Inter 2nd Year

Students get through AP Inter 2nd Year Physics Important Questions 1st Lesson Waves which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 1st Lesson Waves

Very Short Answer Questions

Question 1.
Write the formula for the speed of sound in solids and gases.
Answer:
Speed of sound in solids,
V_s = \(\sqrt{Y / \rho}\) [y = Young’s modulus of solid, ρ = density of solid]
Speed of sound in gases,
V_s = \(\sqrt{\gamma P / \rho}\) [γP = Adiabatic Bulk modulus of gas, ρ = density of gas] .

Question 2.
What does a wave represent?
Answer:
A wave represents the transport of energy through a medium from one point to another without translation of the medium.

Question 3.
Distinguish between transverse and longitudinal waves.
Answer:
Transverse waves

1. The particles of the medium vibrate perpendicular to the direction of wave propagation.
2. Crests and troughs are formed alternatively.

Longitudinal waves

1. The particles of the medium vibrate parallel to the direction of wave propagation.
2. Compressions and rare factions are formed alternatively.

Question 4.
What are the parameters used to describe a progressive harmonic wave ?
Answer:
Progressive wave equation is given y = a sin (ωt – kx)
Where ω = 2πv = \(\frac{2 \pi}{T}\); k = \(\frac{2 \pi}{\lambda}\)
Parameters:

1. a = Amplitude
2. λ = Wavelength
3. T = Time period
4. v = Frequency
5. k = Propagation constant
6. ω = Angular frequency.

Question 5.
What is the principle of superposition of waves ? .
Answer:
When two or more waves are acting simultaneously on the particle of the medium, the resultant displacement is equal to the algebraic sum of individual displacements of all the waves. This is the principle of superposition of waves.
If y₁, y₂, ……………… y_n be the individual displacements of the particles,then resultant displacement
y = y₁ + y₂ + ……………… + y_n.

Question 6.
Under what conditions will a wave be reflected ?
Answer:

1. When the medium ends abruptly at any point.
2. If the density and rigidity modulus of the medium changes at any point.

Question 7.
What is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary ?
Answer:
Phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary is radian or 180°.

Question 8.
What is a stationary or standing wave ?
Answer:
When two identical progressive (Transeverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are superimposed then the resultant wave is called stationary waves or standing wave.

Question 9.
What do you understand by the terms node’ and ‘antinode’ ?
Answer:
Node : The points at which the amplitude is zero, are called nodes.
Antinodes: The points at which the amplitude is maximum, are called antinodes.

Question 10.
What is the distance between a node and an antinode in a stationary wave ?
Answer:
The distance between node and antinode is \(\frac{\lambda}{4}\).

Question 11.
What do you understand by ‘natural frequency’ or ‘normal mode of vibration’ ?
Answer:
When a body is set into vibration and then left to itself, the vibrations made by it are called natural or free vibrations. Its frequency is called natural frequency or normal mode of vibration.

Question 12.
What are harmonics ?
Answer:
The frequencies in which the standing waves can be formed are called harmonics..
(Or)
The integral multiple of fundamental frequencies are called harmonics.

Question 13.
A string is stretched between two rigid supports. What frequencies of vibration are possible in such a string ?
Answer:
The possible frequencies of vibrations in a stretched string between two rigid supports is given by
v_n = (n + \(\frac{1}{2}\)) \(\frac{v}{2l}\) where n = 0, 1, 2, 3, ……….

Question 14.
If the air column in a long tube, closed at one end, is set in vibration, what harmonics are possible in the vibrating air column ?
Answer:
The possible harmonics in the vibrating air column of a long closed tube is given by
v_n = [2n +1]\(\frac{v}{4 l}\) where n = 0, 1, 2, 3,

Question 15.
If the air column in a tube, open at both ends, is set in vibration; what harmonics are possible ?
Answer:
The possible harmonics in vibrating air column of a long open tube is given by no
v_n = \(\frac{\mathrm{nv}}{21}\)
where n = 1, 2, 3, ……………….

Question 16.
What are ‘beats’ ?
Answer:
Beats : When two sound notes of nearly frequency travelling in the same direction and interfere to produce waxing and waning of sound at regular intervals of time is called “Beats”.

Short Answer Questions

Question 1.
What are transverse waves ? Give illustrative examples of such waves.
Answer:
Transverse waves: In a wave motion, the vibration of the particles and the direction of the propagation of the waves are perpendicular to each other, the waves are said to be transverse waves.
Illustration:

1. Waves produced in the stretched strings are transverse.
2. When a stretched string is plucked, the waves travel along the string.
3. But the particles in the string vibrate in the direction perpendicular to the propagation of the wave.
4. They can propagate only in solids and on the surface of the liquids.
5. Ex : Light waves, surface water waves.

Question 2.
What are longitudinal waves ? Give illustrative example of such waves.
Answer:
Longitudinal waves : In a wave motion, the direction of the propagation of the wave and vibrations of particles are in the same direction, the waves are said to be longitudinal waves.
Illustration:

1. Longitudinal waves may be easily illustrated by releasing a compressed spring.
2. A series of compressions and rarefactions (expansions) propagate along the spring.
   
   C = Compression; R = Rarefaction.
3. They can travel in solids, liquids and gases.
4. Ex : Sound waves.

Question 3.
What are ‘beats’ ? When do they occur ? Explain their use, if any.
Answer:
Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing (maximum) and waning (minimum) in the intensity of the resultant sound waves at regular intervals of time is called beats.
It two vibrating bodies have slightly difference in frequencies, beats can occur.
No. of beats can be heard, ∆υ = υ₁ ~ υ₂
Importance:
1. It can be used to tune musical instruments.
2. Beats are used to detect dangerous gases.
Explanation for tuning musical instruments with beats:
Musicians use. the beat phenomenon in tuning their musical instruments. If an instrument is sounded against a standard frequency and tuned until the beats disappear. Then the instrument is in tune with the standard frequency.

Question 4.
What is ‘Doppler effect’ ? Give illustrative examples.
Answer:
Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between .the observer and the source of sound is called doppler effect.
Examples:

1. The frequency of whistling, engine heard by a person standing on the platform appears to increase, when the engine is approaching the platform and it appears to decrease when the engine is moving away from the platform.
2. Due to Doppler effect the frequency of sound emitted by the siren of an approaching ambulance appears to increase. Similarly the frequency of sound appears to drop when it is moving away.

Sample Problem on Doppler effect:
Two trucks heading in opposite direction with speeds of 60 kmph and 70 kmph respectively, approach each other. The driver of the first truck sounds his horn of frequency 400Hz. What frequency does the driver of the second truck hear ? (Velocity of sound = 330m/s). After the two trucks have passed each other, what frequency does the driver of the second truckhear?
Answer:
Speed of first truck = 60 kmph
= 60 × \(\frac{5}{18}\) = 16.66 m/s;
Speed of second truck = 70 kmph 5
= 70 × \(\frac{5}{18}\) = 19.44 m/s
Frequency of horn of first truck = 400 Hz;
Velocity of sound, (V) =330 m/s
Frequency of sound heard by the driver of the second truck when approaching each other,
v¹ = \(\left(\frac{V+V_0}{V-V_s}\right) v=\left(\frac{330+19.44}{330-16.66}\right)\) × 400 = 446 Hz
Frequency of sound heard by the driver of the second truck when approaching each other,
V¹¹ = \(\left(\frac{\mathrm{V}-\mathrm{V}_0}{\mathrm{~V}+\mathrm{V}_{\mathrm{s}}}\right) \mathrm{v}=\left(\frac{330-19.44}{330+16.66}\right)\) × 400 = 358.5 Hz

Long Answer Questions

Question 1.
Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse wave in stretched strings. [IPE]
Answer:
A string is a metal wire whose length is large when compared to its thickness. A stretched string is fixed at both ends, when it is plucked at mid point, two reflected waves of same amplitude and frequency at the ends are travelling in opposite direction and overlap along the length. Then the resultant waves are known as the standing waves (or) stationary waves. Let two transverse progressive waves of same amplitude a, wave length λ and frequency v, travelling in opposite direction be given by

y₁ = a sin (kx – ωt) and y₂ = a sin (kx + ωt)
where ω = 2πv and k = \(\frac{2 \pi}{\lambda}\)
The resultant wave is given by y = y₁ + y₂
y = a sin (kx – ωt) + a sin (kx + ωt)
y = (2a sin kx) cos ωt
2a sin kx = Amplitude of resultant wave.
It depends on ‘kx’. If x = 0, \(\frac{\lambda}{2}, \frac{2 \lambda}{2}, \frac{3 \lambda}{2}\) ……………… etc, the amplitude = zero
These positions are known as “Nodes”.
If x = \(\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}\) …………… etc., the amplitude = maximum (2a)
These positions are called “Antinodes”.
If the string vibrates in ‘P’ segments and T is its length, then length of each segment = \(\frac{l}{\mathrm{P}}\)
Which is equal to \(\frac{\lambda}{2}\)
∴ \(\frac{l}{\mathrm{P}}=\frac{\lambda}{2} \Rightarrow \lambda=\frac{2 l}{\mathrm{P}}\)
Harmonic frequency v = \(\frac{v}{\lambda}=\frac{v \mathrm{P}}{2 l}\)
v = \(\frac{v \mathrm{P}}{2 l}\) ………………. (1)
If’ T’ is tension (stretching force) in the string and ‘μ’ is linear density then velocity of transverse wave (v) in the string is
v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\) …………… (2)
From the Eqs (1) and (2) :
Harmonic frequency v = \(\frac{\mathrm{P}}{2 l} \sqrt{\frac{\Gamma}{\mu}}\)
If P = 1 then it is called fundamental frequency (or) first harmonic frequency
∴ Fundamental Frequency v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\) …………….. (3)

Laws of Transverse Waves Along Stretched String:
Fundamental frequency of the vibrating string v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\)
First Law : When the tension (T) and linear density (μ) are constant, the fundamental frequency (v) of a vibrating string is inversely proportional to its length.
∴ v ∝ \(\frac{1}{l}\) ⇒ vl = constant, when ‘T’ and ‘μ’ are constant. .

Second Law: When the length (l) and its, linear density (m) are constant the fundamental frequency of a vibrating string is directly proportional to the square root of the stretching force (T).
∴ v ∝ \(\sqrt{T}\) ⇒ \(\frac{v}{\sqrt{T}}\) = constant, when ‘l’ and ‘m’ are constant.
JT .
Third Law: When the length (l) and the tension (T) are constant, the fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (m).
∴ v ∝ \(\frac{1}{\sqrt{\mu}}\) ⇒ \(v \sqrt{\mu}\) = constant, when ‘l’ and T are constant.

Question 2.
Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equations for the frequencies of the harmonics produced. [A.P. 17; IPE 2015, 2016 (TS)]
Answer:
A pipe, which is opened at both ends is called open pipe. When a sound wave is sent through a open pipe, which gets reflected by the earth. Then incident and reflected waves are in same frequency, travelling in the opposite directions are super-imposed stationary waves are formed.

Harmonics in open pipe : To form the stationary wave in open pipe, which has two anti nodes at two ends of the pipe with a node between them.

∴ The vibrating length (l) = half of the wavelength \(\left(\frac{\lambda_1}{2}\right)\)
l = \(\frac{\lambda_1}{2}\) ⇒ λ₁ = 2l
fundamental frequency v₁ = \(\frac{\mathrm{v}}{\lambda_1}\) where v is velocity,of sound in air, v₁ = \(\frac{v}{21}\) = v
For second harmonic (first overtone) will have one more node and antinode than the fundamental.
If λ₂ is wavelength of second harmonic l = \(\frac{2 \lambda_2}{2}\) ⇒ λ₂ = \(\frac{21}{2}\)
If ‘v₂’ is frequency of second harmonic then v₂ = \(\frac{v}{\lambda_2}=\frac{v \times 2}{2 l}\) = 2v
v₂ = 2v ……………… (2)
Similarly for third harmonic (second overtone) will have three nodes and four antinodes as shown in above figure.
If λ₃ is wave length of third harmonic l = \(\frac{3 \lambda_3}{2}\)
λ₃ = \(\frac{2l}{3}\)
If ‘v₂‘ is frequency of third harmonic then
v₃ = \(\frac{v}{\lambda_3}=\frac{v \times 3}{2 l}\) = 3V
v₃ = 3v …………… (3)
Similarly we can find the remaining or higher harmonic frequencies i.e., v3, v4 etc., can be determined in the same way.
Therefore the ratio of the harmonic frequencies in open pipe can be written as given below.
v : V₁ : v₂ = 1 : 2 : 3

Question 3.
How are stationary waves formed in closed pipes ? Explain the various modes of vibrations and obtain relations for their frequencies. [IPE 2015, 2016(A.P.), (T.S) A.P. & T.S. Mar. 15]
Answer:
A pipe, which is closed at one end and the other is opened is called closed pipe. When a sound wave is sent through a closed pipe, which gets reflected at the closed end of the pipe. Then incident and reflected waves are in same frequency, travelling in the opposite directions are superimposed stationary waves are formed.

To form the stationary wave in closed pipe, which has atleast a node at closed end and antinode at open end of the pipe, it is known as first harmonic in closed pipe. Then length of the pipe (l) is equal to one fourth of the wave length.

∴ l = \(\frac{\lambda_1}{4}\) ⇒ λ₁ = 4l
If ‘v₁‘ is fundamental frequency then
v₁ = \(\frac{v}{\lambda_1}\) where ‘υ’ is velocity of sound in air
v₁ = \(\frac{v}{4 l}\) = v …………….. (1)
To form the next harmonic in closed pipe, two nodes and two antinodes should be formed. So that there is possible to form third harmonic in closed pipe. Since one more node and antinode should be included.
Then length of the pipe is equal to \(\frac{3}{4}\) of the wavelength.
∴ l = \(\frac{3 \lambda_3}{4}\) where ‘λ₃‘ is wave length of third harmonic
λ₃ = \(\frac{4l}{3}\)
where ‘X3’ is wave length of third harmonic.
If ‘v₃‘ is third harmonic frequency (first overtone)
∴ v₃ = \(\frac{v}{\lambda_3}=\frac{3 v}{41}\)
v₃ = 3v …………….. (2)
Similarly the next overtone in the close pipe is only fifth harmonic, it will have three nodes and 3 antinodes between the closed end and open end.
Then length of the pipe is equal to \(\frac{5}{4}\) of wave length (λ₅)
∴ l = \(\frac{5 \lambda_5}{4}\) where ‘λ₅‘ is wave length of fifth harmonic.
λ₅ = \(\frac{4l}{5}\)
If ‘v₅‘ is frequency of fifth harmonic (second overtone)
v₅ = \(\frac{v}{\lambda_5}=\frac{5 v}{4 l}\)
v₅ = 5v …………….. (3)
∴ The frequencies of higher harmonics can be determined by using the same procedure. Therefore from the Eq (1), (2) and (3) only odd harmonics are formed.
Therefore the ratio of the frequencies of harmonics in closed pipe can be written as
v₁ : v₃ : v₅ = v : 3v : 5v
v₁ : v₃ : v₅ = 1 : 3 : 5

Question 4.
What is Doppler effect ? Obtain an expression for the apparent frequency of sound heard when the source is in motion with respect to an observer at rest. [Mar. 17, BMP, 2016 (AP) Mar. 14, (TS)]
Answer:
Doppier effect: The apparent change in the frequency heard by the observer due to the relative motion between the observer and the source of sound is called doppier effect. When a whistling railway engine approaches an observer standing on the platform, the frequency of sound appears to increase. When it moves away the frequency appear to decrease.

Expression for apparent frequency when source is in motion and listener at rest:
Let S = Source of sound
O = Listener
Let ‘S’ be the source, moving with a velocity ‘υ_s‘ towards the stationary listener.
The distance travelled by the source in time period T = υ_s. T Therefore the successive compressions and rarefactions are drawn closer to listener.
∴ Apparent wavelength λ’ = λ – υ_sT.
λ’ = λ – \(\frac{v_s}{v}\) [∵ υ = \(\frac{1}{T}\)]
= \(\frac{\lambda v-v_s}{v}=\frac{v-v_s}{v}\) [∵ υ = vλ]
If “v'” is apparent frequency heard by the listener then v’ = \(\frac{v}{\lambda^{\prime}}\) where ‘υ’ is Velocity of sound in air
v’ = \(\frac{v . v}{v-v_S}\)
Therefore the apparent frequency is greater than the actual frequency.
Similarly, if the source is away from the stationary listener then apparent frequency v’ = \(\frac{v . v}{v+v_s}\), which is less than the actual frequency.

Limitation : Doppler effect is applicable when the velocities of the source and listener are much less than that of sound velocity.

Problems

Question 1.
A stretched wire of length 0.6 m is observed to vibrate with a frequency of 30 Hz in the fundamental mode. If the string has a linear mass of 0.05 kg / m find (a) the velocity of propagation of transverse waves in the string (b) the tension in the string. [IPE 2016 (T.S)
Answer:
v = 30 Hz; I = 0.6 m; μ = 0.05 kg m^-1
υ = ?; T = ?
a) υ = 2vl = 2 × 30 × 0.6 = 36 m/s
b) T = υ²μ = 36 × 36 × 0.05 = 64.8 N .

Question 2.
A string has a length of 0.4m and a mass of 0.16g. If the tension in the string is 70N, what are the three lowest frequencies it produces when plucked ?
Solution:
I = 0.4 m; M = 0.16g = 0.16 × 10^-3 kg;
μ = \(\frac{\mathrm{M}}{1}=\frac{0.16 \times 10^{-3}}{0.4}\) = 0.4 × 10^-3 kg/m;
T = 70 N; v_n = \(\frac{\mathrm{P}}{21} \sqrt{\frac{\mathrm{T}}{\mu}}\)
v₁ = \(\frac{1}{21} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{2 \times 0.4} \sqrt{\frac{70}{0.4 \times 10^{-3}}}\) = 523 Hz
v₂ = 2v₁ = 2 × 523 = 1046 Hz
v₃ = 3v₁ = 3 × 523 = 1569 Hz

Question 3.
A closed organ pipe 70 cm long is sounded. If the velocity of sound is 331 m/s, what is the fundamental frequency of vibration of the air column ?
Solution:
l = 70 cm = 70 × 10^-2m; v = 331 m/s ;
v = ?
v = \(\frac{v}{4 l}=\frac{331}{4 \times 70 \times 10^{-2}}\) = 118.2 Hz.

Question 4.
A steel cable of diameter 3 cm is kept under a tension of lOkN. The density of steel is 7.8 g/ cm3. With what speed would transverse waves propagate along the cable ?
Solution:
T = 10 kN = 10⁴ N
D = 3 cm; r = \(\frac{D}{2}=\frac{3}{2}\) cm
= \(\frac{3}{2}\) × 10^-2m;

Question 5.
A train sounds its whistle as it approaches and crosses a level crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as it leaves. If the speed of sound is taken to be 340 m/s, find the speed of the train and the frequency of its whistle. [T.S. Mar. 17]
Solution:
When a whistling train away from rest observer,
v’ = \(\left[\frac{v}{v-v_s}\right] v\) ……………… (1)
When a whistling train away from rest observer,
v” = \(\left[\frac{v}{v+v_s}\right] v\) ……………… (2)
Here v’ = 219 Hz; v” = 184 Hz;
v = 340 m/s

Question 6.
A rocket is moving at a speed of 200 m s-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. [A.P. Mar. 16]
Solution:
1) The observer is at rest and the source is moving with a speed of 200 m s^-1. Since this is comparable with the velocity of sound 330 ms^-1, we must use Eq.
v = v₀ \(\left[\frac{1+v_{\mathrm{S}}}{v}\right]^{-1}\) and not the approximate
v = v₀ [1 – \(\frac{v_s}{v}\)]
Since the source is approaching a stationary target, υ₀ = 0, and υ_s must be replaced by -υ_s. Thus, we have
v = v₀ [1 – \(\frac{v_s}{v}\)]^-1
v = 1000 Hz × [1 – 200 m s^-1/330 m s^-1]^-1
≃ 2540Hz

2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, υ_s = 0 and υ₀ has a positive value. The frequency of the sound emitted by the source (the target) is v, the frequency intercepted by the target and not v₀, Therefore, the frequency as registered by the rocket is
v’ = \(v\left(\frac{v+v_0}{v}\right)\)
= 2540 Hz × \(\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right)\)
≃ 4080Hz

Textual Examples

Question 1.
Given below are some examples of wave motion. State in each case if the wave motion is transverse, longitudinal or a combination of both.
a) Motion of kink in a longitudinal spring produced by displacing one end of the spring sideways.
b) Waves produced in a cylinder containing a liquid by moving its piston back and forth.
c) Waves produced by a motorboat sailing in water.
d) Ultrasonic waves in air produced by a vibrating quartz crystal.
Solution:
a) Transverse and longitudinal
b) Longitudinal
c) Transverse and longitudinal
d) Longitudinal.

Question 2.
A wave travelling along a string is des-cribed by, y(x, t) = 0.005 sin (80.0 x – 3.01), in which the numerical constants are in SI untis (0.005 m, 80.0 rad m^-1, and 3.0 rad s^-1). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s ?
Solution:
On comparing this displacement equation with Eq. y (x, t)n = a sin(kx – ωt + Φ)
y(x, t) = a sin (kx – ωt).
We find
a) the amplitude of the wave is
0. 005 m = 5 mm.

b) the angular wave number k and angular frequency ω are k = 80.0 m^-1 and ω = 3.0 s^-1
We then relate the wavelength λ to k through Eq.
λ = \(\frac{2 \pi}{K}\)
= \(\frac{2 \pi}{80.0 \mathrm{~m}^{-1}}\) = 7.85 cm

c) Now we relate T to ω by the relation
T = \(\frac{2 \pi}{\omega}\)
= \(\frac{2 \pi}{3.0 \mathrm{~s}^{-1}}\)
= 2.09 s
and frequency, v = \(\frac{1}{T}\) = 0.48 Hz
The displacement y at x = 30.0 cm and time t= 20s is given by
y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20)
= (0.005 m) sin (-36 + 12π)
= (0.005 m) sin (1.699)
= (0.005 m) sin (97°) ≃ 5 mm

Question 3.
A steel wire 0.72 m long has a mass of 5.0 × 10^-3 kg. If the .wire is under a tension of 60 N, what is the speed of transverse waves on the wire ? [A.P. Mar. 19]
Solution:
Mass per unit length of the wire,
μ = \(\frac{5.0 \times 10^{-3} \mathrm{~kg}}{0.72 \mathrm{~m}}\) = 6.9 × 10^-3 kg m^-1
Tension, T = 60 N
The speed of wave on the wire is given by
υ = \(\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{60 \mathrm{~N}}{6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}}}\) = 93 m s^-1

Question 4.
Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 × 10^-3 kg.
Solution:
We know that 1 mole of any gas occupies 22.4 litres at STP Therefore, density of air at STP is:
ρ₀ = (mass of one mole of air) / (Volume of one mole of air at STP)
= \(\frac{29.0 \times 10^{-3} \mathrm{~kg}}{22.4 \times 10^{-3} \mathrm{~m}^3}\) = 1.29 kgm^-3
According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP,
υ = \(\left[\frac{1.01 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}}{1.29 \mathrm{~kg} \mathrm{~m}^{-3}}\right]^{1 / 2}\) = 280 m s^-1

Question 5.
A pipe, 30.0 cm long is open at both ends. Which harmonic mode of the pipe resonates a 1.1 kHz source ? Will resonance with the same source be observed if one end of the pipe is closed ? Take the speed of sound in air as 330 m s^-1.
Solution:
The first harmonic frequency is given by
v¹ = \(\frac{v}{\lambda_1}=\frac{v}{2 L}\) (open pipe)
Where L is the length of the pipe. The frequency of its n^th; harmonic is
v_n = \(\frac{n v}{2 L}\) for n = 1, 2, 3, ……………… (open pipe)
First few modes of an open pipe are shown in Fig.
For L = 30.0 cm. υ = 330 m s^-1
v_n = \(\frac{\mathrm{n} \times 330\left(\mathrm{~m} \mathrm{~s}^{-1}\right)}{0.6(\mathrm{~m})}\) = 550 s^-1

Clearly, for a source of frequency 1.1 kHz the air column will resonate at υ₂, i.e. the second harmonic.
Now if one end of the pipe is closed (Fig.), the fundamental frequency is

and only to odd numbered harmonics are present :
v₃ = \(\frac{3 v}{4 L}/latex], v₅ = [latex]\frac{5 v}{4 L}\) and s0 0n.
For L = 30 cm and υ = 330 m s^-1;, the fundamental frequency of the pipe closed at one end is 275 Hz and the source frequency corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed.

Question 6.
Two sitar strings A and B playing the note ‘Dha’ are slightly put of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if the frequency of A is 427 Hz ?
Solution:
Increase in the tension of a string increases its frequency. It the original frequency of B (v_B) were greater than that of A(v_A) .further increase in v_B should have resulted in an increase in the beat frequency. But the beat frequency is found to decrease. This shows that v_B < v_A. Since v_A – v_B = 5 Hz, and v_a = 427 Hz, we get v_B = 422 Hz.

Question 7.
A rocket is moving at a speed of 200 m s^-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. [A.P. Mar. 16]
Solution:
1) The observer is at rest and the source is moving with a speed of 200 ms^-1. Since this is comparable with the velocity of sound 330 ms^-1, we must use Eq.
v = v₀ \(\left[\frac{1+\mathrm{v}_{\mathrm{S}}}{\mathrm{v}}\right]^{-1}\) and not the approximate
v = v₀ \(\left[1-\frac{v_s}{v}\right]\)
Since the source is approaching a stationary target, υ₀ = 0, and υ_s must be replaced by -υ_s. Thus, we have
v = v₀ \(\left(1-\frac{v_{\mathrm{S}}}{v}\right)^{-1}\)
v = 1000 Hz × [1 – 200 m s^-1/330 m s^-1]^-1
≃ 2540Hz

2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, υ_s = 0 and υ₀ has a positive value. The frequency of the sound emitted by the source (the target) is v₀, the frequency intercepted by the target and not v0, Therefore, the frequency as registered by the rocket is
v’ = \(v\left(\frac{v+v_0}{v}\right)\)
= 2540 Hz × \(\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right)\)
≃ 4080Hz

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination

Very Short Answer Questions

Question 1.
Name the blood vessels that enter and exit the kidney.
Answer:
Renal artery enters kidney and renal vein comes out of the kidney.

Question 2.
What are renal pyramids and renal papillae?
Answer:
The conical shaped medullary regions of kidney are the renal pyramids. Tips of the renal pyramids which open into pelvis are renal papillae.

Question 3.
What are the columns of Bertin?
Answer:
Columns of Bertin are the medullary extensions of the renal cortex in between the renal pyramids.

Question 4.
Name the structural and functional unit of kidney. What are the two main types of structural units in it?
Answer:
The structural and functional unit of kidney is ‘Nephrons’. The two main parts are
1) Malpighian body (renal corpuscle),
2) Convoluted tube.

Question 5.
Distinguish between cortical and juxta medullary nephrons.
Answer:

1. Cortical nephrons have renal corpuscle in the superficial renal cortex. They have short loop of Henle but without vasa recta.
2. Juxta medullary nephrons re located near the renal medulla. They have loop of Henle and vasa recta.

Question 6.
Define glomerular Alteration.
Answer:
The process of Alteration of blood, which occur between glomerulus and lumen of the Bowman’s capsule due to difference in net pressure is called glomerular Alteration. The filtered fluid which entered the Bowman’s capsule is primary urine or glomerular filtrate which is hypotonic.

Question 7.
Define glomerular Alteration rate (GFR)?
Answer:
The amount of filterate formed by both the kidneys, per minute is called glomerular Alteration rate (GFR). GFR in a healthy individual is approximately 125 ml 11 minute.

Question 8.
What is meant by mandatory reabsorption? In which parts of nephron does it occur?
Answer:
In a healthy individual the GFR is 125 ml/1 minute or 180 ltr per day. About 85% of the filterate formed is reabsorbed in a constant, unregulated fashion by the proximal convoluted tubule and descending limb of Henle’s loop, called mandatory reabsorption.

Question 9.
Distinguish between juxtaglomerular cells and macula densa.
Answer:

1. The cells of the distal convoluted tubule are crowded in the region where distal convoluted tubule makes contact with afferent arteriole. These cells are known as Macula densa.
2. Along side of macula densa, the wall of the afferent arteriole contains the modified smooth muscle fibers called juxtaglomerular cells.

Question 10.
Whait is Juxtaglomerular apparatus?
Answer:
Macula densa along with juxtaglomerular cells form juxtaglomerular apparatus which releases an enzyme Called renin.

Question 11.
Distinguish between the enzymes reniri and rennin.
Answer:
Renin :
Renin is an enzyme produced by the JG cells. This enzyme catalyses the conversion of angiotensinogen into angiotensin -I.

Rennin :
Rennin is also an enzyme found in the gastric juice of infants. It acts on the milk protein casein in the presence of calcium ions and convert it into calcium para caseinate and proteoses.

Question 12.
What is meant by the term osmoregulation?
Answer:
The process of maintaining the quantity of water and-dissolved solutes in balance is referred to as osmoregulation.

Question 13.
What is the role of atrial-natriuretic peptide in the regulation of urine formation?
Answer:
A large increase in blood volume promotes the release of atrial natriuretic peptide from the heart. Atrial natriuretic peptide decreases the absorption of water, Na⁺ from proximal convoluted tubule.

Short Answer Questions

Question 1.
Terrestrial animals are generally either ureotelic or uricotelic and not ammonotelic. Why?
Answer:
Ammonia is highly toxic and readily soluble in water, hence it should be eliminated from the body quickly in a very dilute solution.

Aquatic animals are surrounded by water, so water conservation is not a problem for them. In this manner, they are continuously eliminating ammonia.

On the other hand, terrestrial animals have to conserve water. They cannot waste water. So ammonia in diluted form can’t be eliminated continuously. Since ammonia is highly toxic, it has to be converted to less toxic form, like urea or uric acid.

Urea is 1,00,000 times less toxic than ammonia and requires less water for their excretion. Uric acid is less toxic than urea and being insoluble in water can be excreted as semi solid waste or pellets with very little water. This is the great advantage for animals with little access to water.

Question 2.
Differentiate vertebrates on the basis of the nitrogenous waste products they excrete, giving example.
Answer:
Vertebrates are divided into three categories based on nitrogenous waste excretory products. They are :
1) Ammonotelic animals:
The animals which excrete ammonia as nitrogenous waste products are called ammonotelic animals. These are aquatic organisms.
Ex : Some bony fishes.

2) Ureotelic animals:
The animals which excrete urea as their chief nitrogenous waste are called ureotelic animals.
Ex : Earth worms, cartilaginous fishes, most of the amphibians and mammals.

3) Uricotelic animals:
These animals excrete their nitrogenous waste products in the form of uric acid.
Ex : Reptiles, birds.

Question 3.
Draw labelled diagram of the V.S of kidney.
Answer:

Question 4.
Describe the internal structure of kidney of man.
Answer:

1. Kidney is bean shaped structure, the outer surface of kidney is convex and inner surface is concave where it has a deep notch called hilum.
2. A longitudinal sections of the human kidney shows two distinct regions namely the outer cortex and the inner medulla.
3. Medulla is divided into multiple cone shaped masses of tissue called renal pyramids. The renal pyramids are separated by the projections of the cortex called columns of Berlin.
4. The tips of the pyramids are renal papilla.
5. Renal papilla projects into cup like calyces, formed by the funnel shaped pelvis, which continues out as the ureter. Ureter carries urine into urinary bladder.
6. In cortex and medulla, nearly one million nephrons are present. They are structural and functional units of kidney. They are embedded in the loose connective tissue of cortex and medulla.
7. In addition, kidney contains a network of blood capillaries, lymph sinuses and intestitial fluid in intra cellular spaces.
8. The kidney gets blood supply through renal artery and blood from kidney is carried out by renal vein.

Question 5.
Explain micturition.
Answer:
The process of passing out of urine is called micro nutrition and the neural mechanism involved is called micturition reflex.

Urine is formed by the nephrons is ultimately carried to the urinary bladder where it is stored till a voluntary signal is given by the central nervous system (CNS). This sighal is initiated by the stretching of the urinary bladder as it gets filled with urine. In response, the stretch receptors on the walls of the bladder send signals to the CNS. The CNS passes on motor messages to initiate the contraction of smodth muscles of the bladder and simultaneous relaxation of the urethral sphincter, causing the release of urine.

Question 6.
What is the significance of juxta glomerular apparatus (JGA) in kidney function?
Answer:
Macula densa together with JG cells form juxtaglomerular apparatus (JAG). JAG plays a complex regulating role. A fall in glomerular blood flow or glomerular blood pressure or GFR can activate JG cells to release an enzyme called renin into the blood. This catalyses the conversion of angiotensinogen into angotensin-I which is further converted into angiotensin-II by angiotensin converting enzyme. Angiotensin-II, being a powerful vasoconstrictor increase the glomerular blood pressure and there by GFR.

Angiotensin-II also activates the adrenal cortex to release aldosterone. Aldosterone causes reabsorption of Na⁺ and water from distal convoluted tubule and collecting duct. To reduce loss through urine, and also promote secretion of K⁺ ions into distal convoluted tubule and collecting duct. It leads to increase in the blood pressure and GFR. This complex mechanism is generally known as renin – angiotensin- aldosterone system (RAAS).

Question 7.
Give a brief account of the counter current mechanism.
Answer:
Mammals have the ability to produce concentrated urine. The Henle’s loop and vasa recta plays an important role in this. The flow of the renal filterate in the two limbs of Henle’s loop is in opposite directions and thus form counter current. The flow of blood through vasa recta is also in counter current pattern. The proximity between the Henle’s loop and vasa recta, as well as the counter currents of renal fluid and blood in them help in maintaining an increasing osmolarity towards the inner medullary interstitium.

This gradient is mainly caused by NaCl and urea. NaCl passes out the ascending limb of Henle’s loop, and it enters the blood of the descending limb of vasa recta. NaCl is returned to the intestitium from the ascending portion of the vasa recta. Similarly small amounts of urea enter the thin segment of ascending limb of Henle’s loop which is transported back to the interstitium, from the collecting duct. Transport of these substances facilitated by the special arrangement of Henle’s loop and vasa recta.is called the counter current mechanism.

This mechanism helps to maintain a concentration gradient .in the medullary interstitium. Presehce of such interstitium gradient help easy passage of water from the collecting duct, there by concentrating the urine.

Question 8.
Explain the auto regulatory mechanism of GFR.
Answer:
Auto Regulation of GFR: *The kidneys have built in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxta glomerular apparatus. Juxta glomerular apparatus is a special region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact.

A fall in GFR can activate the juxta glomerular cells to release an enzyme called renin, which catalyses the conversion of angiotensinogen into angiotensin-I and further converted to angiotensin-II by action of an enzyme angiotensin converting enzyme. Angiotensin-II’ stimulate the adrenal cortex to secrete aldosterone. Aldosterone causes reabsorption of Na+ and water from DCT and collecting duct to reduce loss through urine and also promotes the secretion of K+ ions into the DCT and CD (collecting duct). It leads.an increase in the blood pressure and GFR.

Question 9.
Describe the role of liver, lungs and skin in excretion.
Answer:
In addition to the kidneys, liver, lungs and skin also play an important role in the elimination of excretory wastes.

a) Liver :
Liver is the largest gland in our body, secretes bile, containing substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drags. Most of these substances ultimately pass out along with digestive wastes.

b) Lungs :
Lungs remove large amounts of C02 (18 litres 1 day), various, volatile materials and significant quantities of water.

c) Skin :
Human skin possesses two types of glands namely sweat and sebaceous glands for the elimination of certain substances through their secretions.

• Sweat produced by the sweat glands is a watery fluid containing NaCl, small amount of urea, lactic acid etc.,
• Sebaceous glhnds eliminate certain substances like sterols, hydrocarbons and waxes through sebum. This secretion provides a protective oily covering for the skin.

Question 10.
Name the following.
Answer:
a) A chordate animal having protonephridial type excretatory structures Cephalo chordate.
b) Cortical portions projecting between the medullary pyramids in the human kidney. Columns of Bertini
c) Capillary network paralleing the loop of Henle. Vasa recta.
d) A non chordate animals having green glands as excretory structures. Crustaceans.

Long Answer Questions

Question 1.
Describe the excretory system of man, giving the structure of a nephron.
Answer:
In humans, the excretory system consists of a pair of kidney, a pair of ureters, a urinary bladder and urethra.

Kidney :
Kidneys are reddish brown, bean shaped structures, situated on either side of the vertebral column between the levels of last thoracic and third lumbar vertebrae in a retroperitoneal position. The right kidney is slightly lower than the left one due to the presence of liver.

The outer surface of the kidney is convex and the inner surface is concave, where it has a deep notch called hilum, the point at which the renal artery and nerves enter and renal vein and ureter leave. Each kidney is surrounded by a tough, fibrous tissue, called renal capsule.

Ureter :
These are slender whitish tubes, which emerges from the pelvis of the kidney. The ureter rundown and open into the urinary bladder.

Urinary bladder :
Urinary bladder is a pear shaped like muscular organ. It tempirarily stores the urine, situated in the lower abdominal cavity. The neck of the bladder leads into the urethra. Urethra opens near the vaginal orifice in the female and through the penis in the male.

Structure of nephron:
Each kidney has nearly one million nephrons. These are structural and functional units of kidney, embedded in the loose connective tissue of cortex and medulla. Nephron consist of malpighian body and renal tubule.

I) Malphigian body :
It begins in the cortex of the kidney. It contains Bowman’s capsule and glomerulus.

a) Bowman’s capsule :
It is a thin walled, double layered cup. The inner wall of the Bowman’s capsule has certain unique cells called podocytes.

b) Glomerulus :
It is a dense network of capillaries in the cup of Bowman’s capsule. Afferent arteriole of renal artery enter the cavity of Bowman’s capsule and split into five branches.

They unite and come out of the Bowman’s capsule as an afferent arteriole.

The podocytes of inner wall of Bowman’s capsule wrap around each capillary. The podocytes are arranged in an intricate manner so as to leave some minute spaces called filteration slits. The endothelium cells of capillaries have numerous pores called fenestrations.

II) Renal tubule:
It is narrow, delicate tubule arises from the posterior part of Bowman’s capsule known as neck. It opens into along narrow convoluted tubule with three parts like proximal convoluted tubule, Loop of Henle and Distal convoluted tubule.

a) Proximal convoluted tubule :
It is a lined by simple cuboidal epithelium with brush border to increase area of absorption.

b) Loop of Henle :
It is a hairpin like tubule present in medulla region. It consist of a descending limb and an ascending limb. The proximal part of the ascending limb is thin and the distal part is thick. The thick ascending limb continuous into the distal convoluted tubule.

c) Distal convoluted tubule (DCT) :
It is present in cortex. It is lined by simple cuboidal epithelium. The DCT continuous as the initial collecting duct in the cortex.

Collecting system :
Some initial collecting ducts unite to form straight collecting duct, which passes through the medullary pyramid. In the medulla, the tubes of each pyramid join and form duct of Bellini, which finally opens into tip of the renal papilla.

Capillary network of nephron :
The efferent arteriole emerging from the glomerulus forms a fine capillary network called the peritubular capillaries, around the renal tubule. The portion of the peritubular capillaries that surrounds the loop of Henle is called the vasa recta. The vasa recta is absent or highly reduced in the cortical nephrons. The juxta medullary nephrons possess well developed yasa recta.

Question 2.
Explain the physiology of urine formation.
Answer:
The formation of urine involves three main processes namely

1. Glomerular Alteration
2. Selective reabsorption
3. Tubular secretion.

1) Glomerular Alteration :
It is Arst step in urine formation. The process of Alteration of blood, which occurs between glomerulus and lumen of the Bowman’s capsule due to difference in netpressure is called glomerular Alteration.

The hydrostatic pressure of blood while Aowing in the glomerulus is 60 mm Hg. It is opposed by glomerular colloidal osmotic pressure of 32 mm Hg and Bowman’s capsule hydrostatic pressure of 18 mm Hg.

The net filterate pressure is 10 mm Hg ( 60 – 32 + 18 = 10). This causes the Alteration of blood through the 3 layered filterate membrane formed by endothelium cells of glomerular capillary together with the basement membrane and podocytes of the Bowman’s cup. By the result of glomerular Alteration primary urine or renal Auid is collected in lumen of the Bowman’s capsule.

The primary urine contains almost all the constituents of plasma, except the proteins. The primary urine is hypotonic to the cortical Auid, it passes into the next part of renal tubule.

2) Selective reabsorption:
During the process of glomerular Alteration 125 ml/minute df primary urine is formed. Nearly 99% of which and essential substances are reabsorbed* by renal tubules called selective reabsorption. About 85% of filterate formed (primary urine) is reabsorbed in a constant unregulated manner called obligatory reabsorption.

3) Tubular Secretion :
During the formation of urine, the tubular cells secrete substances such as H⁺, K⁺ and NH₃⁺ into the filterate. Tubular secretion is also an important step in the formation of urine as it helps in maintenance of ionic and acid base balance of the body fluids.

Mechanism of selective reabsorption and secretion takes place is different parts of nephrons.

a) In the proximal convoluted tubule :
Nearly all the essential nutrients and 70-80% of electrolytes and water are reabsorbed by this segment. Na⁺, glucose, amino acids, Cl^– and other essential substances are reabsorbed into blood.

PCT also helps to maintain the pH and ionic balance of body fluids by selective secretion of H⁺ and NH₃ into the filterate and by the absorption of HCO₃^– from it.

b) In the Henle’s loop :
Reabsorption in this segment is minmium.

• The descending loop of Henle is permeable to water and almost impermeable to electrolytes results the filterate concentration gradually increases.
• The ascending limb has two specialized regions, a proximal thin segment in which NaCl diffuses out into interstitial fluid passively, and distal thick segment, in which NaCl is actively pumped out.

The ascending limb is impermeable to water. Thus the filterate becomes progressively more dilute as it moves up to the cortex i.e., towards the DCT.

In the Distal convoluted tubule (DCT) :
It is permeable to water and ions. The reabsorption of water is variable depending on several conditions and is regulated by ADH. DCT is also capable of reabsorption of HCO₃^– and selective secretion of H⁺ and K⁺ ions and NH₃⁺ into DCT from peritubular network, to maintain the pH and sodium – potassium balance in tHe blood.

In the collecting duct (CD) :
Considerable’amount of water could be reabsorbed from this region to produce concentrated urine. This segment allows passage of small amount of urea to medullary interstitium to keep up its osmolarity. It also plays a role in the maintenance of pH and ionic balance of blood by the selective secretion of H⁺ and K⁺ ions.

The renal fluid after the process of facultative reabsorption in the CD, influenced by ADH, constitute the urine, that is sent out. Urine in the CD is hypertonic to the plasma of blood.

Andhra Pradesh BIEAP AP Inter 2nd Year Commerce Study Material 5th Lesson Consumer Protections Textbook Questions and Answers.

AP Inter 2nd Year Commerce Study Material 5th Lesson Consumer Protections

Essay Answer Questions

Question 1.
Explain the composition and jurisdiction of the state commission.
Answer:
The state commission settles consumer disputes at the state level. The state commission is headed by the judge of a high court and comprises other members not less than two and not more than as prescribed, one of whom shall be a woman.

The state commission shall have jurisdiction to entertain consumer complaints where the value of goods and services for which the compensation claimed exceeds ₹ 20 lakhs and less than ₹ 1 crore. The state commission is empowered to call for the records and pass appropriate orders in respect of any consumer dispute within the state jurisdiction. The state commission is empowered to transfer any complaint pending before on the district forum to another district forum within the state. The state commission has circuit Benches.

In case the aggrieved party is not satisfied with the order of the state commission, he can appeal to the national commission within 30 days of passing the order.

Question 2.
Describe the rights of a consumer as per CPA 1986.
Answer:
Although a businessman is aware of his social responsibilities even then we come across many cases of consumer protection. Hence Government of India provided the following six rights to all the Consumers under Consumer Protection Act.
1) Right to safety:
According to this right, the consumers have right to be protected against the marketing of goods and services which are hazardous to life and property. The right is important for safe and secure life.

2) Right to information :
According to this right, the consumer has right to get information about the quality, quantity, purity standard and piece of goods or services. The producer must supply all the relevant information at a suitable place.

3) Right to choice :
According to this right, every consumer has a right to choose the goods or services of his or her likings. The supplier should not force the consumer to buy a particular brand only. Consumer should be free to choose the most suitable product from his view point.

4) Right to consumer education:
According to this right, it is the right of the consumer to acquire knowledge and skill to be informed to customer. It is easier for the literate consumers to know their rights and take actions.

5) Right to seek redressal:
According to this right, the consumer has the right to get compensation or seek redressal against unfair trade practices or any other exploitation. The right assures justice to consumer against exploitation.

6) Right to heard / Right to represent:
According to this right, the consumer has the right to represents himself or to be heard or right to advocate his interest. In case a consumer has been exploited or has any complaint against the product or service then he has a right to be heard.

Question 3.
What are the responsibilities of a consumer?
Answer:
Various efforts have been made by government or non – government organisations to protect the interest of consumer, but exploitation of consumer will stop only when the consumer will come forward to safeguard his own interest. Consumer has bear the following responsibilities.

1) Be quality conscious :
To put to stop to adulteration and corrupt practices of the manufacturers and traders, it is the duty of every consumer to be conscious of the quality of the products they buy. They should look for the standard quality certification marks like ISI, Agmark, Wool mark, Ecomark, Hallmark, etc. While making the purchases.

2) Beware of misleading advertisements :
The advertisement often exaggerates the quality of the products. Hence the consumers should not rely on the advertisement and carefully check the product or ask the users before making a purchase.

3) Responsibility to inspect a variety of goods before making selection :
The consumer should inspect a variety of goods before buying the goods and services. For this purpose, he / she should compare their quality, price, durability, after sales service etc.

4) Collect proof of transaction :
The consumer should insist a valid documentary evidence (Cash memo / invoice) relating to purchase of goods or availing of any services and preserve it carefully. Such proof of purchase is required for filing a complaint. In case of durable goods the manufacturers generally provide the warrantee / guarantee card with the product. It is the duty of the consumers to obtain these documents and ensure that these are duly sighed, stamped and dated. The consumer must preserve them till the warrantee / guarantee period is over.

5) Consumers must aware of their rights:
The consumers must aware of their rights as stated above and exercise them while buying goods and services. For example, it is the responsibility of a consumer to insist on getting all information about the quality of the product and ensure himself / herself that it is free from any kind of defect.

6) Complaint for genuine grievances:
As a consumer, if you are dissatisfied with the product, you can ask for redressal of yoifr grievances. In this regard, you must file a proper claim with the company first. The manufacturer / company do not respond, then you can approach the forums. But your claim must state actual loss and the compensation claim must be reasonable. At no cost fictious complaints should be filed otherwise the forum may penalise you.

7) Proper use of product / service :
It is expected from consumers that they use and handle the product / service properly. It has been noticed that during guarantee period, people tend to reckless use of the product, thinking that it will be replaced during guarantee period. This practice should be avoided.

Question 4.
Explain the redressal mechanism available to consumers under the Consumers Proction Act, 1986.
Answer:
The judicial machinery set up under Consumer Protection Act (C.RA) 1986 consists of consumer courts (Forums) at the District, State and National levels. These are known as District Forum, State consumer disputes redressal commission (State Commission) and National consumer disputes redressal commission (National Commission).

1. District forum :
This is established by the state government in each of its districts.
a) Composition :
The district forum consists of a chairman and two other members one of whom shall be a woman. The district forms are headed by the person of the rank of a District Judge.

b) Jurisdiction:
A written complaint can be filed before the district forum where the value of goods or services and the compensation claimed does not exceed ₹ 20 lakhs.

c) Appeal:
If a consumer is not satisfied by the decision of the District Forum, he can challenge the same before state commission, within 30 days of the order.

2. State commission:
This is established by the state governments in their respective states.
a) Composition :
The state commission consists of a president and not less than two and not more such number of members as may be prescribed, one of whom shall be a woman. The commission is headed by a person of the rank of High Court Judge.

b) Jurisdiction :
A written complaint can be filed before the state commission where the value of goods or services and the compensation claimed exceeds ₹ 20 lakhs but does not exceed ₹ 1 Crore.

c) Appeal :
In case the aggrieved party is not satisfied with the order of the state commission he can a appeal to National Commission within 30 days of passing the order.

3. National Commission :
The national commission was constituted in 1988 by the central government; It is the apex body in the three tier judicial machinery set up by the government for redressal of consumer grievances. Its office is situated Janpath Bhawan in New Delhi.

a) Composition :
It consists of a president and not less than four and not more than such members as may be prescribed, one of whom shall be a woman. The National Commission is headed a sitting or retired judge of supreme court.

b) Jurisdiction:
All complaints pertaining to those goods or services and compensation whose value is more than ₹ 1 Crore can be filed directly before the National Commission.

c) Appeal:
An appeal can be filed against the order of the National Commission to the supreme court within 30 days from the date of order passed.

Question 5.
Who can file a complaint, what complaints can be filed, where to file the complaint, how to tile the complaints redressal of grievances under the Consumer Protection Act 1986?
Answer:
For redressal of consumer grievances a complaint must be filed with the appropriate form.

Who can complaint?
The following persons can file a complaint under Consumer Protection Act 1986.
a) a consumer.
b) Any recognised voluntary consumer association whether the consumer is a member of that association or not;
c) The central or any state government;
d) One more consumers where there are numerous consumers having same interest;
e) Legal heir or representative in case of death of consumer what complaints can be filed?

What complaints can be filed?
A consumer can complaint relating to any one or more of the following;
a) An unfair trade practice or a restrictive trade practice adopted by any trader or service provider;
b) Goods bought by him or agreed to bought by him suffer from one more defects ;
c) Services hired or availed of or agreed to be hired or availed of, suffer from deficiency in any respect;
d) Price charged in excess of the price
i) fixed by or under law for the time being in force
ii) displayed on the goods or the package
iii) displayed in the price list or
iv) agreed between the parties and
e) goods or services which are hazardous or likely to be hazardous to life and safety when used.

Where to file a complaint?
If the value of goods and services and the compensation claimed does not exceed ? 20 lakhs, the complaint can be filed in the district forum ; If it exceeds ₹ 20 lakhs but does not exceed ₹ 1 crore, the complaint can be filed before the State Commission ; and if it exceeds ₹ 1 Crore, the complaint can be filed before the National Commission.

How to file a complaint?
A complaint can be made in person or by any authorised agent or by post. The complaint can be written on a plain paper supported by documentary evidence in support ’of the allegation contained in the complaint. The complaint should clearly specify the relief sought. It should also contain the nature, description and address of the complaint as opposite party, and so also the facts relating to the complaint and when and where it arose.

Very Short Answer Questions

Question 1.
Give the meaning of consumer.
Answer:
Under the Consumer Protection Act 1986, The word consumer has been defined separately for the purpose of goods and services.

For the purpose of goods, a consumer is one who buys any goods for consideration and any user of such goods other than the person who actually buys it, provided such use is made with the approval of buyer.

For the purpose of services, a consumer is one who has any service or services for consideration ; and any benificiary of such services provided the service is availed with the approval of the person who had hired the service for a consideration.

Question 2.
What is consumerism?
Answer:
Consumerism is defined as a social force designed to protect consumer interest in the market place by organising consumer pressure on business. By consumerism we mean the process of realising the rights of the consumer as enrises in the Consumer Protection Act, 1986 and ensuring right standards for the goods and services for which one makes payment.

Question 3.
What is meant by consumer protection?
Answer:
Consumer protection means safe guarding the interest and rights of consumers. In other words, it refers to the measures adopted for the protection of consumers from redressal of their grievances. The most common business malpractices are sale of adulterated, spurious, substandard and duplicate goods, false and under weighting, hoarding and black marketing, charging more than MRP Price etc. .

Question 4.
District Forums.
Answer:
The state government in each district establishes District forum by notification. The district forum consists of a president nominated by the state government. The forum also comprises two other members who shall have atleast 10 years of experience in dealing problems of economics, law commerce and industry. Every member of the form shall have tenure of 5 years or 65 years whichever is earlier. The District collector acts as the chairman of the District Forum. The District forum shall have jurisdiction to entertain consumers complaints where the value of goods and services which the compensation claimed, should ₹ 20 lakhs.

Question 5.
State commission.
Answer:
The state commission settles the consumer dispute at state level. The state commission is headed by the judge of High Court and comprised of other members not less than two and not more than such members as prescribed. The state commission is empowered to call for the records and appropriate orders in respect of any consumer dispute within the state jurisdiction. The state commission shall have jurisdiction to entertain consumer complaints where the value of goods and services for which compensation claimed exceeds ₹ 20 lakhs.

Question 6.
National commission.
Answer:
National commission operates at National level. It settle the consumer disputes at in the country. The National Commission has a President, who should be a serving or retained Supreme Court Judge the commission also comprises other members of not less than four. The president and all the members of the commission are appointed by central government. The National Commission shall have jurisdiction to entertain consumers complaints where the value of goods and services and compensation exceeds ₹ 1 crore.

Question 7.
Who is consumer? In the opinion of Mahatma Gandhi.
Answer:
Mahatma Gandhi, the father of nation, attached great importance to what he described as the ‘poor consumer’, who according to him should be the principle benificiary of the consumer movement. He said “A consumer is the most important visitor on our premises. He is not dependent on us, we are on him. He is not an interruption to our work; he is the purpose of it. He is not an outsider to our business ; he is a part of it. We are not doing him a favour by serving him ; he is doing us a favour by giving an opportunity to do so.

Andhra Pradesh BIEAP AP Inter 2nd Year Commerce Study Material 3rd Lesson Business Services Textbook Questions and Answers.

AP Inter 2nd Year Commerce Study Material 3rd Lesson Business Services

Essay Answer Questions

Question 1.
Define banking. Explain the functions of banking.
Answer:
The word Bank is derived from the French word ‘Bancus’ means a bench. According to Banking Regulation Act of 1949, banking is defined as “Accepting for the purpose of lending or investment of deposits of money from the public, repayable on demand or otherwise and withdrawable by cheque, draft, order or otherwise”.

The basic functions of banks are classified as primary functions and secondary functions.

A) Primary functions:
i) Accepting Deposits :
Bank accept various types of deposits, such as

Fixed deposits :
Fixed deposits are also called as time deposits or term deposits. In this deposit amount cannot be withdrawn until the maturity. Interest rate is also high.

Current Deposits :
Current accounts bears no interest companies, institutions, government and business men hold the current account. The greatest advantage of having current account is that there is no restriction on the withdrawls.

Savings Deposits :
The aim of these accounts is to encourage small savings from the public. Certain restrictions are imposed on the depositors regarding the number of withdrawals and the amount to be withdrawn in a given period of time.

Recurring deposits :
The purpose of these accounts is to encourage regular savings particularly by the fixed income group. Generally money is deposited in these accounts on monthly instalments for a fixed period. It is repaid to the depositors along with interest on maturity.

ii) Advancing of Loans :
Lending is carried out purely on profit motive. Banks lend the amount which is mobilised through the deposits. The different forms of lending are :

Loans :
A specified amount sanctioned by bank is called a ‘Loan’. A loan is granted against the security of property or personal security. The loan may be repaid in lumpsum or in instalments. The loan may be classified into i) Demand loan ii) Term loan. Demand loan is repayable on demand. It is repayable at short notice. Medium and long term loans are called term loans. It is granted for more than year and repayment is done on longer period.

Cash credit:
Cash credit is an arrangement where by the bank agrees to lend money to the borrower upto a certain limit. The amount is credited to the borrowers account. The borrower draws the money as and when he needs. Interest will be charged only on the amount actually drawn. Banks may impose commitment charges on unutilised portion.

Overdraft:
Banks grants overdraft to current account holders by which he is allowed to draw an amount in excess of the balance held in his account. Interest is charged on the overdrawn amount.

Discounting of bills of exchange :
A holder of a bill of exchange may be in urgent need of cash before the due date. He may sell or discount the bill with the bank. He will receive lesser amount than the actual amount. On maturity, the bank gets it payment from the debtor.

B) Secondary Functions :
These services include agency services and general utility services.

1. Agency services :
Banks perform some agency services on behalf of their customers.

• Banks helps their customers in transferring funds from one place to another place through cheques, drafts etc.
• Banks collect and pay various credit instruments like cheques, bills of exchange, promissory notes etc.
• Banks undertake to purchase and sale of various securities like shares, bonds, debentures etc., on behalf of their customers.
• Banks preserve the wills of their customers and execute them after death.

2. General utility services :
These services are

• Letters of credit are issued by the banks to their customers certifying their credit worthiness.
• Banks issue travellers cheques to help to travel without fear of theft or loss of money.
• Banks provide safe deposit locker facilities to the public at selected branches.
• Accepting or collecting foreign bills of exchange.

Question 2.
Discuss the principles of insurance. [A.P. Mar. 17]
Answer:
Insurance means protection against risk of loss. It provides compensation against any loss or damage due to the happening of an event. It is a contract between the two parties by which one of them undertake to indemnity the other person against a loss which may arise due to some events.

Principles of insurance :
1. Insurable interest:
A person cannot enter into a contract of insurance unless he has insurable interest in the subject matter of insurance. It is essential feature of insurance. Without this insurable interest, the contract of insurance will be treated as a wager or gambling contract. A person has insurable interest on his own life or the life of his wife and a creditor has insurable interest in the debtor.

2. Utmost good faith :
Insurance is based on the principle of utmost good faith. It means both the parties of the contract must disclose all the facts relating to the subject matter of insurance. If the insured does not disclose all material facts, the contract between them is void.

A person who had suffered from T.B. in the past had not disclosed it in the proposal form. Later on the insurer comes to know of this fact. He may declare the contract as void.

3. Indemnity :
This is the chief principle of insurance. Indemnity means security against risk of loss. Under this principle, the insured gets only the loss suffered from the insurer but not profits out of the contract of insurance. The principle of indemnity applies to contracts of fire and marine insurances only, but not to life insurance contracts.

4. Contribution :
Sometimes, goods are insured with more than one company. It is double insurance. The insured can get compensation only for the total loss from all insurance companies put together, but not total loss from each company. The insurance companies will pay the compensation on prorata basis.

5. Subrogation :
According to this principle, the insurer after compensating the loss of insured, the right of ownership of the damaged goods is shifted from insured to insurance company. Ex : Mr. X owns a scooter worth ₹ 36,000 and it was insured with a insurance company for full value. Later it was met with an accident and damaged beyond repairs. The insurance company paid the full value as compensation. Then all the rights on the scooter will pass on to insurance company.

6. Causa proxima :
According to this principle, the loss is caused by nearest and direct factor, then only the insurer will have to bear the loss. Ex : Biscuits in a ship are insured and are destroyed because of the sea water entered through a hole made by the mouse in the bottom of the ship and water entered into the ship. The nearest and direct cause is sea water. Hence, the insurer will have to bear the loss.

7. Mitigation of loss :
It is duty of the insured to take necessary steps to minimise the loss happened due to some event. He should not act carelessly and negligently at the time of loss to the insured property.

Question 3.
Define Life Assurance Policy. What are the kinds of life assurance policies?
Answer:
A life insurance contract may be defined as “a contract whereby the insurer, in consideration of premium paid either in lumpsum or in periodical instalments, undertakes to pay an annuity or a certain sum of money, either on the death of the insured or the expiry of certain number of years”.

Kinds of life assurance policies:
The following are some of the popular life assurance policies.
1) Whole life policy :
Under this policy the premium is paid through out the life of the insured. The sum assured is payable only after the death of the assured. The premium payable is low and it is meant for the family.

2) Endowment policy:
This policy is taken up for a specific period called endowment period. The policy will mature at the end of the specified period or at the attainment of particular age or on the death of insured, whichever is earlier. This policy offers the advantage of both protection and investment.

3) With or without profits policies :
When the policy is insured with profits, the policy holders share the profits of the company. The profit is called Bonus. The amount of the policy and bonus is paid on the maturity of the policy. When the policy is insured without profits, the insured does not share any profits and the amount of policy only is paid on maturity.

4) Joint life policies:
A policy may be taken jointly on the lives of two or more persons. On the death of any one person, the policy is paid to other surviving policy holder as the case may be. This type of policy may be taken by husband and wife.

5) Convertible whole life policy :
The policy is insured for whole life policy with a provision to convert it into endowment policy after a specified period. The conversion is done at the request of the assured. The rate of premium is increased after conversion.

6) Janata policy:
Janata policy was introduced by life insurance corporation of India in May 1957. This policy was introduced for the benefit of lower income group people. It can be issued for a term of 5,10, 15, 20 and 25 years subject to the condition that it should mature at the age of 60 years. No loans are granted on such policies.

7) Annuity policy :
Under this policy, the insured would deposit a lumpsum amount with the insurance company. The amount of policy would be paid to the insured for a specified number of years or until the death of the insured.

8) Children endowment policy:
This policy is taken by a person for his / her children to meet the expenses of their education or marriage. The agreement states that a certain sum will be paid by the insurer when the children attain a particular age.

9) Group insurance policy :
Members of a family or the employees of a business concern can take this insurance policy.

Question 4.
What do you understand by the word Transport? Discuss the benefits and limitations of Transport.
Answer:
Transport is the physical means of moving goods and persons from one place to another. Transport creates place utility of goods by moving them from different centres of production to the places of consumption. Goods are now produced thousand miles away from places where the consumer resides. Transportation only help the business men to reach consumer. Not only does transport give place utility, but it also renders time utility in various ways. Transportation, in simple language can be defined as “a means through which goods are transferred from one place to another”.’

Benefits or Functions of transport:
1. Movement of goods:
The first and important function of transport is the movement of goods. The raw materials have to move from their sources to the factory. The manufactured goods have to move from the factory to the consuming areas.

2. Transport enhances mobility of labour and capital :
An efficient network of transport services encourages the movement of people from one place to another labour can migrate to the place where they can get better job opportunities which reduces exploitation of labour.

3. Creation of place utility:
It moves goods from places where they are abundant to the places where they are scarce and thus creates place utility.

4. Specialisation and division of labour :
Transportation facilitates optimum utilisation of natural resources of a country. For example, petroleum resources of Arab countries, watches of Switzerland etc.

5. Creation of time utility :
With the advancement of technology, transportation time is being shortened. So, it creates time utility.

6. Stability in prices :
Goods can be transported from the place where the goods are abundant, to the places where scarcity exists. In this way, prices are equalised throughout the country.

7. Contribution to national income :
The transportation also contributes national income of a nation. For example, our railways.

8. Economies of large scale production :
Transport has helped the development of large scale industries. Transport procure raw material, labour and sells the finished goods.

9. Improve standard of living :
Availability of wide variety of goods at reasonable prices improves standard of living.

10. National defence :
Transport strengthens the national defence transport system. During war period, all the personnel, material and equipment can be moved rapidly to the boarder areas.

Limitations of transport :
1. Cottage and small scale industries lost their glory :
With the development of transport, labour is showing interest to work in big factories. This has led to shortage of workers in tiny and small scale industries.

2. Accidents :
Improvement in transport facility has given rise to new problem viz. accidents.

3. High urbanisation:
Improved means of transport has helped in creating big cities, which have further resulted into concentration of population in these cities. This has given rise to many new problems such as housing, health and pollution.

Question 5.
Describe the road transportation. Explain the kinds of roads in India. [A.P. Mar. 17]
Answer:
Road transport is the oldest form of transport. The Indian road network is one of the largest in the world. Road transport plays an important role in trade and commerce. Road transport is very good for short distance. Door – to – door collection and delivery is possible in road transport. It is most suitable for perishable goods. In road transport both men and animals are used to carry goods and people.

Modes of road transport are bullock carts, Tonga, rickshaws and motor vehicles such as jeeps, buses, motor vans, trucks and other vehicles. Road transport is suitable for the goods such as paper goods, clothing, computers, livestock, cement etc.

Indian roads are classified into three types – National highway, State highway, District and Rural roads.

a) National highway:
These roads are meant for inter state transport and movement of defence men. These also connect the state capitals and major cities. The National Highway Authority of India (NHAI) has the responsibility of development, maintenance and operation of national highways. The national highways have a road length of about 65,000 kms or 2% of the length of the total road system but they carry nearly 40% of goods and passenger traffic.

b) State highways:
These are constructed and maintained by state government. They connect the state capital with district head quarters and other important towns. State highways constitute 4% of total road length in the country.

c) District roads :
These roads are the connecting link between district head quarters and the other important roads of the district. The account for 14 % of the total road length of the country.

d) Rural roads :
These roads provide link to the rural areas. There are about 80% of total length in India are categorised as rural roads.

e) Boarder roads :
These roads are in the northern and northern – eastern boundary of the country. The Boarder Road organisation constructs and maintains boarder roads. They construct roads in high altitude areas and undertakes snow clearance.

f) International highways :
There are meant td promote harmonious relationship with the neighbouring countries by providing effective links with India.

Question 6.
Explain the warehouse concept and its significance.
Answer:
The term warehousing is a combination of two terms, ‘ware’ and ’housing’. The word ware refers to goods. Therefore, warehousing can be defined as the ‘ place suitable for preserving the goods and warehousing is the activity involving storage of goods. In common parlance warehouse means a godown. Warehousing facilitates other marketing functions such as assembling, grading and transportation.

Warehousing performs two important functions with regarding to finished goods. They are movement function and storage function. Movement function refers to the receipt of the products from the manufacturing plant, their transfer into warehouse and transferring them to common carriers on their way to consumers. The storage function is performed by retaining storing products in the warehouse until they are sold because production and consumption cycles differ. Warehousing functions creates time utility at minimum cost.

Significance :

1. Some commodities are produced in a particular season only. To ensure their off – season availability, warehousing is needed.
2. Some products are produced throughout the year but their demand is seasonal. Warehousing is important in such cases.
3. For the companies which opt for large scale production and bulk supply, warehousing is unavoidable factor.
4. Warehousing help companies to ensure quick supply of goods in demand.
5. Production of goods and their movement of goods are important for the companies for continuous production of goods.
6. Warehousing is also important for price stabilisation. For necessary goods, the Government store them in the warehouse and control its supply in the market as per the price fluctuations.
7. Another important need of warehousing is for bulk breaking.

Short Answer Questions

Question 1.
Define services and goods.
Answer:
Services are those separately identifiable, essentially intangible, activities that provide satisfaction of wants and are not necessarily linked to the sale of a product or other services. Services are intangible as they are not seen or touched. Service is inconsistant, since there is no standard tangible product. Service is the simultaneous activities of production and consumption. Services cannot be stored for future. Service is the participation of the customer in the service delivery process.

Goods are physical objects and are homogeneous in nature. They are tangible. Ex: Medicine Different customers get standardised demands fulfilled. Ex: Mobile phones. There will be separation of production and consumption. Ex : Purchasing ice cream from a store. Goods can be kept in stock. Ex : Train journey ticket. Involvement at the time of delivery is not possible. Ex : Manufacturing a vehicle.

Question 2.
What are the advantages of E-Banking?
Answer:
E-Banking brings certain advantages.
1. It reduces costs :
The cost of banking transactions is considerably reduced. It increases the profitability of the banks.

2. Prompt in services :
There is high degree of personalisation and fast and flexible execution. Thus E-Banking prompt service and there is greater customer satisfaction.

3. Anywhere and any time banking :
It is 24 hours in a day and 7 days in a week banking service. Bank account can be accessed from anywhere. So the customer can obtain information his account and conduct transactions from his home or office.

4. Cashless banking :
Handling of cash is not necessary in E-Banking.

5. Global coverage :
It provides global network coverage of bank services. NRIs can monitor their bank account in Indian banks, from abroad.

6. Central data base :
The data base of each branch is centralised. Customer can deposit, withdraw or remit money from any branch of his bank.

7. Internet banking helps banks to reduce the workload of their branches, such as generation of statements, balance of enquiry etc.

Question 3.
What is Mobile Banking? What are the services can be obtained through mobile banking?
Answer:
The delivery of bank services to customers through mobile (cell) phone is called mobile banking. When compared to telephone banking the scope of mobile banking is more and effective also. Mobile banking can take the form of SMS banking, GSM SIM Toolkit and WAP.

a) SMS banking :
Short messages are sent to the customers mobile phones. SMS messages can be used for both passive and active banking operations. A client automatically receives information about his account balance after a certain operation is performed.

b) GSM SIM Tool kit :
The GSM SIM Toolkit service can be only be used from a mobile phone supporting this technology. GSM SIM Toolkit is a software that evolves arbitrary changes in the mobile phone menu. Mobile phones now on the market support GSM SIM Tool kit after buying a special sim card and activating the permanent bank branch. The client can use this service.

c) WAP (Wireless Application Protocol) :
WAP is often compared to web pages although it is simplified. Unlike pages appearing on computer monitor, WAP presents it output on a small mobile phone. WAP Banking is not very popular. Only few banks are providing this service. .

Question 4.
What are the facets of electronic banking?
Answer:
The following are the different facets of E-Banking.
1. ATM :
ATM is popularly known as Any Time Money Machine. The customer gets cash fast, withdrawal, transfer, payment of bills or cash deposit through ATM.

2. Tele banking (home banking) :
Customers can perform number of transactions from their telephone such as checking the balance in the accounts, transfer funds from one account to another, pay certain bills and order statements or cheque book etc.

3. E-mail banking :
Customers may communicate with bank by electronic mail or E-mail. The most frequently used service is sending account statement periodically to the clients mail box.

4. Network banking or online banking:
Internet or online banking is a facility provided by banks to enable the user to execute bank related transactions through internet. The people sitting at home can transact business and they need not visit bank.

5. Mobile banking:
The delivery of bank services to a customer through mobile (cell) phone is called mobile banking.

Question 5.
Explain the term Insurance? Explain the functions of Insurance.
Answer:
The method of sharing of risk through economic cooperation is called insurance. Insurance may described as a social device to reduce or eliminate risk of loss to life and property. Insurance renders valuable services to commerce as well as to the society. Insurance covers many risks and uncertainties in the world of business and act as a boon to business firms.

Functions of Insurance :
1. Providing certainty :
Insurance provides payment of the risk of loss. There are uncertainties of happening of time and amount of loss. Insurance removes these uncertainties and the assured receives payment of loss. The insurer charges premium for providing certainty.

2. Protection:
The second function of insurance is to provide protection from probable chances of loss. Insurance cannot stop the happening of a risk or event but can compensate the losses arising out of it.

3. Risk sharing:
On the happening of a risk event, the loss is shared by all the persons exposed to it. The share is obtained from every insured member by way of premiums.

4. Assist in capital formation:
The accumulated funds of the insurer received by way of premium payments made by the insured are invested in various income generating schemes.

Question 6.
Explain the costs and benefits of Insurance.
Answer:
The following are the benefits of Insurance :
i) Provides certainly :
Insurance helps the insured to convert his uncertainties into certainities by entering into contract of insurance. The payment of premium by insured enables to reduce the risk.

ii) Distribution of losses :
Insurance helps to distribute losses of any uncertain events among the large number of insurer’s. It enables to transfer the risks and spreads the financial loss of insured members over the whole insurers.

iii) Provides security:
It provides security to the insured against the risk of uncertain events. The insurance company guarantees the insured to compensate or indemnify the loss on the occurrence of an event in consideration for payment of premium.

iv) Generates capital :
Insurance reduces financial risks and losses by providing facilities of capital investment in various organisations.

v) Increases efficiency :
Insurance reduces the risk and increases the efficiency in business. It provides security for business community which in turn paves the way for growth and diversification of industry.

vi) Earns foreign exchange :
Insurance provides security to the international traders, shippers and banking institutions, thus paves the way for expansion of foreign trade. The increased foreign trade activities lead to securing foreign exchange which makes the country to become economically strong.

vii) Social security :
Insurance acts as an instrument to fight against evils of poverty, unemployment, disease, old age, sickness, disability, accidents, fire and similar other calamities of nature.

viii) Promotes thrift:
Insurance encourages the people to go for savings. It alter the people in their spending habits and makes them to save a certain sum of money regularly.

Costs of disadvantages of Insurance :
i) Sharing of loss:
The loss of one person should be shared by all other policy holders. But the sharing of loss is opposed by many people as their return on investment is reduced.

ii) Real value of money :
The maturity value of the policy after the specific period may be more but the real value of money is going to be less.

iii) Lack of confidence :
Many of the investors, who propose to save their money, prefer banks and other financial institutions. It is due to lack of confidence in the insurance companies and its policies.

Question 7.
What are the advantages of Life insurance policies?
Answer:
In life insurance contract the policy amount is definitely, it is a question of time. The policy may mature during the life time of the assured or it may be paid on his death.

Advantages of life assurance policies :
1. Encourages savings:
The insured has to pay premium to insurance company every year. Otherwise, the policy will be cancelled. So, the insurance is helpful in creating the habit of saving money.

2. Exemption from income tax :
The amount paid as premium on a life insurance policy is allowed as deduction from income for calculating income tax.

3. Protection:
Life insurance provides protection to the family members if the policy holder dies suddenly. Life insurance builds a fund for the benefit of the dependents.

4. Credit facilities :
Insured can get loans against their policies to meet emergency needs. Life Insurance Corporation itself gives a loan against the policy to the insured at a lower rate of interest.

5. Surrender:
The life insurance company can surrender the life policy if the insured is unable to continue it. The insurance company can return some premium known as’surrender value’.

6. Meets the future needs:
An insurance policy can be helpful in providing funds for marriage and educational needs of insured children.

Question 8.
Explain the characteristics of marine Insurance.
Answer:
Marine insurance is a contract whereby the insurer agrees to indemnify the insured against marine losses.

The following are the characteristics of Marine Insurance.
1. Fundamentals of general contract:
Marine insurance must have the fundamentals ieraj insurance i.e. insurable interest, utmost good faith, indemnity, subrogation, contribution, warranties, causa proxima etc.

2. Consideration :
Marine insurance is a contract between the insured and insurer. Hence, insured is under an obligation to pay certain amount periodically to the insurer in consideration for accepting risk.

3. Coverage for insurance:
In marine insurance, cargo ship and freight can be insured. It covers large number of risks such as sinking of the ship, burning of the ship, standing of the ship, collision of ships, sea decoits etc.

4. Mode of insurance :
In marine insurance, the insurance may be for a single journey or number of journeys or for specific period of time. Insurance must be renewed once the specific condition is lapsed.

5. Indemnify the losses :
In marine insurance, the insurers guarantees to indemnify the losses caused by sea perils only.

6. Condition for compensation :
In marine insurance the insured is compensated only when the loss is occurred to ship or cargo. It also includes third party insurance.

Question 9.
Define Fire Insurance. Explain characteristics.
Answer:
According to Asbury, Fire insurance is defined as “It is a contract of insurance by which the insurer agrees for consideration to indemnify the insured upto a certain extent and subject to certain terms and conditions against loss or damage by fire which may happen to the property of the insured during a specified period”.

Features of fire insurance :
1. Contract of indemnity:
The fire insurance contract is a contract of indemnity and insured cannot claim more than the value of goods lost or damaged by fire or the amount of policy whichever is less.

2. Lawful consideration :
There must be consideration in fire insurance contract. The consideration is paid by the insured, which is called premium. Thus the essential element of fire insurance contract is premium received from the insured.

3. Insurable interest:
The insured must have insurable interest in the property or goods insured against fire. He must have insurable interest at the time of taking the policy and also at the time when the loss occurs and claim is filed for compensation.

4. Claim over residue :
The scrap or damaged goods after the fire accident automatically pass on to the insurer after the payment of claim under fire insurance.

5. Cause of accident:
The loss must be the out come of fire or ignition. No other reason for loss of property is accepted for settlement of claim.

6. Utmost good faith:
In fire insurance contract, both insured and insurer must have utmost good faith on each other.

Question 10.
Briefly state the advantages and disadvantages of road transportation.
Answer:
Advantages of Road transport: The following are some of the merits of Road transport.
1. Low capital:
It requires lesser capital for constructing roads. Roads are maintained by government and local authorities.

2. Low maintenance :
The maintenance charges of the road carriers are much less than the cost of railways.

3. Flexible :
Road transport is flexible. The route and timing can be adjusted to the individual requirements.

4. Suitable for short distance :
It is more economical and quicker for carrying goods and people over short distance.

5. Door-to-door delivery:
Road transport provides door-to-door delivery service for industries. Goods can be loaded at sellers doors and unload at buyers door.

6. Service to rural areas :
Exchange of goods between villages and towns are made possible by road transport.

7. Feeder to other modes of transport:
All the movement of goods begin and ultimately end by making use of roads.

8. High speed :
Road transport reduces the effective duration of the transit.

Disadvantages :
1. Less reliable :
Road vehicles are less reliable for long distance to travel because of the breakdowns and road congetions.

2. Accidents and breakdown :
There are more chances of accidents and breakdown in case of motor transport.

3. Lesser speed :
The speed of motor transport is comparatively slow.

4. Limited carrying capacity :
Load carrying capacity of road transport is limited.

5. More expensive :
The road is more expensive than railway transport for long distance travel.

6. Instable rates:
The rates charged by the road carriers are not stable. They need to charge with market.

Question 11.
State various advantages and disadvantages of Railway transportation.
Answer:
Advantages of railway transport:

1. Railway transport facilitate long distance travel and transport bulky and heavy goods.
2. It is quick and more regular form of transport because it helps in the transportation of goods with speed and certainty.
3. It helps in the industrialisation process of a country by easy transportation of coal and raw materials at cheaper rates.
4. It helps in the quick movement of goods from one place to another at the time of emergencies like famines and scarcity.
5. It encourages mobility of labour and thereby provides a great scope for employment.
6. Railway is the safest form of transport. The chances of breakdown and accidents of railways are less as compared to other modes of transport.
7. The carrying capacity of the railways is extremely large. Moreover, its capacity is elastic which can be easily be increased by adding more wagons.

Disadvantages :

1. The railways require a large investment of capital. The cost of construction, maintenance* and overhead expenses are very high as compared to other modes of transport.
2. Railway transport is not flexible. The routes and timings cannot be adjusted to the individual requirements.
3. Rail transport cannot provide door-to-door service as it is tide to a particular track.
4. Railway transport is unsuitable and uneconomical for short distance and small tariff of goods.
5. It involves much time and labour for booking and taking delivery of goods through railways.

Very Short Answer Questions

Question 1.
ATM.
Answer:
ATM means Automatic Teller Machine. An ATM is an unmanned device located on or off the bank premises. The operation mechanism is that ATM is inserted into ATM, the terminal reads and transmits the tape data to a processor which activate the account. It works 24 hours a day, 7 days a week. ATM are being used to withdraw, deposit and transfer of funds.

Question 2.
Online Banking.
Answer:
It is also called as internet banking. It is a facility provided by banks that enable the user to execute bank related transactions through internet. People sitting at home can transact business.

Question 3.
Tele Banking.
Answer:
Customers can perform a number of transactions from their telephone such as they can check balances and statement information, transfer funds from one account to another, pay certain bills and order statements, cheque book etc.

Question 4.
Mobile Banking. [A.P. Mar 17]
Answer:
This type of service is provided free of cost to all the customers. Under this the customer can access his bank account on the mobile screen for the services such as checking the balance, ordering a demand draft, stop payment or viewing last five transactions. To avail this facility the customer requires to have a mobile phone with WAN facility.

Question 5.
Electronic Banking.
Answer:
The concept of E-banking will enable anyone to transact with bank from anywhere such as home or office at any time convenient to him, which can be beyond the banking hours. Electronic banking is banking with the use of electronic tools and facilities and through electronic delivery channels.

Question 6.
Differentiate Insurer and Insured.
Answer:
The party who agrees to pay money on the happening of an event is called insurer. The party who seeks protection against the risk by paying premium is called insured.

Question 7.
What is premium?
Answer:
It is the money which is paid periodically by the insured to the insurer in consideration for which the insurer gives protection to the insured.

Question 8.
Define Insurance.
Answer:
Insurance is the pooling of fortuitous losses by transfer of such risks to insurers, who agrees to indemnify insured for such losses, to provide other precuniary benefits on the occurrence or to render service connected with the risk”.

Question 9.
Re-Insurance.
Answer:
Insurance company undertakes the risk according to its capacity. If a company undertakes more risks than its capacity, then it tries to share the risks with some other insurance company. When insurance company insurer complete or part of the risk with other insurance company, then it is called re-insurance.

Question 10.
Double Insurance.
Answer:
When more than one insurance policy is taken on the same subject matter, it is called double insurance. In life insurance, any number of policies can be taken by the insured upon his life. He can collect full amount on all the policies. But this is not the case with fire and marine insurance. He is entitled to the compensation of the actual loss only.

Question 11.
What is subrogation?
Answer:
According to this principle, the insurer after compensating the loss of insured, the right of ownership on damaged goods is shifted from insured to insurer, i.e., insurance company.

Question 12.
What is proximate cause?
Answer:
According to this principle, the loss is caused by nearest and direct factor, then only the insurer will have to bear the loss. The principle is useful in deciding the actual cause of loss when number of causes have contributed for occurrence of loss.

Question 13.
What is insurable Interest?
Answer:
A person cannot enter into contract of insurance unless he has insurable interest. It is essential feature of insurance. Without the insurable interest the contract of insurance will be treated as a gambling contract. A person has insurable interest on own life or life of his wife.

Question 14.
Endowment polity.
Answer:
This policy is taken up for a specific period. The policy will mature at the expiry of specific period or attainment of particular age or on the death of the insured whichever is earlier.

Question 15.
Whole life policy.
Answer:
This policy runs throughout the life of the insured. The sum assured under this policy is payable only after the death of the insured. The premium is low and it is meant to protect the family. The insured will have to pay the premium throughout his life even at the old age when he is not earning.

Question 16.
Name the subject matters of marine insurance.
Answer:
Subject matter of marine insurance are

1. Cargo or the goods on transhipment
2. Hull
3. Freight

Question 17.
What is cargo insurance?
Answer:
The cargo while being transported by ship is subject to many risks. These may be at ports i.e., risk of theft, loss of goods on voyage etc. Thus an insurance policy can be issued to cover against such risk to cargo.

Question 18.
What is freight insurance?
Answer:
If the cargo does not reach the destination due to damage or loss in transit, the shipping company is not paid freight charges. Freight insurance is for reimbursing the loss of freight to the shipping company i.e., insured.

Question 19.
Essential of fire insurance.
Answer:
Essentials of fire insurance are :

1. It is a contract of insurance.
2. There must be consideration.
3. The object of the contract should be indemnify the assured for the loss caused by damage or destruction of property by fire.

Question 20.
National Highway.
Answer:
National highways are meant for internal transport. These roads also connect state capitals, major cities etc. The National Highway Authority has the responsibility of development, maintenance and operation of the national highways. These roads encompass a road length of about 65,000 kms.

Question 21.
Pipe Lines.
Answer:
Pipe line transport is used for the movement of liquid commodities. Crude oil, natural gas and other petroleum products are transported through pipe lines. Pipe lines offer continuous movement at a relatively low cost. They fuel are efficient, dependable and involves less losses and damage. It can be operated all around the clock (24 x 7).

Question 22.
Bonded warehouse.
Answer:
Bonded warehouses are owned and operated by port trust authorities. It is located near the port. It is a place where importers store goods till customs duties are paid or goods are re-shipped to other destination without being brought into the country.

Question 23.
Two significance of warehouse.
Answer:
Advantages of warehouses.

1. It serves the business men who have very limited space.
2. Some warehouses indirectly offer financial assistance.

Question 24.
Cash credit.
Answer:
A cash credit is an agreement where by a bank agrees to lend money to the borrower upto a certain limit. The amount is credited to the account of the borrower. The borrower draws money as and when he needs. Interest is charged on the amount actually drawn.

Question 25.
Bill discounting.
Answr:
The holder of a bill or drawer may be in urgent need of cash before the due date. In such circumstances, he can sell or discount the bill to the bank at lesser amount than the actual.

Question 26.
Recurring deposit.
Answer:
In Recurring deposit, the depositor is required to deposit a fixed amount of money every month for a specific period. After the completion of the specific period, the consumer gets back the deposited amount along with the cumulative interest accrued on the deposit.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 10th Lesson ఏకాంతర విద్యుత్ ప్రవాహం Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 10th Lesson ఏకాంతర విద్యుత్ ప్రవాహం

అతిస్వల్ప సమాధాన ప్రశ్నలు

ప్రశ్న 1.
10 ప్రాథమిక తీగచుట్లు ఉన్న ఒక పరివర్తకం (transformer) 200 Vac ని 2000 Vac కి మార్చగలిగితే, దాని గౌణ తీగచుట్లను లెక్కించండి. [TS. Mar. 16]
జవాబు:

ప్రశ్న 2.
6 Vబెడ్ లాంప్ ఎటువంటి పరివర్తకాన్ని ఉపయోగిస్తారు?
జవాబు:
6 V ల బెడ్ంప్ లో అవరోహణ పరివర్తకంను ఉపయోగిస్తారు.

ప్రశ్న 3.
పరివర్తకం పనిచేయడంలో ఏ దృగ్విషయం ఇమిడి ఉంది?
జవాబు:
అన్యోన్య ప్రేరణపై పరివర్తకం పనిచేస్తుంది.

ప్రశ్న 4.
పరివర్తక నిష్పత్తి అంటే ఏమిటి?
జవాబు:
గౌణ వి.చా.బ నికి, ప్రాథమిక వి.చా.బ గల నిష్పత్తిని (లేదా) గౌణ తీగ చుట్టలో సంఖ్యకు, ప్రాథమిక తీగచుట్టలో చుట్ల సంఖ్యకు గల నిష్పత్తిని పరివర్తకం నిష్పత్తి అంటారు.
పరివర్తకం నిష్పత్తి = \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\).

ప్రశ్న 5.
i) ప్రేరకం, ii) క్షమశీలి (కెపాసిటర్) ప్రతిరోధకానికి సమీకరణాలు వ్రాయండి.
జవాబు:
i) ప్రేరకం ప్రతిరోధకం (X_L) = ωL,
ii) క్షమశీలి (కెపాసిటర్) ప్రతిరోధకం (X_c) = \(\frac{1}{\omega \mathrm{C}}\)

ప్రశ్న 6.
ఏకాంతర విద్యుచ్ఛాలక బలం, విద్యుత్ ప్రవాహాల మధ్య దశాభేదం కింది వాటిలో ఏవిధంగా ఉంటుంది : శుద్ధ నిరోధం, శుద్ధ ప్రేరకం, శుద్ధ కెపాసిటర్.
జవాబు:

1. శుద్ధ నిరోధములో A.C. వి. బా.బ మరియు విద్యుత్ ప్రవాహం ఒకే దశలో ఉంటాయి.
2. శుద్ధ (ప్రేరకంలో వి.బా.బ కన్నా విద్యుత్ ప్రవాహం \(\frac{\pi}{2}\) (లేదా) 90°.
3. శుద్ధ కెపాసిటర్ వి. బా.బ కన్నా విద్యుత్ ప్రవాహం \(\frac{\pi}{2}\) (లేదా) 90° ముందు ఉంటుంది.

ప్రశ్న 7.
సామర్థ్య కారకాన్ని నిర్వచించండి. సామర్థ్య కారకం ఏ కారకాలపై ఆధారపడుతుంది? [TS. Mar. 15]
జవాబు:
నిజ సామర్థ్యానికి, దృశ్య సామర్థ్యానికి (మిథ్యా సామర్థ్యం) గల నిష్పత్తిని సామర్థ్య కారకం అంటారు.
సామర్థ్య కారకంcosΦ) = \(\frac{P}{P_{rms}}\) (∵ P_rms = V_rms . I_rms)

సామర్థ్య కారకం, r.m.s వోల్టేజి, rm.s విద్యుత్ ప్రవాహం మరియు సగటు సామర్థ్యంపై ఆధారపడుతుంది.

ప్రశ్న 8.
విద్యుత్ ప్రవాహం యొక్క వాట్లెస్ అంశ అంటే అర్థం ఏమిటి? [TS. Mar.’17]
జవాబు:
సగటు సామర్థ్యం (P) = V_rms (I_rms sin Φ) cos \(\frac{\pi}{2}\)

విద్యుత్ ప్రవాహం అంశము (I_rms sin Φ) వల్ల వలయంలో వినియోగించే సగటు సామర్థ్యం సున్నా. ఈ విద్యుత్ ప్రవాహ అంశాన్ని వాట్లెస్ విద్యుత్ ప్రవాహం అంటారు. ఇక్కడ (Irms sin థ) విద్యుత్ ప్రవాహం యొక్క వాట్లెస్ అంశం అంటారు.

ప్రశ్న 9.
LCR శ్రేణి వలయం కనిష్ఠ అవరోధం ఎప్పుడు కలిగి ఉంటుంది?
జవాబు:

ఈ పౌనఃపున్యమును అనునాద పౌనఃపున్యము అంటారు.

ప్రశ్న 10.
LCR శ్రేణి వలయం సామర్థ్య కారకం విలువ ఏకాంకం అయినప్పుడు వోల్టేజి, విద్యుత్ ప్రవాహాల మధ్య దశాభేదం ఎంత ఉంటుంది?
జవాబు:
LCR శ్రేణి వలయంలో సామర్థ్య కారకం (cos Φ) = 1
∴ వోల్టేజి మరియు విద్యుత్ ప్రవాహము మధ్య దశాభేదం సున్నా (Φ = 0).

స్వల్ప సమాధాన ప్రశ్నలు

ప్రశ్న 1.
ఏకాంతర విద్యుచ్ఛాలక బలం అనువర్తింపచేసిన ప్రేరకంలోని విద్యుత్ ప్రవాహానికి సమీకరణాన్ని పొందండి.
జవాబు:
వలయంలో శుద్ధ ప్రేరకం యొక్క ప్రేరకత L. దీనికి ac వి.చా.బ V = V_m sin ωt ని అన్వర్తించామనుకొనుము. i అనునది తాత్కాల విద్యుత్ ప్రవాహము.

ప్రశ్న 2.
ఏకాంతర విద్యుచ్ఛాలక బలం అనువర్తింపచేసిన కెపాసిటర్లోని విద్యుత్ ప్రవాహానికి సమీకరణాన్ని పొందండి.
జవాబు:

వలయంలో శుద్ధ కెపాసిటర్ యొక్క కెపాసిటీ C. దీనికి A.C వి. చా. బ V = V_m sin ωt ని అన్వర్తించామనుకొనుము. i మరియు q అనునది తాత్కాల విద్యుత్ ప్రవాహము మరియు ఆవేశాలు అనుకొనుము.
కెపాసిటర్ వద్ద పొటెన్షియల్ తేడా = –\(\frac{q}{C}\)
మొత్తం వి.చా. బ = V_m sin ωt – \(\frac{q}{C}\)
ఓమ్ నియమం ప్రకారం ఇది iR = 0 కు సమానం
V_m sin ωt – \(\frac{q}{C}\)= 0

i₀ అనునది శిఖర విద్యుత్ ప్రవాహము. ఇక్కడ విద్యుత్ ప్రవాహము వి.చా. బ కన్నా \(\frac{\pi}{2}\) (లేదా) 90° ముందుంటుంది.

ప్రశ్న 3.
పరివర్తకం (ట్రాన్స్ఫార్మర్) ఏ సూత్రంపై ఆధారపడి పనిచేస్తుందో తెలపండి. పరివర్తకం పనిచేసే విధానాన్ని తగిన సిద్ధాంతంతో వర్ణించండి.
జవాబు:
ఎక్కువ వోల్టేజి, తక్కువ విద్యుత్ ప్రవాహమున్న ఏకాంతర విద్యుత్ ప్రవాహంను తక్కువ వోల్టేజి, ఎక్కువ విద్యుత్ ప్రవాహంగా (లేదా) విపర్యంగా మార్చే పరికరాన్ని పరివర్తకం అంటారు.

నియమం :
పరివర్తకం రెండు తీగచుట్ల మధ్య అన్యోన్య ప్రేరణపై ఆధారపడి పనిచేస్తుంది.

పనిచేయు విధానం :
ఏకాంతర వి. చా. బను ప్రాథమిక తీగచుట్టకు అన్వర్తిస్తే నివేశ వోల్టేజి కాలంతోపాటు మారుతుంది. అందువలన కాలంతో పాటు అయస్కాంత అభివాహం కూడా మారుతుంది.

ఈ మారే అయస్కాంత అభివాహం గౌణ తీగచుట్టలో అనుసంధానం చెంది ఉంటుంది. కావున గౌణ తీగచుట్టలో వి. చా. బ ప్రేమవుతుంది.

సిద్ధాంతం :
N₁ మరియు N₂ అనునవి ప్రాధమిక మరియు గౌణ తీగచుట్టలలో చుట్ల సంఖ్య అనుకొనుము. V_P మరియు V_S అనునవి ప్రాథమిక మరియు గౌణ చుట్టలలో విద్యుత్చ్చాలక బలాలు అనుకొనుము.

దీర్ఘ సమాధాన ప్రశ్నలు

ప్రశ్న 1.
LCR శ్రేణి వలయంలో అవరోధానికి, విద్యుత్ ప్రవాహానికి సమీకరణాన్ని పొందండి. LCR శ్రేణి అనునాద వలయం పౌనఃపున్యానికి సమాసాన్ని రాబట్టండి.
జవాబు:
వలయంలో నిరోధకం యొక్క నిరోధం R, ప్రేరకం యొక్క ప్రేరకత L, కెపాసిటర్ యొక్క కెపాసిటి C లను శ్రేణిలో కలిపి, దానికి V = V_m sin ωt A.C వోల్టేజిని అన్వర్తించామనుకొనుము.

i మరియు q అనునవి తాత్కాల విద్యుత్ ప్రవాహము మరియు ఆవేశము అనుకొనుము.

అనునాద పౌనఃపున్యము (f₀) :
ఈ పౌనఃపున్యము వద్ద LCR శ్రేణీ వలయంలో అవరోధం కనిష్ఠంగా ఉంటుంది. ఇది R కు సమానం. ఈ పౌనః పున్యము వద్ద విద్యుత్ ప్రవాహం గరిష్ఠం.

అనునాద పౌనఃపున్యము (f₀) వద్ద విద్యుత్ ప్రవాహం గరిష్ఠం. ఈ శ్రేణీ అనునాద వలయాన్ని గ్రహీత వలయం

లెక్కలు Problems

ప్రశ్న 1.
20 mH ప్రేరకత్వం ఉన్న ఒక ఆదర్శ ప్రేరకాన్ని (తీగచుట్ట అంతర్నిరోధం శూన్యం) AC అమ్మీటర్కు శ్రేణిలో కలిపి, దీన్ని విద్యుచ్ఛాలకు బలం e = 20√2 sin(200t + π/3)V ఉన్న AC జనకానికి కలిపారు. ఇక్కడ t సెకనులలో గలదు. అమ్మీటర్ రీడింగును కనుక్కోండి.
సాధన:
L = 20 mH = 20 × 10^-3H

ప్రశ్న 2.
నిరోధకం, ప్రేరకం ఉన్న శ్రేణీవలయం చివరల తాక్షణిక విద్యుత్ ప్రవాహం, వోల్టేజి విలువలు 1 = √2 sin (100t – π/4)A, υ = 40 sin (100t) V అయితే నిరోధాన్ని లెక్కించండి.
సాధన:
i = √2 sin (100t – π/4)A. (∵ i = i₀ sin(ωt – Φ)
V = 40 sin(100t) V (∵ V = V₀sin(ωt))

ప్రశ్న 3.
ఒక AC వలయంలో ఒక కండెన్సర్, ఒక నిరోధకం, ఒక ప్రేరకంలు ఒక AC జనరేటర్, ఏకాంతరకానికి (alternator) అడ్డంగా శ్రేణిలో కలిపారు. వాటి చివరల వోల్టేజిలు వరసగా 20 V, 35 V, 20 V అయితే ఆల్టర్నేటర్ సరఫరా చేసిన వోల్టేజిని కనుక్కోండి
సాధన:
V_C = 20 V, V_R = 35 V, V_L = 20 V

ప్రశ్న 4.
ఒక ఏకాంతర విద్యుత్ వలయంలో నిరోధం R, ప్రేరకం L, కెపాసిటెన్స్ C లను శ్రేణిలో స్థిర వోల్టేజి, చర పౌనః పున్యం ఉన్న ఏకాంతరకం కొనల మధ్య కలిపారు. అనునాద పౌనఃపున్యం వద్ద ప్రేరకత్వ ప్రతిరోధం, క్షమత్వ ప్రతిరోధం, నిరోధం సమానం మరియు వలయంలోని విద్యుత్ ప్రవాహం i₀ అయితే, అనునాద పౌనఃపున్యానికి రెట్టింపు పౌనఃపున్యం వద్ద వలయంలోని విద్యుత్ ప్రవాహాన్ని కనుక్కోండి.
సాధన:

ప్రశ్న 5.
ఒక శ్రేణీ అనునాద వలయం L, R,, C లను కలిగి ఉంది. అనునాద పౌనఃపున్యం f. మరొక శ్రేణి అనునాద వలయం Lz, Rz, C, అను కలిగి ఉంది. దీని అనునాద పౌనఃపున్యం కూడా f. ఈ రెండు వలయాలను శ్రేణిలో కలిపితే అనునాద పౌనఃపున్యాన్ని లెక్కించండి.
సాధన:
అనునాద పౌనఃపున్యము (f)

ప్రశ్న 6.
ఒక LCR శ్రేణి వలయంలో నిరోధం R = 200 Q, ప్రధాన సరఫరా వోల్టేజి 200V, పౌనఃపున్యం 50 Hz. వలయం నుంచి కెపాసిటెన్స్ను బయటకు తీసినప్పుడు విద్యుత్ ప్రవాహం వోల్టేజి కంటే 45° వెనుకబడి ఉంది. ప్రేరకాన్ని వలయం నుంచి బయటకు తీసినప్పుడు విద్యుత్ ప్రవాహం వోల్టేజి కంటే 45° ముందు ఉంది. అయితే LCR వలయంలో దుర్వ్యయమైన సామర్థ్యాన్ని లెక్కించండి.
సాధన:

ప్రశ్న 7.
ప్రాథమిక, గౌణ చుట్ల నిష్పత్తి 1:2 ఉన్న ట్రాన్స్ఫార్మర్ ప్రాథమికాన్ని వోల్టేజి 200 ఉన్న ఏకాంతరానికి కలిపారు. ప్రాథమిక తీగచుట్ట ద్వారా 4A విద్యుత్తు ప్రవహిస్తున్నది. ట్రాన్స్ఫార్మర్ ఎటువంటి నష్టాలు కలిగి లేవనుకొన్నట్లయితే గౌణ వోల్టేజి, విద్యుత్ ప్రవాహాలను కనుక్కోండి.
సాధన:

అభ్యాసాలు Textual Exercises

ప్రశ్న 1.
100Ω నిరోధాన్ని 220V, 50 Hz ఉన్న ac సరఫరాకు కలిపారు.
(a) వలయంలో విద్యుత్ ప్రవాహ వర్గ మధ్యమ ‘మూలం (rms) విలువ ఎంత?
(b) ఒక పూర్తి చక్రంలో వినియోగమైన నికర సామర్థ్యం ఎంత?
సాధన:
నిరోధము R = 100Ω
V_rms = 220V
పౌనఃపున్యము (f) = 50Hz

a) వలయంలో విద్యుత్ ప్రవాహం.
(I_rms) = \(\frac{V_{rms}}{R}=\frac{220}{100}\) = 2.2A

b) వినియోగించిన మొత్తం సామర్థ్యం
(P) = V_rms × I_rms
= 220 × 2.2
= 484W

ప్రశ్న 2.
(a) ఒక ac సరఫరా శిఖర వోల్టేజి 300 V. వర్గ మధ్యమ మూలం వోల్టేజి ఎంత?
(b) ఒక ac వలయంలో విద్యుత్ ప్రవాహ rms విలువ 10 A. శిఖర విద్యుత్ ప్రవాహం ఎంత?
సాధన:
a) గరిష్ఠ వోల్టేజి విలువ (V₀) = 300V.

ప్రశ్న 3.
44 mH ప్రేరకాన్ని 220V, 50 Hz ac సరఫరాకి కలిపారు. వలయంలో విద్యుత్ ప్రవాహ rms విలువను నిర్ధారించండి.
సాధన:
ప్రేరకత (L) = 44 mH = 44 × 10^-3H
V_rms = 220 V

పౌనఃపున్యము (f) = 50 Hz
ప్రేరక ప్రతిరోధం (X_L) = 2πfL
= 2 × 3.14 × 50 × 44 × 10^-3
X_L = 13.83Ω
r.m.s విద్యుత్ ప్రవాహం

ప్రశ్న 4.
110V, 60Hz ఉన్నacసరఫరాకి 60 µF కెపాసిటర్ను కలిపారు. వలయంలో విద్యుత్ ప్రవాహ rms విలువను నిర్ధారించండి.
సాధన:
కెపాసిటర్ యొక్క కెపాసిటి
C = 60 µF = 60 × 10^-6F
V_rms = 110 V
పౌనఃపున్యం (f) = 60Hz
క్షమత్వ ప్రతిరోధం

ప్రశ్న 5.
3, 4 అభ్యాసాలలో ఒక పూర్తి చక్రంలో ప్రతీ వలయం శోషణం చేసుకొనే నికర సామర్థ్యం ఎంత? మీ సమాధానాన్ని వివరించండి.
సాధన:
i) సగటు సామర్థ్యము (P) = V_rms × I_rms × cos Φ
P = V_rms × I_rms × cos 90° = 0 (∴ Φ = 90°)
P = 0

ii) P = V_rms × I_rms × cos Φ
విద్యుత్ ప్రవాహము, వోల్టేజి మధ్య దశాభేదం క్షమత్వంలో 90°
P = V_rms × I_rms × cos 90° = 0

ప్రశ్న 6.
L = 2.0H, C = 32 µF, R = 10Ω లు ఉన్న శ్రేణి LCR వలయం అనువాద పౌనఃపున్యం అని పొందండి. ఈ వలయం Q-విలువ ఎంత?
సాధన:
L = 2H, C = 32µF, R = 10Ω
అనునాద కోణీయ పౌనఃపున్యము

ప్రశ్న 7.
30µF ఉన్న ఆవేశిత కెపాసిటర్ను 27 mH ఉన్న ప్రేరకానికి కలిపారు. వలయం స్వేచ్ఛా డోలనాల కోణీయ పౌనఃపున్యం ఎంత ?
సాధన:
C = 30µF = 30 × 10^-6F
ప్రేరకత (L) = 27mH = 27 × 10^-3H
అనువాద కోణీయ పౌనఃపున్యము

ప్రశ్న 8.
ఒకవేళ అభ్యాసం 7 లో కెపాసిటర్ మీద తొలి ఆవేశం 6 mC అనుకోండి.. తొలుత ఆ వలయంలో నిల్వ ఉండే మొత్తం శక్తి ఎంత ? ఆ తరవాత సమయంలో మొత్తం శక్తి ఎంత?
సాధన:
కెపాసిటర్పై ఆవేశం (Q) = 6mC = 6 × 10^-3C
C = 30µF = 30 × 10^-6F
వలయంలో నిల్వయున్న శక్తి

కొంత సేపటికిLమరియు Cలు శక్తిని పంచుకుంటాయి. కాని మొత్తం శక్తి స్థిరంగా ఉంటుంది. కావున శక్తి నష్టం ఉండదు.

ప్రశ్న 9.
R = 20Ω, L = 1.5 H, C = 35µF లు ఉన్న LCR శ్రేణి వలయాన్ని చర పౌనఃపున్యం ఉన్న 200 V ac సరఫరాకు కలిపారు. సరఫరా పౌనఃపున్యం వలయం సహజ పౌనఃపున్యానికి ఎప్పుడు సమానం అవుతుంది? ఒక పూర్తి చక్రంలో వలయానికి బదిలీ అయిన సగటు సామర్థ్యం ఎంత?
సాధన:
నిరోధం (R) = 20Ω,
ప్రేరకత (L) = 1.5H,
కెపాసిటెన్స్ (C) = 35 × 10^-6F
V_rms = 200V
అనునాదము వద్ద
Z = R = 20Ω
rms విద్యుత్ ప్రవాహము
I_rms = \(\frac{V_{rms}}{Z}=\frac{200}{20}\) = 10A
Φ = 0° (అనునాదం)
ఒక పూర్తి చక్రానికి వలయానికి బదిలీ అయిన సామర్థ్యం
P = I_rms × V_rms × cos Φ
= 10 × 200 × cos 0° = 2000 W
P = 2KW

ప్రశ్న 10.
ఒక రేడియో MW (Medium Wave – మధ్యంతర తరంగం) ప్రసార అవధిలోని భాగమైన 800 kHz నుంచి 1200 kHz ఉన్న పౌనఃపున్య అవధికి శృతి చేయగలదు. దాని వలయం 200 µH ప్రభావాత్మక ప్రేరకత్వాన్ని కలిగి ఉంటే, దానిలోని చర కెపాసిటర్ అవధి ఎంత ఉండాలి?
(Hint : శృతి చేయడానికి, సహజ పౌనఃపున్యం, అంటే LC వలయం స్వేచ్ఛా డోలనాల పౌనఃపున్యం, రేడియో తరంగ పౌనఃపున్యానికి సమానం కావాలి.)
సాధన:
కనిష్ఠ పౌనఃపున్యము
f₁ = 800KHz = 800 × 10³Hz
ప్రేరకత (L) = 200 µH = 200 × 10^-6H
గరిష్ట పౌనఃపున్యము
f₂ = 1200 KHz = 1200 × 10³Hz

కెపాసిటర్ల వ్యాప్తి 87.8pf నుండి 197.7pF వరకు ఉంటుంది.

ప్రశ్న 11.
I = 5.0 H, C = 80µE, R = 40Ω విలువలు ఉన్న శ్రేణి LCR వలయాన్ని చర పౌనఃపున్యం ఉన్న 230 v జనకానికి పటంలో చూపినట్లుగా కలిపారు.

(a) వలయంలో అనునాదం ఏర్పడాలంటే జనక పౌనఃపున్యం ఎంత ఉండాలి?
(b) అనునాద పౌనఃపున్యం వద్ద వలయం అవరోధం, విద్యుత్ ప్రవాహ కంపనపరిమితిని పొందండి.
(c) వలయంలోని మూడు ఘటకాల కొనల మధ్య ఉండే rms పొటెన్షియల్ను కనుక్కోండి. అనువాద పౌనః పున్యం వద్ద LC సంయోగం కొనల మధ్య ఉండే పొటెన్షియల్ శూన్యం అని చూపండి.
సాధన:
r.m.s వోల్టేజి (V_r.m.s) = 230V
ప్రేరకత (L) = 5H
కెపాసిటెన్స్ C = 80µF = 80 × 10^-6F
నిరోధం (R) = 40Ω
a) అనువాద కోణీయ పౌనఃపున్యం

అనునాద పౌనఃపున్యం

b) అనునాద పౌనఃపున్యము వద్ద X_L = X_C
వలయంలో అవరోధం Z = R = 40Ω
r.m.s విద్యుత్ ప్రవాహము

అదనపు అభ్యాసాలు Additional Exercises

ప్రశ్న 12.
ఒక LC వలయం 20 mH ఉన్న ఒక ప్రేరకం, 10 mC తొలి ఆవేశాన్ని కలిగి ఉన్న 50 µF ఉన్న ఒక కెపాసిటర్ను కలిగి ఉంది. వలయ నిరోధం ఉపేక్షించ దగింది. t = 0 సమయం వద్ద వలయం మూసి ఉందనుకోండి.
(a) తొలుత నిల్వ ఉన్న మొత్తం శక్తి ఎంత? LC డోలనాలు చేసేటప్పుడు ఇది నిత్యత్వమవుతుందా?
(b) వలయం సహజ పౌనఃపున్యం ఎంత?
(c) (i) ఏ సమయం వద్ద నిల్వ ఉన్న శక్తి పూర్తిగా విద్యుత్ శక్తిగా ఉంటుంది. (అంటే కెపాసిటర్లో నిల్వ ఉన్నది), (ii) ఏ సమయం వద్ద నిల్వ ఉన్న శక్తి పూర్తిగా అయస్కాంత శక్తి (అంటే ప్రేరకంలో నిల్వ ఉన్నది) గా ఉంటుంది?
(d) ఏ సమయం వద్ద మొత్తం శక్తి ప్రేరకం, కెపాసిటర్లలో సమానంగా పంచబడుతుంది?
(e) వలయంలో నిరోధాన్ని ఉంచినప్పుడు ఎంత శక్తి ఉష్ణ రూపంలో దుర్వ్యయమవుతుంది?
సాధన:
ప్రేరకత (L) : = 20mH = 20 × 10^-3H
కెపాసిటీ C = 50µf = 50 × 10^-6 F
కెపాసిటర్ మీద తొలి ఆవేశం
(Q₁) = 10mc = 10 × 10³C
a) కెపాసిటర్లో తొలిగా నిల్వయున్న శక్తి (E) = \(\frac{Q^2}{2C}\)

ఈ శక్తి నిత్యత్వముగా ఉండును.

b) అనునాద పౌనఃపున్యము

వలయంలో సహజ పౌనఃపున్యము
ω = 2πυ = 2л × 159.2
999.78 ≈ 1000 = 10³rad/s

c) i) కెపాసిటర్ మీద ఆవేశం Q = Q₀ cos ωt
Q = Q₀ cos \(\frac{2 \pi}{T}\) .t …………. (1)
Q = Q₀ అయితే,
cos \(\frac{2 \pi}{T}\)t = ±1 = cos nл (లేదా) t = \(\frac{nT}{2}\)
ఎక్కడ n = 0, 1, 2, 3…………
t = 0, T/2, T, 3T/2, ………….

ii) ఏదైనా కాలం వద్ద, నిల్వయున్న శక్తి పూర్తిగా అయస్కాంత శక్తి

e) వలయంలో నిరోధాన్ని చేర్చితే, మొత్తం శక్తి ఉష్ణ రూపంలో కోల్పోతుంది. డోలనాలు అవరుద్ధమైతే, కొంత సేపటికి కోల్పోయిన శక్తి ఉష్ణరూపంలోకి మారుతుంది.

ప్రశ్న 13.
0.50 H ప్రేరకత్వం ఉన్న ఒక తీగచుట్ట, 100Ω నిరోధాన్ని 50 Hz పౌనఃపున్యం ఉన్న 240 Vac సరఫరాకు కలిపారు.
a) తీగచుట్టలో గరిష్ఠ విద్యుత్ ప్రవాహం ఎంత?
b) గరిష్ట విద్యుత్ ప్రవాహం, గరిష్ఠ వోల్టేజిల మధ్య కాల విలంబనం (time lag) ఎంత?
సాధన:
ప్రేరకత (L) : = 0.50H
నిరోధము (R) = 100Ω
r.m.s వోల్టేజి V_rms
a) అవరోధము

ప్రశ్న 14.
అభ్యాసం 13లోని వలయాన్ని అధిక పౌనఃపున్యం ఉన్న సరఫరాకు (240 V, 10kHz) కలిపినప్పుడు (a), (b) లను కనుక్కోండి. అందువల్ల అధిక పౌనఃపున్యం వద్ద వలయంలో ప్రేరకం తెరచిన వలయంలాగా ఉంటుందనే ప్రవచనాన్ని వివరించండి. నిలకడ స్థితి తరవాతdcవలయంలో ప్రేరకం ఏవిధంగా ప్రవర్తిస్తుంది?
సాధన:
పౌనఃపున్యము (f) = 10kHz = 10⁴Hz
r.m.s వోల్టేజి (V_r.m.s) = 240V
నిరోధము R = 100Ω,
ప్రేరకత (L) = 0.5H

అల్ప పౌనఃపున్యం వద్ద I₀ = 1.82A, అధిక పౌనః పున్యం వద్ద I₀ = 0.0108 A కావున అధిక పౌనః పున్యము వద్ద ప్రేరకం అధిక నిరోధాన్ని కలిగించి వివృత వలయం అవుతుంది.

వలయంలో ω = 0 అందువలన X_L = ω_L = 0.

ప్రశ్న 15.
100µF కెపాసిటర్ను 40Ω నిరోధానికి శ్రేణిలో కలిపి 110V, 60 Hz సరఫరాకి కలిపారు.
(a) వలయంలో గరిష్ఠ విద్యుత్ ప్రవాహం ఎంత?
(b) గరిష్ఠ విద్యుత్ ప్రవాహం, గరిష్ఠ వోల్టేజిల మధ్య కాల విలంబనం ఎంత?
సాధన:
కెపాసిటర్ యొక్క కెపాసిటీ

C = 100µF = 100 × 10^-6F
నిరోధము (R) = 40Ω,
V_rms = 110 V
పౌనఃపున్యము (f) = 60 Hz
(a) అవరోధము (Z)

గరిష్ఠ వోల్టేజి మరియు గరిష్ఠ విద్యుత్ ప్రవాహం మధ్య కాల అవధి = 1.55 × 10^-3s.

ప్రశ్న 16.
సమస్య 15లోని వలయాన్ని 110 V, 12 kHz సరఫరాకు కలిపినప్పుడు (a), (b) లకు సమాధానాలను కనుక్కోండి. అందువల్ల అధిక పౌనఃపున్యాల వద్ద కెపాసిటర్ ఒక వాహకం లాగా ఉంటుందనే ప్రవచనాన్ని వివరించండి. ఈ ప్రవర్తనను నిలకడ స్థితి తరవాత dc వలయంలో కెపాసిటర్ ప్రవర్తనతో పోల్చండి.
సాధన:
r.m.s వోల్టేజి (V_rms) = 110V.
పౌనఃపున్యము (f) = 12kHz = 12000 Hz.
కెపాసిటీ (C) = 10^-4
నిరోధము (R) = 40Ω
క్షమత్వ ప్రతిరోధం

కావున ఇది వివృత వలయంలాగ పనిచేస్తుంది.

ప్రశ్న 17.
LCR శ్రేణి వలయంలోని అనునాద పౌనఃపున్యానికి సమాన పౌనఃపున్యాన్ని జనకానికి ఉంచి, L,C, R మూడు ఘటకాలను సమాంతరంగా అమర్చినప్పుడు ఈ పౌనఃపున్యం వద్ద LCR సమాంతర వలయంలో మొత్తం విద్యుత్ ప్రవాహం కనిష్ఠం అని చూపండి. ఈ పౌనఃపున్యానికి అభ్యాసం 11 వలయంలో నిర్దేశించిన జనకానికి ప్రతి శాఖలో (ఘటకాలకు) విద్యుత్ ప్రవాహ rms విలువలను పొందండి.
సాధన:
సమాంతర సంధానంలో ఉన్నాయి, కాబట్టి

దీనర్థం \(\frac{1}{Z}\) = కనిష్టం
Z = గరిష్ఠం అయితే, విద్యుత్ ప్రవాహం కనిష్టం.
ప్రేరకం ద్వారా విద్యుత్ ప్రవాహం

మొత్తం విద్యుత్ ప్రవాహం = I_rms
= నిరోధంలో విద్యుత్ ప్రవాహం = 5.75A

ప్రశ్న 18.
ఒక 800mH ప్రేరకం, 60 µF కెపాసిటర్లను శ్రేణిలో 230, 50 Hz సరఫరాకి కలిపారు. వలయం నిరోధం ఉపేక్షించదగినంతగా ఉంది.
(a) విద్యుత్ ప్రవాహ కంపన పరిమితి, rms విలువలను పొందండి.
(b) ప్రతీ మూలకం కొనల మధ్య ఉండే పొటెన్షియల్ భేదం rms విలువలను పొందండి.
(c) ప్రేరకానికి బదిలీ అయ్యే సగటు సామర్థ్యం ఎంత?
(d) కెపాసిటర్కు బదిలీ అయ్యే సగటు సామర్థ్యం ఎంత?
(e) వలయం శోషించుకొనే మొత్తం సగటు సామర్థ్యం ఎంత? (సగటు అంటే ఒక పూర్తి చక్రానికి తీసుకొన్న సగటు).
సాధన:
ప్రేరకత్వం L = 80mH = 80 × 10^-3H
కెపాసిటీ C = 60µF = 60 × 10^-6F
V_rms = 230V

పౌనఃపున్యము (f) = 50Hz, నిరోధము (R) = 0
a) వలయం యొక్క అవరోధము

c) ప్రేరకానికి బదిలీ అయిన సగటు సామర్థ్యం
(P) = I_rms · V_rms· cos Φ
దశాభేదం 90°, కాబట్టి P = 0

d) కెపాసిటర్ బదిలీ అయిన సగటు సామర్థ్యం
(P) = I_rms · V_rms . cos Φ
దశాభేదం 90°, P = 0

e) వలయంలోని నిరోధం లేదు కాబట్టి మొత్తం సామర్థ్యం, ప్రేరకం మరియు కెపాసిటర్ సగటు సామర్థ్యాల మొత్తానికి సమానం. కావున వినియోగించిన మొత్తం సామర్ధ్యం శూన్యం.

ప్రశ్న 19.
అభ్యాసం 18లోని వలయం 15Ω నిరోధం కలిగి ఉందనుకోండి. వలయంలో ప్రతి ఘటకానికి బదిలీ అయ్యే సగటు సామర్థ్యాన్ని శోషించుకొనే మొత్తం సామర్థ్యాన్ని పొందండి.
సాధన:
r.m.s వోల్టేజి (V_rms) = 230V
నిరోధము (R) = 15Ω
పౌనఃపున్యము (f) = 50Hz
విద్యుత్ ప్రవాహము వోల్టేజి మధ్య దశాభేదం 90°
మొత్తం సామర్థ్యం (P_av) = V_rms . I_rms

ప్రశ్న 20.
L = 0.12 H, C = 480 nE, R = 23Ω లతో ఉన్న LCR శ్రేణి వలయం 230 V చర పౌనఃపున్యం ఉన్న సరఫరాకి కలిపారు.
a) విద్యుత్ ప్రవాహ కంపనపరిమితి గరిష్ఠమవడానికి జనక పౌనఃపున్యం ఎంత ఉండాలి? ఈ గరిష్ఠ విలువను పొందండి.
b) వలయ శోషణం చేసుకొనే సగటు సామర్థ్యం గరిష్ఠం అవడానికి జనక పౌనఃపున్యం ఎంత ఉండాలి? ఈ గరిష్ఠ సామర్థ్యం విలువ పొందండి.
c) జనకం యొక్క ఏ పౌనఃపున్యాలకు, వలయానికి సరఫరా అయిన సామర్థ్యం అనునాదం వద్ద ఉండే సామర్థ్యంలో సగం ఉంటుంది. ఈ పౌనఃపున్యాల వద్ద విద్యుత్ ప్రవాహ కంపనపరిమితి ఎంత?
d) ఇచ్చిన వలయం Q-కారకం ఎంత?
సాధన:
ప్రేరకత్వం (L) = 0.12H,
కెపాసిటెన్స్ (C) = 480 = 480 × 10^-9 F
నిరోధము (R) = 23Ω
V_rms = 230V

a) అనునాదము వద్ద విద్యుత్ ప్రవాహం గరిష్ఠం
Z = R = 23Ω
గరిష్ఠ విద్యుత్ ప్రవాహం
(I₀) = √2 I_rms = 1.414 × 10 = 14.14A.
సహజ పౌనఃపున్యము వద్ద, విద్యుత్ ప్రవాహ కంపన పరిమితి గరిష్టం

b) అనునాదం వద్ద సగటు సామర్థ్యం గరిష్టం
P_av = I²_rms .R = 10 × 10 × 23 = 2300W.
R = 10 x 10 x 23 = 2300W.

c) బదిలీ అయిన సామర్థ్యం, అనునాదం వద్ద సామర్థ్యంలో సగం ఉంటుంది.

ప్రశ్న 21.
L = 3.0 H, C = 27µE, R = 10. 4Ω లు ఉన్న సాధన. Xc = 2xfc XL = 2xfL LCR శ్రేణి వలయం అనువాద పౌనఃపున్యం మరియు Q-కారకాలను పొందండి. వలయం యొక్క అర్ధ గరిష్ఠం వద్ద మొత్తం వెడల్పు (FWHM) ను తగ్గించడం ద్వారా వలయం అనునాద నైశిత్యాన్ని రెండు రెట్లు మెరుగుపరచాలని ఆశించడమైనది. తగిన విధానాన్ని సూచించండి.
సాదన:
ప్రేరకత్వం (L) = 3H
కెపాసిటీ (C) = 27 × 10^-6F
నిరోధము (R) = 7.4Ω
అనునాద పౌనఃపున్యము

ప్రశ్న 22.
క్రింది ప్రశ్నలకు సమాధానాలు ఇవ్వండి.
a) ఏదైనా acవలయంలో అనువర్తిత తాక్షణిక వోల్టేజి శ్రేణి వలయంలోని మూలకాల కొనల మధ్య ఉండే వోల్టేజిల బీజీయ మొత్తానికి సమానమవుతుందా? rms వోల్టేజి విషయంలో ఇది నిజమవుతుందా?
సాధన:
అవును. అనువర్తిత వోల్టేజి, వలయంలో శ్రేణిలో కలిపిన మూలకాల తాత్కాల వోల్టేజిల మొత్తానికి సమానం. లేదు r.m.s వోల్టేజికి సంబంధించి సరియైనది. కాదు. అందుకు కారణం వలయంలో మూలకాల వద్ద కొంత దశాభేదం ఉంటుంది.

b) ప్రేరక తీగచుట్ట ప్రాథమిక వలయంలో ఒక కెపాసిటర్ను ఉపయోగించారు.
సాధన:
ప్రేరక తీగచుట్టలో ప్రాథమిక వలయంలో కెపాసిటర్ను కలపాలి. కారణం వలయాన్ని ఛేదించినప్పుడు అధిక ప్రేరిత వోల్టేజి, కెపాసిటర్ను ఆవేశితం చేస్తారు. కాబట్టి ఎలాంటి నష్టం లేకుండా ఉంటుంది.

c) అనువర్తిత వోల్టేజి సంకేతం (సిగ్నల్) dc వోల్టేజి, అధిక పౌనఃపున్యంతో ఉన్న ac వోల్టేజిల అధ్యారోపణంగా ఉంది. వలయంలో ప్రేరకం, కెపాసిటర్లు శ్రేణిలో ఉన్నాయి. కెపాసిటర్ కొనల మధ్య de సంకేతం, ప్రేరకం కొనల మధ్య ac సంకేతం కనిపిస్తుందని చూపండి.
సాధన:

DC వద్ద కెపాసిటర్ ఆగిపోతుంది. C వద్ద AC ఉంటుంది.
అధిక AC పౌనఃపున్యం వద్ద, X_L బాగా అధికం, X_c = 0

d) ఒక బల్బుకు శ్రేణిలో ఉన్న చౌక్ కాయిల్ dc లైనుకు కలిపారు. అప్పుడు బల్బు చాలా ప్రకాశవంతంగా వెలిగినట్లు అనిపించింది. చౌక్ లో ఇనుప కోర్ను ఉంచినప్పుడు బల్బు ప్రకాశవంతతలో ఎటువంటి మార్పు రాలేదు. ఒకవేళ ac లైనుకు కలిపితే అనురూప పరిశీలనలను ఊహించండి.
సాధన:
చోక్ తీగచుట్టకు dc ని కలిపితే, వెలుగులో ఎలాంటి మార్పుండదు.
కారణం f = 0, X_L = 0.
Ac చోక్ అవరోధం (X) ను కలిగిస్తుంది. కాబట్టి డిమ్ గా వెలుగుతుంది.
ఇనుప కోర్ను చొప్పిస్తే అయస్కాంత క్షేత్రం పెరుగుతుంది. కాబట్టి ప్రేరకత పెరుగుతుంది.
BA = LI = Φ
L ∝ B
కాబట్టి X_L పెరిగి, బల్బు వెలుగు తీవ్రత తగ్గుతుంది.

e) ac మొయిన్స్కు కలిపిన ప్రతిదీప్తి బల్బులలో (fluore- scent bulbs) చౌక్ కాయిల్ బదులుగా సాధారణ నిరోధకాన్ని ఎందుకు ఉపయోగించరాదు?
సాధన:
నిరోధం బదులు చోక్ తీగచుట్టను ఉపయోగిస్తే నిరోధం వద్ద సామర్ధ్యం నష్టం గరిష్ఠం. చోక్ వద్ద సామర్థ్య నష్టం శూన్యం
నిరోధానికి Φ = 0
P = I_rms . V_rms . cos 0°
= I_rms . V_rms = గరిష్ఠం
ప్రేరకానికి Φ = 90°
P = I_rms . V_rms. cos 90° = 0

ప్రశ్న 23.
ఒక విద్యుత్ ప్రసార లైన్ 2300 V వద్ద నివేశ సామర్థ్యాన్ని 4000ప్రాథమిక తీగచుట్లు ఉన్న అవరోహణ పరివర్తకానికి అందిస్తుంది.230Vవద్ద నిర్గమ సామర్థ్యాన్ని పొందడానికి గౌణ తీగచుట్ల సంఖ్య ఎంత ఉండాలి?
సాధన:
ప్రాథమిక వోల్టేజి (V_P) = 2300V
N_P = 4000 చుట్లు
గౌణ వోల్టేజి (V_S) = 230v.

గౌణ చుట్ల సంఖ్య (N_S) = 400

ప్రశ్న 24.
జల విద్యుచ్ఛక్తి కేంద్రంలో నీటి పీడన స్తంభం 300 m ఎత్తులో ఉన్నది. అందుబాటులో ఉన్న నీటి ప్రవాహం 100 m³s^-1 గా ఉంది. టర్బైన్ జనరేటర్ దక్షత 60% అయితే కేంద్రం నుంచి అందుబాటులో ఉండే విద్యుత్ సామర్థ్యాన్ని అంచనా వేయండి. (g = 9.8 ms^-2).
సాధన:
నీటియొక్క ఎత్తు (h) = 300m
నీటి ప్రవాహ రేటు (V) = 100m³/s
దక్షత η = 60%
g = 9.8m/².

ప్రశ్న 25.
440 V వద్ద విద్యుచ్ఛక్తిని ఉత్పత్తి చేసే విద్యుదుత్పాదన కేంద్రం నుంచి 15 km దూరంలో ఉన్న చిన్న పట్టణంలో 220 V చొప్పున 800 kW విద్యుచ్ఛక్తి వినియోగ డిమాండ్ ఉన్నది. రెండు తీగలు ఉన్న ప్రసారిత వ్యవస్థ 1kmకి 0.5Ω చొప్పున నిరోధాన్ని కలిగి ఉంది. 4000-220 V అవరోహణ పరివర్తకాన్ని కలిగి ఉన్న ఉపకేంద్రం (సబ్-స్టేషన్) నుంచి తీగల ద్వారా పట్టణానికి విద్యుచ్ఛక్తి సరఫరా అవుతుంది.
(a) తీగలలో ఉష్ణరూపంలో నష్టపోయే శక్తిని అంచనా వేయండి.
(b) లీకేజి వల్ల నష్టం ఉపేక్షించదగినంతగా ఉన్నదనుకొని, విద్యుత్ కేంద్రం ఎంత శక్తిని సరఫరా చేయాల్సి ఉంటుంది?
(c) విద్యుత్ కేంద్రం వద్ద ఆరోహణ పరివర్తకం అభి లక్షణాలను తెలపండి.
సాధన:
విద్యుత్ ప్లాంట్లో జనించిన సామర్థ్యం = 800 kW
V = 220V
= 15km,
జనించిన వోల్టేజి = 440V
నిరోధము/పొడవు = 0.5Ω/km
ప్రాథమిక వోల్టేజి (V_p) = 4000V
గౌణ వోల్టేజి (V_s) = 220V

(a) సామర్థ్యం – I_p . V_p
800 × 1000 = I_p × 4000
I_p = 200A
ఉష్ణరూపంలో కోల్పోయిన సామర్థ్య నష్టం
= (I_p)² × నిరోధం × 2
= (200)² × 0.5 × 15 × 2
= 60 × 10⁴ W = 600KW

(b) లీకేజీ ద్వారా సామర్థ్య నష్టం లేకపోతే
= 800 + 600 = 1400KW

(c) లైన్ వద్ద వోల్టేజి = I_p. R (2 లైన్లు)
= 200 × 0.5 × 15 × 2
= 3000V

ప్రసారం నుండి వోల్టేజి = 3000 + 4000 = 7000
స్టెప్-అప్ ట్రాన్స్ ఫార్మర్ 440 7000 వరకు అవసరం.

ప్రశ్న 26.
పైన ఇచ్చిన అభ్యాసంలోని ట్రాన్స్ఫార్మర్ బదులు 40,000-220V అవరోహణ పరివర్తకాన్ని ఉపయోగించి అన్ని అభ్యాసాలను చేయండి. (ఇంతకు ముందు లాగా లీకేజి నష్టాలను ఉపేక్షించండి. లీకేజి నష్టాలను ఉపేక్షించడమనే ఊహన ఇక ఏ మాత్రం మంచిది కాదు. ఎందుకంటే ప్రసారం చాలా అధిక వోల్టేజితో కూడుకొని ఉంది.) అయితే, అధిక వోల్టేజి ప్రసారం ఎందుకు ప్రాధాన్యం గలదో వివరించండి.
సాధన:
ప్రాధమిక వోల్టేజి (V_P) = 40,000V
ప్రాథమిక విద్యుత్ ప్రవాహం = I_p
∴ V_P . I_P = P
800 × 1000 = 40000 × I_P
I_P = 20A

a) సామర్ధ్య నష్టం = (I_P)² × R (2 లైన్లు)
= (20)² × 2 × 0.5 × 15
= 6000W = 6KW

b) సామర్థ్యం = 800 + 6 =806KW
వోల్టేజి = I_P.R (2 లైన్లు)
= 20 × 2 × 0.5 × 15
= 300V

ప్రసార వోల్టేజి = 40000 + 300 = 40300V
ప్లాంట్ వద్ద స్టెప్ అప్ ట్రాన్స్ఫార్మర్
= 440 – 40300V అవసరం
= \(\frac{6}{100}\) × 100 = 0.74%

అల్ప వోల్టేజి వద్ద సామర్ధ్య నష్టం
\(\frac{600}{1400}\) × 100 = 42.8%

కాబట్టి అధిక వోల్టేజి వద్ద సామర్ధ్య నష్టం కనిష్టం. అందువలన ప్రసారం చేయుటకు అధిక వోల్టేజికి ప్రాధాన్యం ఇవ్వబడును.

సాధించిన సమస్యలు Textual Examples

ప్రశ్న 1.
220.V సరఫరాకు ఒక విద్యుత్ బల్బు రేటింగ్ 100W అయితే (a) బల్బు నిరోధం, (b) జనక శిఖర వోల్టేజి, (c) బల్బు ద్వారా ప్రవహించే rms విద్యుత్ ప్రవాహాలను చెక్కించండి. AP (Mar. ’15)
సాధన:
P = 100 W, V = 220 V అని ఇచ్చారు.

ప్రశ్న 2.
25.0mH ప్రేరకత్వం ఉన్న ఒక శుద్ధ ప్రేరకాన్ని 220V ఉన్న జనకానికి కలిపారు. జనకం పౌనఃపున్యం 50Hz అయితే వలయంలో ప్రేరకత్వ ప్రతిరోధం, rms విద్యుత్ ప్రవాహాన్ని కనుక్కోండి.
సాధన:
ప్రేరకత్వ ప్రతిరోధం X_L = 2πνL
= 2 × 3.14 × 50 × 25 × 10^-3 = 7.85Ω
వలయంలో rms విద్యుత్ ప్రవాహం

ప్రశ్న 3.
ఒక బల్బును కెపాసిటర్కు శ్రేణిలో కలిపారు. dc, ac సంధానాలకు ఏమిజరుగుతుందో ఆ పరిశీలనను తీసుకోండి. కెపాసిటర్ క్షమత్వాన్ని (కెపాసిటెన్స్) తగ్గించి నప్పుడు ప్రతీ సందర్భంలో ఏమి జరుగుతుంది?
సాధన:
కెపాసిటర్కు dc జనకాన్ని కలిపినప్పుడు కెసాసిటర్ ఆవేశితమవుతుంది. ఆవేశితమయిన తరవాత వలయంలో ఎటువంటి విద్యుత్ ప్రవాహం ఉండదు కాబట్టి బల్బు వెలగదు. క్షమత్వాన్ని తగ్గించినప్పటికీ ఎటువంటి మార్పు ఉండదు. ac జనకానికి కెపాసిటర్ని కలిపినప్పుడు కెపాసిటర్ క్షమత్వ ప్రతిరోధం (1/ωC)ని అందిస్తుంది. అందువల్ల వలయంలో విద్యుత్ ప్రవాహం ఉంటుంది. తద్వారా బల్బు వెలుగుతుంది. కెపాసిటన్స్ C విలువ తగ్గిస్తూ పోయేకొద్దీ ప్రతిరోధం పెరుగుతుంది. అందువల్ల బల్బు ఇంతకుముందు కంటే తక్కువ ప్రకాశవంతంగా వెలుగుతుంది.

ప్రశ్న 4.
220, 50 Hz ఉన్న ఏకాంతర జనకానికి 15.0 µF కెపాసిటర్ను కలిపారు. అయితే వలయంలో క్షమత్వ ప్రతిరోధం rms, శిఖర విద్యుత్ ప్రవాహాలను కనుక్కోండి. పౌనఃపున్యాన్ని రెట్టింపు చేసినట్లైతే క్షమత్వ ప్రతిరోధం, విద్యుత్ ప్రవాహాలు ఏమవుతాయి?
సాధన:

ఈ విద్యుత్ ప్రవాహం + 1.47A – 1.47 Aల మధ్య డోలనాలు చేస్తూ వోల్టేజి కంటే 7/2 రేడియన్లు ముందు ఉంటుంది.

ఒకవేళ పౌనఃపున్యం రెట్టింపయితే క్షమత్వ ప్రతిరోధం సగం అవుతుంది. తదనుగుణంగా విద్యుత్ ప్రవాహం రెట్టింపు అవుతుంది.

ప్రశ్న 5.
ఒక విద్యుత్ బల్బు, ఒక వివృత తీగచుట్ట ఉన్న ప్రేరకాన్ని కీ ద్వారా ఏకాంతర జనకానికి పటంలో చూపిన విధంగా కలిపారు.

కీ ని మూసి కొంత సమయం తరవాత ప్రేరకంలోపల ఒక ఇనుప కడ్డీని అమర్చారు. అప్పుడు విద్యుత్ బల్బు దీప్తి (a) పెరుగుతుంది; (b) తగ్గుతుంది; (c) మారదు. తగిన కారణాలతో మీ సమాధానాన్ని ఇవ్వండి.
సాధన:
తీగచుట్టలో ఇనుప కడ్డీని అమర్చగానే తీగచుట్టలోని అయస్కాంతక్షేత్రం ఇనుప కడ్డీని అయస్కాంతీకరిస్తుంది. తద్వారా లోపల అయస్కాంత క్షేత్రం పెరుగుతుంది. అందువల్ల తీగచుట్ట ప్రేరకత్వం పెరుగుతుంది. పర్యవసానంగా తీగచుట్ట ప్రేరకత్వ ప్రతిరోధం పెరుగుతుంది. ఫలితంగా అనువర్తిత ఏకాంతర వోల్టేజిలో అధికభాగం ప్రేరకం కొనల మధ్యే ఉండి వోల్టేజిలోని కొంత భాగం మాత్రమే బల్బుకు అందచేయబడుతుంది. అందువల్ల బల్బు దీప్తి తగ్గుతుంది.

ప్రశ్న 6.
220V, 50 Hz ఏకాంతర వోల్టేజి జనకానికి శ్రేణిలో 200 Ω నిరోధకం, 15.0 µF కెపాసిటర్ను కలిపారు. (a) వలయంలోని విద్యుత్ ప్రవాహాన్ని లెక్కించండి; (b) కెపాసిటర్, నిరోధకాల చివర వోల్టేజి (rms) ని లెక్కించండి. ఈ వోల్టేజిల జీయ మొత్తం జనక వోల్టేజి కంటే ఎక్కువా? అయితే విరోధాభాసాన్ని (paradox) విశ్లేషించండి.
సాధన:
R = 2002. C = 15.0pF = 15.0 × 10^-6F
V = 220V, ν = 50Hz అని ఇచ్చారు.

(a) విద్యుత్ ప్రవాహాన్ని లెక్కించడానికి వలయం అవరోధం తెలుసుకోవాలి.

(b) వలయం అంతటా విద్యుత్ ప్రవాహం సమానం కాబట్టి,
V_R = IR = (0.755 A) (200Ω) = 151V
V_C = IX_C = (0.755A) (212.3Ω) = 160.3V.

V_R, V_C ల బీజీయ మొత్తం 311.3 V. ఈ విలువ జనక వోల్టేజి 220 V కంటే ఎక్కువ. ఈ విరోధాభాసాన్ని ఏవిధంగా విశ్లేషణ చేయవచ్చు ? పాఠంలో ఇదివరకు చదివిన ప్రకారం రెండు వోల్టేజిలు 90° తో దశలో విభేధిస్తున్నాయి. రెండు వోల్టేజిలు ఒకే దశలో లేవు కాబట్టి వీటిని సాధారణ సంఖ్యల లాగా కలపలేం. కాబట్టి రెండు వోల్టేజీల మొత్తాన్ని పైథాగరియన్ సిద్ధాంతం ఉపయోగించి పొందవచ్చు.
V_{R + C} = \(\sqrt{\mathrm{V}_{\mathrm{R}}^2+\mathrm{V}_{\mathrm{C}}^2}\) = 220V
కాబట్టి, రెండు వోల్టేజిల మధ్య దశాభేదాన్ని తగినట్లుగా లెక్కలోకి తీసుకొన్నట్లయితే నిరోధకం, కెపాసిటర్ చివరల ఏర్పడే వోల్టేజిల మొత్తం జనక వోల్టేజికి సమానం అవుతుంది.

ప్రశ్న 7.
a) విద్యుత్ శక్తిని రవాణా చేయడానికి ఉపయోగించే వలయాలకు తక్కువ సామర్థ్యకారకం ఉంటే ప్రసారంలో సామర్థ్య నష్టం అధికం, వివరించండి.
b) వలయంలో తగిన క్షమత్వం ఉన్న కెపాసిటర్ను ఉపయోగించితరచుగా సామర్థ్యకారకాన్ని మెరుగు చేస్తారు. వివరించండి.
సాధన:
a) P = IV cosΦ అని మనకు తెలుసు. ఇందులో cos ని సామర్థ్యకారకం అంటారు. ఇచ్చిన వోల్టేజి వద్ద సామర్ధ్యాన్ని సరఫరా చేయాలంటే, coso తక్కువగా ఉన్నట్లయితే, దానికి అనుగుణంగా విద్యుత్ ప్రవాహాన్ని పెంచాలి. కాని ఇలా చేయడం వల్ల సరఫరాలో ఎక్కువ సామర్థ్య నష్టం (I²R) జరుగుతుంది.

b) వలయంలో విద్యుత్ ప్రవాహం వోల్టేజి కంటే కోణంతో వెనుకబడి ఉన్నదనుకోండి. అప్పుడు సామర్థ్యకారకం cos Φ = R/Z.

Z ను Rకి సమీపింపచేసి సామర్థ్య కారకాన్ని (1కి సమీపించే విధంగా) మెరుగుపరచవచ్చు. దీన్ని ఏవిధంగా సాధించవచ్చో పటంలో చూపిన ఫేజర్ పటం ఆధారంగా అర్ధం చేసుకోవచ్చు. ఇప్పుడు I ని రెండు అంశాలుగా విభజిద్దాం. ఇందులో I_P అనేది అనువర్తిత వోల్టేజి V వెంబడి ఉండే అంశ, I_q అనేది అనువర్తిత వోల్టేజికి లంబదిశలో ఉండే అంశ. 10.7 సెక్షన్ లో చదివిన ప్రకారం I_q ని వాట్ లెస్ అంశ అంటారు. ఎందుకంటే ఈ అంశానికి అనురూపంగా విద్యుత్ ప్రవాహంలో ఎటువంటి సామర్థ్య నష్టం ఉండదు. అనువర్తిత వోల్టేజి I_P అంశదిశ ఉండటంవల్ల దీన్ని సామర్థ్య అంశం అంటారు. వలయంలోని సామర్థ్య నష్టాన్ని ఈ అంశ ఇస్తుంది.

ఈ విశ్లేషణ నుంచి స్పష్టమవుతున్నదేమంటే, సామర్థ్యకారకాన్ని మెరుగు పరచాలంటే వెనుకబడిన వాట్స్ విద్యుత్ ప్రవాహం I_q ని, దీనితో సమానంగా ఉండి, ముందుండే వాట్లెస్ విద్యుత్ ప్రవాహం I’_q తో తటస్థపరచాలి. దీనికోసం తగిన విలువ కలిగిన కెపాసిటర్ను సమాంతరంగా ఏర్పాటుచేయడం వల్లI_q I’_q లు ఒకదానికొకటి రద్దయి ప్రభావాత్మక సామర్థ్యం P విలువ I_p V అవుతుంది.

ప్రశ్న 8.
50 Hz పౌనఃపున్యం, జావక్రీయ శిఖర వోల్టేజి 283V ఉన్న జ్యావక్రీయ వోల్టేజిని R = 3Ω L = 25.48 mH. and C = 796µF విలువ ఉన్న ఒక LCR శ్రేణి వలయానికి అనువర్తించారు. (a) వలయం అవరోధం; (b) జనక వోల్టేజి, విద్యుత్ ప్రవాహాల మధ్య దశాభేదం; (9వలయంలో దుర్వ్యయమయిన సామర్థ్యం: (బి) సామర్ధ్య కారకాలను కనుక్కోండి.
సాధన:
a) వలయం అవరోధాన్ని కనుక్కోవడానికి ముందుగా X_L X_C లను లెక్కించాలి.
X_L = 2πνL
= 2 × 3.14 × 50 × 25.48 × 10^-3Ω = 8Ω

Φ రుణాత్మకం కాబట్టి జనక వోల్టేజి కంటే విద్యుత్ ప్రవాహం వెనుకబడి ఉంటుంది.
(c) వలయంలో దుర్వ్యయమయిన సామర్ధ్యం

d) సామర్థ్య కారకం = cos o = cos 53.1° = 0.6.

ప్రశ్న 9.
ఇంతకు ముందు సమస్యలో జనక పౌనఃపున్యాన్ని మార్చగలిగినట్లయితే (a) అనునాదం సంభవించాలంటే “జనక పౌనఃపున్యం ఎంత ఉండాలి? (b) అనునాద నిబంధన వద్ద అవరోధం, విద్యుత్ ప్రవాహం, దుర్వ్యయమైన సామర్థ్యాన్ని లెక్కించండి.
సాధన:
(a) అనునాదం జరిగే పౌనఃపున్యం

b) అనునాద నిబంధన వద్ద అవరోధం, నిరోధానికి సమానం. అంటే Z = R = 3Ω
అనునాదం వద్ద rms విద్యుత్ ప్రవాహం

అనునాదం వద్ద దుర్వ్యయమైన సామర్థ్యం
P = I² × R = (66.7)² × 3 = 13.35 kW
ఉదాహరణ 8లో కంటే ఈ సందర్భంలో ఎక్కువ సామర్థ్యం దుర్వ్యయం అవడాన్ని మీరు గమనించగలరు.

ప్రశ్న 10.
ఒక విమానాశ్రయంలో భద్రతా కారణాల దృష్ట్యా ఒక మనిషిని లోహ శోధక (metal detector) ద్వారంలో నుంచి నడిచేట్లు చేస్తారు. ఒకవేళ అతడు/ఆమె ఏదైనా లోహంతో తయారైన వస్తువులను కలిగి ఉన్నట్లయితే. లోహ శోధకం శబ్దాన్ని ఉద్గారం చేస్తుంది. లోహ శోధకం ఏ సూత్రంపై ఆధారపడి పనిచేస్తుంది?
సాధన:
ఏకాంతర వలయాల్లో అనునాదం అనే సూత్రంపై ఆధారపడి లోహ శోధకం పనిచేస్తుంది మీరు ఒక లోహ శోధకం ద్వారా నడుస్తున్నారంటే నిజానికి మీరు అనేక చుట్లు ఉన్న తీగచుట్టలో నుంచి నడుస్తున్నారని అర్థం. ఈ తీగచుట్ట ఒక కెపాసిటర్కు కలపబడి ఉంటుంది.

కాబట్టి వలయం అనునాదంలో ఉంటుంది. మీ జేబులో లోహాన్ని కలిగి లోహ శోధక ద్వారంలో నుంచి నచినట్లయితే వలయం అవరోధం మారడంవల్ల వలయంలోని విద్యుత్ ప్రవాహంలో చెప్పుకోదగినంత మార్పు కలుగుతుంది. విద్యుత్ ప్రవాహంలోని ఈ మార్పు గుర్తించబడి వలయం ఒక శబ్దాన్ని అలారం లాగా ఉద్గారం చేసేలా చేస్తుంది.

ప్రశ్న 11.
LC వలయంలోని స్వేచ్ఛా డోలనాలలో ఒక సమయం వద్ద కెపాసిటర్, ప్రేరకాలలో నిల్వ ఉండే శక్తుల మొత్తం స్థిరం అని చూపండి.
సాధన:
కెపాసిటర్పై తొలి ఆవేశం q₀ అనుకోండి. ఆవేశిత కెపాసిటర్ L ప్రేరకత్వం ఉన్న ప్రేరకానికి కలిపా మనుకోండి. సెక్షన్ 8లో చదివిన విధంగా ఈ LC వలయం ω(2πν = \(\frac{1}{\sqrt{LC}}\)) పౌనఃపున్యంతో కలపనాలు కొనసాగిస్తుంది.

ఏదైనా సమయం t వద్ద కెపాసిటర్పై ఉన్న ఆవేశం q విద్యుత్ ప్రవాహం లను
q(t) = q₀ cos ωt,
i(t) = -q₀ sin ωt గా రాయవచ్చు.
t సమయంవద్ద కెపాసిటర్లో నిల్వ ఉండే శక్తి,

q₀, Cలు కాలంపై ఆధారపడవు కాబట్టి మొత్తం శక్తి స్థిరం. ఈ విలువ కెపాసిటర్ తొలి శక్తికి సమానం అని గమనించగలరు. ఎందుకిలా అవుతుంది? ఆలోచించండి.

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Quadratic Expressions Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 1.
Form quadratic equation whose root 7 ± 2\(\sqrt{5}\) (Mar. ’11, ’05)
Solution:
α + β = 7 + 2\(\sqrt{5}\) + 7 – 2\(\sqrt{5}\) = 14
αβ = (7 + 2\(\sqrt{5}\)) (7 – 2\(\sqrt{5}\)) = 49 – 20 = 29
The required equation is
x² – (α + β)x + αβ = 0
x² – 14x + 29 = 0

Question 2.
Form quadratic equation whose root -3 ± 5i. (Mar. ’07)
Solution:
α + β = -3 + 5i – 3 – 51 = -6
αβ = (-3 + 5i)(-3 – 5i)
= 9 + 25 = 34
The required equation is
x² – (α + β)x + αβ = 0
x² + 6x + 34 = 0

Question 3.
For what values of x, 15 + 4x – 3x² expressions are negative? (AP Mar. ’15)
Solution:
The roots of 15 + 4x – 3x² = 0 are
\(\frac{-4 \pm \sqrt{16+180}}{-6}\) i.e., \(\frac{-5}{3}\), 3
∴ The expression 15 + 4x – 3x² is negative if
x < \(\frac{-5}{3}\) or x > 31 ∵ a = -3 < 0

Question 4.
If α, β are the roots of the equation ax² + bx + c = 0, find the value \(\frac{1}{\alpha^{2}}\) + \(\frac{1}{\beta^{2}}\) expressions in terms of a, b, c. (AP & TS Mar. ‘16, 08)
Solution:

Question 5.
Form quadratic equation whose root
\(\frac{p-q}{p+q}\), \(\frac{-p+q}{p-q}\), (p ≠ ±q) (Mar. ’06)
Solution:
α + β = \(\frac{p-q}{p+q}\) – \(\frac{p+q}{p-q}\)

Question 6.
Find the values of m for which the following equations have equal roots?
i) x² – 15 – m(2x – 8) = 0. (AP Mar. ’17) (TS Mar. ’15 13)
Solution:
Given equation is x² – 15 – m(2x – 8) = 0
x² – 2mx + 8m – 15 = 0
a = 1, b = -2m, c = 8m – 15
b² – 4ac = (-2m)² – 4(1) (8m – 15)
= 4m² – 32m + 60
= 4(m² – 8m + 15)
= 4(m – 3)(m – 5)
Hint: If the equation ax² + bx + c = 0 has equal roots then its discriminant is zero.
∵ The roots are equal b² – 4ac = 0
⇒ 4(m – 3) (m – 5) = 0
⇒ m – 3 = 0 or m – 5 = 0
∴ m = 3 or 5

Question 7.
(m + 1)x² + 2(m + 3)x + (m + 8) = 0.
Solution:
Given equation is
(m + 1)x² + 2(m + 3)x + (m + 8) = 0
a = m + 1, b = 2(m + 3), c = m + 8
b² – 4ac = (2(m + 3)]² – 4(m + 1) (m + 8)]
= 4(m² + 6m + 9) – 4(m² + 8m + m + 8)
= 4m² + 24m + 36 – 4m² – 36 m – 32
= -12m + 4
= -4(3m – 1)
∵ The roots are equal ⇒ b² – 4ac = 0
⇒ -4(3m – 1) = 0
⇒ 3m – 1 = 0
⇒ 3m = 1
∴ m = \(\frac{1}{3}\)

Question 8.
If x is real, prove that \(\frac{x}{x^{2}-5 x+9}\) lies between 1 and \(\frac{-1}{11}\). (Mar. ‘14, 13, ‘08, ‘02; May 11, ‘07)
Solution:
Let y = \(\frac{x}{x^{2}-5 x+9}\) ⇒ yx² – 5yx + 9y = x
⇒ yx² + (-5y – 1)x + 9y = 0
x ∈ R ⇒ (-5y – 1)² – 4y(9y) ≥ 0
⇒ 25y² + 1 + 10y – 36y² ≥ 0
⇒ -11y² + 10y + 1 ≥ 0 —— (1)
⇒ -11y² + 10y + 1 = 0 ⇒ -11y² + 11y – y + 1 = 0
⇒ 11y(-y + 1) + 1(-y + 1) = 0
⇒ (-y + 1)(11y + 1) = 0 ⇒ y = 1, \(\frac{-1}{11}\)
-11y² + 10y + 1 ≥ 0
∴ y² coeff is be, but the exp is ≥ 0 from (1)
⇒ \(\frac{-1}{11}\) ≤ y ≤ 1 ⇒ y lies between 1 and \(\frac{-1}{11}\)

Question 9.
Theorem : The roots of ax² + bx + c = 0 are

(Mar. ’02)
Proof:
Given quadratic equation is ax² + bx + c = 0
⇒ 4a(ax² + bx + c) = 0
⇒ 4a²x² + 4abx + 4ac = 0
⇒ (2ax)² + 2(2ax) (b) + b² – b² + 4ac = 0
⇒ (2ax + b)² = b² – 4ac

Question 10.
Find the maximum value of the function \(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R.
Solution:
Let y = \(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\)
⇒ yx² + 2yx + 3y = x² + 14x + 9
⇒ (y – 1)x² + 2(y – 7)x + 3y – 9 = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 7)]² – 4(y – 1) (3y – 9) ≥ 0
⇒ 4[(y² – 14y + 49) [(3y² – 12y + 9)] ≥ 0
⇒ -2y² – 2y + 40 ≥ 0
⇒ y² + y – 20 ≤ 0
⇒ (y + 5) (y -4) ≤ 0
⇒ -5 ≤ y ≤ 4
⇒ y ∈ [-5, 4]
⇒ Maximum value of y = 4
∴ Maximum value of the function
\(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R is 4.

Question 11.
If x² – 6x + 5 = 0 and x² – 12x + p = 0 have a common root, then find p. (TS Mar. ’17)
Solution:
Given x² – 6x + 5 = 0, x² – 12x + p = 0 have a common root.
If α is the common root then
α² -6α + 5 = 0, α² – 12α + p = 0
α² – 6α + 5 = 0 ⇒ (α – 1) (α – 5) = 0
⇒ α = 1 or 5
If α = 1 then α² – 12α + p = 0
⇒ 1 – 12 + p = 0 ⇒ p = 11
If α = 5 then α² – 12α + p = 0
⇒ 25 – 60 + p = 0 ⇒ p = 35
∴ p = 11 or 35

Question 12.
If x₁, x₂ are the roots of the quadratic equation ax₂ + bx + c = 0 and c ≠ 0, find the value of (ax₁ + b)^-2 + (ax₂ + b)_-2 in terms of a, b, c. (TS Mar. ’17)
Solution:
x₁, x₂ are the roots of the equation
ax² + bx + c = 0

Question 13.
Prove that \(\frac{1}{3 x+1}\) + \(\frac{1}{x+1}\) – \(\frac{1}{(3 x+1)(x+1)}\) does not lie between 1 and 4, if x is real. (AP & TS Mar. ’16, AP Mar. 15, ’11) (AP Mar. ‘17)
Solution:

⇒ 3yx² + 4yx + y = 4x + 1
⇒ 3yx² + (4y – 4) x + (y – 1) = 0
x ∈ R ⇒ (4y – 4)² – 4(3y)(y – 1) ≥ 0
⇒ 16y² + 16 – 32y – 12y² + 12y ≥ 0
⇒ 4y² – 20y + 16 ≥ 0
4y² – 20y + 16 = 0
⇒ y² – 5y + 4 = 0
⇒ (y – 1)(y – 4) = 0
⇒ y = 1, 4
⇒ 4y² – 20y + 16 ≥ 0.
⇒ y ≤ 1 or y ≥ 4
⇒ y does not lie between 1 and 4
Since y² coeff is the and exp ≥ 0.

Question 14.
Solve the following equations: (T.S Mar. ‘15)
2x⁴ + x³ – 11x² + x + 2 = 0
Solution:
Dividing by x²

Substituting in (1)
2(a² – 2) + a – 11 = 0
⇒ 2a² – 4 + a – 11 = 0
⇒ 2a² + a – 15 = 0
⇒ (a + 3) (2a – 5) = 0
⇒ a + 3 = 0 or 2a – 5 = 0
a = -3 or 2a = 5 a = \(\frac{5}{2}\)

Case(i) : a = -3
x + \(\frac{1}{x}\) = -3
x² + 1 = -3
x² + 3x + 1 = 0
x = \(\frac{-3 \pm \sqrt{9-4}}{2}\) = \(\frac{-3 \pm \sqrt{5}}{2}\)

Case (ii) : a = \(\frac{5}{2}\)
x + \(\frac{1}{x}\) = \(\frac{5}{2}\)
⇒ \(\frac{x^{2}+1}{x}\) = \(\frac{5}{2}\)
⇒ 2x² + 2 = 5x
⇒ 2x² – 5x + 2 = 0
⇒ (2x – 1) (x – 2) = 0
⇒ 2x – 1 = 0 or x – 2 = 0
⇒ x = \(\frac{1}{2}\), 2
∴ The roots are \(\frac{1}{2}\), 2, \(\frac{-3 \pm \sqrt{5}}{2}\)

Question 15.
For what values of x, the following expressions are positive? (May ’11)
i) x² – 5x + 6
Solution:
x² – 5x + 6 = (x – 2) (x – 3)
Roots of x² – 5x + 6 = 0 are 2, 3 which are real.
The expression x² – 5x + 6 is positive if x < 2 or x > 3, ∴ a = 1 > 0.

ii) 3x² + 4x + 4
Solution:
Here a = 3, b = 4, c = 4,
Δ = b² – 4ac
= 16 – 48 = -32 < 0
∴ 3x² + 4x + 4 is positive ∀ x ∈ R,
∵ a = 3 >0 and Δ < 0
Hint: ax² + bx + c and ‘a’ have same sign ∀ x ∈ R, if Δ < 0

iii) 4x – 5x² + 2
Solution:
Roots of 4x – 5x² + 2 = 0 are

iv) x² – 5x + 14
Solution:
Here a = 1, b = -5, c = 14,
Δ = b² – 4ac
= 25 – 56 = -31 < 0
∴ Δ < 0 ∵ a = 1 > 0 and Δ < 0
⇒ x² – 5x + 14 is positive ∀ x ∈ R.

Question 16.
Determine the range of the \(\frac{x^{2}+x+1}{x^{2}-x+1}\) expressions. (Mar ’04)
Solution:
Let y = \(\frac{x^{2}+x+1}{x^{2}-x+1}\)
⇒ x²y – xy + y = x² + x + 1
⇒ x²y – xy + y – x² – x – 1 = 0
⇒ x²(y – 1) – x(y + 1) + (y – 1) = 0
x is real ⇒ b² – 4ac ≥ 0
⇒ (y + 1)² – 4(y – 1)² ≥ 0
⇒ (y + 1)² – (2y – 2)² ≥ 0
⇒ (y + 1 + 2y – 2)(y + 1 – 2y + 2) ≥ 0
⇒ (3y – 1)(-y + 3) ≥ 0
⇒ -(3y – 1) (y – 3) ≥ 0
a = coeff. of y² = -3 < 0., But
The expression ≥ 0
⇒ y lies between \(\frac{1}{3}\) and 3
∴ The range of \(\frac{x^{2}+x+1}{x^{2}-x+1}\) is \(\left[\frac{1}{3}, 3\right]\)

Question 17.
Theorem : Let α, β be the real roots of ax² + bx + c = 0 and α < β. Then
i) x ∈ R, α < x < β ⇒ ax² + bx + c and ’a’ have opposite signs.
ii) x ∈ R, x < α or x > β ⇒ ax² + bx + c and ‘a’ have the same sign. (Apr. ’96, ’93)
Proof:
α, β are the roots of ax² + bx + c = 0
⇒ ax² + bx + c = a(x – α) (x – β)
⇒ \(\frac{a x^{2}+b x+c}{a}\) = (x – α) (x – β)

i) Suppose x ∈ R, α < x < β
α < x < β ⇒ x – α > 0, x – β < 0
⇒ (x – α) (x – β) < 0
⇒ ax² + bx + c, a have opposite signs.

ii) Suppose x ∈ R, x < α
x < α < 13
⇒ x – α < 0, x – β < 0 ⇒ (x – α) (x – β) > 0
⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign.
Suppose x ∈ R, x > β
x > β > α
⇒ x – α > 0, x – β > 0
⇒ (x – α)(x – β) > 0
⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign,
∴ x ∈ R, x < α or x > β
⇒ ax² + bx + c and a have the same sign.

Question 18.
Theorem: Let f(x) = ax² + bx + c be a quadratic function. (Apr. ‘01)
i) If a > 0 then f(x) has minimum value at x = \(\frac{-b}{2 a}\) and the minimum value = \(\frac{4 a c-b^{2}}{4 a}\)
ii) If a < 0 then f(x) has maximum value at x = \(\frac{-b}{2 a}\) and the maximum value \(\frac{4 a c-b^{2}}{4 a}\)
Proof:
ax² + bx + c =

≤ \(\frac{4 a c-b^{2}}{4 a}\),
when a < 0
∴ If a < 0, then \(\frac{4 a c-b^{2}}{4 a}\) is the maximum fór f when x = \(\frac{-b}{2 a}\)
Second Proof : f(x) = ax² + bx + c
⇒ f'(x) = 2ax + b
⇒ f”(x) = 2a
If f'(x) = 0, then 2ax + b = 0 and hence x = \(-\frac{b}{2 a}\)
If a > 0, then f”(x) > 0 and hence ‘f’ has minimum value at x = \(-\frac{b}{2 a}\).
Minimum of

If a < 0,
then f”(x) < 0 and hence ‘f’ has maximum value at x = \(-\frac{b}{2 a}\) maximum of

Question 19.
Theorem : The roots of ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\).
Proof:
Given quadratic equation is ax² + bx + c = 0
⇒ 4a (ax² + bx + c) = 0
⇒ 4a²x² + 4abx + 4ac = 0
⇒ (2ax)² + 2(2ax) (b) + b² – b² + 4ac = 0
⇒ (2ax + b)² = b² – 4ac
⇒ 2ax + b = ±\(\sqrt{b^{2}-4 a c}\)
⇒ 2ax = -b ± \(\sqrt{b^{2}-4 a c}\)
⇒ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
The roots of ax² + bx + c = 0 are
\(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

Question 20.
Theorem : Let a, b, c ∈ R and a ≠ 0. Then the roots of ax² + bx + c = 0 are non-real complex numbers if and only if ax² + bx + c and a have the same sign for all x ∈ R.
Proof :
The condition for the equation
ax² + bx + c = o to have non-real complex roots is b² – 4ac < 0, i.e., 4ac – b² > 0.

Hence 4ac – b² > 0, so that b² – 4ac < 0. Thus b² – 4ac < 0 if and only if ax² + bx + c and a have the same sign for all real x.

Question 21.
Theorem: If the roots of ax² + bx + c = 0 are real and equal to α = \(\frac{-b}{2 a}\); then for α ≠ x ∈ R, ax² + bx + c and ‘a’ have the same sign.
Proof:
The roots are equal ⇒ b² – 4ac = 0
⇒ 4ac – b² = 0

∴ For α ≠ x ∈ R, ax² + bx + c and a have the same sign.

Question 22.
Theorem : Let α, β be the real roots of ax² + bx + c = 0 and α < β. Then
i) x ∈ R, α < x < 13 ⇒ ax² + bx + c and ’a’ have opposite signs.
ii) x ∈ R, x < α or x > β ⇒ ax² + bx + c and ‘a’ have the same sign.
Proof.
α, β are the roots of ax² + bx + c = 0
⇒ ax² + bx + c = a(x – α)(x – β)
⇒ \(\frac{a x^{2}+b x+c}{a}\) = (x – α)(x – β)

i) Suppose x ∈ R, α < x < β
α 0, x – β < 0
⇒ (x – α)(x – β) < 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) < 0
⇒ ax² + bx + c, a have opposite signs

ii) Suppose x ∈ R, x < α
x < α < β ⇒ x – α < 0, x – β < 0 ⇒ (x – α) (x – β) > 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign.
Suppose x ∈ R, x > 3
x > β > α ⇒ x – α > 0, x – β > 0
⇒ (x – α) (x – β) > 0
⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign,
x ∈ R, x < α or x > β
⇒ ax² + bx + c and a have the same sign

Question 23.
Theorem : Let f(x) = ax² + bx+ c be a quadratic function.
i) If a > 0 then f(x) has minimum value at x = \(\frac{-\mathbf{b}}{2 a}\) and the minimum value = \(\frac{4 a c-b^{2}}{4 a}\)
ii) If a < 0 then f(x) has maximum value at x = \(\frac{-\mathbf{b}}{2 a}\) and the maximum value = \(\frac{4 a c-b^{2}}{4 a}\) (Apr. ’01)
Proof.
ax² + bx+ c =

Second Proof: f(x) = ax² + bx + c
⇒ f'(x) = 2ax + b ⇒ f”(x) = 2a
If f'(x) = 0, then 2ax + b = 0 and hence x = \(-\frac{b}{2 a}\)
If a > 0, then f”(x) > 0 and hence ‘f’ has minimum value at x = \(-\frac{b}{2 a}\)
Minimum of

If a < 0 then f”(x) < 0 and hence ‘f’ has maximum value at x = –\(\frac{b}{2 a}\)
maximum of

Question 24.
Theorem : A necessary and sufficient condition for the quadratic equations
a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 to have a common root is (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁) (b₁c₂ – b₂c₁).
Proof:
Necessity
Let α be a common root of the given equations.
Then a₁α² + b₁α + c₁ = 0 ——(1)
a₂α² + b₂α + c₂ = 0 —— (2)
On multiplying euqation (1) by a₂, equation (2) by a₁ and then subtracting the latter from the former, we get

On multiplying euqation (1) by b₂, equation (2) by b₁ and then subtracting the latter from the former, we get
α² (a₁b₂ – a₂b₁) = b₁c₂ – b₂c₁ —— (4)
On squaring both sides of equation (3) and using (4) we obtain


Therefore the given equations have the same roots.

Case (ii): a₁b₂ – a₂b₁ ≠ 0

Similarly we can prove that a₂α² + b₂α + c₂ = 0
Thus α is a common root of the given equations.

Question 25.
Find the roots of the equation 3x² + 2x – 5 = 0.
Solution:
The roots of the quadratic equation
ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Here a = 3, b = 2 and c = -5.
Therefore the roots of the given equation are

Hence 1 and –[/latex]\frac{5}{3}[/latex] are the roots of the given equation.
Another method
We can also obtain these roots in thè following way.
3x² + 2x – 5 = 3x² + 5x – 3x – 5
= x(3x + 5) -1 (3x + 5)
= (x – 1) (3x + 5)
= 3(x – 1) \(\left(x+\frac{5}{3}\right)\)
Since 1 and –[/latex]\frac{5}{3}[/latex] are the zeros of 3x² + 2x – 5, they are the roots of 3x² + 2x – 5 = 0.

Question 26.
Find the roots of the equation 4x² – 4x + 17 = 3x² – 10x – 17.
Solution:
Given equation can be rewritten as x² + 6x + 34 = 0. The roots of the quadratic equation
ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Here a = 1, b = 6 and c = 34
Therefore the roots of the given equation are

Hence the roots of the given equation are -3 + 5i and -3 – 5i

Question 27.
Find the roots of the equation
\(\sqrt{3}\)x² + 10x – 8\(\sqrt{3}\) = 0.
Solution:
Here a = \(\sqrt{3}\), b = 10, c = -8\(\sqrt{3}\)

Question 28.
Find the nature of the roots of 4x² – 20x + 25 = 0
Solution:
Here a =4, b = -20, c = 25
Hence Δ = b² – 4ac
= (-20)² – 4(4) (25) = 0
Since Δ is zero and a, b, c are real, the roots of the given equation are real and equal.

Question 29.
Find the nature of the roots of 3x² + 7x + 2 = 0
Solution:
Here a = 3, b = 7, c = 2
Hence Δ = b² – 4ac
= 49 – 4(3) (2)
= 49 – 24
= 25 = (5)² > 0
Since a, b, c are rational numbers and Δ = 5² is a perfect square and positive, the roots of the given equation are rational and unequal.

Question 30.
For what values of m, the equation x² – 2(1 + 3m)x + 7(3 + 2m) = 0 will have equal roots?
Solution:
The given equation will have equal roots iff its discriminant is 0.
Here Δ = [(-2(1 + 3m)]² – 4(1) [7 + (3 + 2m)]
= 4(1 + 9m² + 6m) – 28(3 + 28m)
= 9m² – 8m – 20
= (m – 2)(9m + 10)
Hence Δ = 0 ⇔ m = 2, \(\frac{-10}{9} .\)

Question 31.
If α and β are the roots of ax² + bx + c = 0, find the value of α² + β² and α³ + β³ in terms of a, b, c.
Solution:
α, β are the roots of ax² + bx + c = 0

Question 32.
Show that the roots of the equation x² – 2px + p² – q² + 2qr – r² = 0 are rational, given that p, q, r are rational.
Solution:
Here a = 1, b = -2p, c = p² – q² + 2qr – r²
∆ = b² – 4ac
= (-2p)² – 4(1) (p² – q² + 2qr – r²)
= 4p² – 4p² + 4q² + 8qr + 4r²
= 4(q – r)².
∵ p, q, r are rational, the coefficient of the given equation are rational and is a square of a rational number 2(q – r).
∴ The roots of the given equation are rational.

Question 33.
Form a quadratic equation whose roots are 2\(\sqrt{3}\) – 5 and -2\(\sqrt{3}\) – 5.
Solution:
Let α = 2\(\sqrt{3}\) – 5 and β = -2\(\sqrt{3}\) – 5
Then α + β = (2\(\sqrt{3}\) – 5) + (-2\(\sqrt{3}\) – 5) = -10
and αβ = (2\(\sqrt{3}\) – 5) + (-2\(\sqrt{5}\) – 5)
= (-5)² – (-2\(\sqrt{3}\))²
= 25 – 4 × 3
= 13
The required quadratic equation is
x² – (α + β)x + αβ = 0
⇒ x² – (-10)x + 13 = 0
⇒ x² + 10x + 13 = 0

Question 34.
Find the quadratic equation, the sum of whose roots is 1 and the sum of squares of the roots is 13.
Solution:
Let α, β be the roots of a required quadratic equation.
Then α + β = 1 and α² + β² = 13
Now αβ = \(\frac{1}{2}\)[(α + β)² – (α² + β²)]
= \(\frac{1}{2}\)[(1)² – (13)]
Therefore x² – (α + β)x + αβ = 0
⇒ x² – (1)x + (-6) = 0
⇒ x² – x – 6 = 0

Question 35.
Let α and β be the roots of the quadratic equation ax² + bx + c = 0, c ≠ 0, then form the quadratic equation whose roots are \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\).
Solution:
From the given equation

Question 36.
Solve x^2/3 + x^1/3 – 2 = 0
Solution:
(x^1/3)² + (x^1/3) – 2 = 0
Let x^1/3 = a, a² + a – 2 = 0
⇒ (a +2)(a – 1) = 0 ⇒ a = 1, -2
Now, x^1/3 = 1 ⇒ x = (1)³ = 1
x^1/3 = -2 ⇒ x = (-2)³ = -8
∴ roots are 1, -8.

Question 37.
Solve 7^{1 + x} + 7^{1 – x} = 50 for real x.
Solution:
The given equation can be written as,
7.7^x + \(\frac{7}{7^{x}}\) – 50 = 0
Let 7^x = a
7a + \(\frac{7}{a}\) – 50 = 0
⇒ 7a² + 7 – 50a = 0
⇒ 7a² – 49a – a + 7 = 0
⇒ 7a(a – 7) – 1(a – 7) = 0
⇒ (a – 7)(7a – 1) = 0
∴ a = 7, \(\frac{1}{7}\)
Now, if a = 7 then 7^x = 7¹ ⇒ x = 1
a = \(\frac{1}{7}\) then 7^x = \(\frac{1}{7}\) = 7^-1
x = -1
∴ x = -1, 1

Question 38.
Solve

Solution:
On taking \(\sqrt{\frac{x}{1-x}}\) = a
a + \(\frac{1}{a}\) = \(\frac{13}{6}\)
⇒ \(\frac{a^{2}+1}{a}\) = \(\frac{13}{6}\)
⇒ \(\frac{a^{2}+1}{a}\) = \(\frac{13}{6}\)
⇒ 6a² + 6 = 13a
⇒ 6a² – 13a + 6 = 0
⇒ 6a² – 9a – 4a + 6 = 0
⇒ 3a(2a – 3) – 2(2a – 3) = 0
⇒ (2a – 3)(3a – 2) = 0
a = \(\frac{3}{2}\), a = \(\frac{2}{3}\)
If a = \(\frac{3}{2}\) then \(\sqrt{\frac{x}{1-x}}\) = \(\frac{3}{2}\)
⇒ \(\frac{x}{1-x}\) = \(\frac{9}{4}\)
⇒ 4x = 9 – 9x
⇒ 13x = 9
⇒ x = \(\frac{4}{13}\)
∴ x = \(\frac{9}{13}\), \(\frac{4}{13}\)

Question 39.
Find all number which exceed their square root by 12.
Solution:
Let the required number be ‘x’
given, x = \(\sqrt{x}\) + 12
⇒ x – 12= \(\sqrt{x}\)
Squaring on both sides
(x – 12)² = (\(\sqrt{x}\))²
⇒ x² – 24x + 144 = x
⇒ x² – 25x + 144 = 0
⇒ (x – 9)(x – 16) = 0
⇒ x = 9, 16
But x = 9 does not satisfy the given condition
x = 16
∴ The required numbér = 16.

Question 40.
If x² + 4ax + 3 = 0 and 2x² + 3ax – 9 = 0 have a common root, then find the values
of a and the common root.
Solution:
The quadratic equations

Substitute in the above equation
[(3) (2) – (-9) (1)]² = [(1) (3a) – (2) (4a)] [(4a) (-9) – (3a) (3)]
(15)² = (-5a) (-45a)
⇒ 225 = 225a² ⇒ a² = 1 ⇒ a = ±1

Case (i) : If a = 1, the given equations become x² + 4x + 3 = 0 and 2x² + 3x – 9 = 0
⇒ (x + 1)(x + 3) = 0 and (2x + 3)(x + 3) = 0
⇒ x = 3, -1 and -3, \(\frac{3}{2}\)
In this case, the common root of the given equations is -3

Case (ii) : If a = -1, the given equations become x² – 4x + 3 = 0 and 2x² – 3x – 9 = 0
⇒ (x – 1) (x – 3) = 0 and (2x + 3) (x – 3) = 0
⇒ x = 1, 3 and x = 3, –\(\frac{3}{2}\)
In this case, the common root of the given equation is 3.

Question 41.
Prove that there is unique pair of consecutive positive odd integers such that the sum of their squares is. 290 and find it.
Solution:
The difference of two consecutive odd integers is 2.
So, we have to prove that there is a unique positive odd integer ‘x’ such that
x² + (x + 2)² = 290 —– (1)
x² + (x + 2)² = 290
⇒ x² + x² + 4x + 4 = 290
⇒ 2x² + 4x – 286 = 0
⇒ x² + 2x – 143 = 0
⇒ (x + 13) (x – 11) = 0
⇒ x = -13, 11
Hence 11 is the only positive odd integer satisfying equation (1).
∴ 11, 13 is the unique pair of integers which satisfies the given condition.

Question 42.
The cost of a piece of cable wire is Rs. 35/-, If the length of the piece of wire is 4 meters more and each meter costs, Rs. 1/— less, the cost would remain unchanged. What is the length of the wire?
Solution:
Let the length of the piece of the wire be ‘l’ meters and the cost of each meter be Rs. x.
given lx. = 35 —— (1)
and (l + 4) (x – 1) = 35
⇒ lx – l + 4x – 4 = 35
⇒ 35 – l + 4x – 4 = 35
⇒ 4x = l + 4
⇒ x = \(\frac{l+4}{4}\)
Substitute x in (1), we get l[latex]\frac{l+4}{4}[/latex] = 35
⇒ l² + 4l = 140
⇒ l² + 4l – 140 = 0
⇒ l² + 14l – 10l – 140 = 0
⇒ l(l + 14) – 10(l + 14) = 0
⇒ (l – 10)(l + 14) = 0
⇒ l = -14, 10
Since length cannot be negative, l = 10
∴ The length of the piece of wire is 10 meters.

Question 43.
One fourth of a herd of goats was seen in the forest. Twice the square root of the number in the herd had gone up the hill and the remaining 15 goats were on the bank of the river. Find the total number of goats.
Solution:
Let the number of goats in the herd be ‘x’.
The number of goats seen in the forest = \(\frac{x}{4}\)
The number of goats gone upto the hill = \(2 \sqrt{x}\)
The number of goats on the bank of a river = 15
∴ \(\frac{x}{4}\) + 2\(\sqrt{x}\) + 15 = x
⇒ x + 8\(\sqrt{x}\) + 60 = 4x
⇒ 3x – 8\(\sqrt{x}\) – 60 = 0
Put \(\sqrt{x}\) = y
⇒ 3y² – 8y – 60 = 0
⇒ 3y² – 18y + 10y – 60 = 0
⇒ 3y(y – 6) + 10(y – 6) = 0
⇒ (3y + 10)(y – 6) = 0
⇒ y = 6, –\(\frac{10}{3}\)
Since y cannot be negative, y = 6
⇒ \(\sqrt{x}\) = 36
∴ x = 36
∴ Total number of goats = 36

Question 44.
In a cricket match Anil took one wicket less than twice the number of wickets taken by Ravi. If the product of the
number of wickets taken by them is 15, find the number of wickets taken by each of them.
Solution:
Let the number of wickets taken by Ravi be ‘x’ and the number of wickets taken by Anil is 2x – 1.
Given x(2x – 1) = 15
⇒ 2x² – x – 15 = 0
⇒ 2x² – 6x + 5x – 15 = 0
⇒ 2x(x – 3) + 5(x – 15) = 0
⇒ (x – 3)(2x + 5) = 0
⇒ x = 3, –\(\frac{5}{2}\)
Since the number of wickets be integer,
x = 3 and 2x – 1 = 2(3) – 1 = 5
∴ The number of wickets taken by Anil and Ravi are 5 and 3 respectively.

Question 45.
Some points on a plane are marked and they are connected pairwise by line segments. If the total number of line
segments formed is 10, find the number of marked points on the plane.
Solution:
Let the number of points marked on the plane be ’x’.
The total number of line segments actually formed is

Given \(\frac{x(x-1)}{2}\) = 10
⇒ x² – x = 20
⇒ x² – x – 20 = 0
⇒ (x – 5)(x + 4) = 0
Since x cannot be negatives x = 5
∴ The number of points marked on the plane is 5.

Question 46.
Suppose that the quadratic equations ax² + bx + c = 0 and bx² + cx + a = 0 have a common root. Then show that a³ + b³ + c³ = 3abc.
Solution:
Let α be the common root of the equations

Question 47.
For what values of x1 the expression x² – 5x – 14 is positive?
Solution:
Since x² – 5x – 14 = (x + 2) (x – 7), the roots of the equation -2 and 7
Here the coefficient of x² is positive.
Hence x² – 5x – 14 is positive when x < -2 or x > 7.

Question 48.
For what values ofx. the expression -6x² + 2x³ is negative?
Solution:
-6x² + 2x – 3 = 0 can be written as 6x² – 2x + 3 = 0

∴ The roots of -6x² + 2x – 3 = 0 are non-real complex numbers.
Hence coefficient of x² is -6, which is negative.
∴ -6x² + 2x – 3 < 0 for all x ∈ R.

Question 49.
Find the value of x at which the following expressions have maximum or minimum.

i) x² + 5x + 6
Solution:
Here a = 1 > 0, the expression has minimum value at x = \(\frac{-b}{2 a}\)
= \(\frac{-5}{2(1)}\)
= \(\frac{-5}{2}\)

ii) 2x – x² + 7
Solution:
Here a = -1 < 0, the expression has maximum value at x
= \(\frac{-b}{2 a}\)
= \(\frac{-2}{2 x-1}\) = 1

Question 50.
Find the maximum or minimum value of the quadratic expression
(i) 2x – 7 – 5x²
(ii) 3x² + 2x + 17 (Mar. ’14)
Solution:
i) Compare the given equation with ax² + bx + c, we get
a = -5, b = 2, c = -7
Here a = -5 < 0, the given expression has maximum value at x = \(\frac{-b}{2 a}\) = \(\frac{-2}{2(-5)}\)
= \(\frac{1}{5}\)
and maximum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{4(-5) \cdot(-7)-(?)^{2}}{4(-5)}\)
= \(\frac{140-4}{-20}\) = \(\frac{-34}{5}\)

(ii) Compare 3x² + 2x + 11 with
ax² + bx + c, we get a = 3, b = 2, c = 11
∵ a = 3 > 0, the given expression has minimum value at

Question 51.
Find the changes in the sign of 4x – 5x² + 2 for x ∈ R and find the extreme value.
Solution:
The roots of 4x – 5x² + 2 = 0
⇒ 5x² – 4x – 2 = 0
roots are = \(\frac{2 \pm \sqrt{14}}{5}\)
∴ \(\frac{2-\sqrt{14}}{5}\) < x < \(\frac{2+\sqrt{14}}{5}\) the sign of 4x – 5x² + 2 is positive.
x < \(\frac{2-\sqrt{14}}{5}\) or x > \(\frac{2+\sqrt{14}}{5}\), the sign of 4x – 5x² + 2 is negative.
Since a = -5 < 0, then maximum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{4(-5)(2)-(4)^{2}}{4(-5)}\)
= \(\frac{-56}{-20}\) = \(\frac{14}{5}\)
Hence extreme value = \(\frac{14}{5}\)

Question 52.
Find the solution set of x² + x – 12 ≤ 0 by both algebraic and graphical methods.
Solution:
Algebraic Method : We have x² + x – 12 = (x + 4) (x – 3).
Hence -4 and 3 are the roots of the equation x² + x – 12 = 0.
Since the coefficient of x² in the quadratic expression x² + x – 12 = 0 is positive, x² + x – 12 is negative if -4 < x < 3 and positive if either x < -4 or x > 3.
Hence x² + x – 12 ≤ 0 ⇔ -4 ≤ x ≤ 3.
Therefore the solution set is
{x ∈ R : -4 ≤ x ≤ 3}.
Graphical Method: Let y = f(x) = x² + x – 12.
The values of y at some selected values of s are given in the following table:

The graph of the function y = f(x) is drawn using the above tabulated values. This is shown in Fig.

Therefore the graph of y = f(x) we observe that
y = x² – x – 12 < 0 if f -4 ≤ x ≤ 3.
Hence the solution set is {x ∈ R : -4 ≤ x ≤ 3}.

Question 53.
Find the set of values of x for which the inequalities x² – 3x – 10 < 0, 10x – x² – 16
> 0 hold simultaneously.
Solution:
∵ x² – 3x – 10 < 0
⇒ x² – 5x + 2x – 10 < 0
⇒ x(x – 5) + 2(x – 5) < 0
⇒ (x – 5)(x + 2) < 0
∴ x² coeff. is the and expression is <0
⇒ x ∈ (-2, 5) ——- (1)
Now 10x – x² – 16 > 0
⇒ x² – 10x + 16 < 0
⇒ (x – 2) (x – 8) < 0
∴ x² coeff is the ana expression is < 0
⇒ x ∈ (2, 8) ——- (2)
Hence x² – 3x – 10 < 0 and 10x – x² – 16 > 0
⇔ x ∈ (-2, 5) ∩ (2, 8)
∴ The solution set is {x/x ∈ R : 2 < x < 5}

Question 54.
Solve the inequation \(\sqrt{x+2}\) > \(\sqrt{8-x^{2}}\).
Solution:
When a, b ∈ R and a ≥ 0, b ≥ 0 then \(\sqrt{a}\) > \(\sqrt{b}\) ⇔ a > b ≥ 0
∴ \(\sqrt{x+2}\) = \(\sqrt{8-x^{2}}\)
⇔ x + 2 > 8 – x² ≥ 0 and x > -2, |x| < 2\(\sqrt{2}\)
We have (x + 2) > 8 – x²
⇔ x² + x – 6 > 0
⇔ (x + 3)(x – 2) > 0
⇔ x ∈ (-∞, -3)(2, ∞) .
We have 8 – x² ≥ 0
⇔ x² ≤ 8 ⇔ |x| < 2 \(\sqrt{2}\) ⇔ x ∈ [-2 \(\sqrt{2}\), 2 \(\sqrt{2}\)] Also x + 2 > 0 ⇔ x > -2
Hence x + 2 > 8 – x² ≥ 0
⇔ x ∈ ((-∞, -3) ∪ (2, ∞)) ∩ [-2 \(\sqrt{2}\), 2 \(\sqrt{2}\)] and x > -2
⇔ x ∈ (2, 2 \(\sqrt{2}\))
∴ The solution set is [x ∈ R : -2 < x ≤ 2 \(\sqrt{2}\)]

Question 55.
Solve the equation
\(\sqrt{(x-3)(2-x)}\) < \(\sqrt{4 x^{2}+12 x+11}\).
Solution:
The given inequation is equivalent to the following two inequalities.
(x – 3)(2 – x) ≥ 0 and
(x – 3)(2 – x) < 4x² + 12x + 11
(x – 3)(2 – x) ≥ 0 ⇔ (x – 2)(x – 3) ≤ 0
⇔ 2 ≤ x ≤ 3
-x² + 5x – 6 < 4x² + 12x + 11
⇔ 5x² + 7x + 17 > 0
The discriminant = b² – 4ac = 49 – 340 < 0 ⇒ x ∈ R : a = 5 > 0 and expressive is the
Hence the solution set of the given inequation is {x ∈ R : 2 ≤ x ≤ 3}

Question 56.
Solve the inequation

Solution:

⇔ either 6 + x – x² = 0, 2x + 5 ≠ 0 and x + 4 ≠ 0 or 6 + x – x² > 0
and \(\frac{1}{2 x+5}\) ≥ \(\frac{1}{x+4}\)
We have 6 + x – x² = -(x² – x – 6)
= -(x + 2) (x – 3)
Hence 6 + x + x² = 0 ⇔ x = -2 or x = 3
2x + 5 = 0 ⇔ x = –\(\frac{5}{2}\)
x + 4 = 0 ⇔ x = -4
∴ 6 + x – 2x² > 0 ⇔ -2 < x < 3 x ∈ (-2, 3), 2x + 5 > -4 + 5 = 1 > 0 and x + 4 > -2 + 4 = 2 > 0
x + 4 > -2 + 4 = 2 > 0
x ∈ (-2, 3), \(\frac{1}{2 x+5}\) ≥ \(\frac{1}{x+4}\)
⇔ 2x + 5 ≤ x + 4,
2x + 5 ≤ x + 4 ⇔ x ≤ 1
Hence 6 + x – x² > 0 and \(\frac{1}{2 x+5}\) ≥ \(\frac{1}{x+4}\)
⇔ -2 < x ≤ -1
∴ The solution set is {-2, 3} ∪ (-2, -1)
= [-2, -1] ∪ {3}

Question 57.
Find the maximum value of the function
\(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R. (May ’05)
Solution:
Let y = \(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\)
⇒ yx² + 2yx + 3y = x² + 14x + 9
⇒ (y – 1)x² + 2(y – 7)x + 3y – 9 = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 7)²] – 4(y – 1) (3y – 9) ≥ 0
⇒ 4[(y² – 14y + 49) [(3y² – 12y + 9)] ≥ 0
⇒ -2y² – 2y + 40 ≥ 0
⇒ y² + y – 20 ≤ 0
⇒ (y + 5) (y – 4) ≤ 0
⇒ -5 ≤ y ≤ 4
⇒ y ∈ [-5, 4]
∴ Maximum value of y = 4
∴ Maximum value of the function
\(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R is 4.

Question 58.
Show that none of the values of the function over \(\frac{x^{2}+34 x-71}{x^{2}+2 x-7}\) over R lies between 5 and 9. (Mar. 2005)
Solution:
Let y = \(\frac{x^{2}+34 x-7}{x^{2}+2 x-7}\)
⇒ x² + 2x – 7 = x² + 34x – 71
⇒ (y – 1)x² + 2(y – 17)x + (-7y + 71) = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 17)]² – 4(y – 1)(-7y + 71)] ≥ 0
⇒ 4[(y² – 34y + 289) – (-7y² + 78y – 71)] ≥ 0
⇒ 8y² – 112y + 360 ≥ 0
⇒ y² – 14y + 45 ≥ 0
⇒ (y – 5) (y – 9) ≥ 0
⇒ y ≤ 5 or y ≥ 9
⇒ y does not lies between 5 and 9, since expression is ≥ 0 and y² coeff is the
∴ None of the values of the function
\(\frac{x^{2}+34 x-7}{x^{2}+2 x-7}\) over R lies between 5 and 9.

Question 59.
Solve the inequation
\(\sqrt{x^{2}-3 x-10}\) > (8 – x)
Solution:
(i) \(\sqrt{x^{2}-3 x-10}\) > (8 – x)
⇒ x² – 3x – 10 ≥ 0 and (i) (8 – x < 0)
or (ii) x² – 3x – 10 > (8 – x)² and (8 – x ≥ 0)
Now x² – 3x – 10 = (x – 5) (x + 2)
Hence x² – 3x – 10 ≥ 0
⇒ x ∈ (-∞, -2] ∪ [5, ∞)
8 – x < 0 ⇒ x ∈ (8, ∞)
∴ x² – 3x – 10 ≥ 0 and 8 – x < 0
⇔ x ∈ (-∞, -2] ∪ [5, ∞) and x ∈ (8, ∞)
⇔ x ∈ (8, ∞)

(ii) x² – 3x – 10 > (8 – x)²
⇔ x² – 3x – 10 > 64 + x² – 16x
⇔ 13x > 74

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(a)

I.

Question 1.
If ⁿP₃ = 1320, find n.
Solution:
Hint: ⁿP_r = \(\frac{n !}{(n-r) !}\) = n(n – 1)(n – 2)…….(n – r + 1)
∵ ⁿP₃ = 1320
= 10 × 132
= 10 × 12 × 11
= 12 × 11 × 10
= ¹²P₃
∴ n = 12

Question 2.
If ⁿP₇ = 42 . ⁿP₅, find n.
Solution:
ⁿP₇ = 42 . ⁿP₅
⇒ n(n – 1) (n – 2) (n – 3) (n – 4) (n – 5) (n – 6) = 42 . n(n – 1) (n – 2) (n – 3) (n – 4)
⇒ (n – 5) (n – 6) = 42
⇒ (n – 5) (n – 6) = 7 × 6
⇒ n – 5 = 7 or n – 6 = 6
⇒ n = 12

Question 3.
If ⁽ⁿ⁺¹⁾P₅ : ⁿP₆ = 2 : 7, find n.
Solution:
⁽ⁿ⁺¹⁾P₅ : ⁿP₆ = 2 : 7
⇒ \(\frac{(n+1)_{P_5}}{n_{P_6}}=\frac{2}{7}\)
⇒ \(\frac{(n+1) n(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)(n-4)(n-5)}=\frac{2}{7}\)
⇒ 7(n + 1) = 2(n – 4) (n – 5)
⇒ 7n + 7 = 2n² – 18n + 40
⇒ 2n² – 25n + 33 = 0
⇒ 2n² – 22n – 3n + 33 = 0
⇒ 2n(n – 11) – 3(n – 11) = 0
⇒ (n – 11) (2n – 3) = 0
⇒ n = 11 or \(\frac{3}{2}\)
Since n is a positive integer
∴ n = 11

Question 4.
If ¹²P₅ + 5 . ¹²P₄ = ¹³P_r, find r.
Solution:
We have
^(n-1)P_r + r . ^(n-1)P_(r-1) = ⁿP_r and r ≤ n
¹²P₅ + 5 . ¹²P₄ = ¹³P₅ = ¹³P_r (given)
⇒ r = 5

Question 5.
If ¹⁸P_(r-1) : ¹⁷P_(r-1) = 9 : 7, find r.
Solution:
¹⁸P_(r-1) : ¹⁷P_(r-1) = 9 : 7

⇒ 18 × 7 = 9(19 – r)
⇒ 14 = 19 – r
⇒ r = 19 – 14 = 5

Question 6.
A man has 4 sons and there are 5 schools within his reach. In how many ways can he admit his sons into the schools so that no two of them will be in the same school?
Solution:
The number of ways of admitting 4 sons into 5 schools if no two of them will be in the same school = ⁵P₄
= 5 × 4 × 3 × 2
= 120

II.

Question 1.
If there are 25 railway stations on a railway line, how many types of single second-class tickets must be printed, so as to enable a passenger to travel from one station to another?
Solution:
Number of stations on a railway line = 25
∴ The number of single second class tickets must be printed so as to enable a passenger to travel from one station to another = ²⁵P₂
= 25 × 24
= 600

Question 2.
In a class, there are 30 students. On New year’s day, every student posts a greeting card to all his/her classmates. Find the total number of greeting cards posted by them.
Solution:
The number of students in a class is 30.
∴ Total number of greeting cards posted by every student to all his/her classmates = ³⁰P₂
= 30 × 29
= 870

Question 3.
Find the number of ways of arranging the letters of the word TRIANGLE so that the relative positions of the vowels and consonants are not disturbed.
Solution:
Vowels – A, E, I, O, U
In a given, word, the number of vowels is 3
number of consonants is 5

Since the relative positions of the vowels and consonants are not disturbed,
the 3 vowels can be arranged in their relative positions in 3! ways and the 5 consonants can be arranged in their relative positions in 5! ways.
∴ The number of required arrangements = (3!) (5!)
= 6 × 120
= 720

Question 4.
Find the sum of all 4 digited numbers that can be formed using the digits 0, 2, 4, 7, 8 without repetition.
Solution:
First Method:
The number of 4 digited numbers formed by using the digits 0, 2, 4, 7, 8 without repetition = ⁵P₄ – ⁴P₃
= 120 – 24
= 96
Out of these 96 numbers,
⁴P₃ – ³P₂ numbers contain 2 in the units place
⁴P₃ – ³P₂ numbers contain 2 in the tens place
⁴P₃ – ³P₂ numbers contain 2 in the hundreds place
⁴P₃ numbers contain 2 in the thousands place

∴ The value obtained by adding 2 in all the numbers = (⁴P₃ – ³P₂) 2 + (⁴P₃ – ³P₂) 20 + (⁴P₃ – ³P₂) 200 + ⁴P₃ × 2000
= ⁴P₃ (2 + 20 + 200 + 2000) – ³P₂ (2 + 20 + 200)
= 24 × (2222) – 6 × (222)
= 24 × 2 × 1111 – 6 × 2 × 111
Similarly, the value obtained by adding 4 is 24 × 4 × 1111 – 6 × 4 × 111
the value obtained by adding 7 is 24 × 7 × 1111 – 6 × 7 × 111
the value obtained by adding 8 is 24 × 8 × 1111 – 6 × 8 × 111
∴ The sum of all the numbers = (24 × 2 × 1111 – 6 × 2 × 111) + (24 × 4 × 1111 – 6 × 4 × 111) + (24 × 7 × 1111 – 6 × 7 × 111) + (24 × 8 × 1111 – 6 × 8 × 111)
= 24 × 1111 × (2 + 4 + 7 + 8) – 6 × 111 × (2 + 4 + 7 + 8)
= 26664 (21) – 666 (21)
= 21 (26664 – 666)
= 21 × 25998
= 545958

Second Method:
If Zero is one among the given n digits, then the sum of the r – digited numbers that can be formed using the given ‘n’ distinct digits (r ≤ n ≤ 9) is
^(n-1)P_(r-1) × sum of the digits × 111 …… 1 (r times) – ^(n-2)P_(r-2) × Sum of the digits × 111 ……… 1 [(r – 1) times]
Hence n = 5, r = 4, digits are {0, 2, 4, 7, 8}
Hence the sum of all 4 digited numbers that can be formed using the digits {0, 2, 4, 7, 8} without repetition is
^(5-1)P_(4-1) × (0 + 2 + 4 + 7 + 8) × (1111) – ^(5-2)P_(4-2) × (0 + 2 + 4 + 7 + 8) × (111)
= ⁴P₃ (21) × 1111 – ³P₂ (21) (111)
= 24 × 21 × 1111 – 6 (21) (111)
= 21 (26664) – 21 (666)
= 21 (26664 – 666)
= 21 (25998)
= 545958

Question 5.
Find the number of numbers that are greater than 4000 which can be formed using the digits 0, 2, 4, 6, 8 without repetition.
Solution:
While forming any digit number With the given digits, zero cannot be filled in the first place.
We can fill the first place with the remaining 4 digits.
The remaining places can be filled with the remaining 4 digits.
All the numbers of 5 digits are greater than 4000.
In the 4-digit numbers, the number starting with 4 or 6 or 8 are greater than 4000.
The number of 4-digit numbers that begin with 4 or 6 or 8 = 3 × ⁴P₃
= 3 × 24
= 72
The number of 5-digit numbers = 4 × 4!
= 4 × 24
= 96
∴ The number of numbers greater than 4000 is 72 + 96 = 168

Question 6.
Find the number of ways of arranging the letters of the word MONDAY so that no vowel occupies an even place.
Solution:
In the word MONDAY, there are two vowels, 4 consonants and three even places, three odd places.
Since no vowel occupies an even place, the two vowels can be filled in the three odd places in ³P₂ ways.
The 4 consonants can be filled in the remaining 4 places in 4! ways.
∴ The number of required arrangements = ³P₂ × 4!
= 6 × 24
= 144

Question 7.
Find the number of ways of arranging 5 different mathematics books, 4 different Physics books, and 3 different chemistry books such that the books of the same subject are together.
Solution:
The number of ways of arranging Mathematics, Physics, and Chemistry books are arranged 3! ways.
5 different Mathematics books are arranged themselves in 5! ways.
4 different Physics books are arranged themselves in 4! ways.
3 different Chemistry books are arranged themselves in 3! ways.
∴ The number of required arrangements = 3! × 5! × 4! × 3!
= 6 × 120 × 24 × 6
= 1,03,680

III.

Question 1.
Find the number of 5-letter words that can be formed using the letters of the word CONSIDER. How many of them begin with “C”, how many of them end with ‘R’ and how many of them begin with “C” and end with “R”?
Solution:

The number of 5 letter words that can be formed using the letters of the word CONSIDER = ⁸P₅
= 8 × 7 × 6 × 5 × 4
= 6720

The number of 5 letter words beginning with ‘C’ = 1 × ⁷P₄
= 7 × 6 × 5 × 4
= 840

The number of 5 letter words end with ‘R’ = 1 × ⁷P₄ = 840

The number of 5 letter words begins with ‘C’ and ends with ‘R’ = 1 × 1 × ⁶P₃
= 6 × 5 × 4
= 120

Question 2.
Find the number of ways of seating 10 students A₁, A₂, ………, A₁₀ in a row such that
(i) A₁, A₂, A₃ sit together
(ii) A₁, A₂, A₃ sit in a specified order
(iii) A₁, A₂, A₃ sit together in a specified order
Solution:
A₁, A₂, A3, ……….., A₁₀ are the ten students.
(i) Consider A₁, A₂, A₃ as one unit and A₄, A₅, A₆, A₇, A₈, A₉, A₁₀ as seven units.
These 8 units can be arranged in 8! ways.
The students A₁, A₂, A₃ in one unit can be arranged among themselves in 3! ways.
∴ The number of ways seating 10 students in which A₁, A₂, A₃ sit together = (8!) (3!)

(ii) To arrange A₁, A₂, A₃ to sit in a specified order,
A₁, A₂, A₃ can arrange in 10 positions in specific order in \(\frac{{ }^{10} P_3}{3 !}\) ways.
The remaining 7 people can be arranged in the remaining places in 7! ways.
∴ The number of ways of A₁, A₂, A₃ sit in a specific order = \(\frac{10 !}{7 ! \times 3 !} \times 7 !\)
= \(\frac{10 !}{3 !}\)
= ¹⁰P₇

(iii) To arrange A₁, A₂, A₃ sit together in a specified order.
Consider A₁, A₂, A₃ in that order as one unit.
Now there are 8 objects, they can be arranged in 8! ways.
∴ The number of ways of A₁, A₂, A₃ sit together in specified order = 8! ways

Question 3.
Find the number of ways in which 5 red balls, 4 black balls of different sizes can be arranged in a row so that
(i) no two balls of the same colour come together.
(ii) the balls of the same colour come together.
Solution:
No. of red balls = 5
No. of black balls = 4
(i) No. two balls of the same colour come together.
For the required arrangements first, we arrange 4 black balls it can be done in 4! ways.

There are 5 places to arrange 5 red balls it can be done in 5! ways.
∴ The number of required arrangements = 4! 5!

(ii) The balls of the same colour come together.
For the required arrangements 4 black balls of different sizes can be considered as one object and 5 red balls can be considered as one object. These can be arranged in 2! ways.
The 4 black balls can be permuted among themselves in 4! ways.
The 5 red balls can be permuted among themselves in 5! ways.
∴ The no. of required arrangements = 2! 4! 5!

Question 4.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 5, 6, 7. How many of them are divisible by
(i) 2
(ii) 3
(iii) 4
(iv) 5
(v) 25
Solution:
The number of 4 digited numbers that can be formed using the digits 1, 2, 5, 6, 7 is ⁵P₄ = 120.
(i) A number is divisible by 2 when its unit place must be filled with an even digit from among the given integers. This can be done in 2 ways.

Now, the remaining 3 places can be filled with the remaining 4 digits in ⁴P₃ = 4 × 3 × 2 = 24 ways.
∴ The number of 4 digited numbers divisible by 2 = 2 × 24 = 48

(ii) A number is divisible by 3 only when the sum of the digits in that number is a multiple of 3.
Sum of the given 5 digits = 1 + 2 + 5 + 6 + 7 = 21
The 4 digits such that their sum is a multiple of 3 from the given digits are 1, 2, 5, 7 (sum is 15)
They can be arranged in 4! ways and all these 4 digited numbers are divisible by 3.
∴ The number of 4 digited numbers divisible by 3 = 4! = 24

(iii) A number is divisible by 4 only when the last two places (tens and units places) of it are a multiple of 4.

Hence the last two places should be filled with one of the following 12, 16, 52, 56, 72, 76.
Thus the last two places can be filled in 6 ways.
The remaining two places can be filled by the remaining 3 digits in ³P₂ = 3 × 2 = 6 ways.
∴ The number of 4 digited numbers divisible by 4 = 6 × 6 = 36.

(iv) A number is divisible by 5 when its units place must be filled by 5 from the given integers 1, 2, 5, 6, 7. This can be done in one way.

The remaining 3 places can be filled with the remaining 4 digits in ⁴P₃ = 4 × 3 × 2 = 24 ways.
∴ The number of 4 digited numbers divisible by 5 = 1 × 24 = 24

(v) A number is divisible by 25 when its last two places are filled with either 25 or 75.

Thus the last two places can be filled in 2 ways.
The remaining 2 places from the remaining 3 digits can be filled in ³P₂ = 6 ways.
∴ The number of 4 digited numbers divisible by 25 = 2 × 6 = 12

Question 5.
If the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the ranks of the words
(i) REMAST
(ii) MASTER
Solution:
(i) The alphabetical order of the letters of the given word is A, E, M, R, S, T
The number of words beginning with A is 5! = 120
The number of words that begins with E is 5! = 120
The number of words begins with M is 5! = 120
The number of words beginning with RA is 4! = 24
The number of words beginning with REA is 3! = 6
The next word is REMAST
∴ Rank of the word REMAST = 3(120) + 24 + 6 + 1
= 360 + 31
= 391

(ii) The alphabetical order of the letters of the given word is A, E, M, R, S, T
The number of words beginning with A is 5! = 120
The number of words that begins with E is 5! = 120
The number of words beginning with MAE is 3! = 6
The number of words that begins with MAR is 3! = 6
The number of words beginning with MASE is 2! = 2
The number of words begins with MASR is 2! = 2
The next word is MASTER.
∴ Rank of the word MASTER = 2(120) + 2(6) + 2(2) + 1
= 240 + 12 + 4 + 1
= 257

Question 6.
If the letters of the word BRING are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the 59th word.
Solution:
By using the letters of the word BRING in alphabetical order the 59th word must start with ‘I’.
Since the words start with B, G sum to 48.
i.e., Start with B = 4! = 24
Start with G = 4! = 24
Start with IB = 3! = 6
Start with IGB = 2! = 2
Start with IGN = 2! = 2
The next word is 59th = IGRBN

Question 7.
Find the sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition.
Solution:
The number of 4 digited numbers formed by using the digits 1, 2, 4, 5, 6 without repetition = ⁵P₄ = 120
Out of these 120 numbers,
⁴P₃ numbers contain 2 in the units place
⁴P₃ numbers contain 2 in the tens place
⁴P₃ numbers contain 2 in the hundreds place
⁴P₃ numbers contain 2 in the thousands place

∴ The value obtained by adding 2 in all the numbers = ⁴P₃ × 2 + ⁴P₃ × 20 + ⁴P₃ × 200 + ⁴P₃ × 2000
= ⁴P₃ (2 + 20 + 200 + 2000)
= ⁴P₃ (2222)
= ⁴P₃ × 2 × 1111
Similarly, the value obtained by adding 1 is ⁴P₃ × 1 × 1111
the value obtained by adding 4 is ⁴P₃ × 4 × 1111
the value obtained by adding 5 is ⁴P₃ × 5 × 1111
the value obtained by adding 6 is ⁴P₃ × 6 × 1111
∴ The sum of all the numbers = ⁴P₃ × 1 × 1111 + ⁴P₃ × 2 × 1111 + ⁴P₃ × 4 × 1111 + ⁴P₃ × 5 × 1111 + ⁴P₃ × 6 × 1111
= ⁴P₃ (1111) (1 + 2 + 4 + 5 + 6)
= 24 (1111) (18)
= 4,79,952

Second Method:
The sum of the r-digited numbers that can be formed using the given ‘n’ distinct non-zero digits (r ≤ n ≤ 9) is ^(n-1)P_(r-1) × sum of all digits × 111 …… 1 (r times)
Hence n = 5, r = 4, digits = {1, 2, 4, 5, 6}
The sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition is ^(5-1)P_(4-1) × (1 + 2 + 4 + 5 + 6) × (1111)
= ⁴P₃ (18) (1111)
= 24 × 18 × 1111
= 4,79,952

Question 8.
There are 9 objects and 9 boxes. Out of 9 objects, 5 cannot fit in three small boxes. How many arrangements can be made such that each object can be put in one box only?
Solution:
No. of objects = 9
No. of boxes = 9
For the required arrangements, out of 9 objects, 5 cannot bit in three small boxes.
These five can be arranged in 6 boxes these can be done in 6P5 ways.
The remaining 4 objects can be arranged in the remaining 4 boxes it can be done in 4! ways.
∴ No. of required arrangements = ⁶P₅ × 4! ways

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 3rd Lesson Union Executive Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 3rd Lesson Union Executive

Long Answer Questions

Question 1.
Explain the powers and functions of the President of India.
Answer:
Introduction:
The President of India is the constitutional head of the Indian Republic. He is the First citizen of India. He administers the affairs of the union Government either himself or through the officers subordinate to him. (Articles 52 and 53)

Qualifications :
A person to be eligible to contest the office of the President shall possess the following qualifications :

1. He should be a citizen of India.
2. He should have completed the age of 35 years.
3. He should be qualified for election as a member of the Lok Sabha.
4. He should not hold any office of profit under the union, state or local Governments (Article 59 (i))
5. Possess such other qualifications as prescribed by the Parliament.

Election Procedure :
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both Houses of Parliament. State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Oath of office :
The person who is elected as President assumes office only after he takes oath of office and secrecy by the Chief Justice of India.

Term of office :
The President continues in office for five years from the date of his assumption of office.

Salary and Allowances :
The President now gets a monthly salary of ₹ 1,50,000/-. His official residence is Rashtrapathi Bhavan at New Delhi. On retirement, he will get a monthly pension of ₹ 75,000/-.

Removal (or) Impeachment :
The President can be removed from the office by a process of impeachment for violation of the constitution. Impeachment is a quasi-judicial procedure adopted by the Parliament.

Powers and Functions :
The President shall exercise his powers with the help of the Council of Ministers headed by the Prime Minister. His powers may be analysed under the following heads :

1. Executive Powers :
An executive action of the Union Government shall be expressed in the name of the President. The President appoints the Prime Minister and other Ministers, Attorney General, Comptroller and Auditor General of India, State Governors, Judges of the Supreme Court and State High Courts, Finance Commission, Chairman and members of U.P.S.C., Election Commission, and Chief Commissioners of Unit Territories. He allocates portfolios to the Ministers.

2. Legislative Powers :
The President is an integral part of Parliament (Art. 79) and as such enjoys extensive legislative powers.
They are :

1. He summons from time to time each House of Parliament, adjours, and prorogues either or both the Houses.
2. He addresses either House separately or both the Houses Jointly.
3. He can dissolve the Lok Sabha on the advice of the Prime Minister.
4. He opens the first session of Parliament after the General Elections and at the commencement of the first session of each year.
5. He can send messages to the Parliament.
6. He arranges a joint session of both the Houses when there is a dead-lock over an ordinary bill.
7. All bills passed by Parliament require his assent for becoming in acts.
8. He nominates 12 members to Rajya Sabha and two Anglo Indian members to Lok Sabha.
9. He promulgates ordinances when the Parliament is not in session.
10. He sends the annual reports of Finance Commission, U.P.S.C etc., for the consideration and approval of Parliament.

3. Financial Powers :
The President also enjoys some financial powers. They are :

1. He recommends the financial bills to be introduced by the members in parliament. The Budget is caused to be laid down before the Parliament by the President.
2. He operates the Consolidated Fund of India.
3. He determines the shares of States in the proceeds of Income Tax.
4. No Money Bill can be introduced in the Parliament except on his recommendations.
5. He constitutes a Finance Commission for every five years etc.

4. Judicial Powers :

1. The President can grant pardons, reprieves, respites or remission of punishments.
2. He appoints the judges of the Supreme Court and State High Courts.
3. He can also remove them on an address by the Parliament.

5. Military Powers:
The President is the Supreme Commander of the Defence Forces of the Union. He appoints the Chiefs on the Staff for Army, Navy, and Air Force. He can declare war and conclude peace. But he has to take the approval of Parliament.

6. Diplomatic Powers :
The president appoints Ambassadors to foreign countries to represent India. He receives the credentials of the Ambassadors appointed in India. He represents our Nation in International affairs. He makes treaties and agreements with other countries subject to the ratification by the Parliament.

7. Emergency Powers :
In extraordinary conditions, the President can proelaim emergency to safeguard the security, integrity, and independence of our country. They are of three types :

• Emergency caused by war or external aggression or armed rebellion (Article 352).
• Emergency due to failure of Constitutional machinery in the States (Article 356)
• Emergency due to threat to the financial stability of India (Article 360).

Question 2.
Write briefly the Emergency powers of the President of India. [Mar. 16]
Answer:
Articles 352 to 360 of part XVIII of Indian constitution deals with three types of emergency powers of the Indian President. They are :

1. National Emergency,
2. Constitutional Emergency,
3. Financial Emergency

They may be explained as follows.
1) National Emergency : (Article 352)
The President exercises this power during the period of war, external aggression or armed rebellion. He declares emergency if he is satisfied that the sovereignty and security of India or any part thereof is threatened by external aggression.

When National Emergency is in force, the federal provisions of our constitution ceases to operate. So far, National Emergency was proclaimed on four occasions in our country. They are : 1. Chinese Aggression (1962), 2. Indo – Pak war (1965), 3. Indo – Pak war in the context of Bangladesh Liberation Movement (1971), 4. Opposition’s call for blocking Parliament (1975).

2) Constitutional Emergency : (Article 356)
Article 356 of Indian constitution empowers the President to proclaim the constitutional emergency. If the President, on receipt of a report from the Governor or other wise; is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency. It is also called as the President’s Rule. So far this type of emergency was proclaimed for over 100 times.

3) Financial Emergency : (Article 360)
If the President is satisfied that a situation has arisen where by the financial stability or credit of India is threatended, then he may proclaim financial emergency in the country. During the period of financial emergency, the President enjoys the following powers.

• The President may reserve all the money bills or other financial bills of the state after they are approved by the state legislature.
• He may reduce the salaries and allowances of all or any person serving in the states.
• The President can reduce the salaries allowances of the persons working at the union level including the judges of the Supreme Court and the State High Courts. But so far the financial emergency has not been yet imposed in the country.

Question 3.
Discuss the powers and functions of the Prime Minister of India. [Mar. 18, 16]
Answer:
The Prime Minister is the real executive head of the Union Government. He occupies an important position in the administration of our country. Since India has a Parliamentary form of Government the real power rests with him. He is the ‘uncrowned king’ and “the keystone of the Cabinet arch in the Union Government”.

Qualifications :

1. He should be citizen of India.
2. He should have completed the age of 25 ytears.
3. He should be qualified for election as a member of the Lok Sabha.
4. He should not hold any office of profit under the union or state or local governments.

Appointment:
The President appoints the Prime Minister. Generally the President has to summon the leader of the majority party in the Lok Sabha to form the Ministry. If no party gets an absolute majority, the President can use his discretion and summon the leader of the party, who in his opinion can manage to form a ministry. Afterwards the Prime Minister will be asked to prove his majority in the Lok Sabha.

Oath of Office :
The President of India will administers the oath of office of the Prime Minister.

Term of Office :
The Prime Minister shall remain in office during the pleasure of the President. But actually he assumes his powers as long as he retains the confidence of the majority members in the Lok Sabha. He resigns when the Lok Sabha accepts a no-confidence motion against his ministry.

Salary and Allowances:
The salary and allowances of Prime Minister are decided by the Parliament from time to time. He gets his salary and allowances that are payable to a member of Parliament. At present the Prime Minister gets a salary and allowances of ₹ 1,60,000/- per month.

Powers and Functions :
The Prime Minister is the head of the union government. He is the real executive. The Council of Ministers cannot exist without the Prime Minister. His powers are explained here under.

1) Leader of the Union Cabinet:
The Prime Minister is the leader of the Union Cabinet and Union Council of Ministers. He selects some eminent members of his party in parliament and sees that they are appointed as ministers by the President. He has a free choice of both allocating portfolios and reshuffling the ministry. All the ministers are personally and politically loyal to the Prime Minister. He decides the agenda, of the cabinet meetings. Further, he presides over the cabinet meetings.

2) Leader of the Union Government:
The Prime Minister acts as the leader of the union government. The union executive (Union Council of Ministers) initiates its business after the swearing in ceremony of the Prime Minister. All the ministers in the union ministry assume their office, owe their position and exercise their powers along with the Prime Minister. Infact, the Prime Minister influences the nature and working of the union government. He not only has a clear understanding but holds complete control over the affairs of the union government. All the high-level officers and the entire ministry in the union government behave and act according to the wishes of the Prime Minister.

3) Leader of the parliament:
The Prime Minister acts the leader of the Parliament in India. He is primarily a member of Parliament. He extends co-operation to the presiding officers in the smooth conduct of the two Houses. He wields complete control over his party members in the Parliament. He ensures that his party members maintain discipline during the sessions of the Parliament. He informs out the cabinet decisions to the Parliament. He communicates the major domestic and foreign policies of the union government to the members of Parliament. He maintains rapport with the opposition leaders and discusses the major issues confronted by the nation with them.

4) Link between the President and the Council of Ministers :
The Prime Minister acts as the main link between the President and the Union Council of Ministers. It is his duty to communicate to the President about the decisions of the Union Council of Ministers. He furnishes the every information required by the President concerning the affairs of union government. All the ministers shall formally meet the President only with the consent of the Prime Minister.

5) Leader of the Majority Party:
The Prime Minister acts as the leader of the majority party or group in the lower House of Parliament. He participates in the meetings of the party and acquaints his party members on various issues and steps taken by his ministry in implementing the party promises. He utilizes the services of the senior party leaders in running the government. He acts as the main link between the part and the government.

6) Leader of the Nation :
The Prime Minister is the leader of the nation. He takes initiative in finding solutions to several problems in the internal matters of the country. He plays an important role in the development of the nation.

7) Maker of Foreign Policy :
The Prime Minister plays a dominant role in shaping the foreign policy of the nation. He keeps in touch with the developments in all countries. He meets Heads of various countries and maintains friendly relations with them.

8) Chairman of NITI Aayog :
The Prime Minister heads the NITI Aayog (National Institution for Transforming India) NITI Aayog means policy commission. It is a policy think tank of government of India that replaces planning commission which aims to involve the states in economic policy making in India. It will provide strategic and technical advice to the central and state governments. It will have a governing council comprising Chief Ministers of all the states and it governors of Union Territories. Union government set up the NITI Aayog on January 1,-2015.

Question 4.
Explain the composition, powers and functions of the Union Council of Ministers. .
Answer:
Article 74 (1) of the Constitution provides for a Council of Ministers at the Centre. It’s main function is to aid and advice the President in the performance of his duties. It consists of Prime Minister and other Ministers. It is this body which runs the entire administration of our country. It is thefreal Executive authority of the country. It functions on the ‘Principle of Collective Responsibility1. It holds office till it continues to enjoy the confidence and support of the Lok Sabha.

Formation of Council of Ministers :
The formation of the Council of Ministers starts with the appointment of the Prime Minister. The President appoints the Prime Minister and on the advice of the Prime Minister, the other Ministers are appointed by the President.

Composition of Council of Ministers :
Our Constitution did not mention the exact size of the Union Council of Ministers. But there are three kinds of Ministers:

1. Cabinet Ministers
2. Ministers of State
3. Deputy Ministers.

The Cabinet Ministers are entrusted with the maintenance of some important ministries. They enjoy independence and decision making powers.

The Ministers of State act as the heads of some important sections in the ministry. They are directly responsible to the Prime Minister for their activities.

The Deputy Ministers have no independent and discretionary powers. They assist the Cabinet Ministers and perform the functions assigned to them.

The Cabinet or the Council of Ministers is the pivot around which the entire administration of our country revolves. “It is the steering wheel of the ship of the State.” It is the hyphen that joins the Executive and Legislative organs of the Government.

Powers and Functions :
1. Executive Powers :
The Union Cabinet is a deliberative and policy formulating body. It discusses and decides all National and International policies of the country. The policies decided by the cabinet are carried out by the Ministers. It directs and leads the Parliament for action and gets its approval for all its policies. It coordinates and guides the activities of departments of the Government. It also plays an important role by suggesting persons for all major appointments. It considers the reports of various committees before they are presented to the Parliament.

2. Legislative Powers :
The Cabinet plans the legislative programme of the Government at the beginning of each session of Parliament. It drafts Bills on all important matters and introduces them in the Parliament. It also decides the time of summoning and prorogation of Parliament. The inaugural speech of the President to the Parliament is prepared by the Cabinet.

3. Financial Powers:
The Cabinet possess important financial powers. It has complete control over national finance. It prepares the Union Budget. It decides what taxes are to be imposed and how much of expenditure is to be incurred. Money Bills are always introduced in the Lok Sabha by the Finance Minister.

4. Foreign Relations :
In the field of foreign relations also the Cabinet plays an important role. It determines and formulates the foreign policy of the country and decides India’s relations with other countries. It considers and approves all international treaties and agreements.

Collective Responsibility :
Article 75(3) of Indian constitution stated that the union council of Ministers shall be collectively responsible to the Lok Sabha, for all their acts of commissions and commissions. They act as a team under the leadership of the Prime Minister. They sail together, they swim together and they sink together.

Conclusion:
It is thus clear that the Council of Minister or Cabinet enjoys far reaching powers both with regard to the internal and external policies of the country. Internally it maintains law and order within the country and externally protects the country from foreign aggression. The progress of the country largely depends upon the ability of the Cabinet.

Short Answer Questions

Question 1.
How is the President of India elected?
Answer:
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both Houses of Parliament. State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Each member of the Electoral College has one vote. But the value of the vote differes from State to State. The value of the vote of a Parliament member also differs from the value of the vote of a member of State Assembly.

The value of the vote of an M.L.A is worked out as follows. The total population of the State is divided by the total number of elected members of the Assembly. The quotient thus obtained is to be divided by 1,000. Fractions of half or more should be counted as one and added to the quotient. If it is less than half, it is ignored. This may be shown as follows.

1. Value of vote of an M.L.A = Population of State / number of elected members of the Assembly / 1,000.
2. Value of vote of M.P. = Total value of votes of all Assembly members / Total number of elected members of both Houses of Parliaments.

This method is followed to keep the election of the President above narrow political considerations.

Question 2.
Write briefly about the procedure of impeachment of President
Answer:
The President may be removed from the office for violation of the constitution by a process of impeachment. Impeachment is a quasi-judicial procedure adopted by the Parliament. Either House of Parliament shall prefer the charge for removal of the President. The other House shall investigate into the charges itself or cause the charge to be investigated.

There are four stages in the impeachment of the President.

Firstly the impeachment resolution has to be moved with a 14 days prior notice in writing signed by not less than 1/4^th of the total members of that House. Such a resolution has to be passed by a majority of not less than 2/3^rds of the total members of the House.

Secondly, the resolution approved by the first House will be sent to the second House for consideration and approval.

Thirdly, the second House investigates into the charges directly or constitutes a committee to enquire into the charges. The President has the right to present his views directly or through a deputy during such enquiry.

Fourthly, if the charges against the President are established and adopted by the second House with 2/3^rds majority of the total members, the President stands removed from the office. With regard to voting on the resolution for impeachment, only the elected members cast their vote. No president has so far been impeached in our country till today.

Question 3.
Mention any two Emergency powers of the Indian President.
Answer:
1) National Emergency : (Article 352)
The President exercises this power during the period of war, External aggression or armed rebellion. He declares emergency if he is satisfied that the sovereignty and security of India or any part thereof is threatened by external aggression.

When National Emergency is in force, the federal provisions of our constitution ceases to operate. So far, National Emergency was proclaimed on four occasions in our country. They are :

1. Chinese Aggression (1962) ‘
2. Indo – Pak war (1965) .
3. Indo – Pak war in the context of Bangaldesh Liberation Movement (1971)
4. Oppositions call for blocking Parliament (1975)

2) Constitutional Emergency : (Article 356)
Article 356 of Indian constitution Empowers the President to proclaim the constitutional emergency. If the President, on receipt of a report from the Governor or other wise, is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency, It is also called as the President’s rule. So far this type of emergency was proclaimed for over 100 times.

Question 4.
Explain the role and position of the President in Union Government.
Answer:
The President of India is the head of the union executive. He is the first citizen of India. He could exercise many powers as enstined in the constitution in the following manner.

Position of the President:
There are different opinions on the actual position of the president of India in the administration of our country. The farmers of the constitution wanted him to be a nominal Head of the state.

Dr. Ambedkar compared his position to that of the British King Sri. M. C. Setalved, the farmer Attorney General of India mentioned that the position of the President of India is like the king in England and the Governor General in a Dominion. Sri Alladi Krishna Swamya Ayyar also said that it was perfectly clear that our presidents position was similar to that of the constitutional Monarch in England.

Jawaharlal Nehru, the first Prime Minister of India, said that “We have not given our President any real power but we have made his position one of great authority and dignity.” These opinions make it clear that our President is only a nominal figure head and he does not have any real powers. This is confirmed by the 42nd Amendment of the constitution of India.

However, the President exercises independent powers under some conditions. He utilises these powers in regard to the appointment of the Prime Minister, dissolving the Lok Sabha and ordering midterm poll to the Lok Sabha. Some presidents like Sanjiva Reddy, Zail Singh, R.Venkata Raman, Dr. S.D. Sharma etc., utilised their discretionary powers where there was political instability or Hung Parliament after the general elections in the country. Like the Monarch of England, he still enjoys three rights the right to be ! consulted the right to encourage and the right to warn.

Question 5.
Write about any two powers of the Vice – President of India.
Answer:
The Vice President of India occupies the second highest position in the union government. He is accorded a rank next to the President.

Qualifications :
A person to be eligible as vice-president should possess the following qualifications.

1. He should be a citizen of India.
2. He should have completed 35 years of age.
3. He should be qualified for election as a member of the eousil of states.
4. He should not hold any office of profit under the union, state or local Governments in India.

Election :
The election of the vice – president like that of the President shall be indirect and in accordance with the system of proportional representation by means of the single transferable vote system. He is elected by the members of an electoral college consisting of the members of both the houses of Parliament.

Term of Office :
The Vice-President holds office for a term of 5 years from the date on which he enters his office.

Removal:
The Vice President may be removed from his office by a resolution of the council of states passed by a majority of all the members of the council and agreed to by the house of the people.

Salary and Allowances :
The Vice President of India receives a monthly salary of ₹ 1,25,000/-in addition, he is entitled to daily allowance, free furnished residence, medical, travel and other facilities.

Powers and Functions :
The Vice – President is the ex-officio Chairman of the Rajya Sabha. As. such he enjoys the same powers like the Speaker of Lok Sabha, such as (1) presiding over the meetings of Rajya Sabha, (2) maintaining discipline, decency and decorum in the House, (3) exercising casting vote in case of a tie, (4) admitting visitors, (5) protecting the privileges and rights of the members. He has no right with regard to money bills.

Acting as President of India :
He discharges the functions of the President during the temporary absertce of the President. He may take over the office of the President under 4 situations like (1) Death of the President, (2) Resignation of the President, (3) Removal of the President, (4) Inability of the President due to absence, illness or any other cause.

Question 6.
How is the Prime Minister appointed?
Answer:
Article 75 (1) of the Indian constitution deals with the appointment of the Prime Minister of India.

Appointment:
The constitution simply lays down that the Prime Minister shall be appointed by the President. After the conduct of General Elections to the Lok Sabha, the President has to invite the majority party leader of the Ldk Sabha to form the Government.

When no single party is able to secure majority seats in Lok Sabha, the President invites the leader of a coalition to form the Government. The president uses his discretionary powers in this regard. The President appoints the leader of the coalition as the Prime Minister on the condition that he has to prove his majority in the Lok Sabha within a specified period. Being the leader of the majority in Lok Sabha to be the Prime Minister, the person has to be a member of Parliament. If he is not a member at the time of appointment, he has to acquire it within six months from the date of his appointment as Prime Minister.

The powers of the President in choosing, inviting and appointing the Prime Minister cannot be questioned in any court of Law.

Question 7.
Explain the role of the Prime Minister in Union Government.
Answer:
The prime Minister plays a predominant role in the affairs of the union government. He will have an indelible impression on every one in the administration of the union government. He is described as the Primus Interparus (first among equals). His role as the leader of the Union Council of Ministers, Union Cabinet, Party in power, Lok Sabha, Nation and as the link between the President and the Union Council of Ministers is unique. He wields tremendous political power and patronage. He enjoys enormous powers and fulfils innumerable tasks. It all depends on the image, influence, stature, and personality of the Prime Minister in the union government.

Jawaharlal Nehru, Dr. Ambedkar, and other eminent leaders of the National Movement and members of the Constituent Assembly described the Prime Minister as the linchpin of the union government. It is in this context that Sir william Harcourt remarked that every one expects from the Prime Minister dignity and authority, firmness to control, tact, practice and firmness, an impartial mind, a tolerant temper, a kind and prudent counsellorship, and accessibility to the people.

Question 8.
Describe the composition and powers of the Union Council of Ministers.
Answer:
Composition of Council of Ministers :
Our Constitution did not mention the exact size of the Union Council of Ministers. But there are three kinds of Ministers: 1) Cabinet Ministers 2) Ministers of State 3) Deputy Ministers.

The Cabinet Ministers are entrusted with the maintenance of some important ministries. They enjoy independence and decision making powers.

The Ministers of State act as the heads of some important sections in the ministry. They are directly responsible to the Prime Minister for their activities.

The Deputy Ministers have no independent and discretionary powers. They a&ist the Cabinet Ministers and perform the functions assigned to them.

The Cabinet Or the Council of Ministers is the pivot around which the entire administration of our country revolves. “It is the steering wheel of the ship of the State”. It is the hyphen that joins the Executive and Legislative organs of the Government.

Powers and Functions :
1) Executive Powers :
The Union Cabinet is a deliberative and policy formulating body. It discusses and decides all National and International policies of the country. The policies decided by the cabinet are carried out by the Ministers. It directs and leads the Parliament for action and gets its approval for all its policies. It co-ordinates and guides the activities of departments of the Government. It also plays an important role by suggesting persons for all major appointments. It considers the reports of various committees before they are presented to the Parliament.

2) Legislative Powers :
The Cabinet plans the legislative programme of the Government at the beginning of each session of. Parliament. It drafts Bills on all important matters and introduces them in the Parliament. It also decides the time of summoning and prorogation of Parliament. The inaugural speech of the President to the Parliament is prepared by the Cabinet.

3) Financial Powers :
The Cabinet possesses important financial powers. It has complete control over national finance. It prepares the Union Budget. It decides what taxes are to be imposed and how much of expenditure is to be incurred. Money Bills are always introduced in the Lok Sabha by the Finance Minister.

4) Foreign Relations :
In the field of foreign relations also the Cabinet plays an important role. It determines and formulates the foreign policy of the country and decides India’s relations with other countries. It considers and approves all international treaties and agreements.

Very Short answers

Question 1.
Composition of the Union Executive.
Answer:
The constitution of India provides for the Union Executive. Articles 52 to 78 in part V of the constitution deal with the union executive. The Union Executive consists.
i) The President
ii) The Vice-President
iii) The Prime Minister and
iv) The Union Council of Ministers

Question 2.
Qualifications required for contesting the Presidential elections.
Answer:
A person to be eligible to contest the office of the president shall possess the following qualifications.

1. He should be a citizen of India.
2. He should have completed the age of 35 years.
3. He should be qualified for election as a member of the Lok Sabha.
4. He should not hold any office of profit under the Union, State or Local Governments.
5. Possess such other qualifications as prescribed by the Parliament.

Question 3.
Election of President.
Answer:
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both houses of Parliament, State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Question 4.
Important appointments of President.
Answer:
The president has the power to appoint the following high dignitaries :

• The Prime Minister of India.
• Members of the union council of Ministers.
• The Affomey General of India.
• The Comptroller and Auditor general of India.
• The judges of the Supreme Court and the High Court.
• State Governors etc.

Question 5.
Judicial Powers of the President.
Answer:

1. The President can grant pardons, reprieves, respites or remission of punishments.
2. He appoints the Judges of the Supreme Court and State High Courts.
3. He can also remove them on an address by the Parliament.

Question 6.
Article 352. [Mar. 18, 16]
Answer:
Article 352 of the Indian constitution empowers the President to impose National Emergency during the period of war, External Aggression, Armed Rebellion or internal disturbance. So far National Emergency was proclaimed on Four occasions. They are :

1. Chinese Aggression (1962)
2. Indo – Pak War (1965)
3. Indo – Pak war in the context of Bangladesh Liberation movement (1971) and
4. Opposition’s call for blocking Parliament (1975).

Question 7.
Article 356.
Answer:
Article 356 of Indian Constitution empowers the President to proclaim the constitutional emergency. If the President on receipt of a report from the governor or otherwise is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency. It is also called as the President’s Rule.

Question 8.
Financial Emergency.
Answer:
If the President is satisfied that a situation has arisen where by the financial stability or credit of India is threatened then he may proclaim financial emergency in the country as per Article 360 of the Indian Constitution.

Question 9.
National Emergency.
Answer:
Article 352 of the Indian constitution empowers the president to impose National Erpergency during the period of war, External Aggression, Armed Rebellion or Internal disturbances. When National Emergency is in force, the Federal provisions of our constitution cease to operate. So far, National Emergency was imposed four times in 1962, 1965, 1971 and 1975.

Question 10.
Qualifications required for contesting as Vice-President. [Mar. 16]
Answer:
A person to be eligible for election as Vice-president should possess the following qualifications : ,

1. He should be a citizen of India.
2. He should have completed 35 years of age.
3. He should be qualified for election as a member of the council of states.
4. He should not hold any office of profit under the Union, State or Local Governments in India.

Question 11.
Chairman of Rajya Sabha,
Answer:
The Vice President is the ex-officio chairman of the Rajya Sabha. As such he enjoys the same powers like the speaker of Lok Sabha such as

1. Presiding over the meetings of Rajya Sabha.
2. Maintaining discipline, decency and decorum in the House.
3. Exercising casting vote in case of a tie.
4. Protecting the privileges and rights of the members.

Question 12.
Appointment of Prime Minister.
Answer:
After the conduct of General elections of the Lok Sabha, the President has to invite the majority party leader of the Lok Sabha to form the Government. When no single party is able to secure majority seats in Lok Sabha, the President invites the Leader of a coalition to form the Government. The president uses his discretionary powers in this regard.

Question 13.
Categories of Union Council of Ministers.
Answer:
There are three kinds of ministers in the union council of ministers. They are

1. Cabinet Minister.
2. Ministers of State.
3. Deputy Ministers.

1) The Cabinet Ministers are entrusted with the maintenance of some important ministries like Finance, Home, Defence etc.

2) The Ministers of state act as the heads of some important sections in the Ministry. They are directly responsible to the Prime Minister for their activities.

3) The Deputy Ministers have no independent and discretionary powers. They assist the Cabinet Ministers.

Question 14.
Any two functions of the Union Cabinet.
Answer:

1. The Union Cabinet formulates the policies of the union government. It finalizes the domestic as well as foreign policies of the nation after having serious deliberations.
2. It pilots several bills in the Parliament at various stages and strives to secure the approval of the latter.

Question 15.
Collective Responsibility. [Mar. 18]
Answer:
Article 75 (3) of Indian constitution stated that the union council of Ministers shall be collectively responsible to the Lok Sabha, for all their acts of omissions and commissions. They act as a team tinder the leadership of the Prime Minister. They sail together, they swim together and they sink together.

Andhra Pradesh BIEAP AP Inter 2nd Year Economics Study Material 6th Lesson Tertiary Sector Textbook Questions and Answers.

AP Inter 2nd Year Economics Study Material 6th Lesson Tertiary Sector

Essay Questions

Question 1.
Define tertiary sector. Explain the importance of tertiary sector in Indian economy.
Answer:
The tertiary sector activities include all other activities like transport communication, banking, insurance and trade. Tertiary sector with its sub-sectors act as complementary sector to the development of primary and secondary sectors. Economists argue that the development of the economy is associated with high proportion of grass domestic product and working population employed in tertiary sector. The importance of the tertiary sector can be illustrated by the following indices.

1) Share in the gross domestic product: A review of the sectoral contribution to the gross domestic product reveals that the contribution of service sector has been increasing from 1950, its share was 27.5 percent of the GDP in 1950 – 51 and increased to around 56.0 percent in 2007 – 2008. The share of services sector in India is continuously increasing. It was only 27.5 percent in 1950 – 51 and increased to 40.59 percent in 1990 – 91 and further to 55.73 percent in 2007 – 08. In 2013 it was 57.0%.

2) Workers employed : Generally there is a tendency for the tertiary sector to expand more rapidly than secondary sector activities as a consequency the percentage of labour force engaged in this sector also registers an increase.

The workers engaged in tertiary sector has increased from 178.56 lakhs (1991) to 182.65 lakhs in 2008. Though 189.60 lakh workers were employed in tertiary sector during 2001 there was a decline in the number of workers employed to 182.65 lakhs in 2008. This was mainly due to the decline in the employment of workers in public sector.

It is evident form the data that 54.1 percent of the workers employed in tertiaiy sector are in rural areas and the proportion of urban areas was 45.9 percent.

3) Exports of services: India has been recording high growth in the export of services during the last few years. Exports of services was US $ 460 billion in 2004 – 05, increased to US $ 61.4 billion in 2005 – 06 and to US $ 90.1 billion in 2007 – 08 growth of exports was particularly repaid in the miscellaneous services which comprise software services, business services, financial services and communication services India’s export of services is expected to touch US $ 310.9 billion powered by the booming software consultancy, engineering and tourism sectors by 2011 -12.

Question 2.
Infrastructure contributes to the economic development of a country. Explain.
Answer:
“Infrastructure” is an umbrella term for many service activities referred to as social overhead. Capital by many development economists infrastructure facilities are also referred to as economic and social overhead often they have also referred to urban and rural infrastructure. Adequate infrastructural facilities help to determine a country’s success and another failure. They diversity production, expand trade, reduce poverty and improve the environmental conditions.
Generally infrastructure is categorised into two groups. They are :

1. Economic infrastructure
2. Social infrastructure

The services and facilities that are used in the economic production and by households is called as Economic infrastructure.

Engineering structures, equipment, power, gas, telecommunications, water supply, sanitation public works for irrigation, roads and railways, urban transport ports, waterways, air ports etc., are grouped as “economic infrastructure”.

Social infrastructure includes education, healthcare, family welfare housing, labour welfare etc. Both economic and social infrastructural serve as the true “engine of growth” and provides the needed impetus to the economy. The services associated with the use of infrastructure are called as infrastructural services, act as wheels of the economic activity. The availability on infrastructure improves the per capita GDP and has the following effects:

1. Raises the productivity of the production process.
2. Increases access to markets.
3. Leads to agricultural expansion.
4. Brings higher yields in agriculture.
5. Lowers the cost of doing business.
6. Provides ability to complete in the international trade, even in traditional commodities.
7. Provides employment opportunities.
8. Achieves cost – saving in inventory and working capital.
9. Delivers the product just-in-time particularly in exports.
10. Facilitates diversification of trade.
11. Leads to modernisation and diversification of production.
12. Provides ability to repond to changes in demand and prices.
13. Facilitates employment intensive growth.
14. Defines welfare and ensures growth with poverty reduction.
15. Helps to identify the poor.
16. Improves the quality of life.
17. Offers non-farm employment opportunities.
18. Encourages the modification of physical surroundings as environment friendly surroundings.
19. Promotes environment sustainability of human settlements.
20. Removes and disposes the liquid and solid wastes.
21. Protects public health and provides health benefits.

Keeping in view all these beneficial impacts of infrastructural services, the Government of India has decided to invest ₹ 4,35,349 crore, to spend exclusively on the improvement of rural infrastructure. This amount is allocated for the development of electricity, roads, telecommunications, irrigation, water supply and sanitation.

Short Answer Questions

Question 1.
Explain the contribution of GDP in service sector.
Answer:
Service sector has become very prominent in world’s economies. The importance may be highlighted in terms of its contribution to GDP, employment and exports.

The share of GDP is high in many countries. The following table shows that the contribution of services sector to the overall GDP those countries is very high in the tuble U.K has highest share of service sector followed by U.S.A, France. In India the share of service sector increase between 2001 and 2013.

Question 2.
What are the activities considered under the India’s Services Sector ?
Answer:
Services in India are emerging as a prominent sector in terms of contribution of national and State incomes, trade flows, foreign direct investment (FDI) inflows and employment. The following activities can be considered to form the part of the service sector.

1. Trade.
2. Hotels and Restaurants.
3. Transport (Including Railways and transport by other means)
4. Storage.
5. Communication.
6. Banking and Insurance.
7. Real Estate and Business Services.
8. Public administration and defence.
9. Construction.
10. Other services including education, medical and health, religious and other community services, legal services, recreation services.

Question 3.
What are the advantages of Roadways? [A.P. Mar. 18, 17, 16]
Answer:
The principle mode of connectivity between places in roadways. India has one of the largest road networks in the world, spread over 48.65 lakh k.m. district and villages road constitutive 95.2% of the total road network in our country.
Advantages of Roadways :

1. Road transport connects all the villages and regions and finally it connects to the railways.
2. Road transport does not required heavy capital expenditure.
3. The chances of delay, damages are less in case of road transport.
4. Road transport provides transports the goods to the railway station.
5. Road transport help the farmers particularly easily and quickly to transport to mandis and towns.
6. Road transport is more flexible when compared to other means of transport. It can provide door to door service.
7. Enables to defence forces to move areas inaccessible by railways in emergencies.

Question 4.
Explain the importance of Railways.
Answer:
Railways provide the principal mode of transportation for freight and passengers. They are the great integrating force for the past 150 years and helping in acclerating the development of industry and agriculture railways made a very modest beginning in 1853. With a route length of 34 kms. Indian railways have grown into a vast network of 7,183 stations spread over a route length of 53,332 kilometers. They have a flat of 8025 locomotives, 44090 passenger service vehicles, 5990 other loading vehicles and 2,07,176 wagons. The Indian railways is the world’s third largest rail network under a single management. Better resource management rational pricing policy have led to a significant improvement in the performance of the railways. In India the development of agriculture and industrial sectors has generated higher level of demand for rail transport. Coal, pig iron, iron ore, cement, food grains, fertilizers sugar, salt, steel, petroleum products and other essential commodities are transported by railways.

Question 5.
What is tourism ? Explain its importance in Indian economy.
Answer:
Tourism is the sub – sector of tertiary sector in general and services industry in particular.

W.T.O defined Tourism as “the activities of persons travelling and staying in places outside their usual environment for not more then one consecutive year for leisure, business and other purposes”.
Importance:

1. Tourism provides revenue to the government.
2. Tourism creates employment facilities for women.
3. It provides regional development.
4. It is a source of foreign exchange earnings.
5. Tourism sells indirectly the environmental resources.
6. It can be used as a means of reducing poverty.
7. It builds partnership with private sector.

Question 6.
Explain the banking system in India.
Answer:
A well – developed banking system is necessary pro -requisite for achievement of economic development. Banks play an important role in mobilization of savings and investments. Banks are the efficient agents of capital formation in the economy and give access to use the resources in a productive way.

Banking system in India has been playing a very important role in the process of economic development. Depending upon the nature of the activity performed by them, banking system in India may be classified into the following categories. They are :
a) Commercial Banks
b) Cooperative Banks
c) Central Bank (Reserve Bank of India)

a) Commercial Banks : The commercial banks were nationalized in phased manner in 1969 and 1980. They are classified as public sector (nationalized) banks and private sector banks. The State Bank of India and its associates along with another 20 banks are the public sector banks. The Indian scheduled banks, which are not nationalized and branches of foreign banks operating in India are called as private sector banks.

b) Cooperative Banks: Under Cooperative Banking System State Cooperative Banks, District Central Cooperative Banks (DCCB) and primary to short term credit. State Cooperative Agriculture and Rural Development Banks and Primary Agriculture and Rural Development Bank provide long term credit.

c) Reserve Bank of India (RBI) : The RBI is India’s Central Bank was established on 1^st April 1935 was nationalized on 1^st January 1949. RBI is Supreme Monetary Authority in the country. It keeps the reserves of all scheduled banks, which were included in 2nd schedule of RBI and which were not included is called as ‘Non-schedule Banks’.

Question 7.
What are the major constituents of insurance industry in India ?
Answer:
A Health and Development Insurance Sector of vital importance to every modem economy. It encourages the savings habits provides a safety net to rural and urban enterprises and individuals and generates long term funds for infrastructure development. Development and insurance is therefore, necessary to support continued economic growth social security and person reforms also benefit from a mature insurance industry.
There are two major constituents of Insurance :

1. Life Insurance
2. Non Life Insurance (General)

1) Life Insurance : LIC offers schemes, policies and plans to investors. Particularly the main objective of UC is giving protecting against risk of death and channalizing the funds for the benefits and the economy in the socially oriented sectors during 2013 -14.

Life Insurance under wrote first – year premium of ₹ 1,19,641 crore as against ₹ 1,07,361 crore during 2012-13 registering a growth of 11.44 percent.

2) Non Life Insurance (General): The general insurance companies deal with non life insurance. The GIC was approved as the Indian Reinsurer on 3^rd November 2000. It offers fire, marine, motor, health and other insurance. During 2013-14, non-life insurance including stand lone health insurance and specialize insurance (Export Credit Guarantee Scheme) and Agriculture Insurance Company (AIQ underwrote premium worth ₹ 77,584 crore as against ₹ 69,089 crore during 2012 -13 registering a growth of 12.23 percent.

Very Short Answer Questions

Question 1.
Service sector [A.P. Mar 17]
Answer:
Service sector is also known as tertiary sector. Service sector is the life, line for the social and economic growth of a country. The service sector activities include trade, transport, communications, banking, insurance, education, health, energy, marketing etc., all these facilities and services constitutes collectively the tertiary sector.

Question 2.
Infrastructure
Answer:
An umbrella term for service activities in the economy. Infrastructure is categorized into two groups. They are economic infrastructure and social infrastructure.

Question 3.
Transport
Answer:
Transport means conveyance of people or property from one place to another. These services provide a link between production distribution and consumption activities. Road ways, railways, airways, waterways are the important means of transport.

Question 4.
Water Transport
Answer:
Water Transport is the another important means of transport. Shipping is an important indicator of both commodity and service trade of any country. Water transport in India is of two types. They are the land water transport and International water transport (Shipping).

Question 5.
Civil Aviation
Answer:
Air transport has a vital role in the economic development of the country. It is the modem and quickest transport. In India the first commercial flight started on February 18^th, 19.11. The real progress in civil aviation started in 1920.

Question 6.
Tourism [A.P. Mar. 17, 16]
Answer:
Tourism is the sub-sector of tertiary sector in general and services industry in particular. Tourism as “the activities of persons travelling and staying in places outside their usual environment for not more than one consecutive year for leisure, business and other purposes.

Question 7.
LIC [A.P. Mar. 18, 16]
Answer:
Life Insurance Corporation of India was set up in 1956. UC has its central office at Mumbai with 7 Zonal offices, 101 divisional offices and 2,048 branch offices. It mobilise savings of the public to invest in the industrial securities.

Question 8.
GIC
Answer:
General Insurance Industry in India was nationalized in 1972 and a government company known as General Insurance Corporation of India (GIC) was established. There are four GIC companies.

1. National Insurance Company Limited
2. New India Assurance Company Limited
3. Oriental Insurance Company Limited
4. United India Insurance Company Limited. The GIC deal with non-life insurance.

Question 9.
Micro – Insurance [A.P. Mar. 18]
Answer:
A system that blends insurance with savings and credit practices. The member of self help groups, farmers, migrant workers and tribals are the target groups for this micro finance. This insurance offers both individuals and group insurance services.

Question 10.
Communication
Answer:
The communication system is an integral part of the development process. Communication means the transmission of information. By providing necessary information about markets and supply of goods. It consists of posts of telegraphs, telecommunications broadcasting, television etc.

Question 11.
Science and Technology
Answer:
Science and technology are ideas and means with which man seeks to change his environment. Science represents accumulation of knowledge, while technology represents ‘refinement in tools. These two have helped to improve the quality of human life.

Question 12.
Performance of software industry
Answer:
The software industry is the main component of the information technology in India. India’s pool of young-aged man power is the key behind this success story. Presently there are more then 500 software forms in the country. Global software gaints like Microsoft, Oracle etc.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 10th Lesson Chemistry In Everyday Life Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 10th Lesson Chemistry In Everyday Life

Very Short Answer Questions

Question 1.
What are drugs ?
Answer:
Drug : The chemicals of low molecular masses ranging from 100 to 500 U that react with macromolecular targets to produce biological response are called drugs.
E.g.: Morphine, Codeine, Heroin etc.,

Question 2.
When are the drugs called medicines ?
Answer:
When the biological response of a drug is therapeutic arid useful then the chemical substances (drugs) are called medicines.

Question 3.
Define the term chemotherapy.
Answer:
Chemotherapy : The use of medicines (chemical substances) in the treatment of diseases is called chemotherapy.
In chemotherapy diagnosis, prevention and treatment of diseases involved.

Question 4.
Name the macromolecules that are chosen as drug targets.
Answer:
The macromolecules that are chosen as drug targets are carbohydrates, lipids, proteins, nucleic acids and enzymes.

Question 5.
What are enzymes and receptors ?
Answer:
Enzymes : The proteins which perform the role of biological catalysts in the body are called enzymes.
Receptors : The proteins which are crucial to communication system in the body are called receptors.

Question 6.
Which forces are involved in holding the drug to the active site of enzymes ?
Answer:
The forces involved in holding the drugs to the active site of enzymes are ionic bonds, vander waal’s forces, hydrogen bonds, dipole-dipole interactions etc.,

Question 7.
What are enzyme inhibitors ?
Answer:
The drugs which inhibits the catalytic activity of enzymes and can block the binding site of the enzyme and prevent the binding of substrate are called enzyme inhibitors.

Question 8.
What is allosteric site ?
Answer:
Some drugs do not bind to the enzyme’s active site and bind to a different site of enzyme. This different site of enzyme is called allosteric site.

Question 9.
What are antagonists and agonists ?
Answer:
Antagonists : The drugs that bind to the receptor site and inhibit its natural function are called antagonists.

• These are useful when blocking of message is required.

Agonists : The drugs that mimic the natural messenger by switching on receptors are called agonists.

• These are useful when there is lack of natural chemical messenger.

Question 10.
Why do we need to classify the drugs in different ways ?
Answer:

1. By knowing the pharmacological effect it makes easy for doctors.
2. Antacid can be used in case of excessive acidity in stomach.

Question 11.
What are antacids ? Give example. [Mar. 14]
Answer:
Antacids : Chemicals that remove the excess of acid in the stomach and maintain the pH to normal level are antacids. E.g. : Omeprazole, Lansoprozole etc.,

Question 12.
What are antihistamines ? Give example.
Answer:
Antihistamines : Chemicals that prevent the interaction of histamines with receptors of the stomach wall thus producing less amount of acid are antihistamines.
E.g. : Dimetapp, Terfenadine (Seldane).

Question 13.
While antacids and antiallergic drugs interfere with the function of histamines why do not these interfere with the function of each other ?
Answer:
Antacids and antiallergic drugs donot interfere with the function of each other because they work on different receptors in the body.

Question 14.
What are tranquilizers ? Give example.
Answer:
Tranquilizers : The drugs which are used in the management (or) treatment of psychoes and neuroses are called tranquilizers.
E.g.: Luminal, Seconal, Barbituric acid etc.,

Question 15.
What are barbiturates ?
Answer:
Barbiturates : Derivatives of barbituric acid which functions as important class of tranquilizers are called barbiturates.
E.g.: Veronal, Amytatt etc.,

Question 16.
What are analgesics ? How are they classified ?
Answer:
Analgesics : These are to reduce or totally abolish pain without causing impairment of consciousness, mental confusion, incoordination, paralysis, disturbances of nervous system etc. Analgesics are classified as

1. Narcotic analgesics : These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics.
   E.g. : Morphine, Codeine etc.
2. Non-narcotic analgesics : These drugs are analgesics but they have no addictive properties. Their analgesic use is limited to mild aches and pains.
   E.g.: Aspirin, Ibuprofen etc.

Question 17.
What are narcotic analgesics ? Give example.
Answer:
Narcotic analgesics: These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics.
E.g. : Morphine, Codeine etc.

Question 18.
What are non-narcotic analgesics ? Give example.
Answer:
Non-narcotic analgesics: These drugs are analgesics but they have no addictive properties. Their analgesic use is limited to mild aches and pains.
E.g.: Aspirin, Ibuprofen etc.

Question 19.
What are antimicrobials ?
Answer:
Antimicrobials : The chemical substances which destroy (or) prevent the developement (or) inhibit the pathogenic action of microbes such as bacteria, fungi, virus are called antimicrobials.
E.g.: Lysozyme, Lactic acid etc.,

Question 20.
What are antibiotics ? Give example. [T.S. Mar. 19; A.P. Mar. 16]
Answer:
Antibiotics: The chemical substances produced by micro organisms and inhibit the growth or destroy microorganisms are called antibiotics.
(Or)
The substance produced totally or partly by chemical synthesis which in low concentration inhibits the growth (or) destroy microorganism by intervening in their metabolic process are called antibiotics.
E.g. : Penicillin, Chloramphenicol etc.

Question 21.
What are antiseptics ? Give example. [T.S. Mar. 19; A.P. Mar. 19, 15]
Answer:
Antiseptics : The chemical compounds that kill (or) prevent the growth of micro organism are called antiseptics.
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
E.g. : Dettol, Bithional etc.

Question 22.
What are disinfectants ? Give example.
Answer:
Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.

• These are applied to inanimate objects like floors, drainage systems etc…
  E.g.:
  • 4% aqueous solution of formaldehyde called formalin is a disinfectant
  • 0.3 ppm chlorine aqueous solution is disinfectant.
  • SO₂ in very low concentration is disinfectant.

Question 23.
Name a substance which can be used as an antiseptic as well as disinfectant.
Answer:
Phenol is used as antiseptic as well as disinfectant.

1. 0.2% phenol is antiseptic.
2. 1% phenol is disinfectant.

Question 24.
What is the difference between antiseptics and disinfectants ?
Answer:
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
Disinfectants are applied to inanimate objects such as floors, drainage system, instruments etc.

Question 25.
What are the main constituents of dettol ?
Answer:
Dettol (antiseptic) is a mixture of chloroxylenol and terpineol.

Question 26.
What is tincture of iodine ? What is its use ?
Answer:
Tincture of iodine (antiseptic) is a mixture of 2 – 3% Iodine solution in alcohol-water.

Question 27.
What are antifertility drugs ? Give example.
Answer:
Antifertility drugs: These are birth control pills and contain a mixture of synthetic estrogen and progesterone derivatives.
E.g. : Norethindrone, Ethynylestradiol (novestrol)

Question 28.
Why chemicals are added to food ?
Answer:
Chemicals are added to food for

1. preservation
2. enhancing their appeal
3. adding nutritive value in them.

Question 29.
Name different categories of food additives.
Answer:
The following are the categories of food additives.

1. Food colours
2. Flavours and sweetners
3. Fat Emulsifiers and stabilising agents
4. Anti oxidants
5. Flour improvers – antistaling agents and bleaches
6. Preservatives .
7. Nutritional supplements such as minerals, vitamins and amino acids.

Question 30.
What are artificial sweetening agents ? Give example. [A.P. Mar. 16; 15]
Answer:
The chemical substances which are used instead of sucrose (or) sugar are called artificial sweetening agents.
E.g.: Aspartame, Alitame, saccharin.
These decrease the calorific intake and at the same time several times sweater than sucrose.

Question 31.
Why do we require artificial sweetening agents ?
Answer:

1. Artificial sweetening agents are very useful to diabetic persons.
2. These decrease the calorific in take and at the same time several times sweeter than sucrose.
3. These are harmless.

Question 32.
Why is the use of aspartame limited to cold foods and drinks ?
Answer:
Aspartame is unstable at cooking temperature so it’s use is limited to cold foods and soft drinks. –

Question 33.
Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
The sweetening agent used in the preparation of sweets for a diabetic patient is saccharin (or) sucralose. It is stable at cooking temperature.

Question 34.
What problem does arise in using alitame as artificial sweetener ?
Answer:
While using alitame as artificial sweetener, the control of sweetness of food is difficult. Alitame is a high potency sweetner.

Question 35.
What are food preservatives ? Give example.
Answer:
The chemical substances which prevent the spoilage of food due to microbial growth are called food preservatives.
E.g.: Sodium benzoate, Salt of sorbic acid etc.

Questions 36.
Name two most familiar antioxidants used as food additives.
Answer:
The most familiar antioxidants are butylated hydroxy toluene (BHT) and butylated hydroxy anisole (BHA)

Question 37.
What is saponification ?
Answer:
The process of formation of soaps Containing sodium salts by heating esters of fatty acid with aq. NaOH solution is called saponification.

Questions 38.
What are soaps chemically ?
Answer:
Chemically soaps are sodium (OF) potassium salts of long chain fatty acids.
E.g.: Sodium stearate.

Question 39.
Why do soaps not work in hard water ?
Answer:
In hard water Ca, Mg-dissolved salts are present. Ca⁺², Mg⁺² ions form insoluble Ca, Mg, soaps respectively when sodium (or) potassium soaps are dissolved in hard water.

1. These insoluble soaps separate as scum in water and are useless as cleansing agent. These are problematic to good washing because the ppt adheres into the fibres of cloth as gummy mass.
2. Hair washed with hard water looks dull.
3. Dye does not absorb evenly on cloth washed with soap using hard water due to this gummy mass.

Question 40.
What are synthetic detergents ?
Answer:
The cleasing agents which are having all the properties of soaps but do not contain any soap are called synthetic detergents.
Synthetic detergents can be used both in soft and hard water as they give foam even in hard water.
E.g.: Sodium dodecyl benzene sulphonate.

Question 41.
What is the difference between a soap and a synthetic detergent ?
Answer:

1. Generally soaps are sodium or potassium salts of long chain fattyacids.
2. Synthetic detergents are cleansing agents having all the properties of soaps and do not contain any soap.
3. Soaps do not work in hard water but synthetic detergents can be used both in soft and hard water as they give foam even in hard water. Some of the detergents give foam even in ice cold water.

Question 42.
How are synthetic detergents better than soaps ? [Mar 14]
Answer:
Soaps do not work in hard water but synthetic detergents can be used both in soft and hard water as they give foam even in hard water. Some of the detergents give foam even in ice cold water.

Question 43.
Name the different categories of synthetic detergents.
Answer:
Synthetic detergents are classified into three types

1. nionic detergents
2. Cationic detergents
3. Non-ionic detergents.

Question 44.
Can you use soaps and synthetic detergents to check the hardness of water ?
Answer:

1. Soaps are used to check the hardness of water because soaps form insoluble precipitate with hard water.
2. Detergents are soluble in both types of water i.e hardwater and soft water, so detergents are not used to check the hardness of water.

Question 45.
If water contains dissolved calcium hydrogen carbonate, out of soaps and synthetic detergents which one will you use for cleaning clothes and why ?
Answer:
Water contains dissolved calcium hydrogen carbonate is called hard water. This water form insoluble precipitate with soap. Synthetic detergents donot form this type of precipitates, so synthetic detergents are used for cleaning clothes with water containing dissolved Ca(HCO₃)₂.

Short Answer Questions

Question 1.
Explain the term target molecules or drug targets as used in medicinal chemistry.
Answer:
Drug Targets (or) Target molecules: Macro molecules like carbohydrates, lipids, proteins, nucleic acids and enzymes which interact with the drugs are ailed drug targets (or) Target molecules.

Question 2.
Explain the catalytic action of enzymes as drug targets.
Answer:
a) Catalytic action of Enzymes : In the enzyme catalytic activity it performs two functions

1. The first function of an enzyme is to hold the substrate for a chemical reaction. Substrate bind to active site of enzyme through ionic bonding, hydrogen bonding, vander waal force etc.
2. The second function of an enzyme is to provide functional groups that will attack the substrate, and carry out the chemical reaction.

b) Drug- Enzyme interaction :

1. Drugs which inhibit the catalytic activity of the enzymes are called Enzyme inhibitors.
2. Drugs which compete with the substrate for their attachment on active sites of enzymes are called competitive inhibitors.
   

Question 3.
Explain the drug – enzyme interaction.
Answer:
Explanation of drug – Target interaction with Enzymes :

1. The proteins which perform the role of biological catalysts in the body are called enzymes.
2. The binding of inhibitor ai allosteric site changes the shape of the active site.
3. If the bond formed between an enzyme and an inhibitor is a strong covalent bond it cannot be broken easily In this case enzyme is blocked permanently and the body then degrades the enzyme-inhibitoi complex and synthesises the new enzyme.
   

Question 4.
Why are cimetidine and ranitidine better antacids than sodium hydrogen carbonate or magnesium hydroxide or aluminium hydroxide.
Answer:
Sodium hydrogen carbonate or magnesium hydroxide (or) aluminium hydroxide treatment as antacids control only symptoms and not the cause. In advanced stages, ulcers become life threatening and only treatment is removal of affected part of stomach.
The drugs cinetidine, ranitidine prevent the interaction of histamine with the receptors present in the stomach wall and results in release of lesser amount of acid.
So cinetidine, ranitidine are better antacids than NaHCO₃ (or) Mg(OH)₂ (or) Al(OH)₃.

Question 5.
Low level of noradrenaline in the cause of depression. What type of drugs are needed to cure this problem ? Name two drugs.
Answer:
If the level of Noradrenaline is low then the signal sending activity becomes low and the person suffers from depression. In this situations antidepressant drugs (tranquilizers) are required, These drugs inhibit the enzymes which catalyse the degradation of noradrenaline. If the enzyme is inhibited, this important neurotransmitter is slowly metabolised and can activate its receptor for longer periods of time.

Two important antidepressant drugs are IProniazid and phenelzine.

Question 6.
What are analgesics ? How are they classified ? Give examples. [T.S. & A.P. Mar. 19]
Answer:
Analgesics : These are to reduce or totally abolish pain without causing impairment of consciousness, mental confusion, incoordination, paralysis, disturbances of nervous system etc.
Analgesics are classified as

1. Narcotic analgesics : These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics.
   E.g. : Morphine, Codeine etc.
2. Non-narcotic analgesics : These drugs are analgesics but they have no addictive properties. Their analgesic use is limited to mild aches and pains.
   E.g.: Aspirin, Ibuprofen etc.

Question 7.
What are different types of microbial drugs ? Give one example for each.
Answer:
Antimicrobials : The chemical substances which destroy or prevent the developement (or) inhibit the pathogenic action of microbes such as bacteria, fungi, virus are called antimicrobials.
E.g.: Lysozyme, Lactic acid etc.,
Different types of antimicrobial drugs are antibiotics, antiseptics, disinfectants.
1) Antibiotics : The chemical substances produced by micro organisms and inhibit the growth or destroy microorganisms are called antibiotics.
(Or)
The substance produced totally or partly by chemical synthesis in low concentration inhibits the growth (or) destroy microorganism by intervening in their metabolic process are called antibiotics.

2) Antiseptics: The chemical compounds that kill (or) prevent the growth of micro organism
are caiied antiseptics. .
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.

3) Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.
These are applied to inanimate objects like floors, drainage systems etc…
E.g.:

• 4% aqueous solution of formaldehyde called formalin is a disinfectant
• 0.3 ppm chlorine aqueous solution is disinfectant.
• SO₂ in very low concentration is disinfectant.

Question 8.
Write the characteristic properties of antibiotics.
Answer:
Characteristic properties of antibiotics :

1. Antibiotic drugs must be the products of metabolism. ,
2. Antibiotic drugs are effective in low concentration.
3. Antibiotic drug retards the growth (or) survival of microorganism.
4. Antibiotic should be a synthetic substance produced as a structural analogue of naturally occurring antibiotic.
5. Antibiotics have either cidal (killing) effect (or) a static (inhibiting) effect on microbes.
   E.g.: Penicillin, Ofloxacin are Bactericidal.
   Erythromycin, Tetracycline are Bacteriostatic.

Question 9.
What is meant by the term broad spectrum antibiotics ? Explain.
Answer:
The range of bacteria (or) other micro organisms that are effected by a certain antibiotic is expressed as its spectrum of action.
Broad spectrum antibiotics : Antibiotics which kill (or) inhibit a wide range of gram-positive and gram-negative bacteria are called broad spectrum antibiotics.

Question 10.
What are broad spectrum and narrow spectrum antibiotics ? Give one example for each.
Answer:
The range of bacteria (or) other micro organisms that are effected by a certain antibiotic is expressed as its spectrum of action.
Broad spectrum antibiotics : Antibiotics which kill (or) inhibit a wide range of gram¬positive and gram-negative bacteria are called broad spectrum antibiotics.

Narrow spectrum antibiotics: Antibiotics which are effective mainly against gram-positive (or) gram-negative bacteria are called narrow spectrum antibiotics,
E.g.: Penicillin – G is a narrow spectrum antibiotic.

Limited spectrum antibiotics : Antibiotics which are effective mainly against a single organism (or) disease are called as limited spectrum antibiotics.

Question 11.
Write notes on antiseptics and disinfectants.
Answer:
Antiseptics: The chemical compounds that kill (or) prevent the growth of micro organism are called antiseptics.
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
E.g.: Dettol, Bithional etc.
Phenol is used as antiseptic as well as disinfectant. 0.2% phenol is antiseptic.
Dettol (antiseptic) is a mixture of chloroxylenol and terpineol.
Tincture of iodine (antiseptic) is a mixture of 2 – 3% Iodine solution in alcohol-water.

Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.

1. These are applied to inanimate objects like floors, drainage systems etc…
   E.g.: 4% aqueous solution of formaldehyde called formalin is a disinfectant.
2. 0.3 ppm chlorine aqueous solution is disinfectant.
3. SO₂ in very low concentration is disinfectant.

Phenol is used as antiseptic as well as disinfectant 1% phenol is disinfectant.

Question 12.
How do antiseptics differ from disinfectants ? Does the same substance be used as both ? Give one example for each.
Answer:
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
Disinfectants are applied to inanimate objects such as floors, drainage system, instruments etc.
Phenol is used as antiseptic as well as disinfectant.
i) 0.2% phenol is antiseptic.
ii) 1% phenol is disinfectant.

1. Examples of Antiseptics :
   • Dettol (antiseptic) is a mixture of chloroxylenol and terpineol.
   • Tincture of iodine (antiseptic) is a mixture of 2.3% Iodine solution in alcohol-water.
2. Examples of disinfectants :
   • 4% aqueous solution of formaldehyde called formalin is a disinfectant.
   • 0.3 ppm chlorine aqueous solution is disinfectant.
   • SO₂ in very low concentration is disinfectant.

Phenol is used as anti in very low concentration is disinfectant.

Question 13.
What are the main categories of food additives ?
Answer:
The following are the categories of food additives.

1. Food colours
2. Flavours and sweetners
3. Fat Emulsifiers and stabilising agents
4. Anti oxidants
5. Flour improvers – antistaling agents and bleaches
6. Preservatives
7. Nutritional supplements such as minerals, vitamins and amino acids.

Question 14.
Write notes on antioxidants in food.
Answer:
Antioxidants :

• Antioxidants are important and necessary food additives.
• Antioxidants help in food preservation by retarding the action of oxygen on food. Antioxidants are more reactive towards, oxygen than the food material which they protecting.
• The most familiar antioxidants are Butylated hydroxy toluene (BHT) and Butylated hydroxy anisole (BHA).
• The addition of BHA to butter increases its shelf life from months to years.
• BHT and BHA along with citric acid are added to produce more antioxidant effect.
• SO₂ and sulphites are useful anti Oxidants for wine and beer, sugar syrups and cut, peeled (or) dried fruits and vegetables.

Question 15.
Name different types of soaps.
Answer:
The following are the different types of soaps.

1. Toilet soaps
2. Soaps that float in water
3. Medicated soaps
4. Shaving soaps
5. Laundry soaps
6. Soap powders and scouring soaps etc.

Question 16.
Explain the following terms with suitable examples.
i) Cationic detergents
ii) Anionic detergents
iii) Non-ionic detergents
Answer:
Synthetic detergents are classified into three types.
i) Cationic detergents : These are synthetic detergents.
a) Cationic detergents are quarternary ammonium salts of amines with acetates, chlorides (or) bromides as anions.

b) Cationic part possess a long hydrocarbon chain and a positive charge on nitrogen atom. Hence these are called catiomc detergents.
E.g. : Cetyl trimethyl ammonium bromide
It is used in hair conditioners.

ii) Anionic Detergents: These are synthetic detergents.
a) Anionic detergents are sodium salts of sulphonated long chain alcohols (or) hydrocarbons.
b) Anionic detergents are formed by the treatment of long chain alcohols with conc. H₂SO₄ followed by the neutralisation with alkali.
E.g.: Sodium lauryl sulphate.

These are used for house hold work and in tooth pastes.

iii) Non-ionic detergents : These are synthetic detergents.
a) Non-ionic detergents do not contain any ion in their constitution.
b) The detergent formed by the reaction of stearic acid with poly ethylene glycol is an example of non ionic detergent.

Non-ionic detergents are used in liquid dish washing purpose.

Question 17.
What are biodegradable and non-bio degradable detergents ? Give one example for each.
Answer:

1. Biodegradable detergents :
   The detergents which are degraded (or) decomposed by micro organisms are called biodegradable detergents. Biodegradable detergents have less branching.
   These do not cause water pollution.
   E.g.: n-dodecyle benzene sulphonate, soap (non synthetic),
2. Non Biodegradable detergents :
   The detergents which are not decomposed (or) degraded by microbes (or) micro organisms are called non-biodegradable detergents. These have more branching.
   These cause water pollution.
   E.g.: ABS detergent.

Question 18.
Explain the cleansing action of soaps.
Answer:
Soap is sodium stearate, C₁₇H₃₅COONa, in water soap gives the ions stearate anion and sodium ion.

Cleansing action of soap : Soap anions form a micelle. The grease or dirt of the cloth are absorbed into the interior of the micelle. The tails of the anion are pegged into micelle and these micelle are washed away with the soap solution.

The main function of the soap is therefore, to convert the oily and greasy dirt on the cloth into large colloidal particles. For this a critical concentration of soap solution is required. Soap thus, functions as an emulsifying agent for the water – dirt emulsion. The emulsified grease or dirt is then washed away with soap solution.

Question 19.
Label the hydrophilic and. hydrophobic parts in the following compounds.
i) CH₃ (CH₂)₁₀ CH₂ OSO^–₃ Na⁺
ii) CH₃(CH₂)₁₅N⁺(CH₃)₃ Br^–
iii) CH₃ (CH₂)₁₆ COO (CH₂ CH₂O)_n CH₂ CH₂ OH
Answer:

Question 20.
Draw the structures of following:
i) Serotonin
ii) Bithionol
iii) Chloramphenicol
iv) saccharin
Answer:

Long Answer Questions

Question 1.
Describe the classification of drugs into different classes.
Answer:
Classification of drugs : Drugs are classified in different ways.

1. On the basis of drug action : Based on the actions of drugs on particular biochemical processes they are classified as
   • Antibacterials
   • Antibiotics
   • Hypnotics
   • Sedatives and tranquilizers
   • Cardio vascular drugs
   • Antiseptics etc.
2. On the basis of molecular targets : Biomolecules like Carbohydrates, lipids, proteins, nucleic acids etc. react with drugs. These biomolecules are called drug targets or simply target molecules.
3. On the basis of chemical structure : The drugs with some common structural features generally show similar pharmacological activity.
   E.g.: Sulfonamides have common structural feature.
   So they have same chemotherapeutic action.
4. On the basis of pharmocological effect:
   This type of classification is useful for doctors because it provides them the whole range of drugs available for the treatment.
   E.g.: Analgesics have pair killing effect.

Question 2.
Describe briefly the therapeutic action of different classes of drugs.
Answer:
The therapeutic action of different classes of drugs discussed below.
Antacids : Chemicals that remove the excess of acid in the stomach and maintain the pH to normal level are antacids.
E.g. : Omeprazole, Lansoprazole etc.,

Antihistamines : Chemicals that prevent the interaction of histamines with receptors of the stomach wall thus producing less amount of acid are anti histamines.
E.g.: Dimetapp, Terfenadine (Seldane)

Tranquilizers : The drugs which are used in the management (or) treatment of psychoes and neuroses are called tranquilizers
E.g.: Luminal, Seconal, Barbituric acid etc.,

Barbiturates : Derivatives of barbituric acid which functions as important class of tranquilizers are called barbiturates
E.g.: Veronal, Amytal etc.,

Analgesics : These are to reduce or totally abolish pain without causing impairment of consciousness, mental confusion, incoordination, paralysis, disturbances of nervous system etc.
Analgesics are classified as

1. Narcotic analgesics : These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics.
   E.g.: Morphine, Codeine etc.
2. Non-narcotic analgesics : These drugs are analgesics but they have no addictive properties. Their analgesic use is limited to mild aches and pains.
   E.g. : Aspirin, Ibuprofen etc.

Antimicrobials : The chemical substances which destroy (or) prevent the developement (or) inhibit the pathogenic action of microbes such as bacteria, fungi, virus are called antimicrobials. E.g.: Lysozyme, Lactic acid etc.,

Antibiotics: The chemical substances produced by micro Organisms and inhibit the growth or destroy microorganisms are called antibiotics. .
(Or)
The substance produced totally or partly by chemical synthesis in low concentration inhibits the growth (or) destroy microorganism by intervening in their metabolic process are called antibiotics.
E.g. : Penicillin, Chloramphenicol etc.

Antiseptics : The chemical compounds that kill (or) prevent the growth of micro organism are called antiseptics.
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.

Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.
These are applied to inanimate objects like floors, drainage systems etc.
E.g.:

1. 4% aqueous solution of formaldehyde called formalin is a disinfectant.
2. 0.3 ppm chlorine aqueous solution is disinfectant.
3. SO₂ in very low concentration is disinfectant.

Antifertility drugs: These are birth control pills and contains a mixture of synthetic estrogen and progesterone derivatives.
E.g. : Norethindrone, Ethynylestradiol (novestrol).

Question 3.
Write an essay on antimicrobials.
Answer:
Antimicrobials : The chemical substances which destroy or prevent the developement (or) inhibit the pathogenic action of microbes such as bacteria, fungi, virus are called antimicrobials
E.g. : Lysozyme, Lactic acid etc.,
Different types of anti microbial drugs are antibiotics, antiseptics, disinfectants.

Question 4.
Write notes on the following : [T.S. Mar. 16]
i) Artificial sweetening agents
ii) Food preservatives
iii) Antioxidants in food.
Answer:
i) Artificial sweetening agents: The chemical substances which are used instead of sucrose (or) sugar are called artificial sweetening agents.
E.g.: Aspartame, Alitame, saccharin.

1. These decrease the calorific intake and at the same time several times sweeter than sucrose.
2. Artificial sweetening agents are very useful to diabetic persons.
3. These are harmless.
4. Aspartame is unstable at cooking temperature so it’s use is limited to cold foods and soft drinks.
5. The sweetening agent used in the preparation of sweets for a diabetic patient is saccharin (or) sucralose. It is stable at cooking tenperature.
6. While using alitame as artificial sweetener, the control of sweetness of food is difficult. Alitame is a high potency sweetner.

ii) Food preservatives : The chemical substances which prevent the spoilage of food due to microbial growth are called food preservatives.
Sodium benzoate, C₆H₅COONa is the most important food preservative used in limited quantities and is metabolised by conversion into hippuricacid. Which is finally excreted in the urine. Salts of sorbicacid, propionic acid are also food presevatives.

iii) Antioxidants :

1. Antioxidants are important and necessary food additives.
2. Antioxidants help in food preservation by retarding the action of oxygen on food.
3. Antioxidants are more reactive towards oxygen than the food material which they protecting.
4. The most familiar anti oxidants are Butylated hydroxy toluene (BHT) and Butylated hydroxy anisole (BHA).
5. The addition of BHA to butter increases its shelf life from months to years.
6. BHT and BHA along with citric acid are added to produce more antioxidant effect.
7. SO₂ and sulphits are useful anti oxidants for wine and beer, sugar syrups and cut, peeled (or) dried fruits and vegetables.

Question 5.
Write notes on the following :
i) Soaps
ii) Synthetic detergents.
Answer:
i) Soaps : 1) Soaps are the detergents which improve cleansing properties of water and help in removal of fats which bind other materials to the fabric (or) skin.
Chemically soaps are sodium (or) potassium salts of long chain fatty acids.
E.g.: Sodium stearate.
The process of formation of soaps containing sodium salts by heating esters of fatty acid with aq. NaOH solution is called saponification.
Only sodium and potassium soaps are soluble in water and are used for cleaning purpose. Potassium soaps are soft to the skin and are prepared by using KOH instead of NaOH.

Types of soaps :
The following are the different types of soaps

1. Toilet soaps
2. Soaps that float in water
3. Medicated soaps
4. Shaving soaps
5. Laundry soaps –
6. Soap powders and scouring soaps etc.

In hard water Ca, Mg-dissolved salts are present. Ca⁺², Mg⁺² ions form insoluble Ca, Mg, soaps respectively when sodium (or) potassium soaps are dissolved in hard water.

1. These insoluble soaps separate as scum in water and are useless as cleansing agent. These are problematic to good washing because the ppt adheres into the fibres of cloth as gummy mass.
2. Hair washed with hard water looks dull.
3. Dye does not absorb evenly on cloth washed with soap using hard water due to this gummy mass.

ii) Synthetic detergents : The cleansing agents which are having all the properties of soaps but donot contain any soap are called synthetic detergents.
Synthetic detergents can be used both is soft and hardwater as they give foam even in hard water.
E.g. : Sodium dodecyl benzene sulphonate.
Soaps do not work in hard water but synthetic detegents can be used both in soft and hard water as they give foam even in hard water. Some of the detergents give foam even in ice cold water.

Synthetic Detergents are classified into three types :
i) Cationic detergents : These are synthetic detergents.
a) Cationic detergents are quarternary ammonium salts of amines with acetates, chlorides (or) bromides as anions.

b) Cationic part possess a long hydrocarbon chain and a positive charge on nitrogen atom. Hence these are called cationic detergents.
E.g. : Cetyl trimethyl ammonium bromide.
It is used in hair conditioners.

ii) Anionic Detergents :
These are synthetic detergents.
a) Anionic detergents are sodium salts of sulphonated long chain alcohols (or) hydro-carbons.
b) Anionic detergents are formed by the treatment of long chain alcohols with cone, H2S04 followed by the neutralisation with alkali.
E.g. : Sodium lauryl sulphate.

These are used for house hold work and in tooth pastes.

iii) Non-ionic detergents : .
These are synthetic detergents.
a) Non-ionic detergents donot contain any ion in their constitution.
b) The detergent formed by the reaction of stearic acid with poly ethylene glycol is an example of non ionic detergent.

Non-ionic detergents are used in liquid dish washing purpose.

Intext Questions

Question 1.
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor. Why ?
• Consider the effect of tranquilizers or antidepressants on the nervous system.
Solution:
Sleeping pills contain drugs that may be tranquilizers or antidepressants. They affect the nervous system, relieve anxiety, stress, irritability or excitement. But they should strictly be used under the supervision of a doctor. If not, the uncontrolled and over dosage can cause harm to the body and mind because in higher doses these drugs act as poisons.

Question 2.
With reference to which classification has the statement, “ranitidine is an antacid” been given ?
Solution:
This statement refers to the classification of drugs according to pharmacological effect because in neutralizes the acidity (excess) of stomach.

Question 3.
Why do we require artificial sweetening agents ?
• Mention the function of natural sweeteners, discuss the same for artificial sweeteners.
Solution:
Natural sweeteners (sucrose etc.) provide calories to the body. Taking extra calories is harmful
for diabetic patients. So, artificial sweeteners are used
(a) to control intake of calories and
(b) as a substitute of sugar for diabetics.

Question 4.
Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structural formulae of these compounds are given below.
i) (C₁₅H₃₁COO)₃ C₃H₅ – Glyceryl palmitate
ii) (C₁₇H₃₂COO)₃ C₃H₅ – Glyceryl oleate
• Consider the following method.

Solution:

Question 5.
Following types of non-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic parts in the molecule. Identify the functional group (s) present in the molecule.

Solution:

(b) Functional group : Ether and alcohol.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material 7th Lesson Organic Evolution Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material 7th Lesson Organic Evolution

Very Short Answer Questions

Question 1.
What is panspermia?
Answer:
According to Cosomozoic theory, life might have existed all over the universe in the form of resistant spores called panspermia. They might have reached the earth accidentally.

Question 2.
Define prebiotic soap. Who coined the term?
Answer:
Prebiotic soup word is coined by J.B.S.Haldane. Haldane called the ocean prebiotic soup. All the reactions which lead to the formation of organic molecules like sugars, amino acids, fatty acids purines and pyramids etc., occurred in the ocean so it was described as the prebiotic soup.

Question 3.
How did-eukaryotes evolve?
Answer:
Eukaryotes evolved probably by two processes.

1. Prokaryotes lived in the ancestral eukaryotes symbiotically and evolved into organelles such as mitrochondria and plastids.
2. The endo membrane system of eukaryotes might have evolved by the infolding of plasma membrane of the ancestral prokaryotes.

Question 4.
What are the components of the mixture used by Urey and Miller in their experiments to simulate the primitive atmosphere?
Answer:
Urey and Miller -used a mixture of water vapour, methane, ammonia and hydrogen in their experiments to simulate the primitive atmosphere.

Question 5.
Mention the names of any four connecting links that you have studied.
Answer:
Connecting links clearly explain the path of evolution.

• Peripatas between annelida and arthropoda.
• Prototherians between reptilia and mammalia.

Question 6.
Define Biogenetic law, giving an example.
Answer:
It was proposed by Ernst Haeckel. It states that ontogeny repeats phylogeny which means the development history of an organism repeats the evolutionary history of its ancestor Eg : Tadpole larva of frog, it resembles fish both externally and internally. It possesses a, tail, gills and two chambered heart like that of fish. Later it metamorphoses into adult frog.

Question 7.
Define atavism with an example.
Answer:
Sudden appearance of some vestigial organs in a better developed condition is called atavism.
Eg : Tailed human baby

Question 8.
Cite two examples to disprove Lamarck’s inheritance of acquired characters.
Answer:

1. Well developed muscles of athletes are not inherited to their children.
2. Making perforations to pinna for wearing ornaments has been in practice in India for past several centuries. However no girl child is bom with readymade perforations . in their pinna.

Question 9.
Who influenced Darwin much in formulating the idea of Natural Selection.
Answer:
Three scientists influenced Darwin, they are :

1. T.R.Malthus – On the principles of population
2. Sir Charles lyell – Principle$ of Geology
3. Alfred Russel Wallace – On the tendency of varieties do depart from original types.

Question 10.
What is common between Darwinism and Lamarckism?
Answer:
Presence of variations is common to Darwinism and Lamarckism.

Question 11.
What is meant by genetic load. Give an example.
Answer:
The existence of deletorious genes within the populations is called genetic load.
Eg: Genes for Sickle cell anaemia – The individuals homozygous for Sickle cell gene usually die early due to anaemia. .

Question 12.
Distinguish between allopatric and sympatric speciations.
Answer:
Allopatric speciation :
Speciation occurring in which first geographical isolation occurs, then secondly reproductive isolation occurs.

Sympatric speciation :
Reproduction isolation occurs without geographical isolation.

Question 13.
Mention the scientific names of ape like and man like earlier primates. Which man like primate first used hides to cover the bodies?
Answer:

• Scientific name of ape like primate – Dryopithecus
• Scientific name of man like primate – Ramapithecus
• Homo neanderthalensis – They used hides to protect their body.

Short Answer Questions

Question 1.
Distinguish between homologous and analogous organs.
Answer:
1. Homologous organs :
The organs which have similar structure and origin but not necessarily the same function are called homologous organs. Eg : The appendages of vertebrates such as the flippers of whale, wings of bat, forelimbs of horse, paw of cat and hands of man have a common pattern in the arrangement of bones even though their external form and functions may vary to suit their mode of life.

2. Analogous organs :
The organs which have dissimilar structure and origin but perform the same function are called the analogous organs. Eg : Wings of butterfly and wings of a bird.

Question 2.
Write a short note on the theory of mutations.
Answer:
Theory of mutation was proposed by Hugo de Varies, who coined the term mutation.

Mutations are sudden, random inheritable changes that occur in organisms. Hugo de Varies observed this phenomenon in the evening primrose plant Oenothera lamarckiana, which shows different forms like

• O. brevistylis – with small style.
• O. levifolia – with smooth leaves.
• O. gigas – with the giant form.
• O. nanella – with dwarf form.

These characters are inherited to the progeny.

• Darwin called mutations as sports of nature or saltations.
• Bateson called them as discontinuous variations

Salient features of mutation theory :

• Mutations occur from time to time in naturally breeding population.
• Mutants differ from their parents.
• Mutations are inheritable.
• Mutations occur in any direction i.e., they are random.
• They are discontinuous and not accumulated over generations.
• They are full-fledged and so there are no intermediate forms.
• They are subjected to natural selection.

Question 3.
Explain Darwin’s theory of natural selection with industrial melanism as an experimental proof.
Answer:
Darwin’s theory of natural selection does not explain what exactly evolution is, but explains how evolution might have occurred in nature. A classical example for natural selection is industrial melanism, exhibited by peppered moth-Biston betularia. These moths were available in two colours grey and black. Grey moths were abundant before industrial revolution in all over England. The reason for the existence of large number of grey moths during that period was camouflage on the trunks of trees.

But after the establishment of industries in England, black coloured moths were more and grey forms were less. This is due to pollution from industries in the form of soot turned barks of trees into black. So grey moths were easily identified and were more predated by birds. Thus grey moths decreased in number, black moths increased in the population.

Thus natural selection favoured the melanic moths (black) to reproduce more successfully. Natural selection of darker forms in response to industrial pollution is known as industrial melanism.

Question 4.
Discuss the role of different patterns of selections in evolution.
Answer:
Natural selection is a process by which the organisms are physically, physiologically and behaviorally better adopted to environment, survive and reproduce.

It is mainly of three types :

1. Stabilizing selection,
2. Directional selection,
3. Disruptive Selection.

1. Stabilizing selection (Centrifetal selection) :
This selection operates in a stable environment. In this process, the organisms with average phenotype are preserved where as the extreme individuals from both the ends are eliminated. Hence it does not promote any evolutionary change that leads to specification, but maintains the phenotypic stability within the population over generations.
Eg : In England weight of newborn babies were studied in a large sample. Greater mortality was found in the babies whose weight was greater or lesser than the average weight of 81bs.

2. Directional selection :
This selection operates in an environment which gradually undergoes changes. It works by constantly removing the individuals from one end and constantly shifting the average value of fitness towards the other end of phenotypic distribution.
Eg : In case of giraffes the average value of the length of neck shifted towards the long-neck. Stabilizing selection takes over the directional selection. Once the average value of the phenotype coincides with the new optimum environmental conditions.

3. Disruptive selection (Centrifugal selection) :
This selection operates when homogenous environment changes into a heterogenous type. In this process, the organism of both the extreme phenotypes are selected while the individuals with average phenotype are eliminated. This can split the population into two or more sub-populations or species populations.
Eg : In California the sunflower population was divided into two sub-populations. One was adapted to dry area and the other was adapted to wet area.

Question 5.
Write a short note on Neo-Darwinism.
Answer:
It was proposed by Fischer, Sewall Wright, Mayr. According to this theory, five basic factors are involved in the process of Organic evolution. They are :

1. Gene mutations
2. Chromosomal mutations
3. Genetic recombinations
4. Reproductive isolation

1. Gene mutations :
Heritable changes in the structure of a gene are called gene mutations or point mutations. They alter the phenotypic character of the individuals. Thus, gene mutations tend to produce Variations in the offspring.

2. Chromosomal mutations:
Heritable changes in the structure of chromosomes called chromosomal mutations. They also bring about variations in the phenotype of organism which lead to the occurrence of variations in the offspring.

3. Genetic recombinations :
Recombinations of genes due to crossing over during meiosis are responsible for bringing genetic variability among the individuals of the same species.

4. Natural selection :
Natural selection does not produce any genetic changes, but it favours some genetic change while rejecting others.

5. Reproductive isolation:
The absence of gene exchange between population is called the reproductive isolation. It plays a great role in giving rise to new species and preserving the species integrity.

Question 6.
In a population of 100 rabbits which is in Hardy-Weinberg equilibrium, 24 are homozygous long eared. Short ears are recessive to long ears. There are only two alleles for this gene. Find out the frequency of recessive allele in the population.
Answer:
Number of rabbits in the population with Hardy-Weinberg equilibrium = 100
Number of dominant homozygous long eared rabbit = 24
Frequency of homozygous dominant long eared rabbits (P²) = \(\frac{1}{100}\) × 24 = 0.24
Frequency of dominent allele (P) = 0.49
Frequency of recessive allele = q = 1 – 0.49 = 0.51

Question 7.
What is meant by genetic drift ? Explain genetic drift citing the example of Founder Effect.
Answer:
The change in the frequency of a gene that occurs merely by chance and not by selection, in small proportion is called genetic drift.

Suppose, for a gene with two alleles, the frequency of a particular allele is 1% (e = 0.01) the probability of losing that allele by chance from small population is more. The end result is either fixation or loss of that allele.

Genetic drift tend to reduce the amount of genetic variation within the population mainly by removing the alleles with low frequencies. It can examplified by Founder and Bottleneck effect.

Founder effect:
If a small group of individuals fro,m a population starts a new colony in an isolated region, those individuals are called the founders of the new population. The allelic frequency of their descendants are similar those of the founders rather than to their ancestral parent population.
Eg : O^+ve blood group is present in nearly 100% of the red Indians. It means the forefathers of the Red Indians tribe were predominantly O^+ve and they isolated themselves reproductively from other population.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 16th Lesson Communication Systems Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 16th Lesson Communication Systems

Very Short Answer Questions

Question 1.
What are the basic blocks of a communication system?
Answer:
Basic blocks in a communication system are

1. Transmitter
2. Receiver
3. Channel.

Question 2.
What is the “World Wide Web” (WWW)?
Answer:
Tern Berners -Lee invented the World Wide Web.
It is an encyclopedia of knowledge accessible to everyone round the clock throughout the year.

Question 3.
Mention the frequency range of speech signals.
Answer:
Speech signals frequency range is 300 Hz to 3100 Hz.

Question 4.
What is sky wave propagation ?
Answer:
In the frequency range from a MHz upto about 30 MHz, long distance communication can be achieved by ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation.

Question 5.
Mention the various parts of the ionosphere ?
Answer:
Parts of ionosphere are

1. D (Part of stratosphere (65-70 km day only),
2. E (Part of stratosphere (100 km day only),
3. F₁ (Part of mesosphere (170 km – 190 km),
4. F₂ (Part of thermosphere [300 km at night 250 – 400 km during day time]).

Question 6.
Define modulation. Why is it necessary ? (T.S. Mar.’19, 16, 15; A.P. Mar. 16, 15) (Mar. ’14)
Answer:
Modulation : The process of combining low frequency audio signal with high frequency carrier wave is called modulation.

Necessary : Low frequency signals cannot transmit directly. To reduce size of the antenna and to avoid mixing up of signal from different transmitters modulation is necessary.

Question 7.
Mention the basic methods of modulation. (A.P. Mar. ’19, ’16, T.S. Mar. ’15)
Answer:
The basic methods of modulation are :

1. Amplitude modulation (AM)
2. Frequency modulation (FM)
3. Phase modulation (PM)

Question 8.
Which type of communication is employed in Mobile Phones ? (A.P. Mar. ’15)
Answer:
Space wave mode of propagation is employed in mobile phones.

Short Answer Questions

Question 1.
Draw the block diagram of a generalized communication system and explain it briefly.
Answer:
Every communication system has three essential elements :

1. Transmitter
2. Medium / Channel
3. Receiver.

The block diagram is shown in Fig.

In Communication system, the transmitter and the receiver are located at two different places separate from the transmitter.

The channel is the physical medium that connects. The purpose of transmitter is to convert the message signal produced by the source of information into a form, suitable for transmission through the channel. If the output of the information source is a non-electrical signal like a voice signal, a transducer converts it to electrical form before giving it as an input to the transmitter. When a transmitted signal propagates along the channel it may get distorted due to channel imperfection. Moreover, noise adds to the transmitted signal and the receiver receives a corrupted version of the transmitted signal. The receiver has the task of operating on the received signal. It reconstructs a recognizable form of the original message signal for delivering it to the user of information.

Question 2.
What is a Ground wave ? When is it used for communication ?
Answer:
Ground Wave : To radiate signals with high efficiency, The antennas should have a size comparable to the wavelength λ of the signal (at least ~ λ/4). At longer wavelengths (i.e., at lower frequencies), the antennas have large physical size and they are located on or very near to the ground. In standard AM broadcast, ground based vertical towers are generally used as transmitting antennas. For such antennas, ground has a strong influence on the propagation of the signal. The mode of propagation is called surface wave propagation and the wave glides over the surface of the earth. A wave induces current in the ground over which it passes and it is attenuated as a result of absorption of energy by the earth. The attenuation of surface waves increases very rapidly with increase in frequency. The maximum range of coverage depends on the transmitted power and frequency (less than 2 MHz). Ground waves will propagate long distances over sea water due to its high conductivity.

Question 3.
What are Sky Waves ? Explain Sky Wave propagation, briefly.
Answer:
Sky Waves : Long distance communication between two points on the earth is achieved through reflection of electromagnetic waves by Ionosphere, Such waves are called sky waves.

This mode of propagation is used by short wave broadcast services. The Ionosphere is so called because of the presence of a large number of ions or charged particles. It extends from a height of ~ 65 Km to about 400 km above the earth’s surface.

The degree of ionisation varies with the height. The density of atmosphere decreases with height. At greater heights the solar radiation is intense but there are few molecules to be ionized. Close to the earth, the radiation intensity is low so that the ionization is again low. However at some intermediate heights, there occurs a peak of ionization density. The ionospheric layer acts as a reflector for a certain range of frequencies (3 to 30 MHz).
Electromagnetic waves of frequencies higher than 30 MHz penetrate ionosphere and escape. This phenomenon is shown in the Fig.

The phenomenon of bending of em waves is so that they are diverted towards the earth which is similar to Total Internal Reflection in optics.

Question 4.
What is Space Wave Communication ? Explain.
Answer:
A spcae wave travels in a straight line from transmitting antenna to the receiving antenna. Space waves are used for line – of – sight (LOS) communication as well as satellite communication. At frequencies above 40 MHz, communication is essentially limited to line-of-sight paths. At these frequencies, the antennas are relatively smaller and can be placed at heights of many wavelengths above the ground. Because of line-of-sight nature of propagation, direct waves get blocked at some point by the curvature of the earth as illustrated in Fig. If the signal is to be received beyond the horizon then the receiving antenna must be high enough to intercept the line-of-sight waves.

If the transmitting antenna is at a height hT then it can be shown that the distance to the horizon d_T is given as d_T = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{T}}\right)}\) where R is the radius of the earth (approximately 6400 km). Similarly if the receiving antenna is at a height hR, the distance to the horizon d_R is d_R = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{R}}\right)}\). With reference to Fig. the maximum line-of sight distance d_M between the two antennas having heights h_T and h_R above the earth is given by d_M = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{T}}\right)}\) = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{R}}\right)}\)

Television broadcast, microwave links and satellite communication are some examples of communication systems that use space wave mode of propogation.

Question 5.
What do you understand by modulation ? Explain the need for modulation.
Answer:
The process of combining audio frequency signal with high frequency signal is called modulation.
To transmit an electronic signal in the audio frequency range (20 Hz to 20 KHz) over a long distance directly the following constraints limit the possibility :

1. Size of the antenna
2. Effective power radiated by the antenna
3. Mixing up of signals from different transmitters.

For 20 KHz signal the height of antenna is about 4km, still a large height. This makes antenna length impractical. Even if we transmit the signal, they may combine with low frequency signals present in the atmosphere and it is impossible to distinguish the signals at the receiving end.

In order to avoid these problems a low frequency audio signal is combined with high frequency signal to translate the audio signal to high frequencies.

Question 6.
What should be the size of the antenna or aerial ? How the power radiated is related to length of the antenna and wavelength ?
Answer:
Size of antenna (or) aerial: For trasmitting a signal, we need an antenna. The size of the antenna comparable to the wavelength of the signal (at least λ/4). So that the antenna properly senses the time variation of the signal. For an e.m waves of frequency 20 kHz, the wavelength λ is 15 km. Obviously such a long antenna is not possible to construct and operate. There is a need of translating the information contained in our original low frequency base band signal into high frequencies before transmission.

Effective power radiated by an antenna : A linear antenna (length l) show that the power radiated is proportional to \(\frac{l}{\lambda^2}\). For the same antenna length, the power radiated increases with decreasing λ. i.e., increasing frequency. Hence the effective power radiated by a long wave length base band signal would be small.

Question 7.
Explain amplitude modulation.
Answer:
Amplitude modulation (AM.) : In amplitude modulation, the amplitude of carrier wave varies, but frequency and phase remains constant.
Here we can explain AM using a sinusoidal signal as a modulating signal.

Let C(t) = A_c sin ω_c t represent carrier wave
m(t) = A_m sin ω_m t represent modulating signal.
The modulating C_m(t) can be written as
C_m(t) = (A_c + A_m sin ω_mt) sin ω_Ct
C_m(t) = A_c (1 + \(\frac{A_m}{A_c}\) sin ω_mt) sin ω_ct —— (1)
Where ω_m = 2πf_m is the angular frequency of message signai
Note that the modulated signal now contains the message signal.
C_m(t) = A_c sin ω_ct + µ A_c sin ω_mt sin ω_c t —– (2)
Where µ = \(\frac{A_m}{A_c}\) = Modulation index.
To avoid distortion keep . µ ≤ 1
C_m(t) = A_c sin ωt + \(\frac{\mu A_c}{2}\) cos (ω_c – ω_m) t – \(\frac{\mu \mathrm{A}_{\mathrm{c}}}{2}\) cos (ω_c + ω_m)t —— (3)
Here (ω_c – ω_m) and (ω_c + ω_m) are lower side and upper side frequencies.
As long as the broadcast frequencies (camer wave) are sufficiently spaced out the side bands donot over lap.

Question 8.
How can an amplitude modulated wave be generated?
Answer:

The modulated signal A_m sin ω_mt is added to the camer wave signal A_c sin ω_ct to produce the signal x(t).
x(t) = A_m sin ω_m t + A sin ω_ct is passed through a square law device produces an output.
y(t) = B x(t) + Cx² (t)
where B and C are constant.
This signal is passed through a band pass filter. The output of band pass filter produces AM wave, In band pass filter dc and the sinusoidal frequencies ω_m, 2ω_m and 2ω_c rejects and retains the frequency ω_c, (ω_c – ω_m) and (ω_c + ω_m)

Question 9.
How can an amplitude modulated wave be detected ?
Answer:

A block diagram of a typical receiver is shown in figure. Detection is the process of recovering the modulating signal from the modulated carrier wave.
We just saw that the modulated carrier wave contains the frequencies ω_c and ω_c ± ω_m.. In order to obtain message signal m(t) of angular frequency ω_m, a simple method is shown in figure.

The modulated signal is passed through a rectifier produces the output message signal. This message signal is passed through envelope detector (RC circuit).

Textual Exercises

Question 1.
Which of the following frequencies will be suitable for beyond the horizon communication using sky waves ?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Solution:
(b) 10 kHz frequencies cannot be radiated due to large antenna size, 1GHz and 1000 GHz will be generated. So option (b) is correct.

Question 2.
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves
(b) Sky waves
(c) Surface waves
(d) Space waves
Answer:
(d). The frequencies in UHF range normally propagate by means.of space waves. The high frequency space does not bend with ground but are ideal for frequency modulation.

Question 3.
Digital signals
(i) do not provide a continuous set of values
(ii) represent values are discrete steps
(iii) can utilize binary system and
(iv) can utilize decimal as well as binary systems.
Which of the above statements are true ?
(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (i), (ii) and (iii) but not (iv)
(d) All of (i), (ii), (iii) and (iv).
Answer:
(c). A digital signal is a discontinuous function of time in contrast to an analogue signal. The digital signals can be stored as digital data and cannot be transmitted along the telephone lines. Digital signal cannot utilize decimal signals.

Question 4.
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication ? A TV transmitting antenna is 81m tall. How much service area can it cover if the receiving antenna is at the ground level ?
Answer:
No, it is not necessary for line of sight communication, the two antennas may not be at the same height.
Given, height of antenna h = 81 m.
Radius of earth R = 6.4 × 10⁶ m.
Area = πd²; Range, d = \(\sqrt{2 \mathrm{hR}}\)
∴ Service area = π × 2πR = \(\frac{22}{7}\) × 81 × 2 × 6.4 × 10⁶ = 3258.5 km².

Question 5.
A carrier wave of peak voltage 12V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75% ?
Answer:
Given, Peak voltage V₀ = 12V
Modulation index μ = 75% = \(\frac{75}{100}\)
We know that Modulation index (μ)

So, peak voltage of modulating signal V_m = μ × Peak voltage = \(\frac{75}{100}\) × 12 = 9V.

Question 6.
A modulating signal is a square wave, as shown in fig.
The carrier wave is given by c(t) = 2 sin (8πt) volts.
(i) Sketch the amplitude modulated waveform.
(ii) What is the modulation index ?

Answer:
Given, equation of carrier wave c(t) = 2 sin (8πt) —– (1)

(i) According to the diagram
Amplitude of modulating signal A_m = 1V
Amplitude of carrier wave A_c = 2V
T_M = 1s; Ωm = \(\frac{2 \pi}{T_m}\) = \(\frac{2 \pi}{1}\) = 2π rad/s —– (2)
From equation (1)
c(t) = 2 sin 8πt = A_c sin ω_c
From equation (2)
so, ω_c = 4ω_m
Amplitude of modulated wave A = A_m + A_c = 2 + 1 = 3V.
The sketch of amplitude modulated waveform is shown below.

(ii) Modulation index μ = \(\frac{A_m}{A_c}\) = 0.5

Question 7.
For an amplitude modulated wave, the maximum amplitude is found to be 10V while the ‘ minimum amplitude is found to be 2V. Determine the modulation index, μ. What would be the value of m if the minimum amplitude is zero volt ?
Answer:
Given, maximum amplitude A_max = 10V
Minimum amplitude A_min = 2V
Let A_c and A_m be amplitudes of carrier wave and signal wave
A_max = A_c + A_m = 10 —— (1)
and A_min = A_c – A_m = 2 —– (2)
Adding the equations (1) & (2) we get 2 A_c = 12; A_c = 6V; A_m = 10 – 6 = 4V
Modulation index μ = \(\frac{A_m}{A_c}\) = \(\frac{4}{6}\) = \(\frac{2}{3}\).
When the minimum amplitude is zero, then i.e., A_min = 0
A_c + A_m = 10 —– (3)
A_c – A_m = 0 —- (4)
By solving (3) & (4) we get 2 A_m = 10; A_m = 5; A_c = 5
Modulation index μ = \(\frac{A_m}{A_c}\) = \(\frac{5}{5}\) = 1.

Question 8.
Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.
Answer:
Let ω_c be the angular frequency of carrier waves & ω_m be the angular frequency of signal waves.
Let the signal received at the receiving station be e = E₁. cos (ω_c + ω_m) t
Let the instantaneous voltage of carrier wave e_c = E₀ cos ω_c t is available at receiving station.
Multiplying these two signals, we get
e × e_c = E₁E_c cos ω_ct. cos (ω_c + ω_m) t
E = \(\frac{E_1 E_c}{2}\) 2. cos ω_ct. cos (ω_c + ω_m) t. (Let e × e_c = E)
= \(\frac{\mathrm{E}_1 \mathrm{E}_{\mathrm{c}}}{2}\) [cos (ω_c + ω_c + ω_m) t + cos (ω_c + ω_m – ω_c) t]
∵ [2 cos A cos B = cos(A + B) + cos (A – B)]
\(\frac{E_1 E_c}{2}\) = [cos (2 (ω_c + ω_m)t + cos ω_mt]

Now, at the receiving end as the signal passes through filter, it will pass the high frequency (2ω_c + ω_m)t + cos ω_m) but obstruct the frequency ω_m. So we can record the modulating signal \(\frac{\mathrm{E}_1 \mathrm{E}_{\mathrm{c}}}{2}\) cos ω_mt which is a signal of angular frequency to ω_m.