Question 8.
Describe the structure of the following with the help of labelled diagrams.
i) Nucleus ii) Centrosome.
Answer:
i) Nucleus :
It was first discovered by Robert Brown. Later the material of the nucleus stained by the basic dyes was given the name chromatin by Flemming. The interphase nucleus has highly extended and elaborate nucleoprotein fibres called chromatin, nuclear matrix and one or two nucleoli. Electron micrsocopy has revealed that the nucleus is covered by double layered nuclear envelope, with a space between called perinuclear space, forms a barrier between materials present inside the nucleus and the cytoplasm.

The outer nuclear envelope remains continuous with the endoplasmic reticulum and also bears ribosomes. At a number of places, the nuclear envelope is interrupted by nuclear pores, which allows the passage of RNA and protein molecules in both directions between the nucleus and cytoplasm. The nucleoplasm contains nucleolus and chromatin. The nucleoli and spherical structures, involved in active ribosomal RNA synthesis. The chromatin, in different stages of cell division, become chromosomes. They contain DNA and some basic proteins called histones and also RNA. Nucleus plays an important role in biogensis of ribosomes. It plays a significant role in mitosis.

ii) Centrosome :
Centrosome is an organelle usually containing two cylindrical structures called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like cartwheel. They are made up of nine evenly spaced peripheral fibrils of tubulin. Each of the peripheral fibril is a triplet. The adjacent triplets are also linked. The central part of the centriol is also proteinaceous and called the ‘hub’, which is connected with tubules of the peripheral triplets by radial spokes made of protein. The centrioles form the basal body of cilia and flagella and spindle fibres that give rise to spindle apparatus during cell divirsion in animal cells.

Question 9.
What is a centromere? How does the position of centromere form the basis of classification of chromosomes. Support your answer with a diagram showing the position of centromere on different types of chromosomes.
Answer:
Centromere is the region of chromosome that becomes attached to spindle fibres. Special proteins surrounded the centromere. They form a disc shaped structures called kinetochores.

Each chromosome shows centromere at a specific position. Basing on the position of centromere, four types of chromosomes are recognised. They are :
1) Metacentric :
“If the centromere is situated at mid point of a choromosome”. It is “V” shaped and consists of two equal arms.

2) Sub-metacentric :
“If the centromere is situated slightly away from mid point of a chromosome”. It is ‘L’ shaped. It consists of two unequal arms.

C) Acrocentric :
If the centromere is situated at the sub terminal position of a chromosome. It is rod shaped or “J” shaped. It consists of very long arm and a very small arm.

D) Telocentric :
“If the centromere is situated at the terminal position of a chromosome”. It is “i” shaped and has only one arm.

Intext Questions

Question 1.
What is a mesosome in a prokaryotic cell? Mention the functions that it performs.
Answer:
Extensions of plasma membrane into the cell in the form of vesicles, tubules and lamellae are called mesosomes. They help in cell wall formation, DNA Replication and its distribution to daughter cells, in respiration, secretion and to increase the surface area of plasma membrane (absorption of nutrients) and enzymatic content.

Question 2.
How do neutral solutes move across the plasma membrane? Can the polar molecules also move across it in the same way? If not, then how are these transported across the membrane?
Answer:
Neutral solutes move across the plasma membrane by the process of simple diffusion along the concentration gradient. Polar molecules cannot pass through the non polar lipid bilayer, they require a carrier protein of the membrane to facilitate their transport across the membrane.

Question 3.
Name the two cell-organelles that are double membrane bound. What are the characteristics of these two organelles? State their functions and draw labelled diagrams of both.
Answer:
Chloroplast and Mitochondria.

1. Chlorplast is a lens shaped, oval or spherical or even ribbon like organelle, involves in photosynthesis.
2. Mitochondria is a sausage shaped or cylindrical cell organelle, involved in Respiration.

Question 4.
What are the characteristics of prokaryotic cells.
Answer:
i) Nuclear envelope is absent. Cell shows a single circular, supercoiled naked DNA called nucleoid.
ii) Nucleous is absent.
iii) Endomembrane system is absent.
iv) Mitochondria, plastids, lysosomes, peroxisomes and cytoskeleton are absent.
v) Respiratory enzymes are located in cell membrane.
vi) 70 s type of ribosomes are present.
vii) Cell is surrounded by a cell wall, made up of polysaccharides, lipids and proteins.
viii) Cells divide amitotically.

Question 5.
Multicellular organisms have division of labour. Explain.
Answer:
Multicellular organisms are made up of millions and trillions of cells. All these cells perform specific functions. All the cells specialised for performing similar functions are grouped together as tissues in the body. Hence a particular function is carried out by a group of cells at a definite plae in the body. Similarly different functions are carried out by different groups of cells in an organism. By dividing labour, multicelled organisms are able to complete more complex tasks.

Question 6.
Cell is the basic unit of life. Discuss in brief.
Answer:
Schleiden explained a large number of plants and observed that all plants are composed Of different kinds of cells which form the tissues of the plant. All living organisms are composed of cells and products of cells. All cells arise from pre-existing cells. A cell is the basic unit of life. It is the smallest unit (or building block) of a living organism that is capable of life. Within cells, there are structures that work togehter to allow the cell to live. Some structures transport materials throughout the cell other structures make food and others release energy for the cell to use.

Question 7.
What are nuclear pores ? State their function.
Answer:
The nuclear membrane is interupted by a minute pores. Which are formed by the fusion of its two membranes. These pores are called Nuclear pores. These regulate the transportation of molecular, between the nucleus and cytoplasm. In eukaryotic cells, the nucleus is separated from the cytoplasm by nuclear envelope. The envelope safeguards the DNA contained in the nucleus. Inspite of this barrier, there is still communication between the nucleus and cytoplasm by nuclear pores.

Each nuclear pore is a large complex of proteins that allows small molecules and ions to freely pass of diffuse into or out of the nucleus. Nuclear pores also allow necessary proteins to enter the nucleus from the cytoplasm ; if the proteins have special sequences that indicate, they belong to the nucleus. These sequences tags are known as nuclear localization signals. Similarly RNA transcribed in the nucleus and proteins are destined to enter the cytoplasm have nuclear export sequences that tag them for release through the nuclear pores.

Question 8.
Both lysosomes and vacuoles are endomembrane structures, yet they differ in terms of their functions. Comment.
Answer:
Lysosomes are the membrane found vesicular structures, filled with hydrolytic enzymes capable of digesting proteins, lipids and nucleic acids. These enzymes are optimally active at the acidic pH. Under starvation, lysosomes digest cellular contents by releasing hydrylyzing enzymes and cause death of cells called acetolysis.

Vacuole is the membrane found space present in the cytoplasm, contains mainly water, metabolic bye products, excretions and other waste materials. In some plant cells, vacuolar sap contains some pigments like anthocyanin which impart colour to the plant part; The vacuole is covered by single membrane called Tonoplast. The tonoplast facilitates the transport of number of ions and other materials against concentration gradients, into the vacuole. Vacuoles play an important role in osmoregulation.

Question 9.
Describe the structure of the following with the help of labelled diagrams.
(i) Nucleus (ii) Centrosome
Answer:
i) Nucleus :
It was first discovered by Robert Brown. Later the material of the nucleus stained by the basic dyes was given the name chromatin by Flemming. The interphase nucleus has highly extended and elaborate nucleoprotein fibres called chromatin, nuclear matrix and one or two nucleoli. Electron micrsocopy has revealed that the nucleus is covered by double layered nuclear envelope, with a space between called perinuclear space, forms a barrier between materials present inside the nucleus and the cytoplasm.

The outer nuclear envelope remains continuous with the endoplasmic reticulum and also bears ribosomes. At a number of places, the nuclear envelope is interrupted by nuclear pores, which allows the passage of RNA and protein molecules in both directions between the nucleus and cytoplasm. The nucleoplasm contains nucleolus and chromatin. The nucleoli and spherical structures, involved in active ribosomal RNA synthesis. The chromatin, in different stages of cell division, become chromosomes. They contain DNA and some basic proteins called histones and also RNA. Nucleus plays an important role in biogensis of ribosomes. It plays a significant role in mitosis.

ii) Centrosome :
Centrosome is an organelle usually containing two cylindrical structures called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like cartwheel. They are made up of nine evenly spaced peripheral fibrils of tubulin. Each of the peripheral fibril is a triplet. The adjacent triplets are also linked. The central part of the centriol is also proteinaceous and called the ‘hub’, which is connected with tubules of the peripheral triplets by radial spokes made of protein. The centrioles form the basal body of cilia and flagella and spindle fibres that give rise to spindle apparatus during cell divirsion in animal cells.

Question 10.
What is a centromere? How does the position of centromere form the basis of classification of chromosomes. Support your answer with a diagram showing the position of centromere on different types of chromosomes.
Answer:
Centromere is the region of chromosome that becomes attached to spindle fibres. Special proteins surrounded the centromere. They form a disc shaped structures called kinetochores.

Each chromosome shows centromere at a specific position. Basing on the position of centromere, four types of chromosomes are recognised. They are :
1) Metacentric :
“If the centromere is situated at mid point of a choromosome”. It is “V” shaped and consists of two equal arms.

2) Sub-metacentric :
“If the centromere is situated slightly away from mid point of a chromosome”. It is ‘L’ shaped. It consists of two unequal arms.

C) Acrocentric :
If the centromere is situated at the sub terminal position of a chromosome. It is rod shaped or “J” shaped. It consists of very long arm and a very small arm.

D) Telocentric :
“If the centromere is situated at the terminal position of a chromosome”. It is “i” shaped and has only one arm.

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 8th Lesson Taxonomy of Angiosperms Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 8th Lesson Taxonomy of Angiosperms

Very Short Answer Questions

Question 1.
What is ‘Omega Taxonomy’?
Answer:
Omega Taxonomy is the Taxonomy based on information from other branches i.e., Embryology, Cytology, Palynology, Phytochemistry, Serology etc., along with Morphological characters.

Question 2.
What is the Natural system of plant classification? Name the scientists who followed it.
Answer:
Plants are classified on the basis of all possible Morphological characters is called the Natural system of classification. It was proposed by ‘Bentham’ and ‘Hooker’.

Question 3.
Explain the scope and significance of ‘Numerical Taxonomy’.
Answer:
Numerical Taxonomy is a branch of taxonomy that use mathematical methods to evaluate observable differences and similarities between taxonomic groups. In this, number and codes are assigned to all the characters and the data are then processed. Each character is given equal importance and at the same time hundreds of charactes can be considered.

Question 4.
What is geocarpy? Name the plant which exhibits this phenomenon.
Answer:
Geocarpy is the development of fruits inside the soil. It is seen in Arachis (ground nut).

Question 5.
Name the type of pollination mechanism found in members of Fabaceae.
Answer:
‘Piston Mechanism’.

Question 6.
Write the floral formula of solanum plant.
Answer:

Question 7.
Give the technical description of ovary in solanum nigrum.
Answer:
Bicarpellarly, Syncarpous, bilocular,superior ovary with many ovules on swollen axile placentation. Carpels are arranged obliquely at 45°.

Question 8.
Give the technical description of anther of Allium cepa.
Answer:
In Allium cepa. Anthers are dithecs, basifixed, Introrse and dehisce longitudinally.

Short Answer Type Questions

Question 1.
Write a brief note on semi technical description of a typical flowering plant.
Answer:
The plant is described beginning with its habit, habitat, vegetative characters (root, stem, leaves) and floral characters (inflorescence, flower and its parts) followed by fruit. Then floral diagram and a floral formula are presented. In floral formula, Br stands for bracteate (bracts are present), Ebr stands for ebracteate (bracts are absent). Brl stands for bracteolate (bracteoles are present), Ebrl stands for ebracteolates

Floral formula also indicates the number of free or united (within brackets) numbers Of the each whorl and also show cohesion or adhesion of stamens. Floral diagram provides information about the number of parts of a flower, their arrangement and the relation they have with one another. The mother axis represents the posterior side of the flower indicates as a dot or circle at the top of the floral diagram. K, C, A and G are drawn in successive whorls. The bract represents the anterior side of the flower and is indicated at the bottom of the floral diagram.

Question 2.
Describe the non-essential floral parts of plants belonging to Fabaceae.
Answer:
In Fabaceae, non-essential floral parts are Calyx and Corolla.

Calyx :
Sepals 5, gamosepalous, imbricate aestivation, odd sepal anterior.

Corolla :
Petals 5, polypetalous, papilionaceous type consists of a large posterior petal (standard) two laterals (wings). Two anterior fused petals (keel) enclosing essential organs. They show vexillary/descendingly imbricate aestivation.

Question 3.
Give an account of floral diagram.
Answer:
A floral diagram representing the number of parts of flower, the structure, arrangement, aestivation, cohesion and adhesion of stamens and position with respect to mother axis. The mother axis represents the posterior side of the flower and is indicated as dot at the top of the floral diagram. K, C, A, G are drawn in successive whorls. Calyx being the outermost and the gynoecium being the centre represented by T.S. of ovary. The bract represents the anterior side of the flower and is indicated at the bottom of the floral diagram.

Question 4.
Describe the essential floral parts of plants belonging to Lilliaceae.
Answer:
The essential floral parts of Lilliaceae are Androecium and Gynoecium.
Androecium :
Six stamens in two whorls, free or Epipyllous, anthers are Dithecus Basifixed, Introrse and dehisce longitudinally.

Gynoecium :
Tricarpellary, Syncarpous, trilocular superior ovary with many ovules on axile placentation. Style is terminal and stigma is trifid and capitate.

Question 5.
Write a brief account on the class of Dicotyledanae of Bentham and Hooker’s classification.
Answer:
In Bentham and Hookers classification, the class Dicotyledonae was divided into three sub classes namely Polypetalae, Gamopetalae and Monochlamydae. Polypetalae, sub class is divided into three series namely Thalamiflorae (6 orders), disciflorae (4 orders) and Calyciflorae (5 orders). Gamopetalae, sub class is divided into three series namely Inferae (3 orders), Heteromerae (3 orders) and Bicarpellatae (4 orders). Monochlamydae was divided into eight series.

Question 6.
Explain floral formula.
Answer:
The floral formula is represented by some symbols of floral parts. In it, Br stands for bracteate (bracts are present), Ebr stands for ebracteate, (bracts are absent), Brl stands for bracteolate (bracteoles are present).

Ebrl stands for ebracteolates (bracteoles are absent).
⊕ stands for actinomorphic flower,
% stands for zygomorphic flower.

ovary. Floral formula also indicates the number of free or united members of the corresponding whorls. It also shows cohesion or adhesion of stamens.

Question 7.
Give economic importance of plants belonging to Fabaceae.
Answer:
Economic importance :

1. Pulses like red gram (Cajanus cajan), black gram (Phaseolus mungo), green gram (Phaseolus aureus), Bengal gram (Cicer aritetinum) are a rich source of proteins.
2. Pods of Dolichos, Glycine are used as vegetables.
3. Seeds of Pisurn and Arachis are edible.
4. Groundnut oil from Arachis hypogea seeds and soyabean oil from seeds of Glycine max are used in cooking.
5. The oil cake from Arachis hypogea is used as fodder.
6. The oil from the seeds of Derris indica is used in the making of medicines.
7. Goldsmiths use the seeds of Abrus precatorius for weighting.
8. Several crops are used in crop rotation due to their nitrogen fixing ability.
9. Seeds of Trigonella are used as condiment and medicine. The leaves are used as vegetable.
10. Sesbania andTephrosia are used as green manure.
11. Crotalaria and Phaseolus are used as fodder.
12. Fibre from Crotalaria is used in making ropes.
13. Indigofera yields blue dye, which is used as a fabric whitener.
14. Wood from pterocarpus is used for making musical instruments.
15. Wood from Da/zberg/a is used for making furniture.

Question 8.
Describe the essential organs of Solanaceae.
Answer:
Androecium :
There are five epipetalous stamens alternating witht he petals and are free. Anthers are dithecous, basifixed and introse. The dehiscence may be longitudinal (Datura) or Porous (So/anum).

Gynoecium :
The ovary is superior, bicarpellary and syncarpous. It is oblique in position and anterior carpel to the left at an angle of 45°. Usually bilocular occassionally unilocular (Capsicum). There are numerous anatropous ovules arranged on axile placeritation on swollen placenta. The style is terminal and stigma is capitate,

Long Answer type Questions

Question 1.
Describe the characteristics of plants belonging to Fabaceae.
Answer:
VEGETATIVE CHARACTERISTICS :
Habitat:
Most of the members of this family are mesophytes.

Habit :
The plants are annual herbs, some are shrubs, some others are trees. Some are weak stemmed that climb as twiners (Dolichos) or with tendrils (Pisum).

Root System :
It is tap root system. The roots bear root nodules in which the symbiotic nitrogen fixing bacteria. Rhizobia are present.

Stem :
It is aerial, erect, herbaceous or woody. In some plants stem is weak and prostrate or twinner (Dolichos) or a tendril Climber (Pisum, Lathyrus).

Leaf:
Leaves are cauline, alternate, stipulate, petiolate and dorsiventral. The leaf base is pulvinous. They are simple or pinnately compound, the entire leaf except stipules in Lathyrus and terminal leaflets in Pisum are modified into tendrils. In ulex, the leaflets are modified into spines. Venation is Reticulate.

Floral characteristics :
Inflorescence :
It is usually an axillary or terminal raceme (Crotalaria).

Flower :
Flowers are bracteate, bracteoles may be present or absent, pedicellate, complete, zygomorphic, bisexual, pentamerous and perigynous. The thalamus is cup shaped.

Calyx :
There are five sepals which are fused. The aestivation is valvate. The odd sepal is anterior in position.

Corolla :
It consists of five petals which are free (Polypetalous). The corolla is papilionaceous. The posterior petal is largest and is called ‘Standard petal or Vexillum’. The two lateral petals are called ‘Wings or Alae’. The two boat shaped petals are called ‘Keel or Carina’. These are fused and encloses the essential organs. The aestivation is ‘descendingly imbricate’.

Androecium :
It consists of 10 stamens occasionally there may be only 9 stamens (Abrus, Datbargis). Usually the filaments of the stamens unite to form two bundles of 9 + 1 (Diadelphous) as in Dofichos, Tephrosia, Pisum etc., or a single bundle (Monadelphous) as in Crotalaria, Arachis etc. Anthers are dithecous, introrse and dehisce longitudinally.

Gynoecium :
It consists of monocarpellary, unilocular, half-inferior ovary. Usually many pendulous ovules are arranged in two vertical rows on marginal placentation. Style is long and curved at the apex. Stigma is capitate.

Pollination :
As flowers are protandrous, usually cross pollination occurs. In lathyrus andpisum, there is self pollination.

Fruit:
Mostly the fruit is a legume or pod (Pisum, Cajanus, Dolichos). In pterocarpus and Dalbergia it is a samara while in Arachis the pods are indehiscent and geocarpic.

Seed :
It is non-endospermic and dicotyledonous. The cotyledons store proteins in large quantities.

Question 2.
Write about the key characteristics of Solanaceae.
Answer:
VEGETATIVE CHARACTERS :

Habitat :
These plants are mostly mesophytes. A few are xerophytes. (Solarium surattense).

Habit :
Mostly annual or perennial herbs. Some are shrubs (Oestrum sps).

Root system :
Tap root system.

Stem :
It is aerial, erect and mostly herbaceous. Stem is an underground tuber in Solanum tuberosum (potato). Bicollateral vascular bundles are present in the stem.

Leaf :
Leaves are exstipulate, petiolate and show alternate phyllotaxy. They are usually or pinnately lobed. Venation is reticulate.

FLORAL CHARACTERS :
Inflorescence :
It is usually cymose type. It may be terminal or axillary in position. In some species of Solarium, it is an axillary. In Datura, it is solitary and terminal, panicle in tobacco.

Flower :
The flowers are bracteate or ebracteate, ebracteolate, pedicellate, actinomorphic, complete, bisexual, pentamerous and hypogynous.

Calyx :
It consists of 5 speals which are fused (gamosepalous) and persistent (Capsicum, Solatium). The aestivation is valvate.

Corolla :
The corolla consists of 5 petals and is gamopetalous. The aestivation is valvate or twisted. (Datura)

Androecium :
There are five epipetalous stamens alternating with the petals and are free. Anthers are dithecous, basifixed and introse. The dehiscence may be longitudinal (Datura) or porous (So/anum).

Gynoecium :
The ovary is superior, bicarpellary and syncarpous. It is oblique in position due to the tilting of posterior carpel to the right and anterior carpel to the left at an angle of 45°. Usually bilocular occasionally unilocular (Capsicum). There are numerous anatropous ovules arranged on axile placentation on swollen placenta. The style is terminal and stigma is capitate.

Pollination :
Flowers are usually protandrous. Some species of So/anum are protogynous. Cross pollination through insects (entomophily) is common.

Fruit:
The fruit is mostly berry (Capsicum, Solatium, Lycopersicon, Physalis etc.). It is septifragal capsule in Datura and Nicotiana.

Seeds :
The seeds are endospermic and dicotyledonous.

Question 3.
Give an account of the family Liliaceae.
Answer:
VEGETATIVE CHARACTERS :
Habitat:
Plants may be mesophytes (Allium, Ulium) as well as xerophytes (Asparagus, Ruscus, Aloe) are found in this family.

Habit :
Plants are mostly perennial herbs. In some plants like Dracaena, Yucca, Aloe etc., Shrubs or trees are also found. Few are climbers (Gloriosa, Smilax).

Root system :
It is adventitious root system. In Asparagus fasciculated tuberous roots are present.

Stem :
In majority of the species the stem is underground and perennial. It may be a bulb (Sci/la, Allium, Ulium), a Rhizome (Gloriosa) or a corm (Colchicum). The aerial stem is weak in tendril climbers like Gloriosa, Smilax etc. Stem is aerial and shows anomalous secondary growth in Dracaena and Yucca. Branches are modified into cladophyllus (Asperagus, Ruscus).

Leaf:
The leaves may be radical (Allium; Ulium) or cauline (Smilax, Gloriosa). Alternat phyllotaxy is common. Leaves are petiolate, simple stipulate or exstipulate.Venation is usually parallel, but exceptionally reticulate in Smilax. Leaves are succulent in yucca and Aloe.

FLORAL CHARACTERS :

Inflorescence : Solitary cyme or umbel or raceme.
Flower :
The flowrs are usually bracteate, ebracteolate, pedicellate, actinomorphic, complete. Bisexual, homochlamydeous, trimerous and hypogynous. Exceptionally flowers are unisexual in Smilax and Ruscus.

Perianth :
It consists of six tepals arranged in two whorls of three each. The odd tepal is anterior in position. The aestivation is valvate.

Androecium :
Stamens are six, arranged intwo whorls of three each. Epiphyllous (Asparagus). Anthers are dithecous, basifixed, introrse and dehiscence is longitudinal.

Gynoecium :
It is tricarpellary and syncarpous. The oary is superior and trilocular with several anatropous ovules on axile placentation. The style is terminal and stigma is trifid and capitate.

Pollination :
It is of entomophilous type. Flowers may be protandrous (Allium) or protogynous (Colchicum) in Glorisa flowers show herkogamy.

Fruit:
It may be a berry (Asparagusr, Smilax) or loculicidal capsule (Lilium) or septicidal capsule (Gloriosa).

Seed :
It is endospermic with straight or sometimes curved embryo and monocotyledonous. Polyembryony is seen in some memebrs (Allium).

Question 4.
Write the characteristics of plants that are necessary for classification. Describe them in brief.
Answer:
VEGETATIVE CHARACTERS :
Habit:
Herbs (Plants grow to a height of 1 – 3 feet)

Shrubs (Plants which grow in the form of a bush)
Trees (Plants with erect, woody branched or unbranched)
Habitat : Hydrophytes : (Plants which grow in water)
Mesophytes : (Plants which grow in Moderate climatic conditions)
Xerophytes : (Plants which grow in dry areas)
Root system : Taproot system : (The main root called Tap root, which inturn forms lateral roots and Root lets).
Ex : Dicots.
Fibrous Root system : (Cluster of roots which arise from the base of the stem)
Ex : Monocots.

Stem :
Aerial (which grows aerially) or underground (which grows into the soil), erect (Stands in upright position) or creeping (which grows on the soil Horizontally) Tendril climbers (Climbing with the help of Tendrils) Stragglers (Woody plants that climb up with the help of Hooks or thorns) Lianes (Large woody perennial twinners) Branched (Stem with Branches) or unbranched (stem without Branches) green or Brown or black in colour.

Leaf : Leaf Base :
Pulvinous (Swollen) or sheathing (Broad and surround the stem as envelope).
Stipules : Stipulate (Leaf with stipules)
Exstipulate (Leaf without stipules).

Petiole : Petiolate (Leaf with petiole)
Sessile (Leaf without petiole).

Lamina :
Shape – Ovate (oval) or Linear (Long and slightly Broader) Reniform (Kidney shaped) Cordate (Heart Shaped) Centric (hollow).

Venation :
Reticulate (Midrib, Laterial veins and veinlats are arranged in the form of net like).

Parallel :
(Mid rib produce lateral veins and veinlets arranged parallely).

Kind :
Simple (Leaf with undivided Lamina) .
Compound (Lamina is divided into leaflets).

Phyllotaxy :
Alternate (only one leaf arises at a node)
Opposite (Two leaves arises at each node)
Whorled (More than 2 leaves arises at each node).

FLORAL CHARACTERS :
Inflorescence : Racemose (Peduncle is long)
Cymose (Peduncle is short) ‘
Special (Verticillester or Hypanthodium or Cyathium).

Flower : Bracteate (The flower with Bracts)
Ebracteate (The flower without Bracts)
Bracteolate (The flower with Bracteoles)
Ebracteolate (The flower without Bracteoles)
Pedicillate (The flower with Pedicel)
Sessile (The flower without Pedicel)
Complete (The flower with all four floral parts)
Incomplete (The flower without an any one of the floral parts).

Arrangement of Floral parts Acyclic : The floral parts are spirally arranged.

Cyclic : The floral parts are arranged in whorls.
Hemicyclic : K & C are in whorls and A & G are in spiral Manner).
Sex : Bisexual : The flower with both sex organs.
Unisexual: The flower with any one of the sex organs.

Gynoecium Position : Hypogynous : The flower with superior ovary.
Epigynous : The flower with Inferior ovary.
Perigynous : The flower with Half superior ovary.
Merosity : Trimerous : Three parts in each whorl
Tetramerous : Four parts in each whorl.
Pentamerous : Five parts in each whorl.

Symmetry :
Actinomorphic: A flower can be cut into two equal halves in any vertical plane.

Zygmorphic :
A flower can be cut into two equal halves in one vertical plane.

Calyx :
Number of sepals 3 or 4 or 5, polysepalous (free) or gamo$epalous (fused) valvate (arranged in a whorl) or twristed Aestivation (arranged in one whole with one margin Inside and one margin outside). Imbricate Aestivation (arranged with overlapping pattern).

Corolla :
Number of petals, polypetalous (free) or gamopetalous (fused), Aestivation (valvate or Twisted or Imbricate).

Androecium :
Number of stamens 4 or 5 or 10 or many.
Monadelphous : All are in one Bundle
Diadelphous : All are in two Bundles
Polyadelphous : All are in. more than two Bundles.
dithrecous : Anther with two Theca.
Monothecous : Anther with one Theca.
Basifixed ; Filament gets attached to the Base of the Anther.
Dorsifixed : Filament gets attached to the Dorsal side of the anther.

Dehiscence : Longitudinal (Breaks vertically)
Transverse (Breaks Transversely)
Porous (Pollengrains are released through Apical pore).

Gynoecium :
Monocarpellary – The ovary with one carpel.
Bicarpellary – The ovary with Two carpels.
Tricarpellary – The ovary with Three carpels.
Tetracarpellary – The ovary with Four carpels.
Pentacarpellary – The ovary with Five carpels. .
Multicarpellary – The ovary with more than five carpels.
Syncarpous : All the carpels are fused.
Apocarpous : All the carpels are free.
Superior ovary : K, C, A develops from the base of the ovary.
Interior ovary : K, C, A develops from the upper part of the ovary.
Half superior ovary : K, C, A develops from the centre (Half) of the ovary.

Placentation :
Marginal (Ovules are arranged on Placenta present along the margins).
Axile (Ovules are arranged on the centre)
Basal (Ovules are arranged at the Base of the ovary).

Style : Terminal (develops above the ovary)
Laferal (develops from the latual side).

Stigma : Capitate (Round) Bifid (divided Hairy (with hairs).