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Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 14th Lesson Nuclei Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 14th Lesson Nuclei

Very Short Answer Questions

Question 1.
What are isotopes and isobars?
Answer:
Isotopes: The nuclei having the same atomic number (Z) but different mass numbers (A) are called isotopes.
E.g.:
\({ }_8^{16} \mathrm{O}\), \({ }_8^{17} \mathrm{O}\), \({ }_8^{18} \mathrm{O}\)

Isobars: The nuclei having the same neutron number (N) but different atomic number (Z) are called isobars.
E.g.: \({ }_6^{14} \mathrm{C}\), \({ }_7^{14} \mathrm{N}\)

Question 2.
What are isotones, and isomers?
Answer:
Isotones : The nuclei having same neutron number (N) but different atomic numbers (Z) are called isotones.
E.g. : \({ }_80^{198} \mathrm{Hg}\), \({ }_79^{197} \mathrm{N}\)

Isomers: Nuclei having the same atomic number (Z) and mass number (A) but with different nuclear properties such as radioactive decay and magnetic moments are called isomers.
E.g. : \({ }_{35}^{80} \mathrm{Br}^{\mathrm{m}}\), \({ }_{35}^{80} \mathrm{Br}^{\mathrm{g}}\)

Question 3.
What is a.m.u. ? What is its equivalent energy ?
Answer:
The mass of \(\frac{1}{12}\)th of the mass of \({ }_6^{12} \mathrm{C}\) atom is called atomic mass unit (a.m.u)
1 a.m.u = \(\frac{1}{12}\) of mass of \({ }_6^{12} \mathrm{C}\) atom = 1.66 × 10^-27 kg
Equivalent energy of a.m.u= 931.5 MeV

Question 4.
What will be the ratio of the radii of two nuclei of mass numbers A₁ and A₂ ?
Answer:
The ratio of the radii of two nuclei of mass numbers A₁ and A₂ will be \(\frac{R_1}{R_2}\) = \(\left[\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right]^{\frac{1}{3}}\) since R = R₀ A^1/3.

Question 5.
Natural radioactive nuclei are mostly nuclei of high mass number why ?
Answer:
Natural radioactivity is displayed by heavy nuclei beyond lead in the periodic table because of relatively low binding energy per nucleon as 7.6 MeV Hence to attain greater stability.

Question 6.
Does the ratio of neutrons to protons in a nucleus increase, decrease or remain the same after the emission of an α – particle ?
Answer:
The ratio of neutrons to protons in a nucleus, increases after the emission of an α – particle.
E.g. : Taking,

Before emission, the ratio of neutrons to protons
= \(\frac{\mathrm{A}-\mathrm{Z}}{\mathrm{Z}}\) = \(\frac{238-92}{92}\) = \(\frac{144}{92}\) = 1.57
After emission, the ratio of neutrons to protons
= \(\frac{234-90}{90}\) = 1.6

Question 7.
A nucleus contains no electrons but can emit them. How?
Answer:
When the nucleus disintegrates and radiates β-rays, it is said to be β-decay. β-particles are nothing but electrons. So the nucleus eject electrons.

Question 8.
What are the units and dimensions of the disintegration constant ?
Answer:
λ = –\(\frac{0.693}{\mathrm{~T}}\)
Units = sec^-1
Dimensions = -1

Question 9.
Why do all electrons emitted during β-decay not have the same energy ?
Answer:
When a neutron is converted into a proton, an electron and neutron are emitted along with it

In β – decay proton remains in the nucleus, but electron and neutron are emitted with constant energy. The energy of neutron is not constant. So, all electrons do not have same energy.

Question 10.
Neutrons are the best projectiles to produce nuclear reactions. Why ?
Answer:
Neutrons are uncharged particles. So they do not deflected by the electric and magnetic fields. Hence Neutrons are considered as best projectiles in nuclear reaction.

Question 11.
Neutrons cannot produce ionization. Why ?
Answer:
Because neutrons are uncharged particles and cannot produce ionization.

Question 12.
What are delayed neutrons ?
Answer:
Neutrons are emitted in the fission products after sometime are called delayed neutrons.

Question 13.
What are thermal neutrons ? What is their importance ?
Answer:
Neutrons having kinetic energies approximately 0.025 eV are called as slow neutrons or thermal neutrons. ²³⁵U undergoes fission only when bombarded with thermal neutrons.

Question 14.
What is the value of neutron multiplication factor in a controlled reaction and in an uncontrolled chain reaction ?
Answer:
In controlled chain reaction K = 1
In uncontrolled chain reaction K > 1

Question 15.
What is the role of controlling rods in a nuclear reactor ?
Answer:
In nuclear reactor controlling rods are used to absorb the neutrons. Cadmium, boron materials are used in the form of rods in reactor. These control the fission rate.

Question 16.
Why are nuclear fusion reactions called thermo nuclear reactions ?
Answer:
Nuclear fusion occurs at very high temperatures. So it is called as thermo nuclear reaction.

Question 17.
Define Becquerel and Curie.
Answer:
Becquerel: 1 disintegration or decay per second is called Becquerel. It is SI unit of activity.

Curie : 3.7 × 10¹⁰ decays per second is called Curie.
1 Curie = 1Ci = \(\frac{3.7 \times 10^{10} \text { decays }}{\text { second }}\) = 3.7 × 10¹⁰ Bq.

Question 18.
What is a chain reaction ?
Answer:
Chain reaction: The neutrons produced in the fission of a nucleus can cause fission in other neighbouring nuclei producing more and more neutrons to continue the fission until the whole fissionable material is disintegrated. This is called chain reaction.

Question 19.
What is the function of moderator in a nuclear reactor ?
Answer:
They are used to slow down the fast moving neutrons produced during the fission process.
e.g.: Heavy water, Berilium.

Question 20.
What is the energy released in the fusion of four protons to form a helium nucleus ?
Answer:
26.7 MeV energy is released.

Short Answer Questions

Question 1.
Why is the density of the nucleus more than that of the atom ? Show that the density of nuclear matter is same for all nuclei.
Answer:

1. Volume of the atom is greater than that of nucleus and it consists of nucleons
2. Since density ∝ \(\frac{1}{\text { volume }}\)
   ∴ Density of the nucleus more than that of the atom.
3. Mass of the nucleus = no.of nucleons (A) × mass of nucleon (m)
   = Am
4. Volume of the nucleus V= \(\frac{4}{3} \pi \mathrm{R}^3\)
   = \(\frac{4}{3} \pi\left(R_0 A^{1 / 3}\right)^3\)
   = \(\frac{4 \pi \mathrm{R}_0^3 \mathrm{~A}}{3}\) = 1.2 × 10^-45m³.A. [∵ R₀ = 1.2 × 10^-15m]
   i. e., the volume of the nucleus is proportional to the mass number A.
5. Density of the nucleus (ρ)
   
6. The above equation represents it clear that the density of the nucleus is independent of the mass number A and is same for all the nuclei.

Question 2.
Write a short note on the discovery of neutron.
Answer:

1. Bothe and Becker found that when beryllium is bombarded with α – particles of energy 5 MeV, which emitted a highly penetrating radiation.
2. The equation for above process can be written as
   
3. The radiations are not effected by electric and magnetic fields.
4. In 1932, James Chadwick, had subjected nitrogen and argon to the beryllium radiation. He interpreted the experimental results by assuming that the radiation is of a new kind of particles which has no charge and its mass is equal to proton. These neutral particles were named as ‘neutrons’. Thus the neutron was discovered.
5. The experimental results can be represented by the following equation.
   

Question 3.
What are the properties of a neutron ?
Answer:

1. Neutron is an uncharged particle and hence it is not deflected by the electric and magnetic fields.
2. It has very high penetrating power and has very low ionization power.
3. Inside the nucleus neutrons appear to be stable. The average life of an isolated neutron is about 1000 seconds. A free neutron is unstable and spontaneously decays into a proton, electron and an antineutrino \((\bar{v})\).
4. If fast neutrons pass through substances like heavy water, paraffin wax, graphite etc., they are slowed down.
5. Neutrons are diffracted by crystals.

Question 4.
What are nuclear forces ? Write their properties.
Answer:
The forces which hold the nucleons together in nucleus are called nuclear forces. Properties of Nuclear forces :

1. Nuclear forces are attractive forces between proton and proton (P – P), proton and neutron (P – N) and neutron and neutron (N – N).
2. Nuclear forces are independent of charge. It was found that force between proton and proton is same as force between neutron and neutron.
3. These forces are short range forces i.e., these forces will act upto a small distance only. Generally the range of nuclear forces is upto few Fermi (10^-15 m).
4. These forces are non central forces, i.e., they do not act along the line joining the two nucleons.
5. These forces are exchange forces. The force between two nucleons is due to exchange of π-mesons.
6. These forces are spin dependent. These forces are strong when the spin of two nucleons are in same direction and they are weak when they are in opposite direction.
7. Nuclear forces are saturated forces i.e., the force between nucleons will extend upto the immediate neighbouring nucleons only.
8. These are the strongest forces in nature. They are nearly 10³⁸ times stronger than gravitational forces and nearly 100 times stronger than Coulombic forces.

Question 5.
For greater stability a nucleus should have greater value of binding energy per nucleon. Why?
Answer:

1. Uranium has a relatively low binding energy per nucleon as 7.6 MeV Hence to attain greater stability Uranium breaks up into intermediate mass nuclei resulting in a phenomenon called nuclear fission.
2. Lighter nuclei such as hydrogen combine to form heavy nucleus to form helium for greater stability, resulting in a phenomenon called nuclear fusion.
3. Iron whose binding energy per nucleon stands maximum at 8.7 MeV is the most stable and will undergo neither fission nor fusion.

Question 6.
Explain α – decay ?
Answer:

1. It is the phenomenon of emission of an a particle from a radioactive nucleus. When a nucleus emits an alpha particle, its mass number decreases by 4 and charge number decreases by 2.
2. In general, alpha decay is represented as
   
   Where Q is the energy released in the decay.
3. Thus the total mass energy of the decay products is less than the mass energy of the original nuclide.
4. The difference between the initial mass energy and the total mass energy of decay products is called disintegration energy (Q) of the process.
5. This can be calculated using Einstein’s mass energy equivalence relation, E = (Δm). c²
   i-e., Q = (m_x – m_y – m_He) c²
   The energy released (Q) is shared by daughter nucleus y and alpha particle.

Question 7.
Explain β – decay ?
Answer:

1.  It is the phenomenon of emission of an electron from a radioactive nucleus.
2. When a parent nucleus emits a β-particle (i.e., an electron), mass number remains same because mass of electron is negligibly low. However, the loss of unit negative charge is equivalent to a gain of unit positive charge. Therefore, atomic number is increased by one.
3. In general, we can write
   
   where Q is the energy released in β-decay.
4. The basic nuclear process underlying p-decay is the conversion of neutron to proton.
   n → P + \(\overline{\mathrm{e}}\) + \(\overline{\mathrm{v}}\)
   While for β+ decay, it is the conversion of proton ino neutron.
   P → n + e⁺ + v
5. The emission of electron in β-decay is accompained by the emission of an anti neutrino \((\bar{v})\) In β, decay instead, a neutrino (v) is generated. Neutrons are neutral particles with very small mass compared to electrons. They have only weak interactions with other particles.

Question 8.
Explain γ – decay ?
Answer:

1. It is the phenomenon of emission of gamma ray photon from a radioactive nucleus.
2. Like an atom, a nucleus has discrete energy levels in the ground state and excited states.
3. When a nucleus in an excited state spontaneously decays to its ground state (or to a lower energy state), a photon is emitted with energy equal to the difference in the two energy levels of the nucleus. This is the so called gamma-decay.
4. The energy (MeV) corresponds to radiation of extremely short wave length, shorter than the hard X-ray region.
5. A Gamma ray is emitted when a or p decay results in a daughter nucleus in an excited
   state.
6. The \(\bar{\beta}\) -decay of ₂₇CO⁶⁰ transforms it into an excited ₂₈Ni⁶⁰ nucleus. This reaches the ground state by emission of γ-rays of energy 1.17 MeV and 1.33 MeV. This is shown in figure.
   

Question 9.
Define half life period and decay constant for a radioactive substance. Deduce the relation between them.
Answer:
Half life period (T) : Time taken for the number of radio active nuclei to disintegrate to half of its original number of nuclei is called Half life period.
Decay constant (λ) : The ratio of the rate of radioactive decay to the number of nuclei present at that instant.
It is a proportional constant and is denoted by ‘λ’.
λ = \(\frac{-\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)}{\mathrm{N}}\)

Relation between half the period and decay constant:

1. The radioactive decay law N = N₀ e^-λt states that the number of radioactive-nuclei in a radioactive sample decreases exponentially with time. Here λ is called decay constant.
2. If N₀ is the number of nuclei at t = 0 and N is the radioactive nuclei at any instant of time’t’.
3. Substituting N = \(\frac{\mathrm{N}_0}{2}\) at t = T in N = N₀e^-λt.
   Where T is half life of the radioactive substance.
4.  \(\frac{\mathrm{N}_0}{2}\) = N₀ e^-λT
   e^λT = 2
   λT = ln2
   T = \(\frac{\ln 2}{\lambda}\) = \(\frac{2.303 \log _{10}^2}{\lambda}\)
   ∴ T = \(\frac{0.693}{\lambda}\)

Question 10.
Define average life of a radioactive substance. Obtain the relation between decay constant and average life.
Answer:
Average life \((\tau)\) : It is equal to the total life time of all the N₀ nuclei divided by the total number of original nuclei N₀. It is denoted by \((\tau)\).

Relation between decay constant and average life :

1. Let N₀ be the radioactive nuclei that are present at t = 0 in the radioactive sample; The no’ of nuclei which decay between t and t + dt is dN.
2. The total life time of these dN nuclei is t dN. The total life time of all the nuclei present initially in the sample = \(\int_0 \mathrm{t} \mathrm{dN}\)
3. Average life time \((\tau)\) is equal to the total life time of all the N₀ nuclei divided by the total number of original nuclei N₀.
4. Average \((\tau)\) = \(\frac{\int \mathrm{tdN}}{\mathrm{N}_0}\)
   But \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = -λN
   dN = -λNdt = N₀e^-λtdt [∵ N = N₀e^-λt]
5. 
   On integrating, we get \(\tau\) = \(\frac{1}{\lambda}\)
   \(\tau\) = \(\frac{T}{0.693}\) [∵ λ = \(\frac{0.693}{\mathrm{~T}}\)]
6. From the above equation ‘the reciprocal of the decay constant gives us the average life of a radioactive sample.

Question 11.
Deduce the relation between half life and average life of a radioactive substance.
Answer:
Relation between half life (T) and average life (\(\tau\)) :

1. We know, the radioactive decay law, N = N₀ e^-λt —– (1)
2. Consider, ‘N₀‘ is the number of nuclei present at t = 0 and after time T, only \(\frac{\mathrm{N}_0}{2}\) are left and after a time ‘2T’, only \(\frac{\mathrm{N}_0}{4}\) remain and soon.
3. Substituting N = \(\frac{\mathrm{N}_0}{2}\) at t = T in eqn. (1) then
   \(\frac{\mathrm{N}_0}{2}\) = N₀ e^-λT ⇒ \(\frac{1}{2}\) = \(\frac{1}{\mathrm{e}^{\lambda \mathrm{T}}}\) ⇒ e^λT = 2
   Taking log_e on both sides, we get
   λT = \(\log _{\mathrm{e}}^2\) = 2.303 \(\log _{\mathrm{e}}^2\) = 0.693
   ∴ T = \(\frac{0.693}{\lambda}\) —— (2)
4. Average life \(\tau\) = \(\frac{\int \mathrm{tdN}}{\mathrm{No}}\)
5. But –\(\frac{\mathrm{dN}}{\mathrm{dt}}\) = λN dN = -λ.N₀ e^-λt dt [∵ from eqn. (1)]
6. 
   on integrating, we get \(\tau\) = \(\frac{1}{\lambda}\) —– (3)
7. From equs (2) and (3) we get \(\tau\) = \(\frac{\mathrm{T}}{0.693}\)
   This is the relation between average life and half life of radioactive substance.

Question 12.
What is nuclear fission ? Give an example to illustrate it.
Answer:
Nuclear fission : The process of dividing a heavy nucleus into two or more smaller and stable nuclei due to nuclear reaction is called nuclear fission.
Ex : The fission reaction is

Where Q is the energy released.
Q = (Total mass of reactants – Total mass of product) C²
= [(Mass of \({ }_{92}^{235} \mathrm{U}\) + Mass of \({ }_0^1 n\)) – (Mass of \({ }_{56}^{141} \mathrm{Ba}\) + Mass of \({ }_{36}^{92} \mathrm{Kr}\) + Mass of three neutrons)]C²
= (235.043933 – 140.9177 – 91.895400 – 2 × 1.008665) amu × C².
= 0.2135 × 931.5 MeV = 198.9 MeV = 200 MeV

Question 13.
What is nuclear fusion ? Write the conditions for nuclear fusion to occur.
Answer:
Nuclear fusion : The process of combining lighter nuclei to produce a larger nucleus is known as nuclear fusion.
E.g : Hydrogen nuclei (₁H¹) are fused together to form heavy Helium (₂He⁴) along with 25.71 MeV energy released.

Conditions for nuclear fusion :

1. Nuclear fusion occurs at very high temperatures such as 10⁷ kelvin and very high pressures. These are obtained under the explosion of an atom bomb.
2. Higher density is also desirable so that collisions between light nuclei occur quite frequently.

Question 14.
Distinguish between nuclear fission and nuclear fusion.
Answer:
Nuclear fission

1. In this process heavy nucleus is divided into two fragments along with few neutrons.
2. These reactions will takes place even at room temperature.
3. To start fission atleast one thermal neutron from out side is compulsory.
4. Energy released per unit mass of participants is less.
5. In this process neutrons are liberated.
6. This reaction can be controlled.
   Ex: Nuclear reactor.
7. Atom bomb works on principle of fission reaction.
8. The energy released in fission çan be used for peaceful purpose.
   Ex : Nuclear reactor and Atomic power stations.

Nuclear fusion

1. In this process lighter nuclei will join together to produce heavy nucleus.
2. These reactions will takes place at very high temperature such as Kelwin.
3. No necessary of external neutrons.
4. Energy released per unit mass of participants is high. Nearly seven times more than fission reaction.
5. In this process positrons are liberated.
6. There is no control on fusion reaction.
7. Hydrogen bomb works on the principle of fusion reaction.
8. The energy released in fusion cannot be used for peaceful purpose.

Question 15.
Explain the terms tchain reaction’ and multiplication factor’. How is a chain reaction sustained?
Answer:
Chain reaction : In nuclear fission nearly three neutrons are produced when one uranium atom is destoryed. If they again participate in fission reaction nine neutrons are produced. In next generation the neutrons becomes 27. In this process the number of neutrons increases in geometric progression and the whole uranium is destroyed in few seconds. This type of self sustained fission reaction is called chain reaction.

Neutron multiplication factor (K) : Neutron multiplilcation factor is defined as the ratio of number of neutrons produced in one generation to the number of neutrons in previous generation.

Neutron multiplication factor is useful to understand the nature of nuclear reactions in a nuclear reactor.

To sustained chain reaction:
1. Neutron multiplication factor K ≥ 1.

Long Answer Questions

Question 1.
Define mass defect and binding energy. How does binding energy per nucleon vary with mass number ? What is its significance ?
Answer:

1. Mass defect (ΔM) : The difference in mass of a nucleus and its constituents is called the mass defect. The nuclear mass M is always less than the total mass, Σm, of its constituents.
   Mass defect, (ΔM) = [Zm_p + (A – Z)m_n – M]
2. Binding energy: The energy required to break the nucleus into its constituent nucleons is called the binding energy.
   Binding Energy, (E_b) = ΔMC² = [Zm_p + (A – Z)m_n – M] 931.5 MeV
   Nuclear binding energy is an indication of the stability of the nucleus.
   Nuclear binding energy per nucleon E_bn = \(\frac{\mathrm{E}_{\mathrm{b}}}{\mathrm{A}}\).
3. The following graph represents how the binding energy per nucleon varies with the mass number A.
   
4. From the graph that the binding energy is highest in the range 28 < A < 138. The binding energy of these nuclei is very close to 8.7 MeV.
5. With the increase in the mass number the binding energy per nucleon decreases and consequently for the heavy nuclei like Uranium it is 7.6 MeV
6. In the region of smaller mass numbers, the binding energy per nucleon curve shows the characteristic minima and maxima.
7. Minima are associated with nuclei containing an odd number of protons and neutrons such as \({ }_3^6 \mathrm{Li}\), \({ }_5^{10} \mathrm{~B}\), \({ }_7^{14} \mathrm{~N}\) and the maxima are associated with nuclei having an even number of protons and neutrons such as \({ }_2^4 \mathrm{He}\), \({ }_6^{12} \mathrm{C}\), \({ }_8^{16} \mathrm{O}\).
   Significance :
8. The curve explains the relationship between binding energy per nucleon and stability of the nuclei.
9. Uranium has a relatively low binding energy per nucleon as 7.6 MeV. Hence to attain greater stability Uranium breaks up into intermediate mass nuclei resulting in a phenomenon called fission.
10. On the other hand light nuclei such as hydrogen combine to form heavy nucleus to form helium for greater stability, resulting in a phenomenon called fusion.
11. Iron is the most stable having binding energy per nucleon 8.7 MeV and it neither undergoes fission per fusion.

Question 2.
What is radioactivity ? State the law of radioactive decay. Show that radioactive decay is exponential in nature.
Answer:

1. Radioactivity : .The nuclei of certain elements disintegrate spontaneously by emitting alpha (α), beta (β) and gamma (γ) rays. This phenomenon is called Radioactivity or Natural radioactivity.
2. Law of radioactivity decay: The rate of radioactive decay \(\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)\) (or) the number of nuclei decaying per unit time at any instant, is directly proportional to the number of nuclei (N) present at that instant is called law of radioactivity decay’.
3. Radioactive decay is exponential in nature : Consider a radioactive substance. Let the number of nuclei present in the sample at t = 0, be N₀ and let N be the radioactive nuclei remain at an instant t.
   \(\frac{\mathrm{dN}}{\mathrm{dt}}\) ∝ N ⇒ \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = -λN
   dN = -λNdt ——— (1)
   The proportionality constant λ is called decay constant or disintegration constant. The negative sign indicates the decrease in the number of nuclei.
4. From eq.(1) \(\frac{\mathrm{dN}}{\mathrm{N}}\) = -λ dt ——- (2)
5. Integrating on both sides
   \(\int \frac{\mathrm{dN}}{\mathrm{N}}\) = \(-\lambda \int \mathrm{dt}\)
   In N = -λt + C ——- (3)
   Where C = Integration constant.
6. At t = O; N = N₀. Substituting in eq. (3), we get in ln N₀ = C
   ∴ ln N – ln N₀ = – λt
   ln \(\left(\frac{\mathrm{N}}{\mathrm{N}_0}\right)\) = – λt
   ∴ N = ₀e^-λt
   The above equation represents radioactive decay law.
7. It states that the number of radioactive nuclei in a radioactive sample decreases exponentially with time.

Question 3.
Explain the principle and working of a nuclear reactor with the help of a labelled diagram. (A.P.Mar.’19,’16, ’15 & T.S. Mar. ‘15) (Mar. ’14)
Answer:
Principle : A nuclear reactor works on the principle of achieving controlled chain reaction in natural Uranium ²³⁸U enriched with ²³⁵U, consequently generating large amounts of heat.
A nucleàr reactor consists of
(1) Fuel
(2) Moderator
(3) Control rods
(4) Radiation shielding
(5) Coolant.

1. Fuel and clad : In reactor the nuclear fuel is fabricated in the form of thin and long cylindrical rods. These group of rods treated as a fuel assembly. These rods are surrounded by coolant, which is used to transfer of heat produced in them. A part of the nuclear reactor which use to store the nuclear fuel is called the core of the reactor. Natural uranium, enriched uranium, plutonium and uranium – 233 are used as nuclear fuels.

2. Moderator : The average energy of neutrons released in fission process is 2 MeV They are used to slow down the velocity of neutrons. Heavy water or graphite are used as moderating materials in reactor.

3. Control Rods : These are used to control the fission rate in reactor by absorbing the neutrons. Cadmium and boron are used as controlling the neutrons, in the form of rods.

4. Shielding : During fission reaction beta and gamma rays are emitted in addition to neutrons. Suitable shielding such as steel, lead, concrete etc are provided around the reactor to absorb and reduce the intensity of radiations to such low levels that do not harm the operating personnel.

5. Coolant: The heat generated in fuel elements is removed by using a suitable coolant to flow around them. The coolants used are water at high pressures, molten sodium etc.

Working: Uranium fuel rods are placed in the aluminium cylinders. The graphite moderator is placed in between the fuel cylinders. To control the number of neutrons, a number of control rods of cadmium or beryllium or boron are placed in the holes of graphite block: When a few ²³⁵U nuclei undergo fission fast neutrons are liberated. These neutrons pass through the surrounding graphite moderator and loose their energy to become thermal neutrons.

These thermal neutrons are captured by ²³⁵U. The heat generated here is used for heating suitable coolants which in turn heat water and produce steam. This steam is made to rotate steam turbine and there by drive a generator of production for electric power.

Question 4.
Explain the source of stellar energy. Explain the carbon – nitrogen cycle, proton – proton cycle occurring in stars.
Answer:
Scientists proposed two types of cyclic processes for the sources of energy in the sun and stars. The first is known as carbon-nitrogen cycle and the second is proton-proton cycle.

1. Carbon-Nitrogen Cycle: According to Bethe carbon – nitrogen cycle is mainly responsible for the production of solar energy. This cycle consists of a chain of nuclear reactions in which hydrogen is converted into Helium, with the help of Carbon and Nitrogen as catalysts. The nuclear reactions are as given below.

2. Proton – Proton Cycle: A star is formed by the condensation of a large amount of matter at a point in space. Its temperature rises to 2,00,000°C as the matter contracts under the influence of gravitational attraction. At this temperature the thermal energy of the protons is sufficient to form a deuteron and a positron. The deuteron then combines with another proton to form lighter nuclei of helium \({ }_2^3 \mathrm{He}\). Two such helium nuclei combine to form a helium nucleus \({ }_2^4 \mathrm{He}\) and two protons releasing a total amount of energy 25.71 MeV The nuclear fusion reactions are given below.

The net result is

Problems

Question 1.
Show that the density of a nucleus does not depend upon its mass number (density is independent of mass)
Solution:
Density of nucleus matter =

No. of nucleons (A) × mass of nucleons (m)
Volume of nucleus V = \(\frac{4}{3} \pi R^3\)
But R = R₀A^1/3
∴ V = \(\frac{4}{3} \pi \mathrm{R}_0^3 \mathrm{~A}\)
∴ Density of nucleus matter = \(\frac{\mathrm{Am}}{\frac{4}{3} \pi \mathrm{R}_0^3 \mathrm{~A}}\) = \(\frac{3 \mathrm{~m}}{4 \pi \mathrm{R}_0^3}\)
∴ Density of nucleus is independent of mass.

Question 2.
Compare the radii of the nuclei of mass numbers 27 and 64.
Solution:
A₁ = 27; A₂ = 64
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = \(\left[\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right]^{1 / 3}\) [∵ R = R₀A^1/3]
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = \(\left[\frac{27}{64}\right]^{\frac{1}{3}}\) = \(\frac{3}{4}\)
∴ R₁ : R₂ = 3 : 4

Question 3.
The radius of the oxygen nucleus \({ }_8^{16} \mathrm{O}\) is 2.8 × 10^-15m. Find the radius of lead nucleus \({ }_{82}^{205} \mathrm{~Pb}\).
Solution:
R₀ = 2.8 × 10^-15 m; A₀ = 16
A_Pb = 205; R_Pb = ?
\(\frac{R_{P b}}{R_0}\) = \(\left[\frac{A_{P b}}{A_0}\right]^{1 / 3}\) = \(\left[\frac{205}{16}\right]^{1 / 3}\)
[∵ R = R₀A^1/3]
\(\frac{R_{P b}}{2.8 \times 10^{-15}}\) = (12.82)^1/3 = 2.34
R_Pb = (2.34) × (2.8 × 10^-15)
= 6.55 × 10^-15m

Question 4.
Find the binding energy of \({ }_{26}^{56} \mathrm{Fe}\). Atomic mass of Fe is 55.9349u and that of Hydrogen is 1.00783u and mass of neutron is 1.00876u.
Solution:
Mass of hydrogen atom
m_p = 1.00876u; m_n = 1.00867u
Z = 26; A = 56
Mass of Iron atom M = 55.9349u

i) Mass defect Δm
= [Zm_p + (A – Z) m_n – M]
= [26 × 1.00876 + (56 – 26) (1.00867) – 55.9349] u
∴ Δm = 0.55296u

ii) BE of nucleus = ΔMC²
= ΔM × 931.5 MeV
= 0.55296 (931.5) MeV
= 515.08 MeV

Question 5.
How much energy is required to separate the typical middle mass nucleus \({ }_{50}^{120} \mathrm{Sn}\) into its constituent nucleons? (Mass of \({ }_{50}^{120} \mathrm{Sn}\) = 119.902199u, and mass of neutron = 1.008665u)
Solution:
m_p = 1.007825u
m_n = 1.008665u
For Sn, Z = 50;

A = 120; M = 119.902199u

i) Mass defect Δm
= [Zm_p + (A – Z)m_n – M]u
= 50 (1.007825) + (120 – 50) [(1.008665) – 119.902 199]
= [150 × 1.007825 + 70 × 1.008665 – 119.902199]u
= [50.39125 + 70.60655 – 119.902199]u
ΔM = [120.9978 – 119.902199]
= 1095601u

ii) Energy required to šeparate the nucleons = B.E of the nucleus
BE = ΔMc² = ΔM × 931.5MeV
= 1.095601 × 931.5 MeV
= 1020.5 MeV

Question 6.
Calculate the binding energy of an α-particle. Given that mass of proton = 1.0073 u, mass of neutron = 1.0087u, and mass of α- particle = 4.0015u.
Solution:
For ₂He⁴, A = 4, Z = 2, m_p = 1.0073u
m_n = 1.0087u, m_n = 4.0015u

i) ΔM
= [Zm_p + (A – Z)m_n – M]
= [2(1.0073) + (4 – 2) (1.0087) – 4.00260]
= [2 × 1.0073 + 2 × 1.0087 – 4.00260]
= (2.0146 + 2.0174) – 4.0015
ΔM = [4.032 – 4.0015] = 0.0305 u

ii) BE = ΔM × c² = ΔM × 931.5 MeV
= 0.0305 × 931.5
∴ BE = 28.41 MeV

Question 7.
Find the energy required to split \({ }_8^{16} \mathrm{O}\) nucleus into four α-particles. The mass of an α-particle is 4.002603u and that of oxygen is 15.994915u.
Solution:
The energy required to split O =
[Total mass of the products – Total mass of the reactants] c²
Mass of four \({ }_2^4 \mathrm{He}\) – Mass of \({ }_8^{16} \mathrm{O}\)] × c²
= [(4 × 4.002603) – 15.994915] u × c²
= [16.010412 – 15.994915] u × c²
= (0.015497) 931.5 MeV
= 14.43 MeV

Question 8.
Calculate the binding energy per nucleon of \({ }_{17}^{35} \mathrm{Cl}\) nucleus. Given that mass of \({ }_{17}^{35} \mathrm{Cl}\) nucleus = 34.98000 u, mass of proton = 1.007825u, mass of neutron = 1.008665u and 1 is equivalent to 931 MeV.
Solution:
For \({ }_{17}^{36} \mathrm{Cl}\), A = 35, Z = 17;
m_p = 1.007825 u
m_n = 1.008665 u,
M = 34.98u

(i) ΔM = [Zm_p + (A – Z) m_n – M]
= [17 × 1.007825 + (35 – 17)(1.008665) – 34.98]
= [17.13303 + 18.15597 – 34.98]
ΔM = [35.289 – 34.98]
= 0.3089 u

(ii) BE = ΔMc²
= 0.3089 × 931 MeV = 287.5859 MeV
∴ BE per nucleon
= \(\frac{B \cdot E}{A}\) = \(\frac{287.5859}{35}\) = 8.21 MeV

Question 9.
Calculate the binding energy per nucleon of \({ }_{20}^{40} \mathrm{Ca}\). Given that mass of \({ }_{20}^{40} \mathrm{Ca}\) nucleus = 39.962589u, mass of a proton = 1.007825 u,; mass of Neutron = 1.008665u and 1u is equivalent to 931 MeV.
Solution:
For \({ }_{20}^{40} \mathrm{C}\), A = 40, Z = 20; m_p = 1.007825 u
m_n = 1.008665 u; M = 39.962589 u

(i) ΔM = [Zm_p + (A – Z) m_n – M]
= [(20) (1.007825) + (40 – 20)(1 .008665) – 39.962589]
[(20 × 1.007825) + (20 × 1.008665) – 39.962589]
= [40.3298 – 39.962589] = 0.3672 u

(ii) BE = ΔMc² = 0.3672 × 931 MeV
= 341.86 MeV
B.E ‘341.86

(iii) B.E per nucleon = \(\frac{B . E}{A}\) = \(\frac{341.86}{40}\)
= 8.547 MeV

Question 10.
Calculate
(i) mass defect,
(ii) binding energy and
(iii) the binding energy per nucleon of \({ }_6^{12} \mathrm{C}\) nucleus. Nuclear mass of \({ }_6^{12} \mathrm{C}\) = 12.000000 u; mass of proton = 1.007825 u and mass of neutron = 1.008665 u.
Solution:
For \({ }_6^{12} \mathrm{C}\), A = 12; Z = 6; m_p = 1.007825u
m_n = 1.008665u; M = 12.000000u

(i) ΔM = [Zm_p + (A – Z) m_n – M]
= [6(1.007825) + (12 – 6)(1.008665) – 12,00]
= [6 × 1.007825 + 6 × 1.008665 – 12]
= [6.04695 + 6.05199 – 12.00]
ΔM = [12.09894 – 12.00] = 0.098944

(ii) BE = ΔM × c² = 0.09894 × 931.5 MeV
= 92.16 MeV

(iii) BE per nucleon
= \(\frac{B . E}{A}\) = \(\frac{92.16}{12}\) = 7.68 MeV

Question 11.
The binding energies per nucleon for deuterium and helium are 1.1 MeV and 7.0 MeV respectively. What energy in joules will be liberated when 10⁶ deuterons take part in-the reaction.
Solution:
\(\left[\frac{B \cdot E}{A}\right]_D\) = 1.1 MeV; \(\left[\frac{B . E}{A}\right]_{H e}\) = 7.0 MeV
For deuterium \(\left({ }_1^2 \mathrm{H}\right)\),
A = 2
For He \(\left({ }_2^4 \mathrm{He}\right)\), A = 4
\(\left[\frac{B \cdot E}{2}\right]_D\) = 1.1 MeV ⇒ [B.E.]_D
= 2 × 1.1 MeV = 2.2 MeV
\(\left[\frac{B . E}{4}\right]_{\mathrm{He}}\) = 70 MeV ⇒ [B.E.]_He
= 4 × 7.0 MeV = 28 MeV
We know ₁H² + ₁H² → ₂He⁴
Energy released = B.E of 10⁶ deuterons – B.E of \(\frac{1}{2}\) × 10⁶ Helium atoms
B.E = 2.2 × 10⁶ × \(\frac{1}{2}\) × 10⁶ × 28
= 10⁶(2.2 – 14)
= -11.8 × 10⁶ MeV
= -11.8 × 10⁶ × 1.6 × 10^-13J
= -18.88 × 10^-7 J
(- ve sign indicates that energy is released)
∴ Energy released = 18.88 × 10^-7 J

Question 12.
Bombardment of lithium with protons gives rise to the following reaction :
\({ }_3^7 \mathrm{Li}\) + \({ }_1^1 \mathrm{H}\) → 2 \(\left[{ }_2^4 \mathrm{He}\right]\) + Q. Find the Q-value of the reaction. The atomic masses of lithium, proton and helium are 7.016u, 1.0084 and 4.004u respectively.
Solution:
Mass of Lithium = 7.0 16 u
m_p = 1.008 u
Mass of Helium = 4004 u;
u = 931.5 MeV
Q = [Total mass of the reactants – Total mass of the products] c²
= [Mass of Lithium + m_p – (2 × mass of Helium)] × 931.5 MeV
= [7.016+ 1.008 – 2(4.004)] × 931.5MeV
= [18.024 – 8.008] × 931.5 MeV
∴ Energy Q = 0.016 × 931.5
= 14.904 MeV

Question 13.
The half life radium is 1600 years. How much time does lg of radium take to reduce to 0.125 g. (T.S. Mar. ’16)
Solution:
Half life of radium = 1600 years
Initial mass = 1g
Final mass 0.125 g = \(\frac{1}{8}\) g
The quantity remaining after ‘n’ half lifes is \(\frac{1}{2^{\mathrm{n}}}\) of the initial quantity.
In this problem,

∴ Time taken = ’n’ half-lifes = 3 × 1600
= 4,800 years

Question 14.
Plutonium decays with a half – life of 24,000 years. If plutonium is stored for 72,000 years, what fraction of it remains ?
Solution:
Half-life of plutonium = 24,000 years
The duration of the time = 72,000 years
Initial mass = Mg
Final mass = mg
Number of half lifes (n)

Fraction of plutonium that remains

Question 15.
A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.
Solution:
Fraction of substance decays

= \(\frac{1}{2^n}\) = \(\frac{1}{32}\) = \(\frac{1}{2^5}\)
∴ n = 5
Duration of time = 25 days

Question 16.
The half-life period of a radioactive substance is 20 days. What is the time taken for 7/8^th of its original mass to disintegrate?
Solution:
Half life period = 20 days
In this problem,

∴ Time taken to disintegrate
= n × Half life time
= 3 × 20 = 60 days

Question 17.
How many disintegrations per second will Occur in one gram of \({ }_{92}^{238} \mathrm{U}\), if its half-life against α-decay is 1.42 × 10^-17s?
Solution:
T = 1.42 × 10¹⁷ sec
Decay constant (λ) = \(\frac{0.693}{\mathrm{~T}}\) = \(\frac{0.693}{1.42 \times 10^7}\)
No. of disintegration (n) in 1 gm
= \(\frac{1}{238}\) × 6.023 × 10²³
∴ Activity A = λN
= \(\frac{0.693}{1.42 \times 10^{17}}\) × \(\frac{1}{238}\) × 6.023 × 10²³
= 1.235 × 10⁴ disintegrations / sec

Question 18.
The half-life of a radioactive substance is 100 years. Calculate in how many years the activity will decay to 1/10^th of its initial value.
Solution:
T = 100 years
λ = \(\frac{0.693}{\mathrm{~T}}\) = \(\frac{0.693}{\mathrm{~T}}\) = 0.693 × 10^-2 years
N = N₀e^-λt ⇒ e^-λt = \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\frac{1}{10}\)
e^λt = 10 ⇒ λt = \(\log _e^{10}\) = 2.303 × \(\log _{10}^{10}\)
t = \(\frac{2.303 \times 1}{0.693 \times 10^{-2}}\) = 3.323 × 10²
= 332.3years

Question 19.
One gram of radium is reduced by 2 milligram in 5 years by a-decay. Calculate the half-life of radium.
Solution:
Initial (original) mass (N₀) = 1 gram
Reduced mass = 2 mg = 0.002 grams
Final mass (N) = 1 – 0.002 = 0.998 grams
t = 5 years
e^-λt = \(\frac{\mathrm{N}}{\mathrm{N}_0}\) ⇒ e^λt = \(\frac{\mathrm{N}_0}{\mathrm{~N}}\) ⇒ λt = \(\log _{\mathrm{e}}\left[\frac{\mathrm{N}_0}{\mathrm{~N}}\right]\)
λt = 2.303 log\(\left[\frac{\mathrm{N}_0}{\mathrm{~N}}\right]\)
λt = 2.303 log \(\left[\frac{1}{0.998}\right]\)
= 2.303 × 0.000868
= 0.001999
λ = \(\frac{0.001999}{5}\) = 0.0003998
T = \(\frac{0.693}{\lambda}\) = \(\frac{0.693}{0.0003998}\) = 1733.3 years

Question 20.
The half-life of a radioactive substance is 5000 years. In how many years, its activity will decay to 0.2 times a its initial value ? Given log₁₀5 = 0.6990.
Solution:
T = 5000 years; t = ?
Activity, A = Nλ = 0.2 times initial value
Initial activity A₀ = N₀λ
In radioactivity,
N = N₀e^-λt ⇒ \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = e^-λt = 0.2
Put 0.2 = \(\frac{1}{5}\) ; \(\frac{1}{5}\) = e^-λt (or) -λt = \(-\log _e^5\)
(or) t = \(\log _{\mathrm{e}}^5 \frac{5}{\lambda}\)
Radioactive decay constant A = \(\frac{\log _{\mathrm{e}} 2}{\mathrm{~T}}\)
= \(\frac{2.303 \log _{10}^2}{5000}\) = \(\frac{0.693}{5000}\)
Time taken to decay
t = \(\frac{\log _{\mathrm{e}}^5}{\lambda}\) × 5000
∴ t = \(\frac{2.303 \times 0.6990 \times 5000}{0.693}\) = \(\frac{8049}{0.693}\)
= 1.161 × 10⁴ years

Question 21.
An explosion of atomic bomb releases an energy of 7.6 × 10¹³J. If 200 MeV energy is released of fission of one ²³⁵U atom calculate
(i) the number of uranium atoms undergoing fission,
(ii) the mass of uranium used in the bomb.
Solution:
Energy released (E) = 7.6 × 10¹³ J
Energy released on fissions (E) = 200 MeV
= 200 × 10⁶ × 1.6 × 10^-19 J
i) No. of fissions (n) = \(\frac{\mathrm{E}^{\prime}}{\mathrm{E}}\)
= \(\frac{7.6 \times 10^{13}}{200 \times 10^6 \times 1.6 \times 10^{-19}}\)
∴ n = 2.375 × 10²⁴ atoms

But Avogadro number N = 6.023 × 10²³ atoms
Mass of uranium (m) = \(\frac{\mathrm{n} \times 235}{\mathrm{~N}}\)
= \(\frac{2.375 \times 10^{24}}{6.023 \times 10^{23}}\)
= 926.66 g

Question 22.
If one microgram of \({ }_{92}^{235} \mathrm{U}\) is completely destroyed in an atom bomb, how much energy will be released ? (T.S. Mar. ’19)
Solution:
m = 1μg = 1 × 10^-6 g = 1 × 10^-6 × 10^-3 kg
= 10^-9 kg
c = 3 × 10⁸ m/s
E = mc² = 1 × 10^-9 × 9 × 10⁶ = 9 × 10⁷ J

Question 23.
Calculate the energy released by fission from 2g of \({ }_{92}^{235} \mathrm{U}\) in kWh. Given that the energy released per fission is 200 MeV.
Solution:
Mass of uranium = 2g
Energy per fission = 200 MeV
No. of atoms in 2g, n = \(\frac{2 \times 6.023 \times 10^{23}}{235}\)
Total energy released (E’) = nE
= \(\frac{2 \times 6.023 \times 10^{23}}{235}\) × 200 × 10⁶ × 1.6 × 10^-19J
= \(\frac{2 \times 602.3}{235}\) × 32 × 10⁹
= \(\frac{1640.3 \times 10^8}{36 \times 10^5}\)
∴ E’ = 45.56 × 10³ kWh
= 4.556 × 10⁴ kWh

Question 24.
200 MeV energy is released when one nucleus of ²³⁵U undergoes fission. Find the number of fissions per second required for producing a power of 1 megawatt.
Solution:
E = 200 MeV
P = 1 × 10⁶W
P = \(\frac{\mathrm{nE}}{\mathrm{t}}\) ⇒ \(\frac{\mathrm{n}}{\mathrm{t}}\) = \(\frac{\mathrm{P}}{\mathrm{E}}\) = \(\frac{10^6}{200 \times 10^6 \times 1.6 \times 10^{-19}}\)
= \(\frac{1}{32}\) × 10¹⁸
= 3.125 × 10⁶

Question 25.
How much ²³⁵U is consumed in a day in an atomic power house operating at 400 MW, provided the whole of mass ²³⁵U is converted into energy ?
Solution:
P = 400 MW = 400 × 106 W, c = 3 × 10⁸ m/s
t = 24 hours = 24 × 60 × 60 sec
E = mc²
\(\frac{\mathrm{Pt}}{\mathrm{c}^2}\) = m [∵ P = \(\frac{E}{t}\)]
m = \(\frac{400 \times 10^{-6} \times 24 \times 60 \times 60}{9 \times 10^6}\)
= 384 × 10^-6 kg
∴ Mass required = 384 × 10^-6 × 10³ g = 0.384 g

Textual Exercises

Question 1.
a) Two stable isotopes of lithium \({ }_3^6 \mathrm{Li}\) and \({ }_3^7 \mathrm{Li}\) have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.
b) Boron has two stable isotopes, \({ }_5^{10} \mathrm{~B}\) and \({ }_5^{11} \mathrm{~B}\). Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811u. Find the abundances of \({ }_5^{10} B\) and \({ }_5^{11} \mathrm{~B}\).
Solution:
a) Atomic weight
= Weighted average of the isotopes.
= \(\frac{6.01512 \times 7.5+7.01600 \times 92.5}{(7.5+92.5)}\)
= \(\frac{45.1134+648.98}{100}\)
= 6.9414

b) Let relative abundance of ₅B¹⁰ be x%
∴ Relative abundance ₅B¹¹ = (100 – x) %
Proceeding as above
10.811 = \(\frac{10.01294 x+i 1.00931 \times(100-x)}{100}\)
x = 19.9% and (100 – x) = 30.1%

Question 2.
The three stable isotopes of neon : \({ }_{10}^{20} \mathrm{Ne}\), \({ }_{10}^{21} \mathrm{Ne}\) and \({ }_{10}^{22} \mathrm{Ne}\) have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99u, 20.99 u and 21.99u, respectively. Obtain the average atomic mass of Neon.
Solution:
The masses of three isotopes are 19.99u, 20.99 u, 21.99u
Their relative abundances are 90.51%, 10.27% and 9.22%
∴ Average atomic mass of Neon is
m = \(\frac{90.51 \times 19.99+0.27 \times 20.99+9.22 \times 21.99}{(90.51+0.27+9.22)}\)
= \(\frac{1809.29+5.67+202.75}{100}\) = \(\frac{2017.7}{100}\) = 20.17u

Question 3.
Obtain the binding energy (in MeV. of a nitrogen nucleus \(\left({ }_7^{14} \mathrm{~N}\right)\), given m \(\left({ }_7^{14} \mathrm{~N}\right)\) = 14.00307 u.
Solution:
₇N¹⁴ Nucleius contains 7 protons and 7 neutrons
Mass defect (ΔM) = 7m_H + 7m_n – m_N
= 7 × 1.00783 + 7 × 1.00867 – 14.00307
= 7.05481 + 7.06069 – 14.00307
= 0.11243μ
Binding energy = 0.11243 × 931 MeV
= 104.67 MeV

Question 4.
Obtain the binding energy of the nuclei \({ }_{26}^{56} \mathrm{~F}\) and \({ }_{83}^{209} \mathrm{Bi}\) in units of MeV from the following data : m\(\left({ }_{26}^{56} \mathrm{Fe}\right)\) = 55.934939 u, m\(\left({ }_{83}^{209} \mathrm{Bi}\right)\) = 208.980388 u
Solution:
(i) ₂₆F⁵⁶ nucleus contains 26 protons and (56 – 26) = 30 neutrons
Mass of 26 protons = 26 × 1.007825
= 26.26345 a.m.u
Mass of 30 neutrons = 30 × 1.008665
= 30.25995 amu
Total mass of 56 nucleons
= 56.46340 a.m.u
Mass of ₂₆Fe⁵⁶ Nucleus
= 55.934939 a.m.u
Mass defect Δm = 56.46340 – 55.934939
= 0.528461 a.m.u
Total binding energy = 0.524861 × 931.5 MeV
= 492.26 MeV
Average B.E per nucleon = \(\frac{492.26}{56}\)
= 8.790 MeV

(ii) ₈₃Bi²⁰⁹ nucleus contains 83 protons and (209 – 83) = 126 neutrons
Mass of 83 protons = 83 × 1.007825
= 83.649475 a.m.u
Mass of 126 Neutrons = 126 × 1.008665
= 127.09170 a.m.u
Total mass of nucleons = 210.741260 a.m.u
Mass of ₈₃Bi²⁰⁹ nucleus = 208.986388 a.m.u
Mass defect Δm = 210.741260 – 208.980388
= 1.760872
Total B.E = 1.760872 × 931.5 MeV
= 1640.26 MeV
Average B.E. per nucleon = \(\frac{1640.26}{209}\)
= 7.848 MeV
Hence ₂₆Fe⁵⁶ has greater B.E per nucleon than ₈₃Bi²⁰⁹

Question 5.
A given coin has a mass of 3.0gi Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 29Cu atoms (of mass 62.92960 u.
Solution:
Number of atoms in 3g coin =
\(\frac{6.023 \times 10^{23} \times 3}{63}\)
= 2.868 × 10²²
Each atom of copper contains 29 protons and 34 neutrons. Therefore, mass defect of each atom= [29 × 1.00783 + 34 × 1.00867] – 62.92960 = 0.59225 u
Total mass defect for all the atoms
= 0.59225 × 2.868 × 10²² u
ΔM = 1.6985 × 10²²u
As, 1u = 931 MeV
Nuclear energy required
= 1.6985 × 10²² × 931 MeV
= 1.58 × 10²⁵ MeV

Question 6.
Write nuclear reaction equations for
i. α-decay of \({ }_{88}^{226} \mathrm{Ra}\)
ii. α-decay of \({ }_{94}^{242} \mathrm{Pu}\)
iii. β-decay of \({ }_{15}^{32} \mathrm{P}\)
iv. β-decay of \({ }_{83}^{210} \mathrm{Bi}\)
v. β⁺-decay of \({ }_6^{11} \mathrm{C}\)
vi. β⁺-decay of \({ }_{43}^{97} \mathrm{Tc}\)
Solution:

Question 7.
A radio active isotope has a half-life of T years. How long will it take the activity to reduce to
a) 3.125%,
b) 1% of its original value ?
Solution:
a) Here \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\frac{3.125}{100}\) = \(\frac{1}{32}\)
From \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\left(\frac{1}{2}\right)^{\mathrm{n}}\) = \(\frac{1}{32}\left(\frac{1}{2}\right)^5\) ∴ n = 5
From t = nT = 5T

b) Here \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\frac{1}{100}\)
From \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = e^-λt = \(\frac{1}{100}\)
-λt = log 1 – \(\log _{\mathrm{e}} 100\)
= 0 – 2.303 \(\log _{10} 100\) = -2.203 × 2
= -4.606
t = \(\frac{4.606}{\lambda}\) = \(\frac{4.606}{0.693 / \mathrm{T}}\) = 6.65 T

Question 8.
The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive \({ }_6^{14} \mathrm{C}\) present with the stable carbon isotope \({ }_6^{12} \mathrm{C}\). When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity, ceases and its activity begins to drop. From the known half-life (5730 years. of \({ }_6^{14} \mathrm{C}\), and the measured activity, the age of the specimen can be approximately estimated. This is the principle of \({ }_6^{14} \mathrm{C}\) dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
Solution:
Here normal activity R₀ = 15 decays/min
Present activity R = 9 decays / min
T = 5730 yrs
Age t = ?
As activity is proportional to the number of radio active atoms, therefore
\(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\frac{\mathrm{R}}{\mathrm{R}_0}\) = \(\frac{9}{15}\)
But \(\frac{N}{N_0}\) = e^-λt
e^-λt = \(\frac{9}{15}\) = \(\frac{3}{5}\)
e^+λt = \(\frac{5}{3}\)
λt \(\log _{\mathrm{e}} \mathrm{e}\) = \(\log _e \frac{5}{3}\) = 2.3023 log 1.6667
λt = 2.3026 × 0.2218 = 0.5109
t = \(\frac{0.5109}{\lambda}\)
λ = \(\frac{0.693}{T}\) = \(\frac{0.693}{5730}\)yt^-1
∴ t = \(\frac{0.5109}{0.693 / 5730}\) = \(\frac{0.5109 \times 5730}{0.693}\)
t = 4224.3 years

Question 9.
Obtain the amount of \({ }_{27}^{60} \mathrm{Co}\) necessary to provide a radioactive source of 8.0 mCi strength. The half-life of \({ }_{27}^{60} \mathrm{Co}\) is 5.3 years.
Solution:
Here, mass of ₂₇Co⁶⁰ =?
Strength of source, \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = 8.0 mci
= 8.0 × 3.7 × 10⁷ disint/sec
Half life, T = 5.3 years
= 5.3 × 365 × 24 × 60 × 60 sec
= 1.67 × 10⁸ sec
λ = \(\frac{0.693}{T}\) = \(\frac{0.693}{1.67 \times 10^8}\) = 4.14 × 10^-9 s^-1
As \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = 2N
∴ N = \(\frac{\mathrm{dN} / \mathrm{dt}}{\lambda}\) = \(\frac{8 \times 3.7 \times 10^7}{4.14 \times 10^{-9}}\)
= 7.15 × 10¹⁶
By definition of Avogadros number, Mass of 6.023 × 10²³ atoms of ₂₇Co⁶⁰ = 60 g
Mass of 7.15 × 10¹⁶ atoms of ₂₇Co⁶⁰
= \(\frac{60 \times 7.15 \times 10^{16}}{6.023 \times 10^{23}}\) = 7.12 × 10^-6g

Question 10.
The half-life of \({ }_{38}^{90} \mathrm{~Sr}\) is 28 years. What is the disintegration rate of 15 mg of this isotope?
Solution:
Here T = 28 years = 28 × 3.154 × 10⁷s
As number of atoms in 90 g of ₃₈Sr⁹⁰
= 6.023 × 10²³
∴ Number of atoms in 15mg of ₃₈Sr⁹⁰
= \(\frac{6.023 \times 10^{23}}{90}\) × \(\frac{15}{1000}\)
i.e., N = 1.0038 × 10²⁰
Rate of disintegration \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = λN
= \(\frac{0.693}{\mathrm{~T}} \mathrm{~N}\)
= \(\frac{0.693 \times 1.0038 \times 10^{20}}{28 \times 3.154 \times 10^7}\)
= 7.877 × 10¹⁰ Bq

Question 11.
Obtain approximately the ratio of the nuclear radii of the gold isotope \({ }_{79}^{197} \mathrm{Au}\) and the silver isotope \({ }_{47}^{107} \mathrm{Ag}\).
Solution:
Here A₁ = 197 and A₂ = 107
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = \(\left(\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right)^{1 / 3}\) = \(\left(\frac{197}{107}\right)^{1 / 3}\) = 1.225

Question 12.
Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) \({ }_{88}^{226} \mathrm{Ra}\) and (b) \({ }_{86}^{220} \mathrm{Rn}\).
Given m \(\left({ }_{88}^{226} \mathrm{Ra}\right)\) = 226.02540 u,
m \(\left({ }_{86}^{222} R n\right)\) = 222.01750 u,
m \(\left({ }_{86}^{222} \mathrm{Rn}\right)\) =220.01137 u, m \(\left({ }_{84}^{216} \mathrm{Po}\right)\) = 216.00189 u.
Solution:
a) ₈₈Ra²²⁶ → ₈₅Rn²²² + ₂He⁴ Q value
[m(₈₈Ra²²⁶) – m(₈₆Rn²²²) – m_α] × 931.5 MeV
= [226.02540 – 222.0 1750 – 4.00260] × 931.5 MeV
Q = 0.0053 × 931.5 MeV = 4.94 MeV
K.E of a particle =
\(\frac{(A-4) Q}{A}\) = \(\frac{226-4}{226}\) × 4.94 = 4.85 MeV

b) Proceeding as above, in case of
Q = 6.41 MeV
K.E of a particle
= \(\frac{(\mathrm{A}-4) \mathrm{Q}}{\mathrm{A}}\) = \(\frac{(220-4)}{220}\) × 6.41 = 6.29 MeV

Question 13.
The radionuclide ¹¹C decays according to The maximum energy of the emitted positron is 0.960 MeV. Given the mass values ; m\(\left({ }_6^{11} \mathrm{C}\right)\) = 11.011434u and \(\left({ }_6^{11} \mathrm{~B}\right)\) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.
Solution:
Mass defect in the given reaction is Δm = m(₆C¹¹)
= [m (₅B¹¹) + Me]
This is in terms of nuclear masses. If we express the Q value interms of atomic masses we have to subtract 6m_e from atomic mass of carbon and 5 m_e from that of boron to get the corresponding nuclear masses
Therefore, we have
Δm = [m(₆C¹¹) – 6 m_e – m(₅B¹¹) + 5m_e – m_e
= [m(₆C¹¹) – m(₅B¹¹) – 2 m_e]
= [11.011434 – 11.009305 – 2 × 0.000548] u
= 0.001033u
As, 1u = 931 MeV
Q = 0.001033 × 931 MeV = 0.961 MeV
Which is the maximum energy of emitted position.

Question 14.
The nucleus \({ }_{10}^{23} \mathrm{Ne}\) decays by β-emission. Write down the β – decay. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
\(m\left({ }_{10}^{23} \mathrm{Ne}\right)\) = 22.994466 u
\(\mathrm{m}\left({ }_{11}^{23} \mathrm{Na}\right)\) = 22.089770 u.
Solution:
The β decay of ₁₀Ne²³ may be represented as
₁₀Ne²³ → ₁₁Na²³ – ₁e⁰ + v + Q
Ignoring the rest mass of antineutrino and v electron
Mass defect, Δm = m(₁₀Ne²³) – m (₁₁Na²³)
= 22.994466 – 22.989770
= 0.004696 amu
Q = 0.004696 × 931 MeV
= 4.372 MeV
As ₁₁Na²³ is very massive, this energy of 4.3792 MeV is shared by e^–v pair. The max K.E of e^– = 4.372 MeV when energy carried by v is zero.

Question 15.
The Q value of a nuclear reaction A + b → C + d is defined by Q = [m_A + m_b – m_c – m_d] c² where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
i) \(\mathrm{H}_{\mathrm{1}}^{\mathrm{1}}\) + \({ }_1^3 \mathrm{H}\) → \({ }_1^2 \mathrm{H}\) + \({ }_1^2 \mathrm{H}\)
ii) \({ }_6^{12} \mathrm{C}\) + \({ }_6^{12} \mathrm{C}\) → \({ }_{10}^{20} \mathrm{Ne}\) + \({ }_2^4 \mathrm{He}\)
Atomic masses are given to be
\(\mathrm{m}\left({ }_{\mathrm{1}}^2 \mathrm{H}\right)\) = 2.014102u
\(\mathrm{m}\left({ }_1^3 \mathrm{H}\right)\) = 3.016049 u
\(\mathrm{m}\left({ }_6^{12} \mathrm{C}\right)\) = 12.000000u
\(\mathrm{m}\left({ }_{10}^{20} \mathrm{Ne}\right)\) = 19.992439 u
Solution:
i) \(\mathrm{H}_{\mathrm{1}}^{\mathrm{1}}\) + \({ }_1^3 \mathrm{H}\) → \({ }_1^2 \mathrm{H}\) + \({ }_1^2 \mathrm{H}\)
Q = Δm × 931 MeV =
[m (₁H¹ + m(₁H³) – 2m (₁H²)] × 931 MeV
= [1.007825 + 3.01604 – 2 × 2.014102] × 931 MeV
= -4.03 MeV
∴ This reaction is endothermic,

ii) ₆C¹² + ₆C¹² → ₁₀Ne²⁰ + ₂He⁴
Q = Δm × 931 MeV =
[2m(₆C¹²) – m(₁₀Ne²⁰) – m(₂He⁴)] × 931 MeV
= [24.000000 – 19.992439 – 4.002603] × 931 MeV
= ±4.61 MeV
∴ The reaction is exothermic.

Question 16.
Suppose we think of fission of a \({ }_{26}^{56} \mathrm{Fe}\) nucleus into two equal fragments, \({ }_{13}^{28} \mathrm{Al}\), Is the fission energetically possible ? Argue by working out Q of the process. Given \(\mathrm{m}\left({ }_{26}^{56} \mathrm{Fe}\right)\) = 55.93494 u and \(\mathbf{m}\left({ }_{13}^{28} \mathrm{~A} l\right)\) = 27.98191 u.
Solution:
Q = [m(₂₆Fe⁵⁶ – 2m (₁₃ Al²⁸.] × 931.5 MeV
= [55.93494 – 2 × 27.9819] × 931.5 MeV
Q = – 0.2886 × 931.5 MeV = – 26.88 MeV
Which is negative.
This fission is not possible energetically.

Question 17.
The fission properties of \({ }_{94}^{239} \mathrm{Pu}\) are very similar to those of \({ }_{92}^{238} U\). The average-energy released per fission is 180 MeV. How much energy, in MeV is released if all the atoms in 1kg of pure \({ }_{94}^{239} \mathrm{Pu}\) undergo fission?
Solution:
No. of atoms in 1kg of pure .
U^P = \(\frac{6.023 \times 10^{23}}{239}\) × 1000 = 2.52 × 10²⁴
As average energy released / fission is 180 MeV, therefore total energy released
= 2.52 × 10²⁴ × 180 MeV = 4.53 × 10²⁶ MeV

Question 18.
A 1000 MW fission reactor consumes half of its fuel In 5.00 y. How much \({ }_{92}^{235} \mathrm{U}\) did it contain initially ? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of \({ }_{92}^{235} \mathrm{U}\) and that this nuclide is consumed only by the fission process.
Solution:
In the fission of one nucleus of ₉₂U²³⁵ energy generated is 200 MeV
∴ Energy generated in fission of 1 kg of
₉₂U²³⁵ = 200 × \(\frac{6 \times 10^{23}}{235}\) × 1000 MeV
= 5.106 × 10²⁶ MeV = 5.106 × 10²⁶ × 1.6 × 10^-13J
= 8.17 × 10³ J
Time for which reactor operates \(\frac{80}{100}\) × 5
years = 4 years.
Total energy generated in 5 years.
= 1000 × 106 × 60 × 60 × 24 × 365 × 4J
∴ Amount of U consumed in 5 years
= \(\frac{1000 \times 10^6 \times 60 \times 60 \times 24 \times 365 \times 4}{8.17 \times 10^{13}} \mathrm{~kg}\)
= 1544 kg
∴ Initial amount of ₉₂U²³⁵ = 2 × 1544 kg
= 3088 kg

Question 19.
How long can an electric lamp. of 100W be kept glowing by fusion of 20 kg of deuterium ? Take the fusion reaction as \({ }_1^2 \mathrm{H}\) + \({ }_1^2 \mathrm{H}\) → \({ }_2^3 \mathrm{He}\) + n + 3.27 MeV
Solution:
Number of deuterium atoms in 2.0 kg
\(\frac{6.023 \times 10^{23} \times 2000}{2}\) = 6.023 × 10²⁶
Energy released when 2 atoms füse = 3.27 MeV
∴ Total energy released
= \(\frac{3.27}{2}\) × 6.023 × 10²⁶ MeV
= 1.635 × 6.023 × 10²⁶ × 1.6 × 10^-13 j
= 15.75 × 10³ J
Enery consumed by the bulb/sec = 100 J
∴ Time for which bulb will glow
= \(\frac{15.75 \times 10^{13}}{100} \mathrm{~S}\)
= \(\frac{15.75 \times 10^{11}}{60 \times 60 \times 24 \times 365}\) = 4.99 × 10⁷ years

Question 20.
Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint : The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.
Solution:
For head on collision distance between centres of two deuterons = r = 2 × radius
r = 4 fm = 4 × 10^-15 m
Charge of each deuteron e = 1.6 × 10^-10 C
Potential energy
\(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\) = \(\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{4 \times 10^{-15}}\) Joule
= \(\frac{9 \times 1.6 \times 1.6 \times 10^{-14}}{4 \times 1.6 \times 10^{-16}}\)KeV
PE = 360 KeV
P.E = 2 × K.E of each deuteron = 360 KeV
K.E of each deuteron = \(\frac{360}{2}\) = 180 KeV
This is a measure of height of Coulomb barrier

Question 21.
From the relation R = R₀A^1/3, where R₀, is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e., independent of A.
Solution:
Density of nuclear matter

ρ = \(\frac{\mathrm{mA}}{\frac{4}{3} \pi \mathrm{R}^3}\), where m is average mass of a nucleon
Using R = R₀A^1/3 we get
ρ = \(\frac{3 \mathrm{~mA}}{4 \pi\left(\mathrm{R}_0 \mathrm{~A}^{1 / 3}\right)^3}\) = \(\frac{3 \mathrm{~m}}{4 \pi \mathrm{R}_0^3}\)
As R₀ is constant, therefore ρ is constant.

Question 22.
For the β⁺ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K – shell, is captured by the nucleus and a neutrino is emitted).

Show that if β⁺ emission is energetically allowed, electron capture is necessarily allowed but not vice-versa.
Solution:
The β⁺ emission from a nucleus _ZX^A may be represented as
_zX^A = _z-1Y^A + ₁e⁰ + v + Q₁ —– (i)
The other competing process of electron capture may be represented as
_-1e⁰ + _ZX^A = _Z-1y^A + v + Q₂ —– (ii)
The energy released in Q₁ in (1. is given by

Note that m_N here denotes mass of nucleus and m denotes the mass of atom similarly from (ii)

Ir Q₁ > 0 then Q₂ > 0
i.e., If positron emission is energetically allowed electron capture is necessarily allowed. But Q₂ > 0 does not necessarily mean Q₁ > 0. Hence the reverse is not true.

Additional Exercises

Question 1.
In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \({ }_{12}^{24} \mathrm{Mg}\) (23.98504u), \({ }_{12}^{25} \mathrm{Mg}\) (24.98584u) and \({ }_{12}^{26} \mathrm{Mg}\) (25.98259u). The natural abundance of \({ }_{12}^{24} \mathrm{Mg}\) is 78.99% by mass. Calculate the abundances of other two isotopes.
Solution:
Let the abundance of \({ }_{12} \mathrm{Mg}^{25}\) by mass be x% therefore, abundance of ₁₂Mg²⁶ by mass
= (100 – 78.99 – x%)
= (21.01 – x%)
Now average atomic mass of magnesium is
24.312 =
\(\frac{23.98504 \times 78.99+24.98584+25.98529(21.01-\mathrm{x})}{100}\)
on solving we get x = 9.303% for ₁₂Mg²⁵ and for ₁₂Mg²⁶ (21.01 – x) = 11.71%

Question 2.
The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei \({ }_{20}^{41} \mathrm{Ca}\) and \({ }_{13}^{27} \mathbf{Al}\) from the following data: .
m(\({ }_{20}^{40} \mathrm{Ca}\)) = 39.962591 u
m(\({ }_{20}^{41} \mathrm{Ca}\)) = 40.962278 u
m({ }_{13}^{26} \mathrm{Ca}) = 25.986895 u
m({ }_{13}^{27} \mathrm{Ca}) = 26.981541 u
Solution:
When a neutron is separated from ₂₀Ca⁴¹ we are left with
₂₀Ca⁴⁰ i.e. ₂₀Ca⁴¹ → ₂₀Ca⁴⁰ + ₀n¹
Now mass defect
ΔM = m(₂₀Ca⁴⁰) + m_n – m (₂₀Ca⁴¹)
= 39.962591 + 1.008665 – 40.962278
= 0.008978 a.m.u
∴ Neutron seperation energy
= 0.008978 × 931MeV
= 8.362 MeV
similarly ₁₃Al²⁷ → ₁₃Al²⁶ + ₀n¹
∴ Mass defect, ΔM = m (₁₃Al²⁶) + m_n – m(₁₃Al²⁷)
= 25.986895 + 1.008665 – 26.981541
= 0.013845 u
∴ Neutron seperation energy. = 0.0138454 × 931MeV
= 12.89 MeV

Question 3.
A source contains two phosphorous radio nuclides \({ }_{15}^{32} \mathbf{P}\) (T_1/2 = 14.3d) and \({ }_{15}^{33} P\) (T1/2 = 25.3d). Initially, 10% of the decays come from \({ }_{15}^{33} \mathrm{P}\). How long one must wait until 90% do so ?
Solution:
Suppose initially the source has 90% ₁₅pt³² and 10% \({ }_{15} \mathrm{P}_{\mathrm{t}}^{32}\), say 9x gram P₂ and x gram of P₁.

After t days, suppose the source has 90% \({ }_{15} \mathbf{P}_2^{33}\) and 10% \({ }_{15} \mathrm{P}_{\mathrm{t}}^{32}\) i.e., y gram of P₂ and 9y gram of P₁
we have to calculate :
from \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\left(\frac{1}{2}\right)^n\) = \(\left(\frac{1}{2}\right)^{t / T}\) = 2^-i/T
N = N₀2^-t/T
y = 9×2^-t/14.3 for P₂ and 9y = x 2^-t/25.3 for P₁
Dividing we get
\(\frac{1}{9}\) = 9 × 2(t/25.3 – t/14.3.)
or \(\frac{1}{81}\) = 2^-11t/25.3 × 14.3
log 1 – log 81 = \(\frac{-11 \mathrm{t}}{25.3 \times 14.3}\) log 2
0 – 1 – 9085 = \(\frac{-11 \mathrm{t}}{25.3 \times 14.3}\) × 0.3010.
t = \(\frac{25.3 \times 14.3 \times 1.9085}{11 \times 0.3010}\) = 208.5 days

Question 4.
Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α – particle. Consider the following decay processes :
\({ }_{88}^{223} \mathrm{Ra}\) → \({ }_{82}^{209} \mathrm{~Pb}\) + \({ }_6^{14} \mathrm{C}\)
\({ }_{88}^{223} \mathrm{Ra}\) → \({ }_{86}^{219} \mathrm{~Pb}\) + \({ }_2^4 \mathrm{He}\)
Calculate the Q-values for these decays and determine that both are energetically allowed.
Solution:
i) For the decay process
88^{Ra 223} → ₈₂pb²⁰⁹ + ₆C¹⁴ + Q
mass defect, ΔM = mass of Ra²²³ – (mass of pb²⁰⁹ + mass of C¹⁴)
= 223.01850 – (208.98107 + 14.00324)
= 0.03419u
Q = 0.03419 × 931 MeV = 31.83 MeV

ii) For the decay process
₈₈Ra²³ → ₈₆Rn²¹⁹ + ₂He⁴ + Q mass defect, ΔM = mass of Ra²²³ – (mass of Rn²¹⁹ + mass of He⁴) = 223.01850 – (219.00948 + 4.00260)
= 0.00642 u
∴ Q = 0.00642 × 931 MeV = 5.98 MeV
As Q values are positive in both the cases, therefore both the decays are energetically possible.

Question 5.
Consider the fission of \({ }_{92}^{238} \mathbf{U}\) by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are \({ }_{58}^{140} \mathrm{Ce}\) and \({ }_{44}^{99} \mathrm{Ru}\). Calculate Q for this fission process. The relevant atomic and particle masses are
m(\({ }_{92}^{238} \mathrm{U}\)) = 238.05079 u
m(\({ }_{58}^{140} \mathrm{Ce}\)) = 139.90543 u
m(\({ }_{44}^{99} \mathrm{Ru}\))= 98.90594 u
Solution:
For this fission reaction,
₉₂U²³⁸ + _on¹ → 58Ce¹⁴⁰ + ₄₄Ru⁹⁹ + Q
mass defect ΔM = mass of U²³⁸ + mass of n – (mass of Ce¹⁴⁰ + mass of Ru⁹⁹.
= 238.05079 + 1.00867 – (139.90543 + 98.90594)
= 0.24809U
∴ Q = 0.24809 × 931 MeV = 230.97 Mev

Question 6.
Consider the D-T reaction (deuterium- tritium fusion)
\({ }_1^2 \mathrm{H}\) + \({ }_1^3 \mathrm{H}\) → \({ }_2^4 \mathrm{He}\) + n
a) Calculate the energy released in MeV in this reaction from the data:
m\(\left({ }_1^2 \mathrm{H}\right)\) = 2.014102 u
m\(\left({ }_1^3 \mathrm{H}\right)\) = 3.016049 u
b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the Coulomb repulsion between the two nuclei ? To what temperature must the gas heated to
initiate the reaction ?
(Hint : Kinetic energy required for one fusion event = average thermal kinetic energy available with the interacting particles = 2(3kt/2); k = Boltzman’s constant, T = absolute temperature..
Solution:
a) For the process ₁H² + ₁H³ + ₂He⁴ + n + Q
Q = [m(₁H²) + m (₁H³) + m(₂He⁴) – m_n] × 931 MeV
= (2.014102 + 3.016049 – 4.002603
1.00867) × 931 MeV
= 0.018878 × 931 = 17.58 MeV

b) Repulsive potential energy of two nuclei when they almost touch each other is
= \(\frac{q^2}{4 \pi \varepsilon_0(2 r)}\) = \(\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)^7}{2 \times 2 \times 10^{-15}}\) Joule
= 5.76 × 10^-14 Joule

Classically KE atleast equal to this amount is required to overcome Coulomb repulsion. Using the relation
K.E. = 2 × \(\frac{3}{2}\) KT
T = \(\frac{\mathrm{K} \cdot \mathrm{E}}{3 \mathrm{k}}\) = \(\frac{5.76 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}\) = 1.39 × 10⁹K
In actual practise the temperature required for trigerring the reaction is somewhat less.

Question 7.
Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. You are given that
m(¹⁹⁸Au) = 197.968233 u
m(¹⁹⁸Hg) = 197.966760 u.

Solution:
Energy corresponding to r₁
E₁ = 1.088 – 0 = 1.088 MeV
= 1.088 × 1.6 × 10^-13 Joule
Frequency v₁ = \(\frac{E_1}{h}\)
= \(\frac{1.088 \times 1.6 \times 10^{-13}}{6.6 \times 10^{-34}}\)
= 2.63 × 10²⁰ HZ
similarly v₂ = \(\frac{\mathrm{E}_2}{\mathrm{~h}}\)
= \(\frac{0.412 \times 1.6 \times 10^{-12}}{6.6 \times 10^{10}}\)
= 9.98 × 10¹³ Hz
and v₃ = \(\frac{E_3}{h}\)
= \(\frac{(1.088-0.412) \times 1.6 \times 10^{-13}}{6.6 \times 10^{20} \mathrm{~Hz}}\)
Maximum K.E. of β₁ particle
K_max (β₁)= [m(₇₉Au¹⁹⁸ – mass of Second excited state of ₈₀Hg¹⁹⁸] × 931 MeV
= [m(₇₉Au¹⁹⁸) – m(₈₂Hg¹⁹⁸) – \(\frac{1.088}{931}\)] × 931 MeV
= 931 [197.968233 – 197.966760] – 1.088 MeV
= 1.371 – 1.088 = 0.283 MeV
similarly k_max (β₂) – 0.957 MeV

Question 8.
Calculate and compare the energy released by a. fusion of 1.0 kg of hydrogen deep within Sun and b. the fission of 1.0 kg of ²³⁵U in a fission reactor.
Solution:
In sun, four hydrogen nuclei fuse to form a helium nucleus with the release of 26 MeV energy.
∴ Energy released by fusion of 1 kg of hydrogen = \(\frac{6 \times 10^{23} \times 26}{4}\) × 10³ MeV
As energy released in fission of one atom of ₉₂U²³⁶ = 200 MeV
Energy released in fission of 1 kg of ₉₂U²³⁸
= \(\frac{6 \times 10^{23} \times 1000}{235}\) × 200 MeV
E₂ = 5.1 × 10²⁶ MeV
\(\frac{\mathrm{E}_1}{\mathrm{E}_2}\) = \(\frac{39 \times 10^{26}}{5.1 \times 10^{26}}\) = 7.65
i.e., Energy released in fusion is 7.65 times the energy released in fission.

Question 9.
Suppose India had a target of producing by 2020 AD, 2,00,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy of thermal energy produced in a reactór was 25%.
How much amount of fissionable uranium would our country need per year by 2020 ? Take the heat energy per fission of ²³⁵U to be about 200 MeV.
Solution:
Total targeted power = 2 × 10⁵ MW
Total Nuclear power = 10% of 2 × 10⁵ MW
= 2 × 10⁴ MW
Energy produced in fission = 200 MeV
Effeciency of power plant =25%
∴ Energy converted into electrical energy per fission = \(\frac{25}{100}\) × 200 = 50 MeV
= 50 × 1.6 × 10^-13 Joule.
Total electrical energy to be produced :
= 2 × 10⁴ MW = 2 × 10⁴ × 10⁶ Watt
= 2 × 10¹⁰ Joule/Sec
= 2 × 10¹⁰ × 60 × 60 × 24 × 365 Joule / year
No. of fissions in one year
= \(\frac{2 \times 10^{10} \times 60 \times 60 \times 24 \times 365}{50 \times 1.6 \times 10^{-13}}\)
= 2 × \(\frac{36 \times 24 \times 365}{8}\) × 10²⁴
Mass of 6.023 × 10²³ atoms of U²³⁵ = 235 gm = 235 × 10^-3 kg
Mass of \(\frac{2 \times 36 \times 24 \times 365}{8}\) × 10²⁴ atoms
= \(\frac{235 \times 10^{-3}}{6.023 \times 10^{23}}\) × \(\frac{2 \times 36 \times 24 \times 365 \times 20^{24}}{8}\)
= 3.08 × 10⁴ Kg
Hence mass of Uranium needed per year = 3.08 × 10⁴ Kg.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 6th Lesson Current Electricity Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 6th Lesson Current Electricity

Very Short Answer Questions

Question 1.
Define mean free path of electron in a conductor.
Answer:
The average distance transversed by an electron during successive collisions in a conductor is called mean free path of electron in a conductor.

Question 2.
State Ohm’s law and write its mathematical form.
Answer:
At constant temperature, the strength of the current (I) in a conductor is directly proportional to the potential difference (V) between its ends.
∴ I ∝ V ⇒ I = \(\frac{\mathrm{V}}{\mathrm{R}}\) ⇒ V = IR (Mathematical form)
where R is constant, it is called the resistance of the conductor.

Question 3.
Define resistivity or specific resistance.
Answer:
Resistivity or specific resistance (ρ) : The resistance of a conductor of unit length and unit area of cross-section is called resistivity.
If l = 1, A = 1 ⇒ ρ = \(\frac{\mathrm{R} \times 1}{1}\) = ρ ⇒ R

Question 4.
Define temperature coefficient of resistance.
Answer:
Temperature coefficient of resistance (α) : The ratio of the change in resistance per 1°C rise in temperature to the resistance at 0°C is called the temperature coefficient of resistance.
α = \(\frac{R_t-R_0}{R_0 t}\)

Question 5.
Under what conditions is the current through the mixed grouping of cells maximum ?
Answer:
The current through the mixed grouping of cells maximum, when

1. Effective emf of all the cells is high.
2. The value of external resistance is equal to the total internal resistance of all the cells.

Question 6.
If a wire is stretched to double its original length without loss of mass, how will the resistivity of the wire be influenced ?
Answer:
Resistivity of the wire remains unchanged as it does not change with change in dimensions of a material without change in its temperature.

Question 7.
Why is manganin used for making standard resistors ?
Answer:
Due to high resistivity and low temperature coefficient of resistance, manganin wire (Cu – 84% + Mn – 12% + Ni – 4%) is used in the preparation of standard resistances.

Question 8.
The sequence of bands marked on a carbon resistor are: Red, Red, Red, Silver. What is its resistance and tolerance ?
Answer:
The resistance of a carbon resistor marked with Red, Red, Red = 22 × 10²Ω = 2.2kΩ = 2200Ω

[∵ Sequence number for Red = 2 and multiplication factor = 10²]
The tolerance of carbon resistor = ± 10%

Question 9.
Write the color code of a carbon resistor of resistance 23 kilo ohms.
Answer:
Color code of a carbon resistor of 23 Kilo Ohms (= 23 × 10³Ω) are Red, Orange, Orange
[∵ Sequence number 2 for Red, 3 for orange, multiplication factor 10³ for orange]

Question 10.
If the voltage V applied across a conductor is increased to 2V, how will the drift velocity of the electrons change ?
Answer:
The drift velocity, V_d = \(\frac{\mathrm{eE}}{\mathrm{m}}\) \(\tau^{\prime}\) = \(\frac{\mathrm{eV}}{\mathrm{mL}} \tau\)
\(\frac{\mathrm{V}_{\mathrm{d}_1}}{\mathrm{~V}_{\mathrm{d}_2}}\) = \(\frac{v_1}{v_2}\)
Here V₁ = V, V₂ = 2V
\(\frac{\mathbf{V}_{\mathrm{d}_1}}{\mathrm{~V}_{\mathrm{d}_2}}\) = \(\frac{V}{2 V}\)
∴ \(\mathrm{V}_{\mathrm{d}_2}\) = \(2 \mathrm{~V}_{\mathrm{d}_1}\)
∴ Drift velocity is increased by twice.

Question 11.
Two wires of equal length, of copper and, manganin, have the same resistance. Which wire is thicker ?
Answer:
R = \(\frac{\rho \mathrm{A}}{l}\) ⇒ A = \(\frac{\mathrm{R} l}{\rho}\)
Since ρ_cu < p_manganin, copper wire is thicker than manganin wire.

Question 12.
Why are household appliances connected in parallel ?
Answer:
In parallel, the voltage (V) across each appliance is same. The current (I) through them depends upon the power (P) of the appliance. The higher power appliance draws more current and lower power appliance draws less current.
(∵ P = VI or I ∝ P)

Question 13.
The electron drift speed in metals is small (~ms^-1) and the charge of the electron is also very small (~10^-19C), but we can still obtain a large amount of current in a metal. Why ?
Answer:
Current through a metal, I = n A eV_d.
A is the area of cross-section of the metal. The electron drift speed, V_d (~10^-5, ms^-1) is small. The charge of electron, e (~1.6 × 10^-19C) is also very small. But we can still obtain a large amount of current in a metal due to presence of large number of free electrons (n) is a conductor (~ 10²⁹ m^-3).

Short Answer Questions

Question 1.
A battery of emf 10V and internal resistance 3Ω is connected to a resistor R.

1. If the current in the circuit is 0.5 A. Calculate the value of R.
2. What is the terminal voltage of the battery when the circuit is closed.

Answer:
Given, E = 10V, r = 3Ω, I = 0.5A, R = ?, V = ?

1. E = I(R + r) or R + r = \(\frac{E}{I}\) = \(\frac{10}{0.5}\) =20Ω ⇒ R = 20 – 3 = 17Ω
2. Terminal voltage, V = IR = 0.5 × 17 = 8.5Ω

Question 2.
Draw a circuit diagram showing how a potentiometer may be used to find Internal resistance of a cell and establish a formula for It.
Answer:
Measurement of internal resistance (r) with potentiometer:

1. Potentiometer to measure internal resistance (r) of a cell (ε) is shown in diagram.
   
2. The cell (emf ε) whose internal resistance (r) is to be determined is connected across a resistance box (RB) through a key K₂.
3. With key K₂ open, balance is obtained at length
   l₁ (AN₁). Then ε = ϕl₁ —– (1)
4. When key K₂ is closed, the cell sends a current (I) through the resistance box (R.B).
5. If V is the terminal potential difference of the cell and balance is obtained at length l₂ (AN₂). Then V = ϕl₂ —– (2)
6. \(\frac{(1)}{(2)}\) ⇒ \(\frac{\varepsilon}{\mathrm{V}}\) = \(\frac{l_1}{l_2}\) —- (3)
7. But ε = I(r + R) and V = IR. This gives
   \(\frac{\varepsilon}{\mathrm{V}}\) = \(\frac{(\mathrm{r}+\mathrm{R})}{\mathrm{R}}\)
   \(\frac{I_1}{I_2}\) = \(\left(\frac{r}{R}+1\right)\) [∵ from (3)]
   ∴ r = R\(\left(\frac{l_1}{l_2}-1\right)\)

Question 3.
Derive an expression for the effective resistance when three resistors are connected in
(i) series
(ii) parallel. (T.S. Mar. ’19)
Answer:
Effective resistance when three resistors are connected:
(i) In series:

1. Three resistors R₁, R₂ and R₃ are connected in series as shown in fig. V₁, V₂, V₃ are the potential differences across R₁, R₂ and R₃. I is the current flowing through them.
2. Applying Ohm’s law to R₁, R₂ and R₃, Then V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
3. In series, V = V₁ + V₂ + V₃
   IR_S = IR₁ + IR₂ + IR₃ [∵ V = IR_S]
   ∴ R_S = R₁ + R₂ + R₃

ii) In parallel:

1. Three resistors, R₁, R₂ and R₃ are connected in parallel as shown in fig. Potential differences across each resistor is V. I₁, I₂, I₃ are the currents flowing through them.
2. Applying Ohmes law to R₁, R₂ and R₃, then
   V = I₁R₁ = I₂R₂ = I₃R₃
   ⇒ I₁ = \(\frac{\mathrm{V}}{\mathrm{R}_1}\), I₂ = \(\frac{\mathrm{V}}{\mathrm{R}_2}\), I₃ = \(\frac{\mathrm{V}}{\mathrm{R}_3}\)
3. In parallel, I = I₁ + I₂ + I₃
   ⇒ \(\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{p}}}\) = \(\frac{\mathrm{V}}{\mathrm{R}_1}\) + \(\frac{\mathrm{V}}{\mathrm{R}_2}\) + \(\frac{\mathrm{V}}{\mathrm{R}_3}\) [∵ I = \(\frac{V}{R_p}\)]
   ∴ \(\frac{1}{\mathrm{R}_{\mathrm{p}}}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\)

Question 4.
‘m’ cells each of emf E and internal resistance ‘r’ are connected in parallel. What is the total emf and internal resistance ? Under what conditions is the current drawn from mixed grouping of cells a maximum ?
Answer:
Cells in parallel:

1. When ‘m’ identical cells each of emf ‘ε’ and internal resistance r are connected to the external resistor of resistance R as shown in fig., then the cells are connected in parallel.
2. As the cells are connected in parallel, their equivalent internal resistance r_p is given by
   \(\frac{1}{\mathrm{r}_{\mathrm{p}}}\) = \(\frac{1}{\mathrm{r}}\) + \(\frac{1}{\mathrm{r}}\) + ….. upto m terms = \(\frac{\mathrm{m}}{\mathrm{r}}\)
   ∴ r_p = \(\frac{\mathrm{r}}{\mathrm{m}}\)
   
3. As R and r_p are in series, so total resistance in the circuit = R + \(\frac{\mathrm{r}}{\mathrm{m}}\).
4. In parallel combination of identical cells, the effective emf in the circuit is equal to the emf due to a single cell, because in this combination, only the size of the electrodes increases but not emf.
5. Therefore, current in the resistance R is given by I = \(\frac{\varepsilon}{\mathrm{R}+\frac{\mathrm{r}}{\mathrm{m}}}\) = \(\frac{\mathrm{m} \varepsilon}{\mathrm{m} R+\mathrm{r}}\)
6. When the external resistance is negligible is comparison to the internal resistance (R<<r), the current drawn from mixed grouping of cells a maximum.

Question 5.
Define electric resistance and write it’s SI unit. How does the resistance of a conductor vary if
(a) Conductor is stretched to 4 times of it’s length
(b) Temperature of a conductor is increased.
Answer:
Electric resistance (R) : The resistance offered by a flow of electrons in a conductor is called electric resistance.
S.l unit of resistance is ohm (Ω).
The resistance of a conductor
R = \(\frac{\rho l}{\mathrm{~A}}\) = \(\frac{\rho l^2}{\mathrm{~V}}\) ⇒ R ∝ l²
a) in first case, R₁ = R, l₁ = l
b) In second case, l₂ = 4l, R₂ =?
\(\frac{R_2}{R_1}\) = \(\frac{l_2^2}{l_1^2}\) ⇒ \(\frac{R_2}{R}\) = \(\left(\frac{4l}{l_4}\right)^2\) ∴ R₂ = 16R

b) Variation of Resistance with temperature is given by R_t = R₀ (1 + α t)
If temperature increases, resistance also increases.

Question 6.
When the resistance connected In series with a cell is halved, the current is equal to or slightly less or slightly greater than double. Why?
Answer:
‘When he resistance R is connected to cell of emf, ε in series, the current is given by
I = \(\frac{\varepsilon}{\mathrm{R}+\mathrm{r}}\)

where r is internal resistance of cell.
When the resistance is halved \(\left(\frac{\mathrm{R}}{2}\right)\), the current flows through the circuit is I’ = \(\frac{\varepsilon}{\frac{R}{2}+r}\)

1. If r is negligible comparison with \(\frac{\mathrm{R}}{2}\), I¹ = \(\frac{2 \varepsilon}{R}\)
   ∴ I¹ = 2 I [∵ \(\frac{\varepsilon}{R}\) also equal to 1]
2. If r < < \(\frac{\mathrm{R}}{2}\), the current I¹ is slightly greater than 2.
3. If r is just slightly greater than R, the current (I¹) is slightly less than 2.

Question 7.
Two cells of emfs 4.5V and 6.0V and infernal resistance 6Ω and 3Ω respectively have their negative terminals joined by a wire of 18Ω and positive terminals by a wire of 12Ω resistance. A third resistance wire of 24Ω connects middle points of these wires. Using Kirchhoffs laws, find the potential difference at the ends of this third wire.
Answer:

1. Let the currents through the various arms, of the network be as shown in fig:
2. Applying KVL to closed mesh ABCDA, we have
   4.5 – 6I₁ – 18I₁ – 24 (I₁ + I₂) = 0
   ⇒ 48I₁ + 24I₂ = 4.5 —–> (i)
3. For a closed mesh CDEFC, we have
   – 24(I₁ + I₂) – 12I₂ + 6 – 3I₂ = 0
   24I₁ + 39I₂ = 6 —–> (ii)
   
4. (ii) × 2 – (i) ⇒
   
   ∴ I₂ = \(\frac{7.5}{54}\) = 0.139 A —–> (iii)
   Substituting (iii) in (i), we get
   48I₁ + 78 × 0.139 = 12
   48I₁ = 12 – 10.84 = 1.158
   I₁ = \(\frac{1.158}{48}\) = 0.024 A
5. Potential difference across third wire = (I₁ + I₂) × 24 = 0.163 × 24 = 3.912 Volt.

Question 8.
Three resistors each of resistance 10 ohm are connected, in turn, to obtain
(i) minimum resistance
(ii) maximum resistance. Compute
(a) The effective resistance in each case
(b) The ratio of minimum to maximum resistance so obtained.
Answer:
Given, Resistance of each resistor R = 10Ω, no. of resistors, n = 3
i) If three resistors are connected in parallel, we get minimum resistance.
∴ Minimum resistance R_min = R_p = \(\frac{\mathrm{R}}{\mathrm{n}}\) = \(\frac{10}{3} \Omega\) = 3.33Ω

ii) If three resistors are connected in series, we get maximum resistance.
∴ Maximum resistance R_max = R_s = n R = 3 × 10 = 30Ω

a) The effective resistance to get minimum resistance,
R_eff = \(\frac{\mathrm{R}}{\mathrm{n}}\) = \(\frac{10}{3}\) = 3.33Ω (In parallel)
The effective resistance to get maximum resistance
R_eff = n R = 3 × 10 = 30Ω (In series)

b) \(\frac{R_{\min }}{R_{\max }}\) = \(\frac{\left(\frac{10}{3}\right)}{(3 \times 10)}\) = \(\frac{10}{90}\) ∴ \(\frac{\mathrm{R}_{\min }}{\mathrm{R}_{\max }}\) = \(\frac{1}{9}\)

Question 9.
State Kirchhoffs law for an elêctrical network. Using these laws deduce the condition for balance in a Wheatstone bridge. (Ã.P. Mar. 19, ‘16 & Mar. 14)
Answer:
1) Kirchhoff s first law (Junction rule or KCL) : The algebraic sum of the currents at any junction is zero. ∴ ΣI = 0
(or)
The sum of the currents flowing towards a junction is equal to the sum of currents away from the junction.

2) Kirchhoffs second law (Loop rule or KVL): The algebraic sum of potential around any closed loop is zero.
∴ Σ(IR) + ΣE = 0
Wheatstone bridge : Wheatstone’s bridge circuit consists of four resistances R₁, R₂, R₃ and R₄ are connected to form a closed path. A cell of emf e is connected between the point A and C and a galvanometer is connected between the points B and D as shown in fig. The current through the various branches are indicated in the figure. The current through the galvanometer is I_g and the resistance of the galvanometer is G.
Applying Kirchhoffs first law at the junction D, I₁ – I₃ – I_g = 0 —— (1)
at the junction B, I₂ + I_g – I₄ = 0 —— (2)
Applying Kirchhoffs second law to the closed path ADBA,
-I₁R₁ + I₂R₂ = 0
or
⇒ I₁R₁ + I_gG = I₂R₂ —– (3)
applying kirchhoffs second law to the closed path DCBD,
-I₃R₃ + I₄R₄ + I_gG = 0
⇒ I₃R₃ – I_gG = I₄R₄ —– (4)
When the galvanometer shows zero deflection the points D and B are at the same potential. So I_g = 0.

Substituting this value in (1), (2), (3) and (4).
I₁ = I₃ ——- (5)
I₂ = I₄ —— (6)
I₁R₁ = I₂R₂ —– (7)
I₃R₃ = I₄R₄ —– (8)
Dividing (7) by (8)
\(\frac{\mathrm{I}_1 \mathrm{R}_1}{\mathrm{I}_3 \mathrm{R}_3}\) = \(\frac{I_2 R_2}{I_4 R_4}\) ⇒ \(\frac{R_1}{R_3}\) = \(\frac{R_2}{R_1}\) [∵ I₁ = I₃ & I₂ = I₄]
∴ Wheatstone’s Bridge principle : R₄ = R₃ × \(\frac{\mathrm{R}_2}{\mathrm{R}_1}\)

Question 10.
State the working principle of potentiometer. Explain with the help of circuit diagram how the emf of two primary cells are compared by using the potentiometer. (T.S. Mar. 19 & A.P. Mar. 16)
Answer:
Working principle of potentiometer : The potential difference across a length of the potentiometer wire is directly proportional to its length (or) when a steady current is passed through a uniform wire, potential drop per unit length or potential gradient is constant,
i.e. ε ∝ l ⇒ ε = ϕl where ϕ is potential gradient.
Comparing the emf of two cells ε₁ and ε₂ :

1. To compare the emf of two cells of emf E₁ and E₂ with potentiometer is shown in diagram.
   
2. The points marked 1, 2, 3 form a two way key.
3. Consider first a position of the key where 1 and 3 are connected so that the galvanometer is connected to ε₁.
4. The Jockey is moved along the wire till at a point N₁ at a distance l₁ from A, there is no deflection in the galvanometer. Then ε₁ ∝ l₁ ⇒ ε₁ = ϕl₁ —– (1)
5. Similarly, if another emf ε₂ is balanced against
   l₂ (AN₂) then ε₂ ∝ l₂ ⇒ ε₂ = ϕl₂ —— (2)
6. \(\frac{(1)}{(2)}\) ⇒ \(\frac{\varepsilon_1}{\varepsilon_2}\) = \(\frac{l_1}{l_2}\)

Question 11.
State the working principle of potentiometer explain with the help of circuit diagram how the potentiometer is used to determine the internal resistance of the given primary cell. (A.P. & T.S. Mar. ’15)
Answer:
Working principle of potentiometer : The potential difference across a length of the potentiometer wire is directly proportional to its length (or) when a steady current is passed through a uniform wire, potential drop per unit length or potential gradient is constant.
i.e. E ∝ l ⇒ e = ϕl
where ϕ is potential gradient.

Measurement of internal resistance (r) with potentiometer:

1. Potentiometer to measure internal resistance (r) of a cell (ε) is shown in diagram.
2. The cell (emf ε) whose internal resistance (r) is to be determined is connected across a resistance box (R.B) through a key K₂.
   
3. With key K₂ open, balance is obtained at length l₁(AN₁). Then ε = ϕl₁ ——-> (1)
4. When key K₂ is closed, the cell sends a current (T) through the resistance box (R.B).
5. If V is the terminal potential difference of the cell and balance is obtained at length l₂ (AN₂).
   Then V = ϕl₂ ——> (2)
6. \(\frac{(1)}{(2)}\) ⇒ \(\frac{\varepsilon}{\mathrm{V}}\) = \(\frac{l_1}{l_2}\) —– (3)
7. But ε = I (r + R) and V = IR. This gives
   \(\frac{\varepsilon}{V}\) = \(\frac{(\mathrm{r}+\mathrm{R})}{\mathrm{R}}\)
   \(\frac{l_1}{l_2}\) = \(\left(\frac{r}{R}+1\right)\) [∵ from (3)]
   ∴ r = \(\mathrm{R}\left(\frac{l_1}{l_2}-1\right)\)

Question 12.
Show the variation of current versus voltage graph for GaAs and mark the
(i) Non-linear region
(ii) Negative resistance region.
Answer:
The relation between V and I is not unique. That is, there is more than one value of V for the same current I. material exhibiting such behaviour is GaAs (i.e., a light emitting diode).

Question 13.
A student has two wires of iron and copper of equal length and diameter. He first joins two wires in series and passes an electric current through the combination which increases gradually. After that he joins two wires in parallel and repeats the process of passing current. Which wire will glow first in each case ?
Answer:
1) In series combination, there will be same current through Iron and as well as copper wire. Since the rate of heat production, P = I² R or P ∝ R (for the given value of I). The resistance of Iron wire is more than that of copper for the given length and diameter. Hence in Iron wire, the rate of heat production increases gradually. In series combination Iron will glow first.

2) In parallel combination of Iron and copper wire, there will be same P.D (V) across them.
Since the rate of heat production, P = \(\frac{\mathrm{V}^2}{\mathrm{R}}\) or P ∝ \(\frac{1}{R}\) (for the given value of V). The resistance of Iron is more than that of copper for the given length and diameter. Hence in copper wire, the rate of heat production is more.
In parallel combination copper will glow first.

Question 14.
Three identical resistors are connected in parallel and total resistance of the circuit is R/3. Find the value of each resistance.
Answer:
Given three resistances are identical.
Hence R₁ = R₂ = R₃ = x (say)
Total resistance in parallel, R_p = \(\frac{\mathrm{R}}{3}\)
If three identical resistances are connected in parallel, then
\(\frac{1}{R_p}\) = \(\frac{1}{\mathrm{R}_1}\) + \(\frac{1}{\mathrm{R}_2}\) + \(\frac{1}{\mathrm{R}_3}\)
\(\frac{1}{\left(\frac{\mathrm{R}}{3}\right)}\) = \(\frac{1}{x}\) + \(\frac{1}{x}\) + \(\frac{1}{x}\) ⇒ \(\frac{3}{\mathrm{R}}\) = \(\frac{1+1+1}{\mathbf{x}}\)
∴ x = R.

Long Answer Questions

Question 1.
Under what condition is the heat produced in an electric circuit
a) directly proportional
b) inversely proportional to the resistance of the circuit ?
Compute the ratio of the total quantity of heat produced in the two cases.
Answer:
Expression of heat produced by electric current:
Consider a conductor AB of resistance R.

Let V = P.D applied across the ends of AB.
I = current flowing through AB.
t = time for which the current is flowing.
∴ Total charge flowing from A to B in time t is q = It. By definition of P.D, work done is carrying unit charge from A to B = V
Total work done in carrying a charge q from A to B is
W = V × q = V It = I² Rt
(∵ V = IR)
This work done is called electric work done. If this electric work done appears as heat, then amount of heat produced (H) is given by H = W = I² Rt Joule.
This is a statement of Joule’s law of heating.
a) If same current flows through an electric circuit, heat is developed.
i.e., H ∝ R.

b) If same P.D applied across the the electric circuit heat is developed.
i.e., H₂ ∝ \(\frac{1}{\mathrm{R}}\).

c) The ratio of H₁ and H₂ is given by
\(\frac{\mathrm{H}_1}{\mathrm{H}_2}\) = \(\frac{\mathrm{R}}{\frac{1}{\mathrm{R}}}\)
∴ \(\frac{\mathrm{H}_1}{\mathrm{H}_2}\) = R²

Question 2.
Two metallic wires A and B are connected in parallel. Wire A has length L and radius r, wire B has a length 2L and radius 2r. Compute the ratio of the total resistance of the parallel combination and resistance of wire A.
Answer:
1) For metal Wire ‘A’
Length = L
Radius = r
Area = πr²
Resistance, R_A = \(\frac{\rho_{\mathrm{A}} \mathrm{L}}{\pi \mathrm{r}^2}\) —– (i)
Where ρ_A is specific resistance.
For metal Wire ‘B’
Length = 2L
Radius = 2r
Area = π(2r)² = 4πr²
Resistance, R_B = \(\frac{\rho_{\mathrm{B}}(2 \mathrm{~L})}{4 \pi \mathrm{r}^2}\) = \(\frac{\rho_{\mathrm{B}} \mathrm{L}}{2 \pi \mathrm{r}^2}\) —– (ii)
Where ρ_B is specific resistance.

2) Total resistance of wire A and wire B in parallel combination is given by

3)

4) The ratio of the total resistance parallel combination to resistance of wire A, is given by

5)
∴ \(\frac{R_p}{R_A}\) = \(\frac{\rho_{\mathrm{B}} \pi \mathrm{r}^2}{\mathrm{~L}\left(2 \rho_{\mathrm{A}}+\rho_{\mathrm{B}}\right)}\)

Question 3.
In a house three bulbs of 100W each are lighted for 4 hours daily and six tube lights of 20W each are lighted for 5 hours daily and a refrigerator of 400W is worked for 10 hours daily for a month of 30 days. Calculate the electricity bill if the cost of one unit is Rs. 4.00.
Answer:
No. of bulbs in a house, N = 3
Rated power on each bulb, P = 100 W
Time of lighted t = 4H
Energy consumption of 3 bulbs per day = \(\frac{\mathrm{Npt}}{1000}\) KWH
Energy consumption of 3 bulbs for 30 days = \(\frac{30 \mathrm{~N} \mathrm{Pt}}{1000}\) KWH
E_B = \(\frac{30 \times 3 \times 100 \times 4}{1000}\) = 36KWH
Similarly, Energy consumption of 6 tube lights for 30 days, ET = \(\frac{30 \times 6 \times 20 \times 5}{1000}\) KWH = 18 KWH.
And similarly, Energy consumption of one Refrigerator for 30 days,
E_R = \(\frac{30 \times 400 \times 10}{1000}\)KWH = 120 KWH
∴ The total energy consumption, E = E_B + E_T + E_R = (36 + 18 + 120) KWH
∴ E = 174 KWH = 174 units [∵ 1KWH = 1 unit]
Cost of 1 unit = Rs. 4.00/-
Cost of 174 units = No. of units × cost of 1 unit
= 174 × 4 = Rs. 696/-
∴ Electricity bill for one month of that house = Rs. 696/-

Question 4.
Three resistors of 4 ohms, 6 ohms and 12 ohms are connected in parallel. The combination of above resistors is connected in series to a resistance of 2 ohms and then to a battery of 6 volts. Draw a circuit diagram and calculate.
a) Current in main circuit.
b) Current flowing through each of the resistors in parallel.
c) p.d and the power used by the 2 ohm resistor.
Answer:
Circuit diagram for the given data is

a) Effective resistance when R₁, R₂ and R₃ are connected in parallel is given by
\(\frac{1}{R_p}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\) ⇒ \(\frac{1}{R_p}\) = \(\frac{1}{4}\) + \(\frac{1}{6}\) + \(\frac{1}{12}\)
⇒ \(\frac{1}{\mathrm{R}_{\mathrm{p}}}\) = \(\frac{3+2+1}{12}\) = \(\frac{6}{12}\) = \(\frac{1}{12}\)
∴ R_p = 2Ω
Total resistance in the circuit R = R_p + R₄ = 2 + 2 = 4Ω
∴ Current in mam circuit I = \(\frac{\mathrm{V}}{\mathrm{R}}\) = \(\frac{6}{4}\) = 1.5A

b) Current flowing through R₁, I₁ = \(\frac{I R_P}{R_1}\) = \(\frac{1.5 \times 2}{4}\) = 0.75A
Current flowing through R₂, I₂ = \(\frac{\mathrm{IR}_{\mathrm{P}}}{\mathrm{R}_2}\) = \(\frac{1.5 \times 2}{6}\) = 0.5A
Current flowing through R₃, I₃ = \(\frac{\mathrm{IR}_{\mathrm{P}}}{\mathrm{R}_3}\) = \(\frac{1.5 \times 2}{12}\) = 0.25A

c) RD across 2Ω resistor (i.e., R₄), V₄ = IR₄ = 1.5 × 2 = 3 Volt.
Power used by 2Ω resistor, P = V₄I = 3 × 1.5 = 4.5 W.

Question 5.
Two lamps, one rated 100 Ω at 220 V and the other 60W at 220 V are connected in parallel to a 220 volt supply. What current is drawn from the supply line?
Answer:
Data for

Since R₁ and R₂ are connected in parallel effective resistance

Question 6.
Estimate the average drift speed of conduction electrons in a copper wire of cross – sectional area 3.0 × 10^-7 m² carrying a current of 5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 10³ kg/m³ and its atomic mass is 63.5 u.
Answer:
Given, Cross-sectional area of copper wire, A = 3 × 10^-7m²
carrying current of copper, I = 5A
Charge of electron, e = 1.6 × 10^-19C
Density of conduction electrons = No. of atoms per cubic meter,

∴ Average drift speed of conduction electrons.
V_d = \(\frac{\mathrm{I}}{\mathrm{neA}}\) = \(\frac{5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 3 \times 10^{-7}}\)
⇒ V_d = \(\frac{5}{8.5 \times 1.6 \times 3 \times 10^2}\) = 0.1225 × 10^-2m/s
∴ V_d = 1.225 mm/s

Question 7.
Compare the drift speed obtained above with
i) Thermal speed of copper atoms at ordinary temperatures.
ii) Speed of propagation of electric field along the conductor which causes the drift motion.
Answer:
i) At a temperature T, the thermal speed of a copper atom of mass M is obtained from

∴ drift speed of electron (V_d) = 1.047 × 10^-8
= 10^-8 times of thermal speed at ordinary temperature.

ii) The electric field travels along conductór with speed of EMW
C = 3 × 10⁸ m/s
V_d = 1.225 × 10^-3m/s
\(\frac{\mathrm{V}_{\mathrm{d}}}{\mathrm{C}}\) = \(\frac{1.225 \times 10^{-3}}{3 \times 10^8}\)
V_d = 0.408 × 10^-11 C
∴ Drift speed is, in compansion of C, extremely smaller by a factor of 10^-11.

Problems

Question 1.
A 10Ω thick wire is stretched so that its length becomes three times. Assuming that there Is no change in its density on stretching, calculate the resistance of the stretched wire.
Solution:
Given R₁ = 10Ω,
l₁ = 1
l₂ = 3l, R₂ ?

R₁ = \(\frac{\rho}{\mathrm{V}} l_1^2\)
R₂ = \(\frac{\rho}{\mathrm{V}} l_2^2\)
R₃ = \(\left(\frac{l_2}{l_1}\right)^2\) ⇒ \(\frac{\mathrm{R}_2}{10}\) = \(\left(\frac{3 l}{l}\right)^2\)
∴ R₂ = 10 × 9 = 90Ω.

Question 2.
A wire of resistance 4R is bent in the form of a circle. What is the effective resistance between the ends of the diameter? (A.P. Mar. ’19 & T.S. Mar. ’16, Mar. ’14)
Solution:
Resistance of long wire = 4R
Hence the resistance of half wire =
\(\frac{4 R}{2}\) = 2R

Now these two wire are connected in parallel. Hence the effective resistance between the ends of the diameter
R_P = \(\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}\) ⇒ R_p = \(\frac{2 \mathrm{R} \times 2 \mathrm{R}}{2 \mathrm{R}+2 \mathrm{R}}\)
∴ R_p = R.

Question 3.
Find the resistivity of a conductor which carries a current of density of 2.5 × 10⁶A m^-2 when an electric field of 15 Vm^-1 is applied across it.
Solution:
Given current density
J = \(\frac{1}{A}\) = 2.5 × 1o⁶ Am^-2
Applied electric field E = 15Vm^-1
Resistivity of conductor,
ρ = \(\frac{E}{J}\) = \(\frac{15}{2.5 \times 10^6}\)
∴ ρ = 6 × 10^-6Ωm.

Question 4.
What is the color code for a resistor of resistance 350mΩ with 5% tolerance ?
Solution:
Resistance of resistor = 350mΩ with 5% tolerance
= 350 × 10^-3Ω
= 35 × 10^-2Ω
First significant figure (3) indicates 1^st band
Second significant figure (5) indicates 2^nd band
Third significant figure (10^-2) indicates 3^rd band
We know that
0 1 2 3 4 5 6 7 8
B B R O Y of Great Britian has Very Good Wife wearing Gold silver Necklace
9 10^-1 ← 10^-2
3 indicates orange
5 indicates green
10^-2 indicates Silver
5% tolerance substance is Gold.
∴ Color code of given resistor is orange, green, silver, gold.

Question 5.
You are given 8Ω resistor. What length of wire of resistivity 120 Ωm should be joined in parallel with it to get a value of 6Ω ?
Solution:
Given, Resistance of resistor R = 8Ω

Resisty of wire ρ = 120
Let l length of the resistance x is to be connected to get effective resistance,
R_p = 6Ω
Then \(\frac{1}{\mathrm{R}}\) + \(\frac{1}{x}\) = \(\frac{1}{\mathrm{R}_{\mathrm{p}}}\)
\(\frac{1}{8}\) + \(\frac{1}{x}\) = \(\frac{1}{6}\) ⇒ \(\frac{1}{x}\) = \(\frac{1}{6}\) – \(\frac{1}{8}\) = \(\frac{2}{48}\)
∴ x = 24Ω
And x l = ρ
24 l = 120
∴ l = 5m

Question 6.
Three resistors 3Ω, 6Ω and 9Ω are connected a battery. In which of them will the power dissipation be maximum if:
a) They all are connected in parallel
b) They all are connected in series ? Give reasons.
Solution:
Given R₁ = 3Ω, R₂ = 6Ω, R₃ = 9Ω
a) Effective resistance in parallel is given by
\(\frac{1}{R_p}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\) = \(\frac{1}{3}\) + \(\frac{1}{6}\) + \(\frac{1}{9}\)
\(\frac{1}{R_p}\) = \(\frac{6+3+2}{18}\)
∴ R_P = \(\frac{18}{11} \Omega\)

∴ Dissipated power in parallel,
P_P ∝ \(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) ⇒ P_P ∝ \(\frac{1}{\left(\frac{18}{11}\right)}\) ∴ P_P ∝ \(\frac{11}{18}\) —– (1)

b) Effective resistance in series is given by R_s = R₁ + R₂ + R₃ = 3 + 6 + 9 = 18Ω
∴ Dissipated power in series,
P_S ∝ R_S ⇒ P_S ∝ 18 —- (2)
From equations (1) and (2) power dissipation is maximum in series and minimum in parallel.

Reasons:

1. In series connection, P ∝ R and V ∝ R. Hence dissipated power (P) and potential difference (V) is more because current is same across each resistor.
2. In parallel connection, P ∝ \(\frac{1}{R}\) and I ∝ \(\frac{1}{R}\). Hence dissipated power (P) and potential difference (V) is less because voltage is same across each resistor.

Question 7.
A silver wire has a resistance of 2.1Ω at 27.5°C and a resistance of 2.7Ω at 100°C. Determine the temperature coeff. of resistivity of silver.
Solution:
For silver wire, R₁ = 2.1Ω, t₁ = 27.5°C
R₂ = 2.7Ω, t₂ = 100°C, α = ?
α = \(\frac{R_2-R_1}{R_1 t_2-R_2 t_1}\) = \(\frac{2.7-2.1}{2.1 \times 100-2.7 \times 27.5}\)
= \(\frac{0.6}{210-74.25}\) = \(\frac{0.6}{135.75}\)
∴ Temperature coefficient of resistivity
∝ = 0.443 × 10^-2/°C

Question 8.
If the length of a wire conductor is doubled by stretching it while keeping the potential difference constant, by what factor will the drift speed of the electrons change ?
Solution:
Taking l₁ = l,
l₂ = 2l
Since V_d ∝ l, \(\frac{\mathrm{v}_{\mathrm{d}_2}}{\mathrm{~V}_{\mathrm{d}_1}}\) = \(\frac{l_2}{l_1}\)
\(\frac{\mathrm{V}_{\mathrm{d}_2}}{\mathrm{~V}_{\mathrm{d}_1}}\) = \(\frac{2 l}{l}\)
∴ \(\mathrm{V}_{\mathrm{d}_2}\) = \(2 \mathrm{~V}_{\mathrm{d}_1}\)
∴ Drift speed of electrons changes by a factor 2.

Question 9.
Two 120V light bulbs, one of 25W and’ another of 200W are connected in series. One bulb burnt out almost instantaneously. Which one was burnt and why ?
Solution:
Given, For first bulb,
P₁ = 25W,
V₁ = 120V

Resistance of first bulb R₁ = \(\frac{\mathrm{v}_1^2}{\mathrm{P}_1}\)
R₁ = \(\frac{(120)^2}{25}\) —– (1)
For second bulb, P₂ = 200W, V₂ = 120V
Resistance of second bulb,
R₂ = \(\frac{(120)^2}{200}\) —— (2)
\(\frac{(1)}{(2)}\) ⇒ \(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = 8 ⇒ R₁ = 8R₂
As R₁ > R₂, 25 W bulb burnt out almost instantaneously, since two bulbs have rated at same voltage.

Question 10.
A cylindrical metallic wire is stretched to increase its length by 5%. Calculate the percentage change in resistance.
Solution:
Given, % change in length, \(\frac{\mathrm{d} l}{l}\) = 5%
Resistance of wire R = \(\frac{\rho l^2}{\mathrm{~V}}\)
% Change in Resistance of wire,
\(\frac{d R}{R}\) = 2\(\frac{\mathrm{d} l}{l}\) = 2 × 5% = 10%

Question 11.
Two wires A and B of same length and same material, have their cross sectional areas in the ratio 1:4. What would be the ratio of heat produced in these wires when the Voltage across each is constant ?
Solution:
Given l_A = l_B, ρ_A = ρ_B, V_A = V_B,
A_A : A_B = 1 : 4
Rate of heat produced in a wire,
H = i²R = \(\frac{\mathrm{V}^2}{\mathrm{R}^2}\) = \(\frac{V^2 A}{\rho l}\)
Since V, ρ, l are same for both wires A and B, H ∝ A (area of crossection)
For two wires A and B,
\(\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}\) = \(\frac{A_A}{A_B}\) = \(\frac{1}{4}\)
∴ H_A : H_B = 1 : 4.

Question 12.
Two bulbs whose resistances are in the ratio of 1:2 are connected in parallel to a source of constant voltage. What will be the ratio of power dissipation in these ?
Solution:
Given, R₁ : R₂ = 1 : 2, In parallel series
Dissipated power P = \(\frac{V^2}{R}\)
⇒ P ∝ \(\frac{\mathrm{I}}{\mathrm{R}}\) [ ∵ V = constant]
The ratio of dissipated powers in two bulbs is given by,
\(\frac{P_1}{P_2}\) = \(\frac{\mathrm{R}_2}{\mathrm{R}_1}\) = \(\frac{2}{1}\)
∴ P₁ : P₂ = 2 : 1

Question 13.
A potentiometer wire is 5m long and a potential difference of 6 V is maintained between its ends. Find the emf of a cell which balances against a length of 180cm of the potentiometer wire. (A.P. Mar. ’16)
Solution:
Length of potentiometer wire L = 5m
Potential difference V = 6 Volt
Potential gradient ϕ = \(\frac{\mathrm{V}}{\mathrm{L}}\) = \(\frac{6}{5}\) = 1.2 V / m
Balancing length l = 180cm
= 1.80m
Emf of the cell E = ϕl
= 1.2 × 1.8 = 2.16V

Question 14.
A battery of emf 2.5 V and internal resistance r is connected in series with a resistor of 45 ohm through an ammeter of resistance 1 ohm. The ammeter reads a current of 50 mA. Draw the circuit diagram and calculate the value of r.
Solution:
Circuit diagram for the given data is shown below.

Given, E = 2.5 V; R = 4 5Ω;
r_A = 1A; I = 50mA;
r = ?
E = I (R + r_A + r)
2.5 = 50 × 10^-3 (45 + 1 + r)
46 + r = \(\frac{2.5}{50 \times 10^{-3}}\) = \(\frac{2.5 \times 10^3}{50}\) = 50
∴ r = 50 – 46 = 4Ω.

Question 15.
Amount of charge passing through the cross section of a wire is q(t) = at² + bt + c. Write the dimensional formula for a, b and c. If the values of a, b and c in SI unit are 6, 4, 2 respectively, find the value of current at t = 6 seconds.
Solution:
Charge passing through wire is given by q(t) = at² + bt + c
According to principle of homogenity, Dimensional formula of q(t) = dimensional formula of at²
IT = aT²
∴ Dimensional formula a = IT^-1
Dimensional formula of q(t) = Dimensional formula of bt
IT = bT
∴ Dimensional formula of b = I
Dimensional formula of q(t) = Dimensional formula of C.
IT = C
∴ Dimensonal formula of C = IT
Current, I = \(\frac{\mathrm{dq}(\mathrm{t})}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(at² + bt + c)
= 2at + b
Here a = 6 and b = 4
⇒ I = 12t + 4
∴ Current at t = 6 sec,
I = 12 × 6 + 4 = 76 A.

Textual Exercises

Question 1.
The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery ?
Solution:
Here E = 12 V, r = 0.4Ω
Maximum Current, I_max = \(\frac{\mathrm{E}}{r}\) = \(\frac{12}{0.4}\) = 30A

Question 2.
A battery of emf 10V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?
Solution:
Here E = 10 V, r = 3Ω, I = 0.5 A, R = ?, V = ?
I = \(\frac{E}{(R+r)}\) or (R + r) = \(\frac{E}{I}\) = \(\frac{10}{0.5}\) = 20 or
R = 20 – r = 20 – 3 = 17Ω
Terminal voltage V = IR = 0.5 × 17 = 8.5 Ω.

Question 3.
a) Three resistors 1Ω, 2Ω, and 3Ω are combined in series. What is the total resistance of the combination ?
Solution:
Here R₁ = 1Ω, R = 2 Ω, R₃ = 3Ω, V = 12V
In series, total resistance R_S = R₁ + R₂ + R₃ = 1 + 2 + 3 = 6Ω.

b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Solution:
Current through the circuit I = V/Rs = 12/6 = 2A
∴ Potential drop across R₁ = IR₁ = 2 × 1 = 2V
Potential drop across R₂ = IR₂ = 2 × 2 = 4V
Potential drop across R₃ = IR₃ = 2 × 3 = 6V

Question 4.
a) Three resistors 2Ω, 4Ω and 5Ω are combined in parallel. What is the total resistance of the combination ?
Solution:
Here R₁ = 2Ω, R₂ = 4Ω, R₃ = 5Ω, V = 20V
In parallel combination total resistance R_P is given by
\(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\) = \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) = \(\frac{10+5+4}{20}\) = \(\frac{19}{20}\) or R_P = \(\frac{20}{19} \Omega\)

b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Solution:
Current through R₁ = \(\frac{\mathrm{V}}{\mathrm{R}_1}\) = \(\frac{20}{2}\) = 10A
Current through R₂ = \(\frac{20}{4}\) = 5 A
Current through R₃ = \(\frac{20}{5}\) = 4A
Total current = \(\frac{20}{\left(\frac{20}{9}\right)}\) = 19A

Question 5.
At room temperature (27.0°C) the resistance of a heating element is 100Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10^-4°C^-1.
Solution:
Here R₂₇ = 100Ω, R₁ = 117Ω, t = ? α = 1.70 × 10^-4°C^-1
We know that
α = \(\frac{R_t-R_{27}}{R_{27}(t-27)}\) or t – 27 = R_t – R₂₇
t = \(\frac{\mathrm{R}_1-\mathrm{R}_2}{\mathrm{R}_{27} \times \alpha}\) + 27 = \(\frac{117-100}{100 \times 1.7 \times 10^{-4}}\) + 27
= 1000 + 27 = 1027°C

Question 6.
A negligibly small current is passed through a wire of length 15m and uniform cross-section 6.0 × 10^-7 m², and its resistance is measured to be 5.0Ω. What is the resistivity of the material at the temperature of the experiment ?
Solution:
Here L = 15 m, A = 6.0 × 10^-7m², R = 5.0 Ω, ρ = ?

Question 7.
A silver wire has a resistance of 2.1 Ω at 27.5°C, and a resistance of 2.7 Ω at 100°C. Determine the temperature coefficient of resistivity of silver.
Solution:
Here R_27.5 = 2.1Ω, R₁₀₀ = 2.7Ω; α = ?
α = \(\frac{\mathrm{R}_{100}-\mathrm{R}_{27.5}}{\mathrm{R}_{27.5} \times(100-27.5)}\) = \(\frac{2.7-2.1}{2.1 \times(100-27.5)}\) = 0.0039°C^-1

Question 8.
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0°C ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^-4°C^-1.
Solution:

Question 9.
Determine the current in each branch of the network shown in Fig.

Solution:
The current’s through the various arms of the circuit have been shown in figure.
According to Kirchhoff’s second law;
-10 + 10 (i₁ + i₂) + 10i₁ + 5(i₁ – i₃) = 0
(or) 10 = 25i + 10i – 5i₃
(or) 2 = 5i₁ + 2i₂ – i₃
(or) 2 = 5i₁ + 2i₂ – i₃ ………. → (i)
In a closed circuit ABDA ~
10i₁ + 5i₃ – 5i₂ = 0
(or) 2i₁ + i₃ – i₂ = 0
(or) i₂ = 2i₁ + i₃ …….. → (ii)
In a closed circuit BCDB
5(i₁ – i₃) – 10 (i₂ + i₃) – 5i₃ = 0
(or) 5i₁ – 10i₂ = 20i₃ =0
i₁ = 2i₁ + 4i₃…….. → (iii)
From (ii) and (iii)
i₁ = 2 (2i₁ – i₃) + 4i₃ = 4i₁ + 6i₃
(or) 3i₁ = -6i₃
(or) i₁ = -2i₃
Putting this value in (ii) : i₂ = 2(-2i₃) i₃ = -3i₃
Putting values in (i)
2 = 5(-2i₃) + 2(-3i₃) – i₃ (or) 2 = -17i₃
(or) i₃ = -2/17A
From (iv) i₁ = -(-2/17) = -4/17A
from (v) i₂ = 3(-2/17) = 6/17A
i₁ + i₂ = (4/17) + (6/17) = (10/17)A
i₁ + i₃ = (4/17) + (-2/17) = (6/17) A
i₂ + i₃ = (6/17) + (-2/17) = 4/17A.

Question 10.
a) In a metre bridge the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips ?

A meter bridge. Wire Ac is 1m long R is a resistance to be measured and S is a standard resistant
Solution:
Here l = 39.5cm, R = X = ?, S = Y = 12.5 Ω
As S = \(\frac{100-l}{l} \times \mathrm{R}\)
∴ 12.5 = \(\frac{100-39.5}{39.5} \times \mathrm{X}\)
or X = \(\frac{12.5 \times 39.5}{60.5}\) = 8.16Ω
Thick copper strips are used to minimise resistance of the connections which are not accounted in the formula.

b) Determine the balance point of the bridge above if X and Y are interchanged.
Solution:
As X and Y are interchanged therefore, l₁ and l₂ (i.e.) lengths are also interchanged.
Hence L = 100 – 39.5 = 60.5 cm.

c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge ? Would the galvanometer show any current ?
Solution:
The galvanometer will show no current.

Question 11.
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5Ω. What is the terminal voltage of the battery during charging ? What is the purpose of having a series resistor in the charging circuit ?
Solution:
Here emf of the the battery = 8.0V; voltage of d.c. supply = 120V
Internal resistance of battery r = 0.5Ω; external resistance R = 15.5Ω
Since a storage battery of emf 8V is charged with a.d.c supply of 120 V the effective emf in the circuit is given by ε = 120 – 8 = 112 V
Total resistance of the circuit = R + r = 15.5 + 0.5 = 16.0Ω
∴ Current in the circuit during charging is given by
I = \(\frac{\varepsilon}{R+r}\) = \(\frac{112}{16}\) = 7.0A
∴ Voltage across R = IR = 7.0 × 15.5 = 108.5 V
During charging the voltage of the d.c supply in a circuit must be equal to the sum of the voltage drop across R and terminal voltage of the battery
∴ 120 = 108.5 V or V= 120 – 108.5 = 11.5V
The series resistor limits the current drawn from the external source of d.c supply. In its absence the current will be dangerously high.

Question 12.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell ?
Solution:
Here ε₁ = 1.25 V, l₁ = 35.0 cm, ε₂ = ?.l₂ = 63.0 cm.
As \(\frac{\varepsilon_2}{\varepsilon_1}\) = \(\frac{l_2}{l_1}\) or
ε₂ = \(\frac{\varepsilon_1 \times l_2}{l_1}\) = \(\frac{1.25 \times 63}{35}\) = 2.25V

Question 13.
The number density of free electrons in a copper conductor estimated in textual example 6.1 is 8.5 × 10²⁸ m^-3. How long does an electron take to drift from one end of a wire 3.0m long to its other end ? The area of cross-section of the wire is 2.0 × 10^-6 m² and it is carrying a current of 3.0 A.
Solution:
Here n = 8.5 × 10²⁸ m^-3; L = 3.0 m; A = 2.0 × 10^-6 m²; I = 3.0A, t = ?
As I = n A eVd
∴ V_d = \(\frac{1}{\mathrm{nAe}}\)
Now, t = \(\frac{1}{\mathrm{v}_{\mathrm{d}}}\)
=
= \(\frac{3.08 \times 8.5 \times 10^{28} \times 2.0 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}\)
= 2.72 × 10⁴ S
= 7hour 33 minutes.

Additional Exercises

Question 1.
The earth’s surface has a negative surface charge density of 10^-9 C m^-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface ? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10⁶m).
Solution:
Here r = 6.37 × 10⁶ m; Q = 10^-9 cm²; I = 1800 A
Area of the globe A = 4πr² = 4 × 3.14 × (6.37 × 10⁶)²
= 509.64 × 10³C
t = \(\frac{Q}{I}\) = \(\frac{509.64 \times 10^3}{1800}\) = 283.1 S

Question 2.
a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015Ω are joined in series to provide a supply to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?
Solution:
Here ε = 2.0V; n = 6; r = 0.015Ω; R = 8.5 Ω
Current I = \(\frac{n E}{R+n r}\) = \(\frac{6 \times 2.0}{8.5+6 \times 0.015}\) = 1.4A
Terminal voltage,V = IR = 1.4 × 8.5 = 11.9V.

b) A secondary cell after long use has an emf 011.9 V and a large internal resistance of 380Ω. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car?
Solution:
Here E = 1.9 V; r = 380Ω
I_max = \(\frac{\varepsilon}{\mathrm{r}}\) = \(\frac{1.9}{380}\) = 0.005A
This amount of current cannot start a car because to start the motor, the current required is 100 A for few seconds.

Question 3.
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρ_Al = 2.63 × 10^-8 Ωm, ρ_Cu = 1.72 × 10^-8 Ωm, Relative density of Al = 2.7, of Cu = 8.9.)
Solution:
Given, for aluminium wire; R₁ = R; l₁ = l
Relative density d₁ = 2.7.
For copper wire R₂ = R, t₂ = 1, d₂ = 8.9
Let A₁, A₂ be the area of cross section for aluminium wire and copper wire.
We know, R₁ = \(\rho_1 \frac{l_1}{\mathrm{~A}_1}\) = \(\frac{2.63 \times 10^{-8} \times l}{A_1}\)
and mass of the aluminium wire m₁ = A₁l₁ × d₁ = A₁l₁ × 2.7
R₂ = ρ₂ = \(\frac{l_2}{\mathrm{~A}_2}\) = \(\frac{1.72 \times 10^{-8} \times l}{\mathrm{~A}_2}\)
Mass of copper wire m₂ = A₂l₂ × d₂ = A₂l × 8.9
Since two wires are of equal resistance R₁ = R₂
\(\frac{2.63 \times 10^{-8} \times 1}{\mathrm{~A}_1}\) = \(\frac{1.72 \times 10^{-8} \times l}{\mathrm{~A}_2}\) or \(\frac{\mathrm{A}_2}{\mathrm{~A}_1}\) = \(\frac{1.72}{2.63}\)
from (ii) and (iv) we have
\(\frac{\mathrm{m}_2}{\mathrm{~m}_1}\) = \(\frac{\mathrm{A}_2 l \times 8.9}{\mathrm{~A}_1 l \times 2.7}\) = \(\frac{8.9}{2.7} \times \frac{\mathrm{A}_2}{\mathrm{~A}_1}\)
= \(\frac{8.9}{2.7} \times \frac{1.72}{2.63}\) = 2.16

It shows that copper wire is 2.16 times heavier than aluminium wire since for the same value of length and resistance aluminium wire has lesser mass than copper wire, therefore aluminium wire is preferred for overhead power cables. A heavy cable may sag down owing to its own weight.

Question 4.
What conclusion can you draw from the following observations on a resistor made of alloy manganin ?

Answer:
Since the ratio of voltage and current for different readings is same so ohm’s law is valid to high accuracy. The resistivity of the alloy manganin is nearly independent of temperature.

Question 5.
Answer the following Questions.
a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed ?
Solution:
Only current through the conductor of non-uniform area of cross section is constant as the remaining quantities vary inversly with the area of cross-section of the conductor.

b) Is Ohm’s law.universally applicable for all conducting elements ? If not, give examples of elements which do not obey Ohm’s law.
Solution:
Ohm’s law is not applicable for non-ohmic elements. For example, vaccum tubes, semi-conducting diode, liquid electrolyte etc.

c) A low voltage supply from which one needs high currents must have very low internal resistance. Why ?
Solution:
As, I_max = emf internal resistance so for maximum current internal resistance should be least.

d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why ?
Solution:
A high tension supply must have a large internal resistance otherwise, if accidently the circuit is shorted, the current drawn will exceed safety limit and will cause damage to circuit.

Question 6.
Choose the correct alternative :
a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10²³).
Solution:
a) Greater
b) Lower
c) Nearly independent
d) 10²²

Question 7.
a) Given n resistors each of resistance R, how will you combine them to get the
i) maximum,
ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance ?
Solution:
For maximum effective resistance the n resistors must be connected in series.
Maximum effective resistance R_S = nR
For minimum effective resistance the n resistors must be connected in parallel.
Maximum effective resistance R_P = R/n
∴ \(\frac{R_S}{n p}\) = \(\frac{\mathrm{nR}}{\mathrm{R} / \mathrm{n}}\) = n²

b) Given the resistances of 1Ω, 2Ω, 3Ω, how will be combine them to get an equivalent resistance of
(i) (11/3) Ω
(ii) (11/5) Ω,
(iii) 6Ω,
(iv) (6/11)Ω?
Solution:
It is to be noted that
a) the effective resistance of parallel combination of resistors is less than the individual resistance and
b) the effective resistance of series combination of resistors is more than individual resistance.

case (i) Parallel combination of 1Ω and 2Ω is connected in series with 3Ω.
Effective resistance of 1Ω and 2Ω in parallel will be given by
RP = \(\frac{1 \times 2}{1+2}\) = \(\frac{2}{3} \Omega\)
∴ Equivalent resistance of \(\frac{2}{3} \Omega\) and 3] and 3Ω in series
= \(\frac{2}{3}\) + 3 = \(\frac{11}{3}\)Ω

Case(ii) : Parallel combination of 2Ω and 3Ω is connected in series with 1Ω.
Equivalent resistance of 2Ω and 3Ω in parallel
= \(\frac{2 \times 3}{2+3}\) = \(\frac{6}{5} \Omega\)
Equivalent resistance of \(\frac{6}{5} \Omega\) and 1Ω in series = \(\frac{6}{5}\) + 1 = \(\frac{11}{5}\)Ω

Case (iii) : All the resistances are to be connected in series now
∴ Equivalent resistance = 1 + 2 + 3 = 6Ω

Case (iv) : All the resistances are to be connected in parallel
∴ Equivalent resistance (R) is given by
\(\frac{1}{\mathrm{R}}\) = \(\frac{1}{1}\) + \(\frac{1}{2}\) + \(\frac{1}{3}\)
= \(\frac{6+3+2}{6}\) = \(\frac{11}{6}\) (Or) r = \(\frac{6}{11} \Omega\)

c) Determine the equivalent resistance of networks shown in Fig.

Answer:
a) The given network is a series combination of 4 equal units. Each unit has 4 resistances in which 2 resistances (1Ω each in series) are in parallel with 2 other resistances (2Ω each in series).
∴ Effective resistances of two resistances (each of 1Ω) in series = 1 + 1 = 2Ω.
Effective Resistance of two resistances (each of 2Ω) in series = 2 + 2 = 4Ω
If R is the resistance of one unit of resistances then
\(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) = \(\frac{1}{2}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\) or R_P = \(\frac{4}{3} \Omega\)
∴ Equivalent resistance in network = 4 R_P = 4 × \(\frac{4 \Omega}{3}\) = \(\frac{16 \Omega}{3}\)

b) Total resistances each of value R are connected in series. Their effective resistance = 5R.

Question 8.
Determine the current drawn from a 12 V supply with internal resistance 0.5Ω by the Infinite network shown in Fig. Each resistor has 1Ω resistance.

Solution:
Let x be the equivalent résistance of infinite network. Since the net work is infinite, therefore, the addition of one more unit of three resistances each of value of 1Ω across the terminals will not alter the total resistance of network i.e. it should remain x.

Therefore, the network would appear as shown in the figure and its total resistance should remain x.
There the parallel combination of x and 1Ω is in series with two resistors of 1Ω each.
The resistance of parallel combination is

\(\frac{1}{\mathbf{R}_{\mathbf{P}}}\) = \(\frac{1}{x}\) + \(\frac{1}{1}\) = \(\frac{1+x}{x}\)
R_p = \(\frac{x}{(1+x)}\)
∴ Total resistance of network will be given by
x = 1 + 1 + \(\frac{x}{x+1}\) = 2 + \(\frac{x}{x+1}\)
x(x + 1) = 2(x + 1) + x
or x² + x = 2x + 2 + x or x² – 2x – 2 = 0
or x = \(\frac{2 \pm \sqrt{4+8}}{2}\) = \(\frac{2 \pm \sqrt{12}}{2}\)
Total resistance of the circuit shows a full scale deflection for a current of 2.5 mA. How will you convert the meter into
= \(\frac{2 \pm 2 \sqrt{3}}{2}\) = 1 ± \(\sqrt{3}\)
The value of resistance cannot be negative, therefore the resistance of network
= 1 + \(\sqrt{3}\) = 1 + 1.73 Ω = 2.73 Ω
Total resistance of the cfrcuit = 2.73 + 0.5
= 3.23Ω
∴ Current draw I = \(\frac{12}{3.23}\) = 3.72 amp

Question 9.
Figure shows a potentiometer with a cell of 2.0’V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

a) What is the value ε ?
Solution:
Here ε₁ = 1.02 V, L₁ = 67.3cm, ε₂ = e = ?, L₂ = 82.3 cm
Since \(\frac{\varepsilon_2}{\varepsilon_1}\) = \(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\)
∴ ε = \(\frac{\mathrm{L}_2}{\mathrm{~L}_1} \times \varepsilon_1\) = \(\frac{82.3}{67.3} \times 1.02\) = 1.247V

b) What purpose does the high resistance of 600 kΩ have ?
Answer:
The purpose of using high resistance of 600k Ω is to allow very small current through the galvanometer when the movable contact is far from the balance point.

c) Is the balance point affected by this high resistance ?
Answer:
No, the balance point is not affected by the presence of this resistance.

d) Is the balance point affected by the internal resistance of the driver cell ?
Answer:
No, the balance point is not affected by the internal resistance of the driver cell.

e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V ?
Answer:
No, the method will not work as the balance point will not be obtained on the potentiometer wire if the e.m.f of the driver cell is less than the emf of the other cell.

f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple) ? If not, how will you modify the circuit ?
Answer:
The circuit will not work for measuring extremely small emf because in that case the balance point will be just close to the end A. To modify the circuit we have to use a suitable high resistance in series with the cell of 2.0V This would decrease the current in the potentiometer wire. Therefore potential difference 1cm df wire will decrease. Hence extremely small emf can be measured.

Question 10.
Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε ?

Answer:
Here L₁ = 58.3cm; L₂ = 68.5 cm; R = 10Ω; I X = ?. Let be the current in the potentiometer wire and ε₁ and ε₂ be the potential drops across R and X respectively when connected in circuit by closing respective keky. Then

If there is no balance point with given cell of emf it means potential drop across R or X greater than the potential drop across the potentiometer wire AB. In order to obtain the balance point, the potential drops across R and X are to be reduced which is possible by reducing the current in R and X for that either suitable resistance should be put in series with R and X or a cell of smaller emf E should be used. Another possible way is to increase the potential drop across the potentiometer wire by increasing the voltage of driver cell.

Question 11.
Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Solution:
Here l₁ = 76.3 cm, l₂ = 64.8 cm.
r = ?, R = 9.5 Ω
Now, r = \(\left(\frac{l_1-l_2}{l_2}\right) R\) = \(\left(\frac{76.3-64.8}{64.8}\right)\) 9.5 = 1.68 Ω

Textual Exercises

Question 1.
a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 × 10^-7 m² carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 10³ kg/m³, and its atomic mass is 63.5 u.
b) Compare the drift speed obtained above with,
i) thermal speeds of copper atoms at ordinary temperatures,
ii) speed of propagation of electric field along the conductor which causes the drift motion.
Solution:
a) The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed ud is given by Eq.
IΔt = + neA /v_d/Δt
V_d = (I/neA)
Now, e = 1.6 × 10^-19 C, A = 1.0 × 10^-7 m², I = 1.5 A. The density of conduction electrons, n is equal to the number of atoms per cubic meter (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of ‘copper has a mass of 9.0 × 10³ kg. Since 6.0 × 10²³ copper atoms have a mass of 63.5 g,
n = \(\frac{6.0 \times 10^{23}}{63.5}\) × 9.0 × 10⁶
= 8.5 × 10²⁸ m^-3 Which gives,
\(v_{\mathrm{d}}\) = \(\frac{1.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10 \times 10^{-7}}\)
= 1.1 × 10^-3 m s^-1 = 1.1 mm s^-1

b) i) At a temperature T, the thermal speed of a copper atom of mass M is obtained from
[<(1/2) Mυ² > = (3/2) K_BT] and is thus typically of the order of \(\sqrt{\mathrm{k}_{\mathrm{B}} \mathrm{T} / \mathrm{M}}\), where K_B is the Boltzmann constant. For copper at 300 K, this is about 2 × 10² m/s. This figure indicates the random vibrational speeds of copper atoms in a conductor. Note that the drift speed of electrons is much smaller, about 10^-5 times the typical thermal speed at ordinary temperatures,

ii) An electric field travelling along the conductor has a speed of an electromagnetic wave, namely equal to 3.0 × 10⁸ m s^-1. The drift speed is, in comparison, extremely small, smaller by a factor of 10^-11.

Question 2.
a) In Textual Example 1, the electron drift speed is estimated to be only a few mm s^-1 for currents in the range of a few amperes ? How then is current established almost the instant a circuit is closed ?
b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed ?
c) If the electron drift speed is so small, and the electron’s charge is small, how can we still obtain large amounts of current in a conductor ?
d) When electrons drift in a metal from lower to higher potential, does it mean that all the ‘free’ electrons of the metal are moving in the same direction ?
e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the i) absence of electric field,
ii) presence of electric field ?
Solution:
a) Electric field is established throughout the circuit, almost instantly (with the speed of light) causing at every point a local electron drift. Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end. However it does take a little while for the current to reach its steady value.
b) Each ‘free’ electron does accelerate, increasing its drift speed after collision but starts to accelerate and increases its drift speed again only to suffer a collision again and so on. On the average, therefore, electrons acquire only a drift speed.
c) Simple, because the electron number density is enormous, ~10²⁹ m^-3.
d) By no means. The drift velocity is superposed over the large random velocities of electrons.
e) In the absence of electric field, the paths are straight lines, in the presence of electric field, the paths are, in general curved.

Question 3.
An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27.0 °C) is found to be 75.3 Ω. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element ? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is 1.70 × 10^-4 °C^-1.
Solution:
When the current through the element is very small, heating effects can be ignored and the temperature T₁ of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a few seconds, a steady state will be reached when temperature will rise no further, and both the resistance of the element and the current drawn will achieve steady values. The resistance R₂ at the steady temperature T₂ is
R₂ = \(\frac{230 \mathrm{~V}}{2.68 \mathrm{~A}}\) = 85.8Ω
Using the relation
R₂ = R₁ [1 + α(T₂ – T₁)]
with α = 1.70 × 10^-4°C^-1, we get
T₂ – T₁ = \(\frac{(85.8-75.3)}{(75.3) \times 1.70 \times 10^{-4}}\) = 820°C
that is, T₂ = (820 + 27.0)°C = 847 °C

Question 4.
The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 Ω and at steam point is 5.39 Ω. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is 5.795 Ω. Calculate the temperature of the bath.
Solution:
R₀ = 5Ω, R₁₀₀ = 5.23 Ω and R_t = 5.795Ω
Now, t = \(\frac{R_t-R_0}{R_{100}-R_0} \times 100\), R_t = R₀ (1 + αt)
= \(\frac{5.795-5}{5.23-5} \times 100\)
= \(\frac{0.795}{0.23} \times 100\) = 345.65°C

Question 5.
A network of resistors is connected to a 16 V battery with internal resistance of 1Ω, a shown in Fig.
a) Compute the equivalent resistance of the network.
b) Obtain the current in each resistor.
c) Obtain the voltage drops V_AB, V_BC and V_CD.

Solution:
a) The network is a simple series and parallel combination of resistors. First the two 4Ω resistors in parallel are equivalent to a resistor
= [(4 × 4)/(4 + 4)]Ω = 2Ω
In the same way, the 12Ω and 6Ω resistors in parallel are equivalent to a resistor of
[(12 × 6)/(12 + 6)]Ω = 4Ω
The equivalent resistance R of the network is obtained by combining these resistors (2Ω and 4Ω) With 1Ω in series, that is,
R = 2Ω + 4Ω + 1Ω = 7Ω

b) The total current I in the circuit is
I = \(\frac{\varepsilon}{R+r}\) = \(\frac{16 \mathrm{~V}}{(7+1) \Omega}\) = 2A
Consider the resistors between A and B. If I₁ is the current in one of the 4 Ω resistors and I₂ the current in the other.
I₁ × 4 = I₂ × 4
That is, I₁ = I₂, which is otherwise obvious from the symmetry of the two arms. But I₁ + I₂ = I = 2A. Thus,
That is, current in each 4Ω resistor is 1 A. Current in lfi resistor between B and C would be 2 A.
Now, consider the resistances between C and D. If I₃ is the current in the 12Ω resistor, and I₄ in the 6Ω resistor,
I₃ × 12 = I₄ × 6 i.e., I₄ = 2I₃
But, I₃ + I₄ = I = 2A
Thus, I₃ = \(\left(\frac{2}{3}\right)\)A, I₄ = \(\left(\frac{4}{3}\right) \mathrm{A}\)
That is, the current in the 12Ω resistor is (2/3)A, while the current in the 6Ω resistor is (4/3) A.

c) The voltage drop across AB is
V_AB = I₁ × 4 = 1 A × 4Ω = 4V
This can also be obtained by multiplying the total current between A and B by the equivalent resistance between A and B, that is,
V_A = 2A × 2Ω = 4V
The voltage drop across BC is
V_BC = 2A × 1Ω = 2V
Finally, the voltage drop across CD is,
V_CD = 12Ω × I₃ = 12Ω × \(\left(\frac{2}{3}\right) \mathrm{A}\) = 8V.
This can alternately be obtained by multiplying total current between C and D by the equivalent resistance between C and D, that is.
V_CD = 2A × 4Ω = 8V
Note that the total voltage drop across AD is 4V + 2V + 8V = 14V. Thus, the terminal voltage of the battery is 14V, while its emf is 16V The loss of the voltage (= 2V) is accounted for by the internal resistance ID of the battery [2A × 1Ω = 2V].

Question 6.
A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1Ω Fig. Determine the equivalent resistance of the network and the current along each edge of the cube.
Solution:

The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.

The paths AA’. AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners A’, B and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem.
Next take a closed loop, say, ABCC’EA, and apply Kirchhoff’s second rule :
-IR – (1/2)IR – IR + ε = 0
where R is the resistance of each edge and ε the emf of battery. Thus, ε = \(\frac{5}{2}\)IR
The equivalent resistance R_eq of the network is R_eq = \(\frac{\varepsilon}{3 I}\) = \(\frac{5}{6}\)R
For R = IΩ, R,sub>eq = (5/6) Ω and for ε = 10V, the total current (=3I) in the network is 3I = 10V/(5/6) p = 12 A, i.e., I = 4 A
The current flowing in each edge can now be read off from the Fig.

Question 7.
Determine the current in each branch of the network shown in Fig.

Solution:
Each branch of the network is assigned an unknown current to be determined by the application of Krichhoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknown I₁, I₂ and I₃ which can be found by applying the second rule of Krichhoff to three different closed loops. Kirchhoff s second rule for the closed loop ADCA gives,
10 – 4(I₁ – I₂) + 2 (I₂ + I₃ – I₁) – I₁ = 0
that is, 7I₁ – 6I₂ – 2I₃ = 10 ——-> (1)
For the closed loop ABCA, we get
10 – 4I₂ – 2 (I₂ + I₃) – I₁ = 0
that is, I₁ + 6I₂ + 2I₃ = 10 —-—> (2)
For the closed loop BCDEB, we get
5 – 2 (I₂ + I₃) – 2 (I₂ + I₃ – I₁) = 0
that is, 2I₁ – 4I₂ – 4I₃ = -5 ——–> (3)
Equations (1, 2, 3) are three simultaneous equations in three unknowns. These can be solved by the usual method to give.
I₁ = 2.5A, I₂ = \(\frac{5}{8}\)A, I₃ = 1\(\frac{7}{8}\)A
The currents in the various branches of the network are
AB : \(\frac{5}{8}\) A, CA : 2\(\frac{1}{2}\) A, DEB : 1\(\frac{7}{8}\) A
AD = 1\(\frac{7}{8}\) A, CD : 0 A, BC : 2\(\frac{1}{2}\) A

It is easily verified that Krichhoffs second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage drop over the closed loop BADEB
5V + \(\left(\frac{5}{8} \times 4\right)\) – \(\left(\frac{15}{8} \times 4\right) \mathrm{V}\)
equal to zero, as required by Krichhoffs second rule.

Question 8.
The four arms of a Wheatstone bridge (Fig.) have the following resistances :
AB = 100Ω, BC = 10Ω, CD = 5Ω, and DA = 60Ω

A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.
Solution:
Considering the mesh BADB, we have
100I₁ + 15I_g – 60I₂ = 0
or 20I₁ + 3I_g – 12I₂ = 0 —–> (1)
Considering the mesh BCDB, we have
10(I₁ – I_g) – 15I_g – 5(I₂ + I_g) = 0
10I₁ – 30I_g – 5I₂ = 0
2I₁ – 6I_g – I₂ = 0 ——> (2)
Considering the mesh ADCEA,
60I₂ + 5(I₂ + I_g) = 10
65I₂ + 5I_g = 10
13I₂ + I_g = 2 —–> (3)
Multiplying equation (2) by 10 ‘
20I₁ + 60I_g – 10I₂ = 0
From Equations. (4) and (1) we have  ——> (4)
63I_g + 2I₂ = 0
I₂ = 31.5I_g
Substituting the value of I₂ into Equation (3) we get.
13(31.5I_g) + I_g = 2
410.5 I_g = 2
I_g = 4.87 mA.

Question 9.
In a metre bridge (Fig.), the null point is found at a distance of 36.7 cm from A. If now a resistance of 12Ω is connected in parallel with S, the null point occurs at 51.9 cm. Determine the values of R and S.

Solution:
From the first balance point, we get
\(\frac{\mathrm{R}}{\mathrm{S}}\) = \(\frac{33.7}{66.3}\) —–> (1)
After S is connected in parallel with a resistance of 12Ω, the resistance across the gap changes from S to S_eq, where
S_eq = \(\frac{12 \mathrm{~S}}{\mathrm{~S}+12}\) —–> (2)
and hence the new balance condition now gives
\(\frac{51.9}{48.1}\) = \(\frac{\mathrm{R}}{\mathrm{S}_{\mathrm{eq}}}\) = \(\frac{R(S+12)}{12 S}\)
Substituting the value of R/S from Equation (1), we get
\(\frac{51.9}{48.1}\) = \(\frac{\mathrm{S}+12}{12}\) . \(\frac{33.7}{66.3}\)
Which gives S = 13.5 Ω. Using the value of R/S above, we get R = 6.86 Ω.

Question 10.
A resistance of R Ω draws current from a potentiometer. The potentiometer has a total resistance R₀ Ω. (Figure) A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

Solution:
While the slide is in the middle of, the potentiometer only half of its resistance (R₀/2) will be between the points A and B. Hence, the total resistance between A and B, say, R₁, will be given by the following expression.
\(\frac{1}{\mathrm{R}_1}\) = \(\frac{1}{R}\) + \(\frac{1}{\left(\mathrm{R}_0 / 2\right)}\)
R₁ = \(\frac{\mathrm{R}_0 \mathrm{R}}{\mathrm{R}_0+2 \mathrm{R}}\)
The total resistance between A and C will be süm of resistance between A and B and B and C, i.e., R₁ + R₀/2
∴ The current flowing through the potentiometer will be
I = \(\frac{\mathrm{V}}{\mathrm{R}_1+\mathrm{R}_0 / 2}\) = \(\frac{2 \mathrm{~V}}{2 \mathrm{R}_1+\mathrm{R}_0}\)
The voltage V₁ taken from the potentiometer will be the product of current I and resistance R₁,

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 5th Lesson Electrostatic Potential and Capacitance Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 5th Lesson Electrostatic Potential and Capacitance

Very Short Answer Questions

Question 1.
Can there be electric potential at a point with zero electric intensity? Give an example.
Answer:
Yes, There can exist potential at a point where the electric intensity is zero.

Ex :

1. Between two similar charges intensity of electric field is zero. But potential is not zero.
2. Inside the charged spherical conductor electric field intensity is zero but potential is not zero.

Question 2.
Can there be electric intensity at a point with zero electric potential ? Give an example.
Answer:
Yes, electic intensity need not be zero at a point where the potential is zero.

Ex :
1) At mid point between two equal opposite charges potential is zero. But intensity is not zero.

Question 3.
What are meant by equipotential surfaces ?
Answer:
Surface at every point of which the value of potential is the same is defined as equipotential surface
For a point charge, concentric spheres centred at a location of the charge are equipotential surfaces.

Question 4.
Why is the electric field always at right angles to the equipotential surface ? Explain.
Answer:
No work is done in moving a charge from one point on equipotential surface to the other. Therefore, component of electric field intensity along the equipotential surface is zero. Hence, the surface is perpendicular to the field lines.

Question 5.
Three capacitors of capacitances 1μF, 2μF and 3μF, are connected in parallel
(a) What is the ratio of charges ?
(b) What is the ratio of potential difference ?
Answer:
When capacitors are connnected in parallel
(a) q₁ : q₂ : q₃ = V: C₂ V: C₃ V = 1μF : 2μF : 3μF

∴ q₁ : q₂ ; q₃ = 1 : 2 : 3
(b) V₁ : V₂ : V₃ = V : V : V = 1 : 1 : 1

Question 6.
Three capacitors of capacitances 1μE, 2μF and 3μF are connected in series
(a) What is the ratio of charges ?
(b) What is the ratio of potential differences ?
Answer:
When capacitors are connected in series

(a) q₁ : q₂ : q₃ = q : q : q = 1 : 1 ; 1
(b) V₁ : V₂ : V₃ = \(\frac{\mathrm{q}}{\mathrm{C}_1}: \frac{\mathrm{q}}{\mathrm{C}_2}: \frac{\mathrm{q}}{\mathrm{C}_3}\) = \(\frac{1}{1}: \frac{1}{2}: \frac{1}{3}\)
∴ V₁ : V₂ : V₃ = 6 : 3 : 2

Question 7.
What happens to the capacitance of a parallel plate capacitor if the area of its plates is doubled ?
Answer:
\(\frac{C_2}{C_1}\) = \(\frac{A_2}{A_1}\) [∵ C₂ = 2C₁]
Given A₂ = 2A₁ ; \(\frac{C_2}{C_1}\) = \(\frac{2 \mathrm{~A}_1}{\mathrm{~A}_1}\) [∴ C₂ = 2C₁]
Therefore capacity increases by twice.

Question 8.
The dielectric strength of air is 3 × 10⁶.Vm^-1 at certain pressure, A parallel plate capacitor with air in between the plates has a plate separation of 1 cm. Can you charge the capacitor
to 3 × 10⁶V?
Answer:
Dielectric strength of air E₀ = 3 × 10⁶ Vm^-1
Electric field intensity between the plates, E = \(\frac{E_0}{K}\) = 3 × 10⁶ Vm^-1 [∵ for air K = 1]
Distance between two plates, d = 1 cm = 102m
Electric potential difference between plates, V = Ed = 3 × 10⁶ × 10^-2
∴ V = 3 × 10⁴ Volt.
Hence we cant charge the capacitor upto 3 × 10⁶ Volt.

Short Answer Questions

Question 1.
Derive an expression for the electric potential due to a point charge. (T.S. Mar. ’16)
Answer:
Expression for the electric potential due to a point charge:

1. Electric potential at a point is defined as the amount of workdone in moving a unit +ve charge from infinity to that point.
   
2. Consider a point P at a distance r from the point charge having charge + q. The electric field at P = E = \(\frac{q}{4 \pi \varepsilon_0 x^2}\)
3. Workdone in taking a unit +ve charge from B to A = dV = -E.dx (-ve sign shows that the workdone is +ve in the direction B to A, Whereas the potential difference is +ve in, the direction A to B.
4. Therefore, potential at P = The amount of workdone in taking a unit +ve charge from infinity to P
   

Question 2.
Derive an expression for the electrostatic potential energy of a system of two point charges and find its relation with electric potential of a charge.
Answer:
Expression for the electrostatic potential energy of a system of two point charges:

1. Let two point charges q₁ and q₂ are separated by distance ‘r’ in space.
2. An electric field will develop around the charge q₁
   
3. To bring a charge q2 from infinity to the point B some work must be done.
   workdone = q₂ V_B
   But V_B = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1}{\mathrm{r}}\)
   W = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\)
4. This amount workdone is stored as electrostatic potential energy (U) of a system of two charged particles. Its unit is joule.
   ∴ U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
5. If the two charges are similar then ‘U’ is positive. This is in accordance with the fact that two similar charges repel one another and positive work has to be done on the system to bring the charges nearer.
6. Conversely if the two charges are of opposite sign, they attract one another and potential energy is negative.

Question 3.
Derive an expression for the potential energy of an electric dipole placed in a uniform electric field.
Answer:
Expression for potential energy of an electric dipole placed in a uniform electric field:

1. Consider a electric dipole of length 2a having charges + q and -q.
2. The electric dipole is placed in uniform electric field E and its axis makes an angle θ with E.
   
3. Force on charges are equal but opposite sign. They constitute torque on the dipole.
   Torque \(\tau\) = one of its force (F) × ⊥^r distance (BC)
   F = qE and sinθ = \(\frac{\mathrm{BC}}{2 \mathrm{a}}\) ⇒ BC = 2a sinθ
   ∴ Torque \(\tau\) = qE × 2a sinθ = PE sin θ [∴ p = 2aq]
4. Suppose the dipole is rotated through an angle dθ, the workdone dw is given by
   dw = tdθ = PE sinθ dθ
5. For rotating the dipole from angle θ₁ to θ₂,
   workdone W = \(\int_{\theta_1}^{\theta_2} \mathrm{PE} \sin \theta d \theta\) = PE(cosθ₁ – cosθ₂)
6. This workdone (W) is then stored as potential energy(U) in the dipole.
   ∴ U = PE(cosθ₁ – cosθ₂)
7. If θ₁ = 90°and θ₂ = 0°, U = -PE cosθ.
   In vector form U = \(-\vec{P} \cdot \vec{E}\)

Question 4.
Derive an expression for the capacitance of a parallel plate capacitor. (Mar.’16 (AP) Mar ’14)
Answer:
Expression for the capacitance of a parallel plate capacitor:

1. P and Q are two parallel plates of a capacitor separated by a distance of d.
2. The area of each plate is A. The plate P is charged and Q is earth connected.
3. The charge on P is + q and surface charge density of charge = σ
   ∴ q = Aσ
   
4. The electric intensity at poiñt x, E = \(\frac{|\sigma|}{\varepsilon_0}\)
5. Potential difference between the plates P and Q,
   
6. Capacitance of the capacitor Farads (In air)
   Note : Capacity of a capacitor with dielectric medium is C = \(\frac{\varepsilon_0 \mathrm{~A}}{\left[\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathbf{k}}\right]}\) Farads.

Question 5.
Explain the behaviour of dielectrics in an external field. (A.P. Mar. ’19)
Answer:

1) When an external field is applied across dielectrics, the centre of positive charge distribution shifts in the direction of electric field and that of the negative charge distribution shifts

opposite to the electric field and induce a net electric field within the medium opposite to the external field. In such situation the molecules are said to be polarised.

2) Now consider a capacitor with a dielectric between the plates. The net field in the dielectric becomes less.

3) If E₀ the external field strength and E₁ is the electric field strength induced, then the net field \(\overrightarrow{\mathrm{E}}_{\text {net }}\) = \(\overrightarrow{\mathrm{E}}_0\) + \(\overrightarrow{\mathrm{E}}_1\)
(E_net) = E₀ – E_i = \(\frac{E}{K}\) where K is the dielectric constant of the medium.

Long Answer Questions

Question 1.
Define electric potential. Derive and expression for the electric potential due to an electric dipole and hence the electric potential at a point (a) the axial line of electric dipole (b) on the equatorial line of electric dipole.
Answer:
Electric potential (V) : The workdone by a unit positive charge from infinite to a point in an electric field is called electric potential.
Expression for the potential at a point due to a dipole:

1. Consider A and B having -q and + q charges separated by a distance 2a.
2. The electric dipole moment P = q × 2a along AB
3. The electric potential at the point ‘P’ is to be calculated.
4. P is at a distance ‘r’ from the point ‘O’. θ is the angle between the line OP and AB.
5. BN and AM are perpendicular to OP.
6. Potential at P due to charge + q at B,
   
7. Potential at P due to charge -q at A, V₂ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{-\mathrm{q}}{\mathrm{AP}}\right]\)
   ∴ V₂ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{-\mathrm{q}}{\mathrm{MP}}\right]\) [∵ BP = NP]
8. Therefore, Resultant potential at P is V = V₁ + V₂
   V = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{N P}-\frac{q}{M P}\right]\) …… (1)
9. In Δle ONB, ON = OB cosθ = a cosθ; ∴ NP = OP – ON = r – a cosθ ….. (2)
10. In Δle AMO, OM = AO cosθ = a cosθ; ∴ MP = MO + OP = r + a cos θ ….. (3)
11. Substituting (2) and (3) in (1), we get
    
12. As r > >a, a² cos²θ can be neglected with comparision of r².
    ∴ V = \(\frac{\mathrm{P} \cos \theta}{4 \pi \varepsilon_0 \mathbf{r}^2}\)
13. Electric potential on the axial line of dipole :

(i) When θ = 0°, point p lies on the side of + q
∴ V = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 0° = 1]

ii) When θ = 180°, point p lies on the side of -q.
∴ V = \(\frac{-\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 180° = -1]

b) Electric potential on the equitorial line of the diopole:
when θ = 90°, point P lies on the equitorial line.
∴ V = o [∵ cos 90° = 0]

Question 2.
Explain series and parallel combination of capacitors. Derive the formula for equivalent capacitance in each combination. (AP. & T.S. Mar. ‘15)
Answer:
Series combination : If a number of condensers are connected end to end between the fixed points then such combination is called series.

In this combination

1. Charge on each capacitor is equal.
2. PD’s across the capacitors is not equal.

Consider three capacitors of capacitanceš C₁, C₂ and C₃ are connected in series across a battery of P.D ‘V’ as shown in figure.

Let ‘Q’ be the charge on each capacitor.
Let V₁, V₂ and V₃ be the P.D’s of three
V = V₁ + V₂ + V₃ —— (1)
P.D. across I^st condenser V₁ = \(\frac{\mathrm{Q}}{\mathrm{C}_1}\)
P.D. across II^nd condenser V₂ = \(\frac{\mathrm{Q}}{\mathrm{C}_2}\)
RD across III^rd condenser V₃ = \(\frac{\mathrm{Q}}{\mathrm{C}_3}\)
∴ From the equation (1), V = V₁ + V₂ + V₃
= \(\frac{\mathrm{Q}}{\mathrm{C}_1}\) + \(\frac{\mathrm{Q}}{\mathrm{C}_2}\) + \(\frac{\mathrm{Q}}{\mathrm{C}_3}\) = Q\(\left[\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\right]\)

For ‘n’ number of capacitors, the effective capacitance can be written as

Parallel Combination : The first plates of different capacitors are connected at one terminal and all the second plates of the capacitors are connected at another terminal then the two terminals are connected to the two terminals of battery is called parallel combination.

In this combination,

1. The PD’s between each capacitor is equal (or) same.
2. Charge on each capacitor is not equal. Consider three capacitors of capacitancé C₁, C₂ and C₃ are connected in parallel across a RD ‘V’ as shown in fig.

The charge on Ist capacitor Q₁ = C₁ V
The charge on IInd capacitor Q₂ = C₂ V
The charge on IIIrd capacitor Q₂ = C₃ V
∴ The total charge Q = Q₁ + Q₂ + Q₃
= C₁ V + C₂ V + C₃ V
Q = V(C₁ + C₂ + C₃) ⇒ \(\frac{Q}{V}\) = C₁ + C₂ + C₃
C = C₁ + C₂ + C₃ [ ∵ C = \(\frac{\mathrm{Q}}{\mathrm{V}}\)]
for ‘n’ number of capacitors connected in parallel, the equivalent capacitance can be written as C = C₁ + C₂ + C₃ + …. + C_n

Question 3.
Derive an expression for the energy stored in a capacitor. What is the energy stored when the space between the plates is filled with a dielectric
(a) with charging battery disconnected?
(b) with charging battery connected in the circuit?
Answer:
Expression for the energy stored in a capacitor : Consider an uncharged capacitor of capacitance ‘c’ and its initial will be zero. Now it is connected across a battery for charging then the final potential difference across the capacitor be V and final charge on the capacitor be ‘Q’
∴ Average potential difference V_A = \(\frac{\mathrm{O}+\mathrm{V}}{2}\) = \(\frac{\mathrm{V}}{2}\)
Hence workdone to move the charge Q = W = V_A × Q = \(\frac{\mathrm{VQ}}{2}\)
This is stored as electrostatic potential energy U’
∴ U = \(\frac{\mathrm{QV}}{2}\)
We know Q = CV then ‘U’ can be written as given below.
U = \(\frac{\mathrm{QV}}{2}\) = \(\frac{1}{2}\) CV² = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)
∴ Energy stored in a capacitor
U = \(\frac{\mathrm{QV}}{2}\) = \(\frac{1}{2}\) CV² = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)

Effect of Dielectric on energy stored :

Case (a) : When the charging battery is disconnected from the circuit:
Let the capacitor is charged by a battery and the disconnected from the circuit. Now the space between the plates is filled with a dielectric of dielectric constant ‘K’ then potential decreases by \(\frac{1}{K}\) times and charge remains constant.
Capacity increases by ‘K times

∴ Energy stored decreases by \(\frac{1}{\mathrm{~K}}\) times.

Case (b): When the charging battery is connected in the circuit:
Let the charging battery is continue the supply of charge. When the dielectric is introduced then potential decreases by \(\frac{1}{\mathrm{~K}}\) times and charge on the plates increases until the potential difference attains the original value = V ,
New charge on the plates Q’ = KQ
Hence new capacity C’ = \(\frac{Q^{\prime}}{V}=\frac{K Q}{V}\) = KC
Energy stored in the capacitor U’ = \(\frac{1}{2}\)Q’V = \(\frac{1}{2}\)(KQ) V = KU
U’ = KU
∴ Energy stored in the capacitor increases by ‘K times.

Problems

Question 1.
An elementary particle of mass ‘m’ and charge +e initially at a very large distance is projected with velocity ‘v’ at a much more massive particle of charge + Ze at rest. The closest possible distance of approach of the incident particle is
Solution:
For an elementary particle, mass = m; charge = +e; velocity = v.
For much more massive particle, charge = + Ze
From law of conservation of energy, we have
K.E of elementary particles = Electrostatic potential energy of elementary particle at a closest distance (d)

∴ The closest possible distance of approach of the incident particle,
d = \(\frac{Z e^2}{2 \pi \varepsilon_0 m v^2}\)

Question 2.
In a hydrogen atom the electron and proton are at a distance of 0.5 A. The dipole moment of the system is
Answer:
In a hydrogen atom the charge of an electron = -1.6 × 10^-19C

In a hydrogen atom the charge of proton,
q_p = +1.6 × 10^-19C
The distance between the proton and an electron
2a = 0.5A = 0.5 × 10^-10 m
The dipole moment of the system,
P = 2a × q_p = 0.5 × 10^-10 × 1.6 × 10
∴ P = 8 × 10^-30cm

Question 3.
There is a uniform electric field in the XOY plane represented by (40\(\hat{i}\) + 30\(\hat{j}\)) Vm^-1. If the electric potential at the origin is 200 V, the electric potential at the point with coordinates (2m, 1m) is
Answer:
Given, uniform Electric field intensity,
\(\overrightarrow{\mathrm{E}}\) = (40\(\hat{i}\) + 30\(\hat{j}\)) Vm^-1
Electric potential at the origin = 200V
Position vector d\(\begin{aligned}
&\rightarrow \\
&\mathrm{r}
\end{aligned}\) = (2\(\hat{i}\) + 1\(\hat{j}\)) m
We know that,

V_p – V_o = -(80 + 30) = -110Volt
V_p = V_o – 110 = (200 – 110) Volt = 90 Volt
∴ potential at point P, V_p = 90Volt.

Question 4.
An equilateral triangle has a side length L. A charge +q is kept at the centroid of the triangle. P is a point on the perimeter of the triangle. The ratio of the minimum and maximum possible electric potentials for the point P is
Answer:
Charge at the centroid of an equilateral triangle = +q
The charge + q divides the line segment in ratió 2 : 1.
That means r_max = 2 and r_min = 1
V_min = \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{r_{\max }}\) and V_max = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}_{\min }}\)

Question 5.
ABC is an equilateral triangle of side 2m. There is a uniform electric field of intensity 100V/m in the plane of the triangle and parallel to BC as shown. If the electric potential at A is 200 V, then the electric potentials at B and C.
Answer:
Given length of side of an equilateral triangle a = 2m
E = 100V/m; V_A = 200V
Let D be the mid point between B and C Potential at D = V_D = 200V

From fig V_B – V_D = Ed
⇒ V_B – 200 = 100 × 1
∴ Potential at B, V_B = 300 V And V_D – V_C = Ed
200 – V_C = 100 × 1
∴ Potential at C, V_C = 100V.

Question 6.
An electric dipole of moment p is placed in a uniform electric field E, with p parallel to E. It is then rotated by an angle q. The work done is
Solution:
Let AB be a electric dipole having charges -q and + q
Electric dipole moment of AB = p
Electric field = E The workdone by a dipole, when it is rotated through an angle q from E,

W = \(\int_0^q p E \sin \theta d \theta\)
⇒ W = pE \([\cos \theta]_0^q\) = pE (cos0° – cosq)
∴ W = pE(1 – cosq)

Question 7.
Three identical metal plates each of area ‘A’ are arranged parallel to each other, ’d’ is the distance between the plates as shown. A battery of V volts is connected; as shown. The charge stored in the system of plates is

Answer:
Area of each plate = A
distance between two plates = d capacity of each, parallel plate capacitor,
C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
Two capacitors are connected in parallel as shown in figure
Equivalent capacity of two capacitors connected in parallel, C_p = 2C = \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
Charge stored in the system of plates,
q = C_pV = \(\frac{2 \varepsilon_0 A}{d} V\)
∴ q = \(\frac{2 A \varepsilon_0 \mathrm{~V}}{\mathrm{~d}}\)

Question 8.
Four identical metal plates each of area A are separated mutually by a distance d and are connected as shown. Find the capacity of the system between the terminals A and B.

Solution:
Area of each plate of a capacitor = A
Distance between two parallel plate capacitors = d
Capacity of each parallel plate capacitor,
C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
Given fig is

The equivalent circuit of the above fig is

Question 9.
In the circuit shown the battery of ‘V’ volts has no internal resistance. All three condensers are equal in capacity. Find the condenser that carries more charge ?

Answer:
The equivalent circuit to the given circuit is as shown

In series combination of capacitors, charge q flows through each capacitor. Then q₁ = q = C₁V₁; q₂ – q = C₂V₂; q₃ = q = C₃V₃
∴ q₁ = q₂ = q₃
Hence three capacitors C₁, C₂ and C₃ carry the same charge.

Question 10.
Two capacitors A and B of capacities C and 2C are connected in parallel and the combination is connected to a battery of V volts. After the charging is over, the battery is removed. Now a dielectric slab of K = 2 is inserted between the plates of A so as to fill the completely. The energy lost by the system during the sharing of charges is
Solution:
i) With battery of parallel combination:
C₁ = C;C₂ = 2C; V = V
C_p = C₁ + C₂ = 3C; q = 3Cv

ii) Without battery of parallel combination : .

Question 11.
A condenser of certain capacity is charged to a potential V and stores some energy. A second condenser of twice the capacity is to store half the energy of the first, find to what potential one must be charged?
Answer:
For first capacitor, C₁ = C; V₁ = V
And U₁ = \(\frac{1}{2}\) C₁V\(V_1^2\) = \(\frac{1}{2}\) CV² …….. (1)
For second capacitor, C₂ = 2C₁ = 2C;
U₂ = \(\frac{\mathrm{U}_1}{2}\) = \(\frac{1}{4} \mathrm{CV}^2\); Let potential difference across the capacitor = V₂

Textual Exercises

Question 1.
Two charges 5 × 10^-8 C and -3 × 10^-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero.
Solution:
Here q1 = 5 × 10^-8C, q₂ = -3 × 10^-8C
Let the potential be zero at a distance x cm from the charge q₁ = 5 × 10^-8C.
∴ r₁ = x × 10^-2m
r₂ = (16 – x) × 10^-2m
Now V = \(\frac{\mathrm{q}_1}{4 \pi \varepsilon_0 \mathrm{r}_1}\) + \(\frac{\mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{r}_2}\)
= \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}\right]\)
∴ \(\frac{\mathrm{q}_1}{\mathrm{r}_1}\) = \(\frac{-\mathrm{q}_2}{\mathrm{r}_2}\)
= \(\frac{5 \times 10^{-8}}{x \times 10^{-2}}\) = \(\frac{-\left(-3 \times 10^{-8}\right)}{(16-x) 10^{-2}}\) or \(\frac{5}{x}\) = \(\frac{3}{16-x}\)
3x = 80 – 5x
8x = 80, x = 10cm

Question 2.
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.
Answer:
In fig. O is centre of hexagon ABCDEFA of each side 10 cm, As it clear from the figure, OAB, OBC etc. are equilateral triangles.
Therefore
OA = OB = OC = OD = OE = OF = r = 10 cm = 10^-1m
As potential is scalar, there for C potential at O is

Question 3.
Two charges 2 μC and -2 μC are placed at points A and B 6 cm apart.
a) Identify an equipotential surface of the system.
b) What is the direction of the electric field at every point on this surface ?
Solution:
a) The plane normal to AB and passing through its middle point has zero potential everywhere.
b) Normal to the plane in the direction AB.

Question 4.
A spherical conductor of radius 12 cm has a charge of 1.6 × 10^-7C distributed uniformly on its surface. What is the electric field.
a) inside the sphere
b) just outside the sphere
c) At a point 18 cm from the centre of the sphere ?
Solution:
a) Here r = 12 cm = 12 × 10^-2m,
q = 1.6 × 10^-7C. Inside the sphere, E = 0

b) Just coincide the sphere (say on the surface of the sphere)
E = \(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}\)
= 9 × 10⁹ × \(\frac{1.6 \times 10^{-7}}{\left(12 \times 10^{-2}\right)^2}\) = 10⁵ N/c

c) At r = irm = 18 × 10^-2
E = \(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) = \(\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{\left(18 \times 10^{-2}\right)^2}\)
= 4.4 × 10⁴ N/C

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^-12F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?
Solution:
C₁ = \(\frac{\varepsilon_0 A}{d}\) = 8pF
C₂ = k\(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}\) = \(\frac{6 \times 2 \varepsilon_0 A}{d}\) = 12 × 8 = 96pF.

Question 6.
Three capacitors each of capacitance 9 pF are connected in series.
a) What is the total capacitance of the combination ?
b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?
Solution:
a) \(\frac{1}{C_S}\) = \(\frac{1}{9}\) + \(\frac{1}{9}\) + \(\frac{1}{9}\) = \(\frac{3}{9}\) = \(\frac{1}{3}\) ; C_s = 3pF
\(\frac{\mathrm{V}}{3}\) = \(\frac{\mathrm{120}}{3}\) = 40V

b) P.d across each capacitor =
\(\frac{\mathrm{V}}{3}\) = \(\frac{\mathrm{120}}{3}\) = 40V

Question 7.
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
a) What is the total capacitance of the combination ?
b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Solution:
a) C_p = 2 + 3 + 4 = 9pF
b) For each capacitor, V is same = 100 Volt
q₁ = C₁V₁ = 2 × 100 = 200pC
q₂ = C₂V = 3 × 100 = 300pC
q₃ = C₃V = 4 × 100 = 400pC

Question 8.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^-3 m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor ?
Solution:
Here A = 6 × 10^-3m², d = 3mm = 3 × 10^-3m, C = ?
V = 100V, q = ?
C₀ = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{\left(8.85 \times 10^{-12}\right)\left(6 \times 10^{-3}\right)}{3 \times 10^{-3}}\)
= 1.77 × 10^-11F
q = C₀V= 1.77 × 10^-11 × 100
= 1.77 × 10^-9C

Question 9.
Explain what would happen if in the capacitor given in Exercise 8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates.
a) While the voltage supply remained connected.
b) after the supply was disconnected.
Solution:
a) Capacity increases to C = KC₀
= 6 × 1.77 × 10^-11F
charge increases to
q¹ = C¹V = 6 × 1.77 × 10^-11 × 10²C

b) After the supply was disconnected new capacity C = KC₀ = 6 × 1.77 × 10^-11F
New voltage V¹ = \(\frac{q}{C^1}\) = \(\frac{1.77 \times 10^{-9}}{6 \times 1.77 \times 10^{-11}}\)
= 16.67V

Question 10.
A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor ?
Solution:
Here C = 12pF = 12 × 10^-12F, V = 50Volt, E = ?
E = \(\frac{1}{2} \mathrm{CV}^2\) = \(\frac{1}{2}\left(12 \times 10^{-12}\right)(50)^2\)
= 1.5 × 10^-8J

Question 11.
A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Solution:
Here C₁ = C₂ = 600pF = 600 × 10^-12
F = 6 × 10^-10F,
V₁ = 200 V, V₂ = o

Additional Exercises

Question 1.
A charge of 8mC is located at the origin. Calculate the workdone in taking a small charge of’ -2 × 10^-9 C from a point P(0, 0, 3 cm) to a point Q(0,4 cm, 0), Via a point R(0,6 cm, 9 cm).
Solution:
From fig. a charge q = 8mc = 8 × 10^-3C is located at the origin O. Charge to be carried is
q₀ = -2 × 10^-9C from P to Q
Where OP = r_p = 3 cm
= 3 × 10^-2m and OQ = r_Q = 4cm = 4 × 10^-2m
As electrostatic forces are conservative forces, workdone is independent of the path. Therefore there is no relevance of point R.

Question 2.
A cube of side b has a charge q at each of its vertices, Determine the potential and electric field due to this charge array at the centre of the cube.
Answer:
We know that the length of diagonal of thè cube of each side b is \(\sqrt{3 b^2}\) = \(\mathrm{b} \sqrt{3}\)
Distance between centre of the cube and each vertex r = \(\frac{b \sqrt{3}}{2}\)
V = \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{r}\)
and 8 charges each of valué q are present at the eight vertices, of the cube therefore

∴ V = \(\frac{1}{4 \pi \varepsilon_0} \frac{8 q}{b \sqrt{3} / 2}\) or V = \(\frac{4 \mathrm{q}}{\sqrt{3} \pi \varepsilon_0 \mathrm{~b}}\)
Further electric field intensity at the centre due to all the eight charges is zero because the fields due to individual charges cancel in pairs.

Question 3.
Two tiny spheres carrying charges 1.5μC and 2.5μC are located 30 cm apart. Find the potential and electric field
a) at the mid-point of the line joining the two charges and
b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
Solution:
Here q₁ = 1.5,C = 1.5 × 10^-6 C,
q₁ = 2.5μC = 2.5 × 10^-6C
Distance between the two spheres = 30cm
from figure

b) Let P be the point in a plane normal to the line passing through the mid point,
where OP 10cm = 0. 1m
From figure,
Now PA = PB

Resultant field intensity at P is
E = \(\sqrt{\mathrm{E}_1^2+\mathrm{E}_2^2+2 \mathrm{E}_1 \mathrm{E}_2 \cos \theta}\)
E = 6.58 × 10⁵ Vm^-1
Let θ be the angle which resultant intensity \(\overrightarrow{\mathrm{E}}\) makes with \(\overrightarrow{\mathrm{E}_1}\).

Question 4.
A spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge Q.
a) A charge q is placed at the centre of the shell. What ¡s the surface charge density on the inner 4nd outer surfaces of the shell?
Answer:
a) The charge of + Q resides on the Outer surface of the shell. The charge q placed at the centre of the shell induces charge -q on the inner surface and charge + q on the outer surface of the shell, from figure.
∴ Total charge on inner surface of the shell is -q and total charge on the outer surface of the shell is (Q + q)
σ₁ = \(\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}\) and σ₁ = \(\frac{\mathrm{Q}_1+\mathrm{q}}{4 \pi \mathrm{r}_2^2}\)

b) Is the electric field inside a cavity (with no charge) zero, even If the shell is not spherical, but has any irregular shape? Explain.
Solution:
Electric field intensity inside a cavity with no charge is zero, eveñ when the shell has

any irregular shape. If we were to take a closed loop part of which is inside the cavity along a field line and the rest outside it then network done by the field in carrying a rest charge over the closed loop will not be zero. This is impossible for an electrostatic field. Hence electric field intensity inside a cavity with no charge is always zero.

Question 5.
a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to
another given by (E₂ – E₁). \(\hat{\mathbf{n}}\) = \(\frac{\sigma}{\varepsilon_0}\)
Where \(\hat{\mathbf{n}}\) is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of \(\hat{\mathbf{n}}\) is from side 1 to side 2.)
Hence show that just out side a conductor, the electric field is σ\(\hat{\mathrm{n}} / \varepsilon_0\)
b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
(Hint: for (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.)
Answer:
a) Normal component of electric field intensity due to a thin infinite plane sheet of charge on left side.
\(\overrightarrow{\mathrm{E}}_1\) = –\(\frac{\sigma}{2 \varepsilon_0} \hat{\mathbf{n}}\)
and on right side 2 = \(\overrightarrow{\mathrm{E}_2}\) = \(\frac{\sigma}{2 \varepsilon_0} \hat{\mathrm{n}}\)
Discontinuity in the normal component from one side to the other is

b) To show that the tangential component of electostatic field is continous from one side of a charged surface to another, we use the fact that workdone by electrostatic field on a closed loop is zero.

Question 6.
A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Solution:
From figure A is a long charged cylinder of linear charge density λ, lengh l and radius a. A hollow co-axial conducting cylinder B of length L and radius b surrounds A.

The charge q = λL spreads uniformly on the outer surface of λ. It induces – q charge on the cylinder B. which spreads on the inner surface of B. An electric field \(\overrightarrow{\mathrm{E}}\) is produced in the space between the two cylinders which is directed radically outwards. Let us consider a co-axial cylindrical Gaussian surface of radius r. The electric flux through the cylindrical Gaussian surface is

The Electric flux through the end faces of the cylindrical Gaussian surface is zero as \(\overrightarrow{\mathrm{E}}\) is parallel to them. According to Gauss’s theorem

Question 7.
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53Å:
a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A separation ?
Solution:
a) Here q₁ = -1.6 × 10^-19C; q₂ = + 1.6 × 10^-19C.
r = 0.53λ = 0.53 × 10^-10m
Potential energy = P.E at ∞ -P. E at r
= 0 – \(\frac{q_1 q_2}{4 \pi \varepsilon_0 r}\) = \(\frac{-9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{0.53 \times 10^{-10}}\)
= -43.47 × 10^-19 Joule
= \(\frac{-43.47 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}\) = -27.16 eV

b) K.E in the Orbit = \(\frac{1}{2}\) (27.16) eV
Total energy = K.E + P.E
= 13.58 – 27.16 = – 13.58 eV
Work required to free the electron = 13.58eV

c) Potential energy at a seperation of r₁ (= 1.06A) is
= \(\frac{\mathrm{q}_1 \mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{r}_1}\) = \(\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)}{1.06 \times 10^{-10}}\)
= 21.73 × 10^-19J = 13.58eV
Potential energy of the system, when zero of P.E is taken at r₁ = 1.06A is
= RE at r₁ -P.E at r = 13.58 – 27.16 = -13.58eV.
By shifting the zero of potential energy work required to free the electron is not affected. It continues to be the same, being equal to + 13.58 eV

Question 8.
If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion \(\mathbf{H}_2^{+}\). In the ground state of an \(\mathbf{H}_2^{+}\), the two protons are separated by roughly 1.5A. and the electron is roughly 1 A from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Solution:
Here q₁ = charge on electron (= -1.6 × 10^-19C)
q₂, q₃ = charge on two protons, each = 1.6 × 10^-19C
r₁₂ = distance between
q₁ and q₂ = 1A = 10^-10m
r₂₃ = distance between
q₂ and q₃ = 1.5A = 1.5 × 10^-10m
r₃₁ = distance between
q₃ and q₁ = 1A = 10^-10m.
Taking zero of potential energy at infinity, we have

Question 9.
Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres ? Use the result obtained to explain why charge density on the sharp and pointed ends’ of a conductor is higher than on its flatter portions.
Solution:
The charge flows from the sphere at higher potential to the other at lower potential till their potentials become equal. After sharing the charges on two spheres would be.
\(\frac{\mathrm{Q}_1}{\mathrm{Q}_2}\) = \(\frac{C_1 V}{C_2 V}\) where C₁. C₂ are the capacities of two spheres.
But \(\frac{\mathrm{C}_1}{\mathrm{C}_2}\) = \(\frac{\mathrm{a}}{\mathrm{b}}\) ∴ \(\frac{\mathrm{Q}_1}{\mathrm{Q}_2}\) = \(\frac{a}{b}\)
Ratio of surface density of charge on the two spheres

Hence ratio of electric fields at the surface of two spheres
\(\frac{\mathrm{E}_1}{\mathrm{E}_2}\) = \(\frac{\sigma_1}{\sigma_2}\) = \(\frac{\mathrm{b}}{\mathrm{a}}\)
A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius. Therefore, charge density on sharp and pointed ends of conductor is much higher than on its flatter portions.

Question 10.
Two charges -q and + q are located at points (0, 0, -a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0)?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> I.
(c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Here -q is at (0, 0, -a) and +q is at (0, 0, a)
a) Potential at (0, 0, z) would be

Potential at (x, y, 0) i.e at a point 1 to z-axis where charges are located is zero

b) we have proved that
V = \(\frac{P \cos \theta}{4 \pi \varepsilon_0\left(r^2+a^2 \cos ^2 \theta\right)}\)
If \(\frac{r}{a}\) >> 1 then a << r ∴ V = \(\frac{p \cos \theta}{4 \pi \varepsilon_0 r^2}\)
∴ V = \(\frac{1}{\mathrm{r}^2}\)
i.e potential is inversely proportional to square of the distance

c) Potential at (5, 0, 0) is

As work done = charge (V₂ – V₁)
W = zero

As work done by electrostatic field is independent of the path connecting the two points therefore work done will
continue to be zero along every path.

Question 11.
Figure shows a charge array known as an electric quadrupole, For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1. and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Answer:
As is clear from figure an electric quadrupole may be regarded as a system of three charges +q, -2q and + q at A, B and C respectively.
Let AC = 2a we have to calculate electric potential at any point P where BP = r, using superposition principle. Potential at p is given by

∴ \(\frac{\mathrm{a}^2}{\mathrm{r}^2}\) is negligibly small V = \(\frac{\mathrm{q} \cdot 2 \mathrm{a}^2}{4 \pi \varepsilon_0 \mathrm{r}^3}\)
clearly V ∝ \(\frac{1}{\mathrm{r}}\)
In case of an electric dipole V ∝ \(\frac{1}{\mathrm{r}^2}\) and in case of an electric monopole
(i.e a single charge), V ∝ \(\frac{1}{\mathrm{r}}\)

Question 12.
An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1kV. A large number of 1μF capacitors are available to him each of which can withstand a potential difference of not more than 400V. Suggest a possible arrangement that requires the minimum number of capacitors.
Answer:
Here total capacitance, C = 2μF
Potential difference v = 1KV = 1000 volt
Capacity of each capacitor C₁ = 1μF
Maximum potential difference across each V = 400 volt

Let n capacitors of 1μF each be connected in series in a row and m such rows be connected in parallel as shown in the figure. As potential difference in each row
= 1000 Volt
∴ Potential difference across each capacitor = \(\frac{1000}{\mathrm{n}}\) = 400
∴ n = \(\frac{1000}{400}\) = 2.5
As n has to be a whole number (not less than 2.5) therefore n = 3
capacitance of each row of 3 condensors of 1μF
Each is series = 1/3
Total capacitance of m such rows in parallel = \(\frac{\mathrm{m}}{3}\)
∴ \(\frac{\mathrm{m}}{3}\) = 2(μf) or m = 6
∴ Total number of capacitors =
n × m = 3 × 6 = 18.
Hence 1μF capacitors should be connected in six parallel rows, each row containing three capacitors in series.

Question 13.
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Answer:
Here A = ? C = 2F,
d = 0.5 cm = 5 × 10^-3m
As C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
∴ A = \(\frac{c d}{\varepsilon_0}=\frac{2 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}}\)
= 1.13 × 10⁹m²
Which is too large.
That is why ordinary capacitors are in the range of μF or less. However in electrolytic capacitors d is too small. Therefore their capacitance is much larger (=0.1F)

Question 14.
Obtain the equivalent capacitance of the network in Fig. For a 300 V supply, determine the charge and voltage across each capacitor.

Answer:
Here C₂ and C₃ are in series
A = \(\frac{c d}{\varepsilon_0}=\frac{2 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}}\)
∴ \(\frac{1}{C_1}\) = \(\frac{1}{200}\) + \(\frac{1}{200}\) = \(\frac{2}{200}\) = \(\frac{1}{100}\)
C_S = 100pF
Now C_S and C₁ are in parallel
∴ C_p = C_s + C₁ = 100 + 100 = 200pF
Again C_p and C₄ are in series
∴ \(\frac{1}{\mathrm{C}_{\mathrm{s}}}\) = \(\frac{1}{\mathrm{C}_{\mathrm{p}}}\) + \(\frac{1}{\mathrm{C}_4}\) = \(\frac{1}{200}\) + \(\frac{1}{100}\) = \(\frac{3}{200}\)
∴ C = \(\frac{200}{3}\)pF = 66.7 × 10^-12F
As C_p and C₄ are in series
∴ V_p + V₄ = 300
Charge on C₄ is q₄
= CV = \(\frac{200}{3}\) × 10^-12 × 300 = 2 × 10^-8C
Potential difference across
C₄ is V₄ = \(\frac{\mathrm{q}_4}{\mathrm{C}_4}\) = \(\frac{2 \times 10^{-8}}{100 \times 10^{-12}}\) = 200V
from (i) V_p = 300 – V₄ = 300 – 200 = 100
Potential difference across
C₁ is V₁ = V_p = 100V
Charge on C₁ = q₁ = C₁V₁
= 100 × 10^-12 × 100 = 10^-8C.
Potential diff across
C₂ and C₃ in series = 100V
charge on C₂;
q₂ = C₂ V₂ = 200 × 10^-12 × 50 = 10^-8C
Charge on C₃;
q₃ = C₃V₃ = 200 × 10^-12 × 50 = 10^-8C

Question 15.
The plates of parallel plate capacitor have an area of 90 cm² each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor ?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u, Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Answer:
a) Here A = 90 cm² = 90 × 1o^-4m²
= 9 × 10^-3m²
d = 2.5mm = 2.5 × 10^-3m

Question 16.
A 4µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?
Answer:
Here C₁ = 4µF = 4 × 10^-6F, V₁ = 200volt. Initial elctrostatic energy stored in C₁ is E₁
= \(\frac{1}{2} C_1 V_1^2\) = \(\frac{1}{2}\) × 4 × 10^-6 × 200 × 200
E₁ = 8 × 10^-2 Joule
When 4µF capacitor is connected to uncharged capacitor of 2µF charge flows and both acquire a common potential.

∴ final electrostatic energy of both capacitors
E₂ = \(\frac{1}{2}\)(C₁ + C₂)V²
= \(\frac{1}{2}\) × 6 × 10^-6 × \(\frac{800}{6}\) × \(\frac{800}{6}\)
E₂ = 5.33 × 10^-2Joule.
Energy dissipated in the form of heat and electro magnetic radiation.
E₁ – E₂ = 8 × 10^-2 – 5.33 × 10^-2
= 2.67 × 10^-2 Joule.

Question 17.
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2.
Answer:
If F is the force on each plates of parallel plate capacitor, then work done in increasing the seperation between the plates by Δx = fΔx

This must be the increase in potential energy of the capacitor Now the increase the volume of capacitor is = A Δx
If U = energy density = energy stored/ volume then the increase in potential energy = U.AΔx
∴ fΔx = U. AΔx

The origin of factor 1/2 in force can be explained by the fact that inside the conductor field is zero and outside the conductor, the field is. E. Therefore the average value of the field (i.e E/2) contributes to the force.

Question 18.
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig.) Show that the capacitance of a spherical capacitor is given by C = \(\frac{4 \pi \varepsilon_0 r_1 r_2}{r_1-r_2}\)

where r₁ and r₂ are the radii of outer and inner spheres, respectively.
Solution:
As is clear from the figure +Q charge spreads uniformly on inner surface of outer sphere of radius r₁. The induced charge – Q spreads uniformly on the outer surface of inner sphere of radius r₂. The outer surface of outer sphere is earthed. Due to electrostatic shielding E = 0 for r < r₂ and E = 0 for r < r₂ and E = 0 for r > r₁

In the space between the two spheres electric intensity E exists as shown. Potential difference between the two spheres.

Question 19.
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere ?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Solution:
Here r_a = 12cm = 12 × 10^-2m
r_b = 13cm = 13 × 10^-2m
q = 2.5μC = 2.5 × 10^-6C E_r = 32
(a) C = ?

(b)
V = ?
V = \(\frac{q}{c}\) = \(\frac{2.5 \times 10^{-6}}{5.5 \times 10^{-9}}\) = 4.5 × 10²Volt

(c) Capacity of an isolated sphere of radius R
R = 12 × 10^-2m is
C¹ = \(4 \pi \varepsilon_0 R\) = \(\frac{1}{9 \times 10^9}\) × 12 × 10^-12
= 1.33 × 10^-11 Farad.

The capacity of an isolated sphere is much smaller because in a capacitor outer sphere is earthed potential difference decreases and capacitance increases.

Question 20.
Answer carefully:
(a) Two large conducting sphers carrying charges Q₁ and Q₂ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q₁ Q₂/\(4 \pi \varepsilon_0 \mathbf{r}^2\). where r is the distance between their centres ?
(b) If Coulomb’s law involved 1/r³ dependence (instead of 1/r²), would Gauss’s law be still true ?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point ?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron ? What if the orbit is elliptical ?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there ?
(f) What meaning would you give to the capacitance of a single conductor ?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
Answer:
a) When the charged spheres are brought close together the charge distributions on them become non-uniform. Therefore, coloumb’s law is not valid hence the magnitude of force is not given exactly by this formula.
b) No Gauss’s law will not be true if coloumb’s law involved 1/r³ dependence instead of 1/r² dependence.
c) The line of force gives the direction of accelaration of charge. If the electric line of force is linear the test charge will move along the line if the line of force is not linear the charge will not go along the line.
d) As force due to the field is discreted towards the nucleus and the electron does not move in the direction of this force, therefore work done is zero when the orbit is circular. This is true even when orbit is elliptical as electric forces are conservative forces.
e) No electric potential is continuous.
f) The capacity of a single conductor implies that the second conductor is infinity.
g) This is because a molecule of water in its normal state has an unsymmetrical shape and therefore it has a permanent dipole moment.

Question 21.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5μC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effets (i.e., bending of field lines at the ends).
Answer:
Here L = 15cms = 15 × 10^-2m
r_a = 1.4cm = 1.4 × 10^-2m,
r_b = 1.5cm = 1.5 × 10^-2m
q = 3.5 μC = 3.5 × 10^-6 coloumb, C = ? V = ?

Question 22.
A parallel plate capacitor is to be designed with a voltage rating 1kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ Vm^-1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e, without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF ?
Answer:
Here V = 1kV= 1000Volt; K = ε_r = 3
Dielectric strength = 10⁷V/m
As electric field at the most should be 10% of dielectric strength due to reasons of safety.
E = 10% of 10⁷V/m = 10⁶V/m A = ?
C = 50pF = 50 × 10^-12F

Question 23.
Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the z – direction.
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction.
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer:
By definition an equipotential surfaces is that every point of which potential is the same. In the four cases given above:
a) Equipotential surfaces are planes parallel to x – y plane. These are equidistant.
b) Equipotential surfaces are planes parallel to x – y plane. As the field increases uniformly distance between the planes decreases.
c) Equipotential surfaces concentric spheres with origin at the centre.
d) Equipotential surfaces have the shape which changes periodically at far off distances from the grid.

Question 24.
In a Van de Graff type generator a spherical metal shell is to be a 15 × 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 10⁷ Vm^-1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build and electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)
Answer:
Here V = 15 × 10⁶ volt .
Dielectric strength = 5 × 10^-7 Vm^-1
minimum rodius, r = ?
max. Electric field E = 10% (dielectric strength)
E = \(\frac{10}{100}\) × 5 × 10⁷ = 5 × 10⁶VM^-1
As E = \(\frac{\mathrm{V}}{\mathrm{r}}\) ∴ r = \(\frac{\mathrm{V}}{\mathrm{E}}\) = \(\frac{15 \times 10^6}{5 \times 10^6}\) = 3m
obviously we cannot build an electrostatic generator, using a very small shell.

Question 25.
A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radium r₂ and charge q₂. Show that if q₁ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q₂ on the shell is.
Answer:
As the charge resides always on the outer surface of the shell therefore, when the sphere and shell are connected by a wire, charge will flow essentially from the sphere to the shell, whatever be the magnitude and sign of charge q₂.

Question 26.
Answer the following :
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm^-1. Why then do we not get an electric shock as we step out of our house into the open ? (Assume the house to be a steel cage so there is no field inside!)
(b) A man fixes outside his house on evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m². Will he get an electric shock if he touches the metal sheet next morning ?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge it self completely in due course and become electrically neutral ? In other words, what keeps the atmosphere charged ?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lighting?
(Hint : The earth has and electric field of about 100 Vm^-1 at its surface in the downward direction, corresponding to a surface charge density = -10^-9Cm^-2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
Answer:
a) since our body and the surface of earth both are conducting therefore our body and the ground form an equipotential surface. As we step out into the open from our house the original equipotential surfaces of open air change, keeping out body and the ground at the same potential that is why we do not get an electric shock.

b) Yes, the man will get a shock This is because the steady discharging current of the atmosphere charges up the aluminium sheet gradually and raises its voltage to an extent depending on the capacitance of the condenser formed by the aluminium sheet and the ground and the insulating slab.

c) The atmosphere is being discharged continuously by understorms and lightning all over the globe. It is also discharging due to the small conductivity of air. The two opposing processes, on an overage, are in equilibrium. Therefore the atmosphere. Keeps charged.

d) During lightning the electric energy of the atmosphere is dissipated in the form of light, heat and sound.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields

Very Short Answer Questions

Question 1.
What is meant by the statement ‘charge is quantized’?
Answer:
The minimum charge that can be transferred from one body to the other is equal to the charge of the electron (e = 1.602 × 10^-19C). A charge always exists as an integral multiple of the charge of the electron (q = ne). Therefore charge is said to be quantized.

Question 2.
Repulsion is the sure test of charging than attraction. Why?
Answer:
A charged body may attract a neutral body and also an oppositely charged body. But it always repels a like-charged body. Hence repulsion is the sure test of electrification.

Question 3.
How many electrons constitute 1 C of charge ?
Answer:
n = \(\frac{q}{e}\) = \(\frac{1}{1.6 \times 10^{-19}}\) = 6.25 × 10¹⁸ electrons

Question 4.
What happens to the weight of a body when it is charged positively ?
Answer:
When a body positively charged it must loose some electrons. Hence weight of the body will decrease.

Question 5.
What happens to the force between two charges if the distance between them is
a) halved
b) doubled ?
Answer:
From Coulombs law, F ∝ \(\frac{1}{\mathrm{~d}^2}\), so
a) When distance is reduced to half, force increases by four times.

b) When distance is doubled, then force is reduced by four times.

Question 6.
The electric lines of force do not intersect. Why ?
Answer:
They do not intersect because if they intersect, at the point of intersection, intensity of electric field must act in two different directions, which is impossible.

Question 7.
Consider two charges + q and -q placed at B and C of an equilateral triangle ABC. For this system, the total charge is zero. But the electric field (intensity) at A which is equidistant from B and C is not zero. Why ?
Answer:
Charges are scalars, but the .electrical intensities are vectors and add vectorially.

Question 8.
Electrostatic field lines of force do not form closed loops. If they form closed loops then the work done in moving a charge along a closed path will not be zero. From the above two statements can you guess the nature of electrostatic force ?
Answer:
It is conservative force.

Question 9.
State Gauss’s law in electrostatics.
Answer:
Gauss’s law : It states that “the total electric flux through any closed surface is equal to – \(\frac{1}{\varepsilon_0}\) times net charge enclosed by the surface”.
\(\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}\) = \(\frac{\mathrm{q}}{\varepsilon_0}\)

Question 10.
When is the electric flux negative and when is it positive ?
Answer:
Electric flux ϕ = \(\vec{E} \cdot \vec{A}\). If angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{A}}\) is 180°, then flux will have a ‘-ve’ sign. We consider the flux flowing out of the surface as positive and flux entering into the surface as negative.

Question 11.
Write the expression for electric intensity due to an infinite long charged wire at a distance radial distance r from the wire.
Answer:
The electric intensity due to an infinitely long charged wire E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\) the conductor.
Where λ = Uniform linear charge density
r = Distance of the point from the conductor.

Question 12.
Write the expression for electric intensity due to an infinite plane sheet of charge.
Answer:
The electric intensity due to an infinite plane sheet of charge is E = \(\frac{\sigma}{2 \varepsilon_0}\).

Question 13.
Write the expression for electric intensity due to a charged conducting spherical shell at points outside and inside the shell.
Answer:
a) Intensity of electric field at any point inside a spherical shell is zero.
b) Intensity of electric field at any point outside a uniformly charged spherical shell is
E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)

Short Answer Questions

Question 1.
State and explain Coulomb’s inverse square law in electricity.
Answer:
Coulomb’s law – Statement: Force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The force acts along the straight line joining the two charges.
Explanation : Let us consider two charges q₁ and q₂ be separated by a distance r.
Then F ∝ q₁q₂ and F ∝ \(\frac{1}{\mathrm{r}^2}\) or F ∝ \(\frac{q_1 q_2}{r^2}\)

∴ F = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}\) where \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10⁹ Nm²C^-2
In vector form, in free space \(\overrightarrow{\mathrm{F}}\) = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2} \hat{\mathrm{r}}\). Here \(\hat{\mathrm{r}}\) is a unit vector.
ε₀ is called permittivity of free space.
ε₀ = 8.85 × 10^-12 C²/N-m² or Farad/meter.
Where ε is called permittivity of the medium.

Question 2.
Define intensity of electric field at a point. Derive an expression for the intensity due to a point charge.
Answer:
Intensity of electric field (E) : Intensity of electric field at any point in an electric field is defined as the force experienced by a unit positive charge placed at that point.
Expression :
1) Intensity of electric field is a vector. It’s direction is along the direction’ of motion of positive charge.
2) Consider point charge q. Electric field will exist around that charge. Consider any point P in that electric field at a distance r from the given charge. A test charge q₀ is placed at R
3) Force acting on q₀ due to q is F = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q q_0}{r^2}\)
4) Intensity of electric field at that point is equal to the force experienced by a test charge q₀.
Intensity of electric field, E = \(\frac{\mathrm{F}}{\mathrm{q}_0}\)

E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\) N/C or V/m

Question 3.
Derive the equation for the couple acting on a electric dipole in a uniform electric field.
Answer:
1) A pair of opposite charges separated by a small distance is called dipole.
2) Consider the charge of dipole are -q and +q coulomb and the distance between them is 2a.
3) Then the electric dipole moment P is given by P = q × 2a = 2aq. It is a vector. It’s direction is from -q to +q along the axis of dipole.
4) It is placed in a uniform electric field E, making an angle 0 with field direction as shown in fig.
5) Due to electric field force on +q is F = +qE and force on -q is F = -qE.

6) These two equal and opposite charges constitute torque or moment of couple.
i. e., torque, \(\tau\) = ⊥^r distance × magnitude of one of force
∴ \(\tau\) = (2a sin θ)qE = 2aqE sin θ = PE sin θ
In vector form, \(\vec{\tau}\) = \(\overrightarrow{\mathrm{P}}\) × \(\overrightarrow{\mathrm{E}}\)

Question 4.
Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.
Answer:
Electric field at a point on the axis of a dipole :
1) Consider an electric dipole consisting of two charges -q and +q separated by a distance ‘2a’ with centre ‘O’.

2) We shall calculate electric field E at point P on the axial line of dipole, and at a distance OP = r.
3) Let E₁ and E₂ be the intensities of electric field at P due to charges +q and -q respectively.
4)

The resultant intensity at P is E = E₁ – E₂ [∵ They are opposite and E₁ > E₂]

If r > > a then a² can be neglected in comparision to r².

In vector form, \(\overrightarrow{\mathrm{E}}\) = \(\frac{2 \overrightarrow{\mathrm{P}}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

Question 5.
Derive an expression for the intensity of the electric field at a point on the equatorial plane of an electric dipole. (A.P. Mar. ’19, ’15)
Answer:
Electric field intensity on equitorial line of electric dipole:
1) Consider an electric dipole consisting of two charges -q and +q separated by a distance ‘2a’ with centre at ‘O’.
2) We shall calculate electric field E at P on equitorial line of dipole and at a distance OP = r.

3) Let E₁ and E₂ be the electric fields at P due to charges +q and -q respectively.
4) The ⊥^r components (E₁ sin θ and E₂ sin θ) cancel each other because they are equal and opposite. The ||^el components (E₁ cos θ and E₂ cos θ) are in the same direction and hence add up.
5) The resultant field intensity at point P is given by E = E₁ cos θ + E₂ cos θ

6) From figure, cos θ = \(\frac{a}{\left(r^2+a^2\right)^{1 / 2}}\)
∴ E = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0} \times \frac{1}{\mathrm{r}^3}\)
7) If r >> a, then a² can be neglected in comparison to r². Then
E = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0} \times \frac{1}{\mathrm{r}^3}\)
In vector form \(\overrightarrow{\mathrm{E}}\) = \(\frac{\overrightarrow{\mathrm{P}}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

Question 6.
State Gauss’s law in electrostatics and explain its importance.
Answer:
Gauss’s law : The total-electric flux through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the net charge enclosed by the surface.
Total electric flux,

Here q is the total charge enclosed by the surface ‘S’, \(\oint\) represents surface integral of the closed surface.

Importance :

1. Gauss’s law is very useful in. calculating the electric field in case of problems where it is possible to construct a closed surface. Such surface is called Gaussian surface.
2. Gauss’s law is true for any closed surface, no matter what its shape or size.
3. Symmetric considerations in many problems make the application of Gauss’s law much easier.

Long Answer Questions

Question 1.
Define electric flux. Applying Gauss’s law and derive the expression for electric intensity due to an infinite long straight charged wise. (Assume that the electric field is everywhere radial and depends only on the radial distance r of the point from the wire.)
Answer:
Electric flux : The number of electric lines of force passing perpendicular to the area is known as electric flux (ϕ). Electric flux ϕ = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\). So flux is a scalar.

Expression for E due to an infinite long straight charged wire :

1) Consider an infinitely long thin straight wire with uniform linear charge density ‘λ’.
2) Linear charge density λ = \(\frac{\text { change q }}{\text { length } l}\) ⇒ λl —– (1)
3) Construct a coaxial cylindrical gaussion surface of length T and radius ‘r’. Due to symmetry we will assume that electric field is radial i.e., normal to the conducting wire.
4) The flat surfaces AB and CD are ⊥^r to the wire. Select small area ds₁ and ds₂ on the surface as AB and CD.
They are ⊥^r to \(\overrightarrow{\mathrm{E}}\). So flux coming out through them is zero.
Since flux ϕ = \(\oint \vec{E} \cdot d \vec{s}\) = Eds cos 90° = 0

5) So flux coming out through the cylindrical surface ABCD is taken into account.

6) From Gauss’s law

7) From (2) and (3), E(2πrl) = \(\frac{Q}{\varepsilon_0}\) = \(\frac{\lambda /}{\varepsilon_0}\) (∵ Q = λl)
∴ E = \(\frac{\lambda l}{2 \pi \varepsilon_0 \mathrm{r} l}=\frac{1}{2 \pi \varepsilon_0} \frac{\lambda}{\mathrm{r}}\)

8) Therefore electric intensity due to an infinitely long conducting wire E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\).

Question 2.
State Gauss’s law in electrostatics. Applying Gauss’s law derive the expression for electric intensity due to an infinite plane sheet of charge.
Answer:
Gauss’s law: The total electric flux through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the net charge enclosed by the surface, i.e.,

Expression for E due to an infinite plane sheet of charge :

1. Consider an infinite plane sheet of charge. Let the charge distribution is uniform on this plane.
2. Uniform charge density on this surface σ = \(\frac{\mathrm{dq}}{\mathrm{dS}}\) where dq is the charge over an infinite small area ds.
3. Construct a horizontal cylindrical Gaussian surface ABCD perpendicular to the plane with length 2r.
4. The flat surfaces BC and AD are parallel to the plane sheet and are at equal distance from the plane.
5. Let area of these surfaces are dS₁ and dS₂. They are parallel to \(\overrightarrow{\mathrm{E}}\). So flux through these two surfaces is
   ——– (1)
   Where S is area of plane surface AD or BC. Both are equal in area and intensity.
6. Consider cylindrical surface of AB and CD. Let their areas are say dS₃ and dS₄. These surfaces are ⊥^lr to electric intensity \(\overrightarrow{\mathrm{E}}\).
7. So angle between \(\overrightarrow{\mathrm{E}}\) and d\(\overrightarrow{\mathbf{s}_3}\) or dS₄ is 90°. Total flux through these, surfaces is zero. Since
   
8. From Gauss’s law total flux, ϕ = \(\oint \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{S}}\) = 2ES = \(\frac{\mathrm{q}}{\varepsilon_0}\)
   ∴ 2ES = \(\frac{\sigma S}{\varepsilon_0}\) [∵ \(\text { Q }\) = σ × S]
9. Therefore intensity of electric field due to an infinite plane sheet of charge E = \(\frac{\sigma}{2 \varepsilon_0}\)

Question 3.
Applying Gauss’s law derive the expression for electric intensity due to a charged conducting spherical shell at
(i) a point outside the shell
(ii) a point on the surface of the shell and
(iii) a point inside the shell.
Answer:
Expression for E due to a charged conducting spherical shell:

1. Consider a uniformly charged spherical shell. Let total charge on it is ‘q’ and its radius is R.
   
2. Since the shell is uniformly charged, the intensity of electric field at any point depends on radial distance ‘r’ from centre ‘O’. The direction of E is away from the centre along the radius.

i) E at a point outside the shell:

1) Consider a point at a distance ‘r’ outside the sphere. Construct a Gaussian surface with ‘r’ as radius (where r > R).

2) Total flux coming out of this sphere is



3) Therefore at any point outside the sphere, E = \(\frac{\sigma}{\varepsilon_0} \frac{\mathrm{R}^2}{\mathrm{r}^2}\)

ii) E at a point on the surface of shell:

1) Construct a Gaussian surface with radius r = R.

2)


3) Therefore intensity at any point on surface of the sphere E = \(\frac{\sigma}{\varepsilon_0}\)

iii) E at a point inside the shell :

1) Consider a point P inside the shell. Construct a Gaussian surface with radius r (where r < R). There is no charge inside the shell. So from Gauss’s law \(\oint_{\mathrm{S}} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{S}}\) = \(\frac{\mathrm{q}}{\varepsilon_0}\)

2) Therefore, intensity of electric field at any point inside a charged shell is zero.

Textual Exercises

Question 1.
Two small identical balls, each of mass 0.20 g, carry identical charges and are suspended by two threads of equal lengths. The balls position themselves at equilibrium such that the angle between the threads is 60°. If the distance between the balls is 0.5 m, find the charge on each ball.
Solution:
Given m = 0.20 g = 0.2 × 10^-3 kg; θ = 60° ⇒ α = \(\frac{\theta}{2}\) = 30°
r = 0.5 m, Let q₁ = q₂ = q

Question 2.
An infinite number of charges-each of magnitude q are placed on x-axis at distance of 1, 2, 4, 8, …….. meter from the origin respectively. Find intensity of the electric field at origin.
Solution:
Let q₁ = q₂ = q₃ = q₄ = ……. = q
r₁ = 1; r₂ = 2; r₃ = 4; r₄ = 8, …….
The resultant electric field at origin ‘O’ is given by
E = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r_1^2}\) + \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_2}{\mathrm{r}_2^2}\) + \(\frac{1}{4 \pi \varepsilon_0} \frac{q_3}{r_3^2}\) + \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_4}{\mathrm{r}_4^2}\) + ……..

Question 3.
A clock face has negative charges -q, -2q, -3q, ….. -12q fixed at the position of the corresponding numerals on the dial. The clock hands do not disturb the net field due to the point charges. At what time does the hour hand point in the direction of the electric field at the centre of the dial ?
Solution:
Let distance of each charge from unit charge at centre ‘O’ = r.
Resultant electric field of each charge, E = \(\frac{1}{4 \pi \varepsilon_0} \frac{6 q}{r^2}\) [∵ -6q – (-12q)]
Let OX be the reference axis. The angles of resultant fields with OX-axis are shown.

Resultant field along OX-axis = \(\left(0+\frac{1}{2}+\frac{\sqrt{3}}{2}+1+\frac{\sqrt{3}}{2}+\frac{1}{2}\right)\)i = (2 + \(\sqrt{3}\))i
Resultant field along OY-axis = \(\left(1+\frac{\sqrt{3}}{2}+\frac{1}{2}+0-\frac{1}{2}-\frac{\sqrt{3}}{2}\right) \hat{\mathrm{j}}\)
= 1\(\hat{\mathrm{i}}\)
∴ Resultant electric field, E_R(OH) = (2 + \(\sqrt{3}\))\(\hat{i}\) + 1\(\hat{j}\)
The direction of resultant field (OH) is given by, tan θ = \(\frac{|\mathrm{OY}|}{|\mathrm{OX}|}\)
⇒ tan θ = \(\frac{1}{2+\sqrt{3}}\) = tan 15°
⇒ θ = 15°, with OX-axis
∴ The hour hand shows at the centre of the dial is at 9.30.

Question 4.
Consider a uniform electric field E = 3 × 10³ N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x – axis ?
Solution:
a) Given E = 3 × 10³ N/C
S = 10² cm² = 10² × (10^-2m)² = 10^-2m²
θ = 0°
ϕ = ES cos θ
= 3 × 10³ × 10^-2 × cos 0°
∴ ϕ = 30 Nm²C^-1

b) If θ = 60°, ϕ = ES cos θ
= 3 × 10³ × 10^-2 × cos 60°
∴ ϕ = 15 Nm²C^-1

Question 5.
There are four charges, each with a magnitude Q. Two are positive and two are negative. The charges are fixed to the comers of a square of side ‘L’, one to each comer, in such a way that the force on any charge is directed toward the center of the square. Find the magnitude of the net electric force experienced by any charge ?
Solution:

Question 6.
The electric field in a region is given by \(\overrightarrow{\mathbf{E}}\) = a\(\hat{\mathbf{i}}\) + b\(\hat{\mathbf{j}}\). Here a and b are constants. Find the net flux passing through a square area of side L parallel to y-z plane.
Solution:
Given \(\overrightarrow{\mathrm{E}}\) = a\(\hat{\mathrm{i}}\) + b\(\hat{\mathrm{j}}\)
\(\vec{S}\) = L²\(\hat{\mathrm{i}}\)

ϕ = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{S}}\) = (a\(\hat{i}\) + b\(\hat{j}\)) .L²\(\hat{i}\)
∴ ϕ = aL² [∴ \(\hat{i}\). \(\hat{i}\) = 1 and \(\hat{i}\). \(\hat{j}\) = 0]

Question 7.
A hollow spherical shell of radius r has a uniform charge density σ. It is kept in a cube of edge 3r such that the centre of the cube coincides with the center of the shell. Calculate the electric flux that comes out of a face of the cube.
Solution:
For spherical shell, charge = q (say)
Radius = r
Charge density = σ = \(\frac{q}{A}\) = \(\frac{\mathrm{q}}{4 \pi \mathrm{r}^2}\)
∴ Charge on spherical shell, q = 4πr²σ
Flux through one of the face of a cube,
ϕ_E = \(\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}\) = \(\frac{1}{6} \times \frac{4 \pi r^2 \sigma}{\varepsilon_0}\) = \(\frac{2 \pi \mathrm{r}^2 \sigma}{3 \varepsilon_0}\)

Question 8.
An electric dipole consists of two equal and opposite point charge +Q and -Q, separated by a distance 2l. P is a point collinear with the charges such that its distance from the positive charge is half of its distance from the negative charge. Calculate electric intensity at P.
Solution:
Distance of P from -Q = d (say)
Distance of P from +Q = d/2

Question 9.
Two infinitely long thin straight wires having uniform linear charge densities λ and 2λ are arranged parallel to each other at a distance r apart. Calculate intensity of the electric field at a point midway between them.
Solution:
Distance between two parallel infinite long thin straight wires = r.
Electric field due to infinite long thin straight wire, E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\)

∴ Electric intensity at mid point, E = E₂ – E₁ = 2E₁ – E₁ = E
∴ E = \(\frac{\lambda}{\pi \varepsilon_0 \mathrm{r}}\)

Question 10.
Two infinitely long thin straight wires having uniform linear charge densities e and 2e are arranged parallel to each other at a distance r apart. Find the intensity of the electric field at a point midway between them.
Solution:
For first infinitely long straight wire, linear charge density λ = e.
For second infinitely long straight wire, linear charge density λ’ = 2e
Distance between two infinite parallel straight wires = r.
Distance of point P from 1^st and 2^nd wire = \(\frac{\mathrm{r}}{2}\)
Electric field intensity at P due 1^st wire, E₁ = \(\frac{\lambda}{2 \pi \varepsilon_0\left(\frac{\mathrm{r}}{2}\right)}=\frac{\mathrm{e}}{\pi \varepsilon_0 \mathrm{r}}\) —— (1)

Electric field intensity at P due 2^nd wire, E₂ = \(\frac{\lambda^{\prime}}{2 \pi \varepsilon_0\left(\frac{r}{2}\right)}=\frac{2 \mathrm{e}}{\pi \varepsilon_0 \mathrm{r}}\)
∴ E₂ = 2E₁ [∵ from(1)]
∴ Electric field intensity at middle point due to second infinitely long wire
E₂ = \(\frac{2 \lambda}{\pi \varepsilon_0 \mathrm{r}}\)

Question 11.
An electron of mass m and charge e is fired perpendicular to a uniform electric field of intensity E with an initial velocity u. If the electron tranverses a distance x in the field in the direction of firing, find the transverse displacement y it suffers.
Solution:
Given m_e = m; q = e; d = x; u_x = u; u_y = 0
Electric field between the plates = E
Time taken travel in the field, t = \(\frac{d}{u_x}\) = \(\frac{\mathbf{X}}{\mathbf{u}}\)

Force on electron F = qE = eE
Acceleration of electron, a = \(\frac{F}{m}\) = \(\frac{\mathrm{eE}}{\mathrm{m}}\)
Transverse displacement of electron y = u_yt + \(\frac{1}{2} \mathrm{at}^2\)
⇒ y = 0 + \(\frac{1}{2}\left(\frac{e E}{m}\right)\left(\frac{x}{u}\right)^2\)
∴ y = \(\frac{\mathrm{eEx}^2}{2 \mathrm{mu}^2}\)

Additiona Exercises

Question 1.
What is the force between two small charged spheres having charges of 2 × 10^-7 C and 3 × 10^-7 C placed 30 cm apart in air ?
Solution:
Given, q₁ = 2 × 10^-7 C; q₂ = 3 × 10⁷ C; d = 30 cm = 30 × 10^-2 m = 3 × 10^-1m

As q₁, q₂ are positive charges, the force between them is repulsive.

Question 2.
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres ?
(b) What is the force on the second sphere due to the first ?
Solution:
a) Given q₁ = 0.4 μc ;
= 0.8 × 10^-6C
q₂ = 0.8 μc; F = 0.2 N = 0.4
= 0.4 × 10^-6m
0.2 = \(\frac{9 \times 10^9 \times 0.4 \times 10^{-6} \times 0.8 \times 10^{-6}}{\mathrm{r}^2}\)
r² = 16 × 9 × 10^-4
r = 4 × 3 × 10^-2 = 12 × 10^-2 m
∴ Distance between two charges, r = 12 cm

b) Electrostatic force between two charges obeys the Newton’s third law. i.e., force on q₁ due to q₂ = force on q₂ due to q₁
f₁₂ = f₂₁ = 0.2N

Question 3.
Check that the ratio ke²/G m_em_p is dimensionless. Look up a table of Physical Constants and determine the value of this ratio. What does the ratio signify ?
Solution:
i) In electrostatics, F_e = \(\frac{\mathrm{Kq}_1 \mathrm{q}_2}{\mathrm{r}^2}\) = \(\frac{\mathrm{Ke}^2}{\mathrm{r}^2}\) ……. (1)
Where q₁ = q₂ = e
In gravitation, F_g = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}\) = \(\frac{\mathrm{Gm}_{\mathrm{e}} \mathrm{m}_{\mathrm{p}}}{\mathrm{r}^2}\) …. (2)
Where m₁ = m_e ; m₂ = m_p

Thus the given ratio is dimensionless.

ii) We know that e = 1.6 × 10^-19 C ; G = 6.67 × 10^-11 N-m²C²

Question 4.
a) Explain the meaning of the statement ‘electric charge of a body is quantized’.
b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e, large scale charges ?
Answer:
a) The electric charge of a body is quantized means that the charge on a body can occur in some particular values only. Charge on any body is the integral multiple of charge on an electron because the charge of an electron is the elementary charge in nature. The charge on any body can be expressed by the formula q = ± ne. Where n = number of electrons transferred and e = charge on one electron. The cause of quantization is that only integral number of electrons can be transferred from one body to other

b) We can ignore the quantization of electric charge when dealing with macroscopic charges because the charge on one electron is 1.6 × 10^-19 C in magnitude, which is very small as compared to the large scale change.

Question 5.
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Answer:
According to law of conservation of charge, “charge can neither be created nor be destroyed but it can be transferred from one body to another body”. Before rubbing the two bodies they both are neutral i.e., the total charge of the system is zero. When the glass rod is rubbed with a silk cloth, some electrons are transferred from glass rod to silk cloth. Hence glass rod attains positive charge and silk cloth attains same negative charge.

Again the total charge of the system is zero, i.e., the charge before rubbing is same as the charge after rubbing. This is consistent with the law of conservation of charge. Here we can also say that charges can be created only in equal and unlike pairs.

Question 6.
Four point charges q_A = 2 µC, q_B = -5 µC, q_C = 2 µC and q_D = -5 µC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square?
Solution:
Let the centre of the square is at O.
The charge placed on the centre is µC
AB = BC = CD = DA = 10 cm; AC = \(\sqrt{2}\) × 10 = 10\(\sqrt{2}\)cm
AO = BO = CO = DO = \(\frac{10 \sqrt{2}}{2}\) = 5\(\sqrt{2}\) cm


Here we observe that, F_A = -F_C and F_D = -F_B
∴ The net resultant force on 1 µC is
F = F_A + F_B + F_C + F_D
= -F_C + F_B + F_C – F_B
= 0.

Question 7.
a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not ?
b) Explain why two field lines never cross each other at any point ?
Answer:
a) An electrostatic field line represents the actual path travelled by a unit positive charge in an electric field. If the line have sudden breaks it means the unit positive test charge Jumps from one place to another which is not possible. It also means that electric field becomes zero suddenly at the breaks which is not possible. So, the field line cannot have any sudden breaks.

b) If two field lines cross each other, then we can draw two tangents at the point of intersection which indicates that (as tangent drawn at any point on electric line of force gives the direction of electric field at that point) there are two directions of electric field at a particular point, which is not possible at the same instant. Thus, two field lines never cross each other at any point.

Question 8.
Two point charges q_A = 3 μC and q_B = -3 μC are located 20 cm apart in vaccum.
a) What is the electric field at the midpoint O of the line AB joining the two charges ?
b) If a negative test charge of magnitude 1.5 × 10^-9 C is placed at this point, what is the force experienced by the test charge ?
Solution:
a) Given q_A = 3 μC = 3 × 10^-6 C; q_B = -3 μC = -3 × 10^-6C

From fig. AO = OB = 10 cm = 0.1 m
Electric field at midpoint ‘O’ due to q_A

The direction of E_A is A to O.
Electric field at midpoint ‘O’ due to q_B at B is

The direction of E_B is O to B.
Now we see that E_A and E_B are in same direction. So, the resultant electric field at O is E. Hence,
E = E_A + E_B = 2.7 × 10⁶ + 2.7 × 10⁶ = 5.4 × 10⁶ N/C :
The direction of E will be from O to B or toward B.

b) Let test charge q₀ = -1.5 × 10^-9 C is placed at midpoint O’.
Electric field intensity at ‘O’ is E = 5.4 × 10⁶

Force F = E_q = 5.4 × 10⁶ × -1.5 × 10^-9 N
= -8.1 × 10³N
The direction of force is from O to A.

Question 9.
A system has two charges q_A = 2.5 × 10^-7 C, and q_B = -2.5 × 10^-7 C located at points A(0, 0, -15 cm) and B(0, 0, +15 cm). What are the total charge and electric dipole moment of the system ?
Solution:
Given A(0, 0, -15 cm) and B(0, 0, 15 cm)
q_A = 2.5 × 10^-7C
q_B = -2.5 × 10^-7 C
AB = 2a = length of the dipole
= 30 cm = 30 × 10^-2 m

The total charge q on the dipole is
q = q_A + q_B = 2.5 × 10^-7C – 2.5 × 10^-7C = 0
The electric dipolemoment
P = Any charge (q_A) × length of dipole (2a)
= 2.5 × 10^-7 × 10 × 10^-2
∴ P = 7.5 × 10^-8 C-m
The direction of P is from negative charge to positive charge that is along B to A.

Question 10.
An electric dipole with dipole moment 4 × 10^-9 Cm is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ NC^-1. Calculate the magnitude of the torque acting on the dipole.
Solution:
Given, P = 4 × 10^-9 C-m; E = 5 × 10⁴ N/C; θ = 30°,
Torque, \(\tau\) = PE sin θ

= 4 × 10^-9 × 5 × 10⁴ sin 30° = \(\frac{20 \times 10^{-5}}{2}\) = 10^-4N-m
The direction of torque is ⊥^r to both electric field and dipole moment.

Question 11.
A polythene piece rubbed with wool is found to have a negative charge 3 × 10^-7 C.
a) Estimate the number of electrons transferred (from which to which ?)
b) Is there a transfer of mass from wool to polythene ?
Solution:
a) Given, charge on Polythene, q = -3 × 10^-7 C
e = -1.6 × 10^-19 C
No. of electrons transferred, n = \(\frac{\mathrm{q}}{\mathrm{e}}\) = \(\frac{-3 \times 10^{-7}}{-1.6 \times 10^{-19}}\)
∴ n = 1.875 × 10¹² [∵ q = ± ne]
Electrons are transferred from wool to polythene.
So wool gets positive charge and polythene gets negative charge.

b) The number of electrons transferred = 1.875 × 10¹²
The mass of one electron, m_e = 9.1 × 10^-3 kg
Mass transferred from wool to polythene M = n × m_e
M = 1.875 × 10¹² × 9.1 × 10^-31 = 1.8 × 10^-18 kg

Question 12.
a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each 6.5 × 10^-7 C ? The radii of A and B are negligible compared to the distance of
separation.
b) What is the force of repulsion if each sphere is charged double the above amount and the distance between them is halved?
Solution:

a) Given, q_A = 6.5 × 10^-7C ; q_B = 6.5 × 10^-7C
r = AB = 50 cm = 50 × 10^-2m

b)

This force is also repulsive in nature because both the charges are similar (positive) in nature.

Question 13.
Suppose the spheres A and B in Exercise – 12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with second and finally removed from both. What is the new force of repulsion between A and B?
Solution:
Given q_A = 6.5 × 10^-7C;
q_B = 6.5 × 10^-7 C; q_C = 0

After contact of A and C, the charges will be divided equally on both of them. Then final charge on A, then
\(\mathrm{q}_{\mathrm{A}}^{\prime}\) = \(\frac{\mathrm{q}_{\mathrm{A}}+\mathrm{q}_{\mathrm{C}}}{2}\) = \(\frac{6.5 \times 10^{-7}+0}{2}\)
= 3.25 × 10^-7C
Similarly charge on C, \(\mathrm{q}_{\mathrm{c}}^{\prime}\) = 3.25 × 10^-7 C
After contact of B and C, the charges will be divided equally on both of them.

Then final charge on B, \(q_B^{\prime}\) = \(\frac{\mathrm{q}_{\mathrm{B}}+\mathrm{q}_{\mathrm{C}}^{\prime}}{2}\) = \(\frac{6.5 \times 10^{-7}+3.25 \times 10^{-7}}{2}\) = 4.875 × 10^-7 C
Similarly final charge one, \(q_C^{\prime \prime}\) = 4.875 × 10^-7 C

Question 14.
Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio ?

Answer:
We know that a positively charged particle is attracted towards the negatively charged plate and a negatively charged particle is attracted towards the positively charged plate.

Here, particle 1 and particle 2 are attracted towards positive plate that means particle 1 and particle 2 are negatively charged. Particle 3 is attracted towards negatively charged plate so it is positively charged. As the deflection in the path of a charged particle is directly proportional to the charge/mass ratio.
y ∝ \(\frac{\mathrm{q}}{\mathrm{m}}\)
Here, the deflection in particle 3 is maximum, so the charge to mass ratio of particle 3 is maximum.

Question 15.
Consider a uniform electric field E = 3 × 10³ \(\hat{\mathbf{i}}\) N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis ?
Solution:
Given \(\overrightarrow{\mathrm{E}}\) = 3 × 10³ \(\hat{\mathbf{i}}\) N/C

a) As the surface is in Y – Z plane, so the area vector (normal to the square) is along X – axis
Area S = 10 × 10 = 100 cm² = 10^-2 m²

Area vector \(\vec{S}\) = 10^-2 \(\hat{\mathbf{i}}\) m²
ϕ = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{S}}\) = (3 × 10³ \(\hat{\mathbf{i}}\)). (10^-2i)
∴ ϕ = 3 × 10³ × 10^-2 = 30N-m²/c

b) \(\overrightarrow{\mathrm{E}}\) = 3 × 10³ \(\hat{\mathbf{i}}\) N/C ; \(\vec{S}\) = \(\hat{\mathbf{i}}\) m² ; θ = 60°
ϕ = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{S}}\) = ES cos 60° = 3 × 10³ × 10^-2 × cos 60°
∴ ϕ = 15 N – m²/C

Question 16.
What is the net flux of the uniform electric field of Exercise -15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes ?
Answer:
As we know that the number of lines entering in the cube is the same as that the number of lines leaving the cube. So, no flux is remained on the cube and hence, the net flux over the cube is zero.

Question 17.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10³ Nm²/C.
(a) What is the net charge inside the box ?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box ? Why or Why not ?
Solution:
a) Given, ϕ = 8.0 × 10³ N – m²/C
ε₀ = 8 × 10³ × 8.854 × 10^-12
∴ q = 0.07 μc
The flux is outward hence the charge is positive in nature

b) Net outward flux = 0
Then, we can conclude that the net charge inside the box is zero. i.e., the box may have either zero charge or have equal amount of positive and negative charges. It means we cannot conclude that there is no charge inside the box.

Question 18.
A point charge +10 μC is a ‘distance 5 cm directly above the centre of a square of side 10 cm, as shown in fig. What is the magnitude of the electric flux through the square ? (Hint: Think of the square as one face of a cube with edge 10 cm).

Solution:
Let the charge q is placed at the centre of cube as shown in fig.
The total flux enclosed through the cube is ϕ = \(\frac{q}{\varepsilon_0}\)
The flux enclosed by one face ϕ = \(\frac{1}{6}\) of total flux.
[∵ Cube has 6 faces]

ϕ = \(\frac{\phi}{6}\) = \(\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}\)
Here q = 10 μC = 10 × 10^-6C ; ε₀ = 8.854 × 10^-12C² – N^-1-m^-2
∴ ϕ = \(\frac{1}{6} \times \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}}\)
= 1.88 × 10⁵ N-m²/C

Question 19.
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface ?
Solution:
Given, q = 2.0 μC = 2.0 × 10^-6C
ε₀ = 8.854 × 10^-12 C²-N^-1 – m^-2
The net flux through the surface,
ϕ = \(\frac{\mathrm{q}}{\varepsilon_0}\) = \(\frac{2 \times 10^{-6}}{8.854 \times 10^{-12}}\) = 2.26 × 10⁵N-m²/C

Question 20.
A point charge causes an electric flux of -1.0 × 10³ Nm²/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge,
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface ?
(b) What is the value of the point charge ?
Solution:
a) From Gauss’s law, ϕ = \(\frac{\mathrm{q}}{\varepsilon_0}\)
Electric flux ϕ depends on charge q.
It is independent of radius of Gaussian surface. Hence the radius of Gaussian surface were doubled, flux does not change.

b) ϕ = – 1.0 × 10³ N-m²/c ; ε₀ = 8.854 × 10^-12 e²-N^-1-m^-2
q = ϕε₀ = -1.0 × 10³ × 8.854 × 10^-12 = -8.85 × 10^-9C.
∴ The value of point charge, q = -8.85 × 10^-9C

Question 21.
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ N/C and points radially inward, what is the net charge on the sphere?
Solution:
E = 1.5 × 10³ N/C; r = 20 cm = 20 × 10^-2m.
E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)
1.5 × 10³ = \(\frac{9 \times 10^9 \times \mathrm{q}}{\left(20 \times 10^{-2}\right)^2}\)
q = \(\frac{1.5 \times 10^3 \times 20 \times 20 \times 10^{-4}}{9 \times 10^9}\) = 6.67 × 10^-9C.

Question 22.
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density, of 80.0 μC/m².
(a) Find the charge on the sphere,
(b) What is the total electric flux leaving the surface of the sphere ?
Solution:
a) Given D = 2.4 m; r = \(\frac{\mathrm{D}}{2}\) = 1.2 m
σ = 80 µc/m² = 80 × 10^-6 C/m²
σ = \(\frac{\mathrm{q}}{4 \pi r^2}\) ⇒ q = σ 4πr²
⇒ q = 80 × 10^-6 × 4 × 3.14 × 1.2 × 1.2
∴ q = 1.45 × 10^-3C

b) ϕ = \(\frac{Q}{\varepsilon_0}\) = \(\frac{1.4 \times 10^{-3}}{8.854 \times 10^{-12}}\) = 1.6 × 10⁸N-m²/C
Thus, the flux leaving the surface of sphere is 1.6 × 10⁸ N – m²/c

Question 23.
An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.
Solution:
Given r = 2 cm = 2 × 10^-2m ; E = 9 × 10⁴ N/C


Thus, the linear charge density is 10^-7 C/m.

Question 24.
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10^-22 C/m². What is E :
(a) in the outer region of the first plate,
(b) in the outer region of the second plate and
(c) between the plates ?
Solution:
Given σ_A = 127.0 × 10^-22 C/m²
σ_B = 17.0 × 10^-22 C/m²

Question 25.
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ NC^-1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm^-3. Estimate the radius of the drop, (g = 9.81 ms^-2; e = 1.60 × 10^-19C).
Solution:
Given n = 12; E = 2.55 × 10⁴ N/C
p = 1.26 g/cm³ = 1.26 × 10³ kg/m³
e = 1.6 × 10^-19C ; g = 9.81 ms^-2

As the oil drop is stationary,
Electrostatic force = Gravitational force
⇒ qE = mg
neE = \(\frac{4}{3} \pi r^3 \mathrm{\rho g}\)
r³ = \(\frac{3 \mathrm{neE}}{4 \pi \rho \mathrm{g}}\) = \(\frac{3 \times 12 \times 1.6 \times 10^{-19} \times 2.55 \times 10^4}{4 \times 3.14 \times 1.26 \times 10^3 \times 9.8}\)
r = 0.94 × 10^-18
r = [0.94 × 10^-18]^{\(\frac{1}{3}\)} = 9.81 × 10^-7m
∴ Radius of the drop = 9.81 × 10^-7 m.

Question 26.
Which among the curves shown in Fig. cannot possibly represent electrostatic field lines ?

Solution:
a) According to the properties of electric lines of force, the lines should be always ⊥r to the surface of a conductor as they starts or they ends. Here, some of the lines are not ⊥r to the surface, thus it cannot represent the electrostatic field lines.

b) According to the property of electrostatic field lines, they never start from negative charge, here some of the lines start from negative charge. So, it cannot represent the electrostatic field lines.

c) As the property of electric field lines that they start outwards from positive charge. Hence, it represents the electrostatic field lines.

d) By the property of electric field lines, two electric field lines never intersect each other. Here, two lines intersect. So it does not represent the electric field lines.

e) By the property of electric field lines that they are not in the form of closed loops. Here, the lines form closed loop. So, it does not represent the electric field lines.

Question 27.
In a certain region of space, electric field is along the Z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive Z-direction, at the rate of 10⁵ NC^-1 per metre. What are the force and torque experienced by a system having a total dipolemoment equal to 10^-7 Cm in the negative Z-direction ?
Solution:
The electric field increases in positive Z – direction. dE
\(\frac{\mathrm{dE}}{\mathrm{dZ}}\) = 10⁵ N/C-m
The direction of dipolemoment is in the negative Z-direction

So the negative charge q is placed at A and positive charge q is placed at B as the direction of dipole moment is from negative charge to positive charge.
P_Z = -10^-7C-m

The negative sign shows its direction in negative Z – axis. According to the basic definition of electric field, F = qdE Now, multiplying and dividing by dz,
F = q\(\frac{\mathrm{dE}}{\mathrm{dz}} \cdot \mathrm{dz}\) .dz = q.dz\(\frac{\mathrm{dE}}{\mathrm{dz}}\)
qdz = dipolement p_z, as the length of the dipole is dz.

∴ F = P_z. \(\frac{\mathrm{dE}}{\mathrm{dz}}\) = -10^-7 × 10⁵ = -10^-2N
Torque, \(\tau\) = PE sin θ (∵ θ = 180° angle between P and E)
\(\tau\) = PE sin 180° = 0
Thus the force is -10^-2 N and the torque is 0.

Question 28.
a) A conductor A with a cavity as shown in Fig. (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor, (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q (Fig. (b)). (c) A sensitive instrument is o he shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Solution:
a) As we know the property of conductor that the net electric field inside a charged conductor is zero, i.e., E = 0.
Now let us choose a Gaussian surface lying completely inside the conductor enclosing the cavity.
So, from Gauss’s theorem \(\oint \text { E. dS }\) = \(\frac{\mathrm{q}}{\varepsilon_0}\)
As E = 0 ⇒ \(\frac{q}{\varepsilon_0}\) = 0 ⇒ q = 0
That means the charge inside the cavity is zero. Thus, the entire charge Q on the conductor must appear on the outer surface of the conductor.

b) As the conductor B carrying a charge +q inserted in the cavity, the charge -q is induced on the metal surface of the cavity and then charge +q induced on the outside surface of the conductor A. Initially the outer surface of A of A has a charge Q and now it has a charge +q induced. So the total charge on the outer surface of A is Q + q.

c) To protect any sensitive instrument from electrostatic field, the sensitive instrument must be put in the metallic cover. This is known as electrostatic shielding.

Question 29.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/εε₀)\(\hat{\mathbf{n}}\), where \(\hat{\mathbf{n}}\) is the unit vector in the outward normal direction and σ is the surface charge density near the hole.
Solution:
Surface charge density near the hole = σ
Unit vector = \(\hat{\mathbf{n}}\) (normal directed outwards)
Let P be the point on the hole.
The electric field at point P closed to the surface of conductor, according to Gauss’s theorem,

\(\oint \mathrm{E} \cdot \mathrm{dS}\) = \(\frac{q}{\varepsilon_0}\)
Where q is the charge near the hole.
E ds cos θ = \(\frac{\sigma \mathrm{dS}}{\varepsilon_0}\) (∴ σ = \(\frac{\mathrm{q}}{\mathrm{dS}}\) ∴q = σ dS) where dS = area

∴ Angle between electric field and area vector is 0°.
EdS = \(\frac{\sigma \mathrm{dS}}{\varepsilon_0}\)
E = \(\frac{\sigma}{\varepsilon_0}\)
E = \(\frac{\sigma}{\varepsilon_0} \hat{\mathrm{n}}\)

This electric field is due to the filled up hole and the field due to the rest of the charged conductor. The two fields inside the conductor are equal and opposite. So, there is no electric field inside the conductor. Outside the conductor, the electric fields are equal and are in the same direction.
So, the electric field at P due to each part = \(\frac{1}{2} \mathrm{E}\) = \(\frac{\sigma}{2 \varepsilon_0} \hat{n}\)

Question 30.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
Solution:
Let us consider a long thin wire of linear charge density λ. We have to find the resultant electric field due to this wire at point P.
Now, consider a very small element of length dx at a distance x from C.

The charge on this elementary portion of length dx
q = λ dx ——- (1)

Electric field intensity at point P due to the elementary portion
dE = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{(\mathrm{OP})^2}\) = \(\frac{1}{4 \pi \varepsilon_0} \frac{\lambda d x}{(\mathrm{OP})^2}\) [∵ from (1)]
Now, in ΔPCO (PO)² = (PC)² + (CO)²
(OP)² = r² + x²
dE = \(\frac{1}{4 \pi \varepsilon_0} \frac{\lambda d \mathbf{x}}{\left(x^2+r^2\right)}\) ——- (2)

The components of dE are dE cos θ along PD and dE sin θ along PF.
Here, there are so many elementary portion. So all the dE sin θ components balance each other. The resultant electric field at P is due to only dE cos θ components.
The resultant electric field due to elementary component, dE’ = dE cos θ

dE’ = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda d x}{\left(x^2+r^2\right)} \cos \theta\) —— (3)
In ΔOCP tan θ = \(\frac{x}{r}\) ⇒ x = r tan θ
Differentiating with respect to θ, we get dx = r sec² θ dθ
Putting in equation (3), we get

As the wire is of infinite length, so integrate within the limits –\(\frac{\pi}{2}\) to \(\frac{\pi}{2}\), we get

Question 31.
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge +(2/3)e and the ‘down’ quark (denoted by d) of charge (-1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.
Solution:
For the protons, the charge on it is +e let the number of up quarks are a, then the number of down quarks are (3 – a) as the total number of quarks are 3.
So, a_x up quark charge + (3 – a) down quark charge = +e
a × \(\frac{2}{3} \mathrm{e}\) + (3 – a)\(\left(\frac{-\mathrm{e}}{3}\right)\) = e
\(\frac{2 \mathrm{ae}}{3}\) – \(\frac{(3-\mathrm{a}) \mathrm{e}}{3}\) = e
2a – 3 + a = 3
3a = 6
a = 2
Thus, in the proton there are two up quarks and one down quark.
∴ Possible quark composition for proton = uud

For the neutron, the charge on neutron is 0.
Let the number of up quarks are b and the number of down quarks are (3 – b)
So, b_x up quark charge + (3 – b) × down quark charge = 0
b\(\left(\frac{2 \mathrm{e}}{3}\right)\) + (3 – b)\(\left(\frac{-\mathrm{e}}{3}\right)\) = 0
2b – 3 + b = 0
3b = 3
∴ b = 1
Thus, in neutron, there are one up quark and two down quarks.
∴ Possible quark composition for neutrons = udd.

Question 32.
a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e, where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Solution:
a) Let us consider that initially the test charge is in the stable equilibrium. When the test charge is displaced from the null point (where, E = 0) in any direction, it must experience a restoring force towards the null point.
This means that there is a net inward flux through a closed surface around the null point According to the Gauss’s theorem, the net electric flux through a surface net enclosing any charge must be zero. Hence, the equilibrium is not stable.

b) The middle point of the line joining two like charges is a null point. If we displace a test Charge slightly along the
line, the restoring force try to bring the test charge back to the centre. If we displace the test charge normal to the line, the net force on the test charge takes it further away from the null point. Hence the equilibrium is not stable.

Question 33.
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed V_x (as in the fig.). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/(2m \(\mathbf{V}_{\mathbf{x}}^2\)).
Compare this motion with motion of a projectile in gravitational field discussed in section 4.10 of 1^st Year Textbook of Physics.
Solution:
Mass of particle = m
Charge of particle = -q
Speed of particle = V_x
Length of plates = L

Electric field between the plates = E (from positive plate to negative plate).
Let the deflection in the path of the charge – q is Y, because the force acting in +Y axis direction. The direction of force is from negative plate to positive plate because the charge is negative in nature.

Let us discuss the motion in Y axis direction. Initial velocity u = 0
Acceleration a = \(\frac{F}{m}\) = \(\frac{+\mathrm{qE}}{\mathrm{m}}\)
Deflection Y = ?
Time = \(\frac{\text { Distance }}{\text { Velocity }}\) = \(\frac{\mathrm{L}}{\mathrm{V}_{\mathrm{x}}}\)
Using second equation of motion,
S = ut + \(\frac{1}{2} \mathrm{at}^2\)at
Putting the values y = 0 + \(\frac{1}{2} \times\left(+\frac{\mathrm{qE}}{\mathrm{m}}\right) \frac{\mathrm{L}^2}{\mathrm{~V}_{\mathrm{x}}^2}\)
Y = \(\frac{\mathrm{qEL}^2}{2 \mathrm{mV}_{\mathrm{x}}^2}\)

In the case of projectile motion y = \(\frac{1}{2} \mathrm{gt}^2\). Thus, it is exactly similar to the projectile motion in the gravitational field.

Question 34.
Suppose that the particle is an electron projected with velocity V_x = 2.0 × 10⁶ ms^-1. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate ? (|e| = 1.6 × 10^-19 C, m_e = 9.1 × 10^-31 kg.)
Solution:
Given V_x = 2 × 10⁶ m/s; E = 9.1 × 10² N/C
q = e = 1.6 × 10^-19 C; m_e = 9.1 × 10^-31 kg
d = 0.5 cm = 0.5 × 10^-2 m = 5 × 10^-3 m
The electron will strike the upper plate at its other end at X = L as it get deflected.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 3rd Lesson Wave Optics Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 3rd Lesson Wave Optics

Very Short Answer Questions

Question 1.
What is Fresnel distance?
Answer:
“Fresnel distance is the minimum distance a beam of light has to travel before its deviation from straight line path becomes significant”.

Fresnel distance (Z_F) = \(\frac{\mathrm{a}^2}{\lambda}\) ; Where a = width of the aperture, λ = wave length.

Question 2.
Give the justification for validity of ray optics.
Answer:
The distances much smaller than Z_F, the spreading due to diffraction is smaller compared to the size of the beam.

When the distance is approximately Z_F, and much more than Z_F, the spreading due to diffraction dominates over that due to ray optics (size of the aperture (a)]
Z = \(\frac{\mathrm{a}^2}{\lambda}\)
From this equation ray optics is valid in the limit of wave length tending to zero.

Question 3.
What is polarisation of light ?
Answer:
The vibrations of the light confined only one direction. This phenomenon is called polarisation.
(or)
The phenomenon due to which the transverse vibrations of electric field vector of a light wave become confined to one plane, is called polarisation.

Question 4.
What is Malus’ law.
Answer:
Malus’ law : It states that the intensity of polarised light transmitted through the analyser varies as a square of cosine of the angle between the plane of transmission of analyser and polariser.
I ∝ cos² θ; I = I₀ cos² θ.

Question 5.
Explain Brewster’s law.
Answer:
Brewster’s law : The tangent of the polarising angle is equal to the refractive index of the medium.
μ = tan i_B, where i_B = polarising angle and μ = refractive index. Note : r + i_B = 90°

Question 6.
When does a monochromatic beam of light incident on a reflective surface get completely transmitted ?
Answer:
Let the light emitted by laser source passes through polariser, and incident on the surface of the reflective surface with Brewster’s angle (i_B). Now rotate the polariser at particular alignment the light incident on the surface is completely transmitted.

Short Answer Questions

Question 1.
Explain Doppler effect in light. Distinguish between red shift and blue shift. (T.S. Mar. ’18, ’16)
Answer:
Dopper effect in light: The change in the apparent frequency of light, due to relative motion between source of light and observer. This phenomenon is called Doppler effect.

The apparent frequency of light increases when the distance between observer and Source of light is decreasing and the apparent frequency of light decreases, if the distance between source of light and observer increasing.

Doppler shift can be expressed as \(\frac{\Delta v}{v}\) = \(\frac{-v_{\text {radial }}}{\mathrm{C}}\)

Applications of Doppler effect in light:

1. It is used in measuring the speed of a star and speed of galaxies.
2. Measuring the speed of rotation of the sun.

Red shift: The apparent increase in wave length in the middle of the visible region of the spectrum moves towards the red end of the spectrum is called red shift.

Blue shift: When waves are received from a source moving towards the observer, there is an apparent decrease in wave length, this is called blue shift.

Question 2.
What is total internal reflection. Explain the phenomenon using Huygen’s principle.
Answer:
Total internal reflection: When a light ray travels from denser medium to rarer medium, the angle of incidence is greater than critical angle then it reflects into the same medium. This phenomenon is called total internal reflection.

Huygen’s principle : According to Huygen’s principle, every point on the wavefront AB is a source of secondary wavelets and time during which wavelets from B reaches at C, Let \(\tau\) be the time taken by the wave front to advance from B to C.

Then distance BC = υ\(\tau\)
In order to construct the reflected wavefront, we draw a sphere of radius ux from point A.
Let CE represent the tangent drawn from C to this sphere.
AE = BC = υ\(\tau\)

Consider the triangles EAC and BAC are congruent.
∴ Angles i and r would be equal. This is the law of reflection.

Question 3.
Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for maximum and zero intensity. (A.P. & T.S. Mar. ’16)
Answer:
Let y₁ and y₂ be the displacements of the two waves having same amplitude a and Φ is the phase difference between them.
y₁ = a sin ωt …… (1)
y₂ = a sin (ωt + ϕ)) …… (2)
The resultant displacement y = y₁ + y₂
y = a sin ωt + a sin (ωt + ϕ)
y = a sin ωt + a sin ωt cos ϕ + a cos ωt sin ϕ
y = a sin ωt [1 + cos ϕ] + cos ωt (a sin ϕ)

Let R cos θ = a(1 + cos ϕ) —– (4)
R sin θ = a sin ϕ —– (5)
y = R sin ωt . cos θ + R cos ωt . sin θ
y = R sin(ωt + θ) —– (6)
where R is the resultant amplitude at P, squaring equations (4) and (5), then adding
R² (cos² θ + sin θ] = a²[1 + cos² ϕ + 2 cos ϕ + sin² ϕ]
R²[1] = a²[1 + 1 + 2 cos ϕ]
I = R² = 2a² [1 + cos ϕ] = 2a² × 2 cos² \(\frac{\phi}{2}\); I = 4a² cos² \(\frac{\phi}{2}\) —— (7)

i) Minimum intensity (I_max)
c0s² \(\frac{\phi}{2}\) = 1
ϕ = 2nπ Where n = 0, 1, 2, 3 …….
ϕ = 0, 2π, 4π, 6π
∴ I_max = 4a².

ii) Minimum intensity (I_min)
cos² \(\frac{\theta}{2}\) = 0
ϕ = (2n + 1)π where n = 0, 1, 2, 3, …….
ϕ = π, 3π, 5π, 7π
I_min = 0

Question 4.
Does the principle of conservation of energy hold for interference and diffraction phenomena ? Explain briefly. (Mar. ’14)
Answer:
Yes, law of conservation of energy is obeyed. In case of constructive interference, intensity becomes maximum. Hence bright fringes are formed on the screen where as in the case of destructive interference, intensity becomes minimum. Hence dark fringes are formed on the screen.

This establishes that in the interference and diffraction pattern, the intensity of light is simply being redistributed i.e., energy is being transferred from dark fringe to bright fringe. No energy is being created (or) destroyed in the process. Hence energy is redistributed.
Thus the principle of conservation of energy is being obeyed in the process of interference and diffraction.

Question 5.
How do you determine the resolving power of your eye ? (A.P. Mar. ’19)
Answer:
Make black strips of equal width separated by white strips. All the black strips having same width, while the width of white strips should increase from left to right.
Now watch the pattern with one eye. By moving away (or) closer to the wall, find the position where you can just see some two black strips as separate strips.

All black strips to the left of this strips would merge into one another and would not be distinguishable on the other hand, the black strips to the right of this would be more and more clearly visible.

Note the width d of the white strips and measure the distance D of the wall from eye.
Then resolution of your eye = \(\frac{\mathrm{d}}{\mathrm{D}}\).

Question 6.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.
Answer:
Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be I = I₀cos²θ.
Where θ is the angle between pass axes P₁ and P₂. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be \(\left(\frac{\pi}{2}-\theta\right)\)
Hence the intensity of light emerging from P₃ will be
I = I₀cos² θ. cos² \(\left(\frac{\pi}{2}-\theta\right)\)
= I₀ cos² θ . sin² θ
I = \(\frac{I_0}{4}\) sin² 2θ
The transmitted intensity will be maximum when θ = \(\frac{\pi}{4}\)

Long Answer Questions

Question 1.
What is Huygen’s Principle? Explain the optical phenomenon of refraction using Huygen’s principle.
Answer:
Huygens principle: Every point on a wave front is the source of secondary wavelets.
Refraction of a plane wave using Huygen’s principle:
Let the surface PP’ separating the two medium of refractive index μ₁ and μ₂. Let υ₁ and υ₂ be the velocities of light in medium 1 and medium 2.

According to Huygen’s principle, every point on incident wave front AB is a source of secondary wavelets. By the time wavelét from point B reaches at point C, the wavelet from point A would have reached at point E. Let t be the time taken from B to C is equal to time taken from A to D.

μ₁ sin i = μ₂ sin r
This is the Snell’s law of refraction.

Second law of refraction: Since incident ray, refracted ray and the normal all the lie on the same plane PP’ at the point of incidence. This proves the second law of refraction.

Question 2.
Distinguish between Coherent and Incoherent addition of waves. Develop the theory of constructive interferences.
Answer:
Coherent sources : The two sources which maintain zero (or) any constant phase relation between themselves are known as Coherent sources.
Incoherent sources : If the phase difference changes with time, the two sources are known as incoherent sources.

Theory of constructive and destructive interference :
Let the waves of two coherent sources be
y₁ = a sin ωt —— (1)
y₂ = a sin (ωt + ϕ) —– (2)
where a is amplitude and ϕ is the phase difference between two displacements.
According to superposition principle, y = y₁ + y₂
y = a sin ωt + a sin (ωt + ϕ)= a sin ωt + a sin ωt cos ϕ + a cos ωt sin ϕ
y = a sin ωt [1 + cos ϕ] + cos ωt [a sin ϕ] —– (3)
Let A cos θ = a(1 + cos ϕ] —– (4)
A sin θ = a sin ϕ —– (5)
Substituting equations (4) and (5) in equation (3)
y = A sin ωt. cos θ + A cos ωt sin θ
y = A sin (ωt + θ) —— (6)
Where A is resultant amplitude. Squaring equations (4) and (5), then adding
A² [cos² θ + sin² θ] = a²[1 + cos² ϕ + 2 cos ϕ + sin² ϕ]
A² [1] = a² [1 + 1 + 2 cos ϕ]
I = A² = 2a² [1 + cos ϕ] ( ∵ I = A²)
I = 2a² × 2 cos² \(\frac{\phi}{2}\)
I = 4a² cos² \(\frac{\phi}{2}\)
I = 4I₀ cos² \(\frac{\phi}{2}\) —— (7) (∵ I₀ = a²)

Case (i) For constructive interference : Intensity should be maximum.
cos \(\frac{\phi}{2}\) = 1 ⇒ ϕ = 2nπ
Where n = 0, 1, 2, 3….. ⇒ ϕ = 0, 2π, 4π, 6π ….. I_max = 4I₀
Case (ii) For destructive interference : Intensity should be minimum
i.e., cos ϕ = 0 ⇒ ϕ = (2n + 1) π ; where n = 0, 1, 2, 3……. ; ϕ = π, 3π, 5π ⇒ I_min = 0.

Question 3.
Describe Young’s experiment for observing interference and hence arrive at the expression for ‘fringe width’. ‘
Answer:
Interference : The modification of intensity obtained by the super position of two (or) more light waves is called interference.

1. Thomas Young experimentally observed the phenomenon of interference of light using two coherent sources.
2. A small pin hole ‘S’ illuminated by monochromatic source of light which produces a spherical wave.
3. S₁ and S₂ are two narrow pin holes equidistant from S.
4. Screen is placed at a distance D.
5. The points at which any two crests (or) any two troughs are superimposed, constructive interference takes place bright fringe will be observed on the screen.
6. The points at which crest of one wave and trough of another wave are super imposed, destructive interference takes place dark fringe will be observed on the screen.
7. Thus on the screen alternately bright and dark frings are observed.

Expression for fringe width :

i) It is the distance between two successive bright (or) dark fringes, denoted by p.

ii) The path difference (δ) = d sin θ

If θ is very small then from figure sin θ ≈ tan θ = \(\frac{x}{D}\)

iii) For bright fringes path difference S₂P – S₁P = nλ
∴ d sin θ = nλ,
d × \(\frac{x}{D}\) = nλ
x = \(\frac{n \lambda \mathrm{D}}{\mathrm{d}}\) —– (1) where n = 0, 1, 2, 3,…….
This equation represents the position of bright fringe.
When n = 0, x₀ = 0
n = 1, x₁ = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) and n = 2, x₂ = \(\frac{2 \lambda \mathrm{D}}{\mathrm{d}}\)

The distance between any two consecutive bright fringes is
x₂ – x₁ = \(\frac{2 \lambda \mathrm{D}}{\mathrm{d}}-\frac{\lambda \mathrm{D}}{\mathrm{d}}\) ⇒ β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) ——- (2)

iv) For dark fringes path difference S₂P – S₁P = (2n + 1) \(\frac{\lambda}{2}\) ∴ d sin θ = (2n + 1)\(\frac{\lambda}{2}\)
d × \(\frac{x}{D}\) = (2n + 1) \(\frac{\lambda}{2}\) = \(\frac{(2 \mathrm{n}+1) \lambda \mathrm{D}}{2 \mathrm{~d}}\) ——- (3) where n =0, 1, 2, 3, ………
This equation (3) represents, position of dark fringe.
When n = 0, x₀ = \(\frac{\lambda \mathrm{D}}{2 \mathrm{~d}}\) ⇒ n = 1, x₁ = \(\frac{3 \lambda \mathrm{D}}{2 \mathrm{~d}}\) ; n = 2, x₂ = \(\frac{5 \lambda \mathrm{D}}{2 \mathrm{~d}}\) ……
The distance between any two consecutive dark fringes is x₂ – x₁ = \(\frac{5 \lambda \mathrm{D}}{2 \mathrm{~d}}-\frac{3 \lambda \mathrm{D}}{2 \mathrm{~d}}\) = \(\frac{5 \lambda \mathrm{D}-3 \lambda \mathrm{D}}{2 \mathrm{~d}}\)
β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) —– (4)
Hence fringe width is same for bright and dark fringes.

Question 4.
What is diffraction ? Discuss diffraction pattern obtainable from a single slit.
Answer:
Diffraction : The phenomenon of bending of light at the edges of an obstacle and light enters into the geometrical shadow is known as diffraction of light.
Example : The silver lining surrounding the profile of a mountain just before sunrise.

Diffraction of light at a single slit:

1. Consider a narrow slit AB of width d. A parallel beam of light of wave length λ falling normally on a single slit.
2. Let the diffracted light be focussed by means of a convex lens on a screen.
3. The secondary wavelets travelling normally to the slit, i.e., along the direction of OP₀.
   Thus P₀ is a bright central image.
   
4. The secondary wavelets travelling at an angle θ with the normal are focussed at a point P₁ on the screen.
5. In order to find out intensity at P₁, draw a perpendicular AC on BR.
6. The path difference between secondary wavelets = BC
   = AB sin θ = a sin θ (∵ sin θ = 0)
   Path difference (λ) ≈ a θ —– (1)
7. Experimental observations shown in figure, that the intensity has a central maximum at θ = 0 and other secondary maxima at θ ≈ \(\left(\mathrm{n}+\frac{1}{2}\right) \frac{\lambda}{\mathrm{a}}\) and has minima at θ = \(\approx \frac{n \lambda}{a}\)
8. From equation (1), θ = \(\frac{\lambda}{\mathrm{a}}\). Now we divide the slit into two equal halves, each of size \(\frac{a}{2}\).
9. We can show that the intensity is zero for θ = \(\frac{n \lambda}{a}\) where n = 1, 2, 3….
10. It is alsó to see why there are maxima at θ = \(\left(\mathrm{n}+\frac{1}{2}\right) \frac{\lambda}{\mathrm{a}}\)
11. Consider θ = \(\frac{3 \lambda}{2 \mathrm{a}}\) which is midway between two of the dark fringes.
12. If we take the first two thirds of the slit, the path difference between two ends is
    \(\frac{2}{3} a \times \theta\) = \(\frac{2 \mathrm{a}}{3} \times \frac{3 \lambda}{2 \mathrm{a}}\) = λ —– (2)
    
13. The first two third of the slit can be divided into two halves which have a \(\frac{\lambda}{2}\) path difference. The contribution of two halves cancel and only remaining one third of the slit contributes to the intensity minima.

Question 5.
What is resolving power of Optical Instruments? Derive the condition under which images are resolved.
Answer:
Resolving power : The resolving power of a lens is its ability to resolve two points that are to each other.

Resolving power of optical instruments:

1. Consider a parallel beam of light falling on a convex lens. Due to diffraction effect, the beam focussed to a spot of finite area.
2. Taking into account the effects duè to diffraction, the pattern on the focal plane would consist of a central bright region (circular) surrounded by a concentric dark and bright rings.
3. The radius of the central bright region is given by r₀ = \(\frac{1.22 \lambda \mathrm{f}}{2 \mathrm{a}}\) = \(\frac{0.61 \lambda \mathrm{f}}{\mathrm{a}}\)
   where f is focal length of the lens
   2a = diameter of the lens.

Derive the condition under which images are resolved : The size of the spot is very small, it plays an important role in determining the limit of resolution.
For the two stars to be Just resolved
f Δ θ ≈ r₀ ≈ \(\frac{0.61 \lambda \mathrm{f}}{\mathrm{a}}\)
Δ θ ≈ \(\frac{0.61 \lambda}{\mathrm{a}}\) —– (1)
Thus Δθ will be small, if the diameter (2a) of the objective is large. This implies that the telescope will have better resolving power if a is large.
In case of microscope, the object is placed slightly beyond f. The corresponding minimum seperation (d_Min) between the object and the objective lens is given by
d_Min = \(\frac{1.22 \lambda}{2 \mu \sin \beta}\)
Where μ = Refractive index
μ sin β = Numerical aperture.

Textual Exercises

Question 1.
Monochromatic light of wavelength. 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light ? Refractive index of water is 1.33.
Solution:
λ = 589 nm = 589 × 10^-9 m

a) Reflected light: (Wavelength, frequency, speed same as incident light)
λ = 589 nm, v = 5.09 × 10¹⁴ Hz
c = 3 × 10⁸ m/s ⇒ υ = \(\frac{c}{\lambda}=\frac{3 \times 10^8}{589 \times 10^{-9}}\) = 5.093 × 10¹⁴ Hz.

b) Refracted light: (frequency same as the incident frequency)
y = 5.093 × 10¹⁴ Hz
υ = \(\frac{\mathrm{c}}{\mu}=\frac{3 \times 10^8}{1.33}\) = 2.256 × 10⁸ m/s ⇒ λ = \(\frac{v}{v}=\frac{2.26 \times 10^8}{5.09 \times 10^{14}}\) = 443 m.

Question 2.
What is the shape of the wavefront in each of the following cases :
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant intercepted by the Earth.
Soution:
a) It is spherical wavefront.
b) It is plane wavefront.
c) Plane wavefront (a small area on the surface of a large sphere is nearly planar.

Question 3.
(a) The refractive index of glass is 1.5. What is the speed of light in glass ? (Speed of light in vacuum is 3.0 × 10⁸ m s^-1)
(b) Is the speed of light in glass independent of the colour of light ? If not, which of the two colours red and violet travels slower in a glass prism ?
Solution:
a) Here, µ = \(\frac{\mathbf{c}}{v}\) ⇒ υ = \(\frac{\mathrm{c}}{\mu}\) = \(\frac{3 \times 10^8}{1.5}\) = 2 × 10⁸ m/s
b) No, the refractive index and speed of light in a medium depend on wavelength i.e. colour of light. We know that µ_v > µ_r.
Therefore v_violet < v_red. Hence violet component of white light travels slower than the red component.

Question 4.
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Solution:
d = 0.28 mm = 0.28 × 10^-3 m, D = 1.4 m, β = 1.2 × 10^-2 m, n = 4

⇒ λ = 600 nm.

Question 5.
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3 ?
Solution:
Let I₁ = I₂ = I. If ϕ is phase difference between the two light waves, then resultant intensity,
I_R = I₁ + I₂ + \(2 \sqrt{\mathrm{I}_1 \mathrm{I}_2}\) . cos ϕ
When path difference = λ, Phase difference ϕ = 0° ∴ I_R = I + I + \(2 \sqrt{I I}\). cos 0° = 4I = k
When path difference = \(\frac{\lambda}{3}\), phase difference ϕ = \(\frac{2 \pi}{3}\) rad.
∴ I_R = I + I + \(2 \sqrt{\mathrm{II}} \cdot \cos \frac{2 \pi}{3}\) ⇒ I’_R = 2I + 2I\(\left(\frac{-1}{2}\right)\) = I = \(\frac{\mathrm{k}}{4}\)

Question 6.
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both’the wavelengths coincide ?
Solution:
Here λ₁ = 650 nm = 650 × 10^-9 m, λ₂ = 520 nm = 520 × 10^-9 m .
Suppose d = Distance between two slits; D = Distance of screen from the slits
a) For third bright fringe, n = 3 ⇒ x = nλ, \(\frac{\mathrm{D}}{\mathrm{d}}\) = 3 × 650 \(\frac{\mathrm{D}}{\mathrm{d}}\) nm
b) Let n^th fringe due to λ₂ = 520 nm coincide with (n – 1)^th bright fringe due to λ₁ = 650 nm
∴ nλ₂ = (n – 1) λ₁ ; n × 520 = (n – 1) 650; 4n = 5n – 5 or n = 5 .
∴ The least distance required, x = nλ₂ \(\frac{\mathrm{D}}{\mathrm{d}}\) = 5 × 520 \(\frac{\mathrm{D}}{\mathrm{d}}\) = 2600 \(\frac{\mathrm{D}}{\mathrm{d}}\) nm.

Question 7.
In a double-slit experiment the angular width of the fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water ? Take refractive index of water to be 4/3.
Solution:
Here, θ₁ = 0.2°, D = 1m, λ₁ = 600 nm, θ₂ = ?, μ = 4/3

Question 8.
What is the Brewster angle for air to glass transition ? (Refractive index of glass = 1.5.)
Solution:
Here, i_p = ? μ = 1.5; As tan i_p = μ = 1.5 ∴ i_p = tan^-1 (1.5); i_p = 56.3

Question 9.
Light of wavelength 5000 A falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light ? For what angle of incidence is the reflected ray normal to the incident ray ?
Solution:
Given λ = 5000 A = 5 × 10^-7 m
The wavelength and frequency of reflected light remains the same.
∴ Wavelength of reflected light, λ = 5000 A
Frequency of reflected light, υ = \(\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{5 \times 10^{-7}}\) = 6 × 10¹⁴ Hz
The reflected ray is normal to incident if angle of incidence i = 45.

Question 10.
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Solution:
Here, a = 4 mm = 4 × 10^-3 m; λ = 400 nm = 400 × 10^-9 m = 4 × 10^-7 m
Ray optics is good approximation upto distances equal to Fresnels’ distance (Z_F)
Z_F = \(\frac{a^2}{\lambda}=\frac{\left(4 \times 10^{-3}\right)^2}{4 \times 10^{-7}}\) = 40 m

Additional Exercises

Question 11.
The 6563 A Hα line emitted by hydrogen in a star is found to be red-shifted by 15 A. Estimate the speed with which the star is receding from the Earth.
Solution:
Given λ’ – λ= 15A = 15 × 10^-10 m; λ = 6563 A = 6563 × 10^-10 m; v = ?
Since λ’ – λ = \(\frac{v \lambda}{c}\) ∴ v = \(\frac{c\left(\lambda^{\prime}-\lambda\right)}{\lambda}\) ⇒ v = \(\frac{3 \times 10^8 \times 15 \times 10^{-10}}{6563 \times 10^{-10}}\) = 6.86 × 10⁵ m/s

Question 12.
Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water ? If not, which alternative picture of light is consistent with experiment ?
Solution:
In Newton’s corpuscular picture of refraction, particles of light incident from a rarer to a denser medium experience a force of attraction normal to the surface. This results in an increase in the normal component of velocity but the component along the surface remains unchanged. This means
c sin i = v sin r or \(\frac{\mathrm{v}}{\mathrm{c}}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\) = μ; Since μ > 1, ∴ v > c
The prediction is opposite to the experimental result: (v < c) . The wave picture of light is consistent with experiment.

Question 13.
You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
Solution:
In figure, P is a point object placed at a distance r from a plane mirror M₁ M₂. with P as centre and OP = r as radius, draw a spherical arc; AB. This is the spherical wave front from the object, incident on M₁ M₂. If mirrors were not present, the position of wave front AB would be A’B’ where PP’ = 2r. In the presence of the mirror, wave front AB would appear as A”PB”, according to Huygen’s construction. As is clear from the fig. A’B’ and

A”B” are two spherical arcs located symmetrically on either side of M₁ M₂. Therefore, A’PB’ can be treated as reflected image of A”PB”. From simple geometry, we find OP = OP’, which was to be proved.

Question 14.
Let us list some of the factors, which could possibly influence the speed of wave propagation :

1. nature of the source
2. direction of propagation
3. motion of the source and/or observer
4. wavelength
5. intensity of the wave On which of these factors, if any, does

(a) the speed of light in vacuum
(b) the speed of light in a medium (say, glass or water), depend ?
Solution:
a) The speed of light in vacuum is a universal constant, independent of all the factorslisted and anything else.
b) Dependence of the speed of light in a medium

1. Does not depend on the nature of the source.
2. Independent of the direction of propagation for isotropic media.
3. Independent of the motion of the source relative to the medium but depends on the motion of the Observer relative to the medium.
4. Depends on wavelength.
5. Independent of intensity.

Question 15.
For sound waves, the Doppler formula for frequency shift differs slightly between the two situations :

(i) source at rest; observer moving and
(ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium ?

Solution:

Sound waves require a material medium for propagation. That is why situation (i) and (ii) are not identical physically though relative motion between the source and the observer is the same in the two cases. Infact, relative motion of the observer relative to the medium is different in two situations. That is why Doppler’s formulae for sound are different in the two cases.

For light waves travelling in vacuum, there is nothing to distinguish between the two situations. That is why the formulae are strictly identical.
For light propagating in a medium, situation (i) and (ii) are not identical. The formulae governing the two situations would obviously be different.

Question 16.
In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits ?
Solution:
Here λ = 600 nm = 6 × 10^-7 m, θ = 0.1° = \(\frac{0.1^{\circ}}{180^{\circ}} \times \pi \mathrm{rad}\), d = ?
Since θ = \(\frac{\lambda}{\mathrm{d}}\) ⇒ d = \(\frac{\lambda}{\theta}\) = \(\frac{6 \times 10^{-7} \times 180^{\circ}}{0.1^{\circ} \times \pi}\) = 343 × 10^-4 m.

Question 17.
Answer the following questions :
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and density of the central diffraction band ?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment ?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why ?
(d) Two students are separated by a 7m partition wall in a room 10 m high. If both light and sound waves Can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line.fDiffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification ?
Solution:
a) The size of centred diffraction band reduces by half according to the relation : size \(\frac{\lambda}{\mathrm{d}}\). Intensity increase four fold.

b) The intensity of interference fringes in a double slit arrangement is modulated by diffraction pattern Of each slit.

c) Waves diffracted from the edge of the circular obstacle interfer constructively at the centre of the shadow producing a bright spot.

d) For diffraction the size of the obstacle should be comparable to wavelength if the size of the obstacle is much too large compared to wavelength, diffraction is by a small angle. Here the size partition of wall is of the order of few metres. The wavelength of light is about 5 × 10^-7 m, while sound waves of say 1 kHZ frequency have wavelength of about 0.3 m. Thus sound waves can bend around the partition while light waves cannot.

e) Typical sizes of apertures involved in ordinary optical instruments are much larger than the wavelength.

Question 18.
Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects ?
Solution:
We want \(\frac{(5.0)^2}{\lambda}\) > > \(\frac{40,000}{2}\) ⇒ i.e. λ = < < \(\frac{(5.0)^2}{20,000}\) = \(\frac{250}{20,000}\) = 0.125 m = 12.5 cm

Question 19.
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
Solution:
Here λ = 500 nm = 5 × 10^-7 m, D = 1 m, y = 2.5 mm = 2.5 × 10^-3 m, d = ?
sin θ = \(\frac{\lambda}{d}=\frac{y}{D}\) ∴ d = \(\frac{\lambda \mathrm{D}}{\mathrm{y}}=\frac{5 \times 10^{-7} \times 1}{2.5 \times 10^{-3}}\) = 2 × 10^-4 m = 0.2 mm

Question 20.
Answer the following questions :
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle ?
Solution:
a) Interference of the direct signal received by the antenna with the (weak) signal reflected by the passing air craft.
b) Super position principle follows from the linear character of the equation governing wave motion. It is true so ions as wave have small amplitude.

Question 21.
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitable dividing the slit to bring out the cancellation.
Solution:
Divide the signal slit into n smaller slits of width a’ = \(\frac{\mathrm{a}}{\mathrm{n}}\). Each of the smaller slits sends zero intensity in the direction θ. The combination gives zero intensity as well.

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

AP Inter 2nd Year Physics Study Material 1st Lesson Waves

Very Short Answer Questions

Question 1.
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 2.
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 fractions are formed alternatively.

Question 3.
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 = \(\frac{2 \pi}{T}\)

Parameters :

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

Question 4.
Obtain an expression for the wave velocity in terms of these parameters.
Answer:
Let ‘v’ be the velocity of a wave, ‘v’ be frequency and ‘λ’ be the wavelength. If T is the time period, then v = \(\frac{1}{\mathrm{~T}}\)
The distance travelled by the wave in the time T = λ.
Distance travelled in one second = \(\frac{\lambda}{T}\)
which is equal to wave velocity v = \(\frac{\lambda}{T}\).
∴ v = vλ

Question 5.
Using dimensional analysis obtain an expression for the speed of transverse waves in a stretched string.
Answer:
Wave velocity v ∝ T^a µ^b ⇒ V = K T^a µ^b ——-> (1)
Dimensions of v = M⁰L¹ T^-1, Tension T = M¹L¹T^-2,
Linear mass µ = M¹L^-1, Constant K = M⁰L⁰T⁰
Now (1) becomes M⁰L¹T^-1 = [M¹L¹T^-2]^a [M¹L^-1]^b
M⁰L¹T¹ = M^{a + b}L^a-bT^-2a Comparing the powers of same physical quantity.
-1 = -2a ⇒ a = \(\frac{1}{2}\)
a + b = 0 ⇒ b = –\(\frac{1}{2}\)
⇒ v = (1)\(T^{\frac{1}{2}} \mu^{\frac{1}{2}}\) [∵ K = 1 Practically]
∴ v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\)

Question 6.
Using dimensional analysis obtain an expression for the speed of sound waves in a medium. .
Answer:
Speed of sound v ∝ B^a ρ^b ⇒ v = K B^a ρ^b ——–> (1)
Dimensions of v = M⁰L¹T^-1,
Elasticity of medium,
B = M¹L^-1T^-2, density ρ = M¹L^-3, constant K = M⁰L⁰T⁰.
Now (1) becomes M⁰L¹T^-1 = M⁰L⁰T⁰ [M¹L^-1T^-2]^a [M¹L^-3]^b
0 = a + b
1 = -a – 3b
-1 = -2a ⇒ a = \(\frac{1}{2}\)
b = –\(\frac{1}{2}\)
v = K \(B^{\frac{1}{2}} \rho^{\frac{1}{2}}\)
∴ v = \(\sqrt{\frac{B}{\rho}}\) [∵ K = 1, practically]

Question 7.
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 8.
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 9.
What is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary.
Answer:
π Radian or 180°.

Question 10.
What is a stationary or standing wave ?
Answer:
When two identical progressive (Transverse 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 11.
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 12.
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 13.
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 14.
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 15.
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 + \(\frac{1}{2}\))\(\frac{v}{21}\) where n = 0, 1, 2, 3, ……

Question 16.
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 17.
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
V_n = \(\frac{n v}{21}\)
where n = 1, 2, 3, ……….

Question 18.
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”.

Question 19.
Write down an expression beat frequency and explain the terms there in.
Answer:
Expression of beat frequency, Δv = v₁ ~ v₂
where v₁ and v₂ are the frequencies of two waves.

Question 20.
What is ‘Doppler effect’? Give an example.
Answer:
Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between source of sound and observer is called “Doppler effect”.

E.g.: When the whistling railway engine approaches the stationary observer on the platform, the frequency of sound appears to increase above the actual frequency. When it moves away from the observer, the apparent frequency decreases.

Question 21.
Write down an expression for the observed frequency when both source and observer are moving relative to each other in the same direction.
Answer:
Apparent frequency of sound heard by an observer,
v’ = \(\left[\frac{v-v_0}{v-v_s}\right] v\)

where v = frequency of sound
v = velocity of sound
v₀ = velocity of observer
v_S = velocity of source

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.
Write an expression for a progressive harmonic wave and explain the various parameters used in the expression.
Answer:
The expression of a progressive harmonic wave is written as y = a sin(ωt – \(\frac{2 \pi}{\lambda} x\))
or y = a sin(ωt – kx) where ω = 2πv, k = \(\frac{2 \pi}{\lambda}\)

Parameters:

1. Amplitude (a) : It is the maximum displacement of a vibrating particle from its mean position.
2. Frequence (v): It is the number of complete vibrations made by a vibrating body in one second.
3. Wave length (λ) : It is defined as the distance covered by a wave while it completes one vibration, (or) It is the distance between two consecutive points in the same phase.
4. Phase of vibration (ϕ) : The phase of vibration of a vibrating particle gives its state of displacement at a given instant. It is generally given by the phase angle.

Question 4.
Explain the modes of vibration of a stretched string with examples.
Answer:
Modes of vibrations of a stretched string :

1. In sitar or Guitar, a stretched string can vibrate, in different frequencies and form stationary waves. This mode of vibrations are known as harmonics.
2. If it vibrates in one segment, which is known as fundamental harmonic. The higher harmonics are called the overtones.
   
3. It vibrates in two segments then the second harmonic is called first overtone. Similarly the pattern of vibrations are shown fig.
4. If a stretched string vibrates with P ’Seg’ ments (loop) then frequency of vibration v = \(\frac{\mathrm{P}}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\) where T = tension in the string, µ = linear density = \(\frac{\text { mass }}{\text { length }}\)
5. In first mode of vibration, P = 1, then v = \(\frac{1}{2l} \sqrt{\frac{T}{\mu}}\) (1^st hamonic (or) fundamental frequency)
6. second mode of vibration, P = 2, then v₁ = \(\frac{2}{2l} \sqrt{\frac{T}{\mu}}\) = 2v (2^nd harmonic (or) 1^st overtone)
7. In third mode of vibration, P = 3, then v₂ = \(\frac{3}{2l} \sqrt{\frac{T}{\mu}}\) = 3v (3^rd harmonic (or) 2^nd overtone)
   The ratio of the frequency of Harmonics are, v : v₁ : v₂ = v : 2v : 3v = 1 : 2 : 3

Question 5.
Explain the modes of vibration of an air column in an open pipe.
Answer:
Modes of vibration of an air column in an open pipe :
1) For a open pipe both the ends are open. So antinodes will be formed at both the ends. But two antinodes cannot exist without a node between them.
2) The possible harmonics in vibrating air column of a open pipe is given by .
Where n = 1, 2, 3
(1^st harmonic or fundamental frequence)

3) In first normal Mode of vibrating air column in a open pipe v₁ = \(\frac{v}{2l}\) = v
(2nd harmonic 1^st overtone)

4) In second normal Mode of vibrating air column in a open pipe, v₂ = \(\frac{2 v}{2l}\) = 2v

5) In third, normal Mode of vibrating air column in a open pipe, v₃ = \(\frac{3 v}{21}\) = 3u
(3^rd harmonic 2^nd overtone)

6) In open pipe the ratio of frequencies of harmonics is
v₁ : v₂ : v₃ = v : 2v : 3v = 1 : 2 : 3

Question 6.
What do you understand by ‘resonance’ ? How would you use resonance to determine the velocity of sound in air ?
Answer:
Resonance: If the natural frequency of a vibrating body is equal to the frequency of external periodic force then the two bodies are said to be in resonance. At resonance the bodies will vibrate with increasing amplitude.

Determination of velocity of sound in air using resonance :

1) In resonance tube, an air column is made to vibrate by means of vibrating fork. At certain length of air column, the air column would have the same frequency as that of the fork. Then the air column vibrates with the maximum amplitude and the intense sound is produced.

2) The vibrating fork of known frequency (v) is placed above the open end of the tube.

3) The length of air column is gradually increased until the booming sound can be heard at two different lengths of air column.

4) In first resonance, l₁ be the length of air column, then \(\frac{\lambda}{4}\) = l₁ + C …….. (1)
Where λ is the wavelength of sound emitted by the fork and C is the end correction of the tube.

5) In second resonance, l₂ be the length of air column, then \(\frac{3 \lambda}{4}\) = l₂ + C …… (2)
(2) – (1) ⇒ \(\frac{\lambda}{2}\) = l₂ – l₁

λ = 2 (l₂ – l₁)
Speed of sound is given by
υ = vλ = v[2(l₂ – l₁)]
∴ υ = 2v (l₂ – l₁)

6) Hence by knowing v, l₁, l₂ the speed of sound is calculated.

Question 7.
What are standing waves ? Explain how standing waves may be formed in a stretched string.
Answer:
Standing wave or stationary: When two identical progressive (Transverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are super-imposed then the resultant wave is called standing wave.

Formation of standing wave in a stretched string : –

1. If a string of length ‘l’ is stretched between two fixed points and set into vibration, a transverse progressive wave begins to travel along the string.
2. The wave is get reflected at the other fixed end.
3. The incident and reflected waves interfere and produce a stationary wave.
4. The stationary wave with nodes and antinodes is shown below.
   

Question 8.
Describe a procedure for measuring the velocity of sound in a stretched string.
Answer:
The velocity of a transverse wave travelling along a stretched string in fundamental mode is given by v = 2vl, where v = frequency, l = resonating length.

Measurement of velocity of sound in a stretched string using sonometer :

1. The wire is subjected to a fixed tension with suitable load.
2. A tuning fork of known frequency (v), is excited and the stem is held against the sono – meter box.
   
3. The distance between the two bridges is adjusted such that a small paper rider at the middle of B₁ B₂ vibrates vigorously and flies off due to resonance.
4. The resonating length ‘l’ can be measured between two bridges with scale.
5. By knowing v and l; we can find the velocity of a wave using υ = 2vl.

Question 9.
Explain, using suitable diagrams, the formation of standing waves in a closed pipe. How may this be used to determine the frequency of a source of sound ?
Answer:
Formation of standing waves in a closed pipe :

1. In closed pipe one end is closed and the other end is open. So antinode is formed at open end and antinode is formed at closed end.
2. The possible harmonics in vibrating air column in a closed pipe v_n = \(\frac{(2 n+1) v}{4 l}\) where v = 0, 1, 2, 3, …….
3. In first normal mode of vibrating air column in a closed pipe, v₁ = \(\frac{v}{41}\)
   [first harmonic (or) fundamental frequency]
   
4. In second normal mode of vibrating air column in a closed pipe,
   v₃ = \(\frac{3 \mathrm{v}}{4l}\) [Third harmonic (or) first overtone]
5. In third normal mode of vibrating air column in a closed pipe,
   v₅ = \(\frac{5 \mathrm{v}}{4 \mathrm{l}}\) [Fifth harmonic (or) second overtone]

Determination of frequency of a source of sound :

1. The vibrating fork of unknown frequency (v) is placed above the open end of the tube.
   
2. Reservoir is slowly lowered, until a large booming sound is heard. Measure 1st resonating, air column length l₁.
3. Further lower the reservoir, until second time a large booming sound is heard. Measure 2^nd resonating air column length l₂.
4. Velocity of a wave at 0°C is v = 331 m/s.
5. By knowing v, l₁ and l₂ we can find unknown frequency of a tuning fork using
   v = \(\frac{v}{2\left(l_2-l_1\right)}\)

Question 10.
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 Δυ = 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 11.
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.

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. (A.P. Mar. ’19)
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 = \(\frac{\lambda}{4}\), \(\frac{2 \lambda}{2}\), \(\frac{3 \lambda}{2}\) ……… etc, the amplitude = zero

The 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 ‘l’ is its length then length of each segment = \(\frac{l}{p}\)
Which is equal to \(\frac{\lambda}{2}\)
∴ \(\frac{l}{\mathrm{p}}\) = \(\frac{\lambda}{2}\) ⇒ λ = \(\frac{2 l}{\mathrm{P}}\)
Harmonic frequency v = \(\frac{v}{\lambda}=\frac{v \mathrm{P}}{2 l}\)
v = \(\frac{v P}{2 l}\) ——- (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 = \(\frac{p}{2 l} \sqrt{\frac{T}{\mu}}\)
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 (I) 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 ∝ \(\sqrt{\mathrm{T}}\) ⇒ \(\frac{v}{\sqrt{T}}\) = constant, when ‘l’ and ‘m’ are constant.

Third Law: WHien the length (J) 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. (T.S. Mar. ’16, A.P. Mar. ’15)
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 antinodes 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 = \(\frac{v}{2 l}\) = v —— (1)
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 = \(\frac{2l}{2}\)
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 = 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. (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 = \(\frac{v}{4l}\) = 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 = \(\frac{4l}{3}\)
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 = \(\frac{v}{\lambda_5}=\frac{5 v}{4 I}\)
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 are beats ? Obtain an expression for the beat frequency ? Where and how are beats made use of ?
Answer:
Beats : Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing and waning in the intensity of the resultant sound waves at regular intervals of time are called Beats.
If v₁ and v₂ are the frequencies of two sound notes superimposed along the same direction, no of beats heard per second = Δv = v₁ – v₂.
Maximum no. of beats heard per sec is 10 due to persistence of hearing.

Expression for the beat frequency :

1. Consider the two wave trains of equal amplitude but of nearly equal frequencies.
2. Let the frequencies of the waves be v₁ and v₂. Say v₁ is slightly greater than v₂.
3. Let the beat period be T seconds.
4. No.of vibrations, made by the first wave train in T seconds – v₁T
   [∵ no.of oscillations in 1 sec = v]
   [∵ no.of oscillations in T sec = vt]
5. No.of vibrations, made by the second wave train in T seconds = v₂T
6. During the time interval T, the first wave train would have completed one vibration more than the second wave train.
7. Hence, v₁T – v₂T = 1 or v₁ – v₂ = \(\frac{1}{\mathrm{~T}}\)
8. Since, T is the beat period, no.of beats per seconds = \(\frac{1}{\mathrm{~T}}\)
9. Hence the beat frequency = \(\frac{1}{\mathrm{~T}}\) = v₁ – v₂ = Δv
10. That is the beat frequency is the difference between the frequencies of the two wave trains.

Practical applications of beats:

1. Determination of an unknown frequency: Out of two tuning forks, one is loaded with wax and the other is filed. The excited tuning forks are close together and no.of beats can be heard. Then after unknown frequencies of them will be found practically.
2. For tuning musical instruments : Musicians use the beat phenomenon in tuning their musical instruments.
3. For producing colourful effects in music: Sometimes, a rapid succession of beats is knowingly introduced in music. This produces an effect similar to that of human voice and is appreciated by the audience.
4. For detection of Marsh gas (dangerous gases) in mines.

Question 5.
What is Doppler effect? Obtain an expression for the apparent frequencý of sound heard when the source is in motion with respect to an observer at rest. (A.P. Mar. ’16, Mar. ’14)
Answer:
Doppler 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 doppler 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.

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

Question 6.
What is Doppler shift? Obtain an expression for the apparent frequency of sound heard when the observer Is In motion with respect to a source át rest.
Answer:
Doppler Shift: Due to the relative motion, when the source comes closer to listener, the apparent frequency is greater than actual frequency and source away from listener; the apparent frequency is less than actual frequency So the difference in apparent and actual frequencies is known as Doppler shift.
Expression for the apparent frequency heard by a moving observer:

Case (1) : When observer Is moving towards source:
Let ‘υ₀’ be velocity of listener ‘O’, moving towards the stationary source ‘s’ as shown in figure. So observer will receive more number of waves in each second.
The distance travelled by observer in one second = υ₀

The number of extra waves received by the observer = \(\frac{v_0}{\lambda}\)
We know v = vλ ⇒ λ = \(\frac{v}{v}\)

Where υ = Velocity of sound
v = Frequency of sound
If ‘v’ is apparent frequency heard by him then

Therefore the apparent frequency is greater than actual frequency.

Case (2) : When observer Is moving away from rest source:
If the observer is moving away from the stationary source then he loses the number of waves \(\frac{v_0}{\lambda}\)
∴ Apparent frequency v’ = v – \(\frac{v_0}{\lambda}\) = v – \(\frac{v_0 \cdot v}{v}\)
v’ = \(\left[\frac{v-v_0}{v}\right] \cdot v\)
Hence the apparent frequency is less than actual frequency.

Problems

Question 1.
A stretched wire of length 0.6m is observed to vibrate with a frequency of 30Hz 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.
Solution:
v = 30 Hz; l = 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 steel cable of diameter 3 cm is kept under a tension of 10kN. 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 = \(\frac{\mathrm{D}}{2}\) = \(\frac{3}{2}\)cm
= \(\frac{3}{2}\) × 10^-2m;
A = πr² = \(\frac{22}{7} \times\left[\frac{3}{2} \times 10^{-2}\right]^2\)

Question 3.
Two progressive transverse waves given by y₁ = 0.07 sinπ(12x-500t) and y₂ = 0.07 sinπ(12x + 500t) travelling along a stretched string from nodes and antinodes. What is the displacement at the (a) nodes (b) antinodes ? What is the wavelength of the standing wave ?
Solution:
A₁ = 0.07; A₂ = 0.07; K = 12π
a) At nodes, displacement
y = A₁ – A₂ = 0.07 – 0.07 = 0
b) At antinodes, displacement
y = A₁ + A₂ = 0.07 + 0.07 = 0.14 m
c) Wavelength λ = \(\frac{2 \pi}{\mathrm{K}}=\frac{2 \pi}{12 \pi}\) = 0.16 m

Question 4.
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;

Question 5.
A metal bar when clamped at its centre resonantes in its fundamental frequency with longitudinal waves of frequency 4 kHz. If the clamp is moved to one end. What will be its fundamental resonance frequency ?
Solution:
When a metal bar of length l is clamped in the middle, it has one node in the middle and two antinodes at its free ends. In the fundamental mode. l = \(\frac{\lambda}{2}\) ⇒ λ = 21

In fundamental mode of frequency of bar
= frequency of wave = 4 kHz.
∴ v = \(\frac{\mathrm{v}}{\lambda}\) = \(\frac{\mathrm{v}}{2l}\) = 4kHz —– (1)
When clamp is moved to one end,
l = \(\frac{\lambda^{\prime}}{4}\) ⇒ λ’ = 4l
∴ V¹ = \(\frac{\mathrm{v}}{\lambda}\) = \(\frac{\mathrm{v}}{4 \mathrm{l}}\) = \(\frac{4 \mathrm{kHz}}{2}\) = 2kHz
[∵ from (1)]

Question 6.
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:
I = 70 cm = 70 × 10^-2m; v = 331 m/s ; v = ?
v = ?
v = \(\frac{v}{4 l}\) = \(\frac{331}{4 \times 70 \times 10^{-2}}\) = 118.2 Hz

Question 7.
A vertical tube is made to stand in water so that the water level can be adjusted. Sound waves of frequency 320 Hz are sent into the top of the tube. If standing waves are produced at two successive water levels of 20 cm and 73 cm, what is the speed of sound waves in the air in the tube ?
Solution:
v = 320 Hz; l₁ = 20cm = 20 × 10^-2
l₂ = 73 cm = 73 × 10^-2m; v = ?
v = 2v(l₂ – l₁)
= 2 × 320 (73 × 10^-2 – 20 × 10^-2)
∴ v = 339 m/s .

Question 8.
Two organ pipes of lengths 65cm and 70cm respectively, are sounded simultaneously. How many beats per second will be produced between the fundamental frequencies of the two pipes ? (Velocity of sound = 330 m/s).
Solution:
l₁ = 65 cm = 0.65 m
₂ = 70 cm = 0.7 m
v = 330 m/s
No. of beats per second ∆υ = υ₁ – υ₂
= \(\frac{v}{2 h}\) – \(\frac{\mathrm{v}}{2 l_2}\) = \(\frac{330}{2 \times 0.65}\) – \(\frac{330}{2 \times 0.7}\)
∴ ∆v = 253.8 – 235.8 = 18Hz

Question 9.
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.
Solution:
When a whistling train approaches to rest observer,

v’ = \(\left[\frac{v}{v-v_s}\right] v\) ——– (1)
When a whistling train away from rest observer

v” = \(\left[\frac{v}{v+v_{\mathrm{S}}}\right] v\) —— (2)
Here v’ = 219 Hz; V” = 184 Hz;
v = 340 m/s
\(\frac{(1)}{(2)}\) ⇒ \(\frac{v^{\prime}}{v^{\prime \prime}}\) = \(\frac{\left(v+v_s\right)}{\left(v-v_s\right)}\)
\(\frac{219}{184}\) = \(\frac{340+v_{\mathrm{s}}}{340-v_{\mathrm{s}}}\)
219(340 – υ_s) = 184(340 + υ_s)
219 × 340 – 219 υ_s = 184 × 340 + 184 υ_s
403 υ_s = 35 × 340
∴ Velocity of train υ_s = 29.5 m/s
Frequency of whistle, v = v’ × \(\left[\frac{v-v_{\mathrm{S}}}{v}\right]\)
= 219 × \(\left[\frac{340-29.5}{340}\right]\)
= 199.98
∴ v = 200 Hz.

Question 10.
Two trucks heading in opposite directions 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 = 330 m/ s). After the two trucks have passed each other, what frequency does the driver of the second truck hear ?
Solution:
v_s = 60 kmph = 60 × \(\frac{5}{18}\) m/s = \(\frac{300}{18}\) m/s
v₀ = 70 kmph = 70 × \(\frac{5}{18}\) m/s = \(\frac{350}{18}\) m/s
v = 400 Hz
When two trucks approach each other

When two trucks crossed each other,

Textual Exercises

Question 1.
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is caused at one end of the string, how long does the disturbance take to reach the other end ?
Answer:
Here M = 2.50 kg, T = 200 N, l = 20.0M
Mass per unit length; μ = \(\frac{\mathrm{M}}{l}\) = \(\frac{2.5}{20.0}\)
= 0.125 kg/m
Velocity V = \(\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{200}{0.125}}\) = 40 m/s
Time taken by disturbance to reach the other end
t = \(\frac{l}{\mathrm{~V}}\) = \(\frac{20}{40}\) = 0.5s.

Question 2.
A stone dropped from the top of a tower of height 300 m high splashes into the pond of water near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s^-1, (g = 9.8m s^-2)
Answer:
Here, h = 300m, g = 9.8 m/s², V = 340 m/s. If t₁ = time taken by stone to strike the surface of water in the pond, then from
S = ut + \(\frac{1}{2}\) at²
300 = 0 + \(\frac{1}{2}\) × 9.8 \(\mathrm{t}_1^2\)
t₁ = \(\sqrt{\frac{300}{4.9}}\) = 7.82s.
Time taken by sound to reach the top of tower t₂ = \(\frac{\mathrm{h}}{\mathrm{v}}\)
= \(\frac{300}{400}\) = 0.88s
Total time after which splash of sound is heard = t₁ + t₂ = 7.82 + 0.88 = 8.70s

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20° C = 343 m s^-1.
Answer:
Here, l = 12.0M, M = 2.10kg, T = ?
V = 343 m/s
Mass per unit length μ = \(\frac{\mathrm{M}}{l}\) = \(\frac{2.10}{12.0}\)
= 0.175 kg/m
As V = \(\sqrt{\frac{\mathrm{T}}{\mu}}\)
T = V² . μ = (343)² × 0.175 = 2.06 × 10⁴N.

Question 4.
Use the formula υ = \(\sqrt{\frac{\gamma P}{\rho}}\) to explain why the speed of sound in air
a) is independent of pressure,
b) increases with temperature,
c) increases with humidity.
Answer:
a) Effect of pressure:
The speed of sound in gases, υ = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\)
At constant temperature, PV = constant
P\(\frac{m}{\rho}\) = constant ⇒ \(\frac{\mathrm{P}}{\rho}\) = constant
If P increases, ρ also increases. Hence speed of sound in air is independent of pressure.

b) Effect of temperature:
As PV = nRT, P\(\frac{\mathrm{m}}{\rho}\) = \(\frac{m}{M} R T\)
⇒ \(\frac{P}{\rho}\) = \(\frac{\mathrm{RT}}{\mathrm{M}}\)
∴ υ = \(\sqrt{\frac{R T}{M}}\)
Since R, M are constants υ ∝ \(\sqrt{\mathrm{T}}\)
∴ Velocity of sound in air depends on temperature.

c) Effect of humidity:
As υ = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\) ∴ υ ∝ \(\frac{1}{\sqrt{\rho}}\)
As the density of water vapour is less than density of dry air at STP. So the presence of moisture in air decreases the
density of air. Since the speed of sound is inversely proportional to the square root of density. So sound travels faster in moist air than dry air. Hence velocity of sound
V ∝ humidity

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f(x, t) where x and t must appear in the combination x – υt or x + υt, i.e. y = f(x ± υ t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:
Answer:
No, the converse is not true. The basic requirement for a wave function to represent a travelling wave is that for all values of x & t, wave function must have a finite value.

Out of the given functions y, no one satisfies this condition therefore, none can represent a travelling wave.

Question 6.
A bat emits ultrasonic sound a frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of
(a) the reflected sound,
(b) the transmitted sound?
Speed of sound in air is 340 m s^-1 and in water 1486 m s^-1.
Answer:
Here V = 100 KHz = 10⁵Hz, V_a = 340m/s, V_w = 1486 m/s^-1
Wavelength of reflected sound, λ_a = \(\frac{\mathrm{V}^{\mathrm{a}}}{\mathrm{V}}\)
= \(\frac{340}{10^5}\) = 3.4 × 10^-3 m
Wavelength of transmitted sound,
λ_w = \(\frac{V_w}{V}\) = \(\frac{1486}{10^5}\) = 1.486 × 10^-2 m

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^-1? The operating frequency of the scanner is 4.2 MHz.
Answer:
λ = ? υ = 1.7 Km/s = 1700 ms^-1
y = 4.2 MHz = 4.2 × 10⁶Hz
λ = \(\frac{v}{v}\) = \(\frac{1700}{4.2 \times 10^6} \mathrm{~m}\) = 0.405 × 10^-3 m
= 0.405 mm

Question 8.
A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right.
a) Is this a travelling wave or a stationary wave?
If it is travelling, what are the speed and direction of its propagation?
b) What are Its amplitude and frequency?
C) What is the initial phase at the origin?
d) What is the least distance between two successive crests in the wave?
Answer:
Compare the given equation with that of plane progressive wave of amplitude r, travelling with a velocity V from right to left.
y(x, t) = rsin\(\left[\frac{2 \pi}{\lambda}(v t+x)+\phi_0\right]\) ……… (1)
We find that
a) The given equation represents a transvërse harmonic wave travelling from right to left. It is ñot a stationary wave.

b) The given equation can be written as
Y(x, t) = 3.0sin[0.018(\(\frac{36}{0.018}\) + x) + \(\frac{\pi}{4}\)] ……… (2)
equating coefficient of t in the two
(1) & (2) we get. :
V = \(\frac{.36}{0.018}\) = 2000 cm/sec.
Obviously, r = 3.0 cm
Also, \(\frac{2 \pi}{\lambda}\) = 0.018
λ = \(\frac{2 \pi}{0.018} \mathrm{~cm}\)
Frequency, v = \(\frac{v}{\lambda}\) = \(\frac{2000}{2 \pi}\) × 0.018
= 5.7351.

c) Intial phase, φ₀ = \(\frac{\pi}{4}\) radian.

d) Least distance between two successive crests of the wave =
Wave length, λ = \(\frac{2 \pi}{0.018 \mathrm{~cm}}\) = 349 cm,

Question 9.
For the wave described in the last problem plot the displacement (y) versus (t) graphs for x = 0.2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
Answer:
The transverse harmonic wave is
y(x,t) = 3.0 sin[36t + 0.018x + \(\frac{\pi}{4}\)]
For x = 0
y(0, t) = 3.0 sin(36t + \(\frac{\pi}{4}\)) —— (i)
Here w = \(\frac{2 \pi}{T}\) = 36, T = \(\frac{2 \pi}{36}\) = \(\frac{\pi}{18}\)-sec.

For different values of t, we calculate y using eq(i). These values are tabulated below.

On plotting y versus t graph, we obtain a sinusoidal curve as shown in fig.

Similar graphs are obtained for x = 2cm & x = 4cm. The oscillary motion in travelling wave differs from one point to another only in terms of phase. Amplitude and frequency of oscillatory motion remain the same in all the three areas.

Question 10.
For the travelling harmonic wave
y(x, t) = 2.0 cos 2 π (10t – 0.0080 x + 0.35)
where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
a) 4 m,
b) 0.5 m,
c) λ/2,
d) 3λ/4,
Answer:
The given equation be written as
y = 2.0 cos[2π(10t – 0.0080x) + 2π × 0.35]
y = 2.0 cos[2π × (0.0080(\(\frac{10 \mathrm{t}}{0.0080}\) – x) + 0.7π]
Compare it with the standard equation of a travelling harmonic, we have
y = r.cos[\(\frac{2 \pi}{\lambda}(v t-x)+\phi_0\)
We get, \(\frac{2 \pi}{\lambda}\) = 2π × 0.0080
Further we know that phase diff. φ = \(\frac{2 \pi}{\lambda} \mathrm{x}\)
a) When x = 4m = 400 cm
φ = \(\frac{2 \pi}{\lambda}\) . x = 2π × 0.0080 × 400
= 6.4 π rad.

b) When x = 0.5 = 50 cm
φ = \(\frac{2 \pi}{\lambda}\) . x = 2π × 0.0080 × 50
= 0.8π rad.

c) When x = \(\frac{\lambda}{2}\)
φ = \(\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) = π rad.

d) When x = \(\frac{3 \lambda}{4}\)
φ = \(\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4}\) = \(\frac{3 \lambda}{2} \mathrm{rad}\)

Question 11.
The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin\(\left(\frac{2 \pi}{3} x\right)\)cos (120 πt)
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10^-2 kg
Answer the following:
a) Does the function represent a travelling wave or a stationary wave?
b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
c) Determine the tension in the string.
Answer:
The given equation is
y(x, t) = 0.06 sin \(\frac{2 \pi}{3} \mathrm{x} \cos 120 \pi \mathrm{t}\)

a) As the equation involves harmonic functions of x and t seperately, it represents a stationary wave.

b) We know that when a wave pulse
y₁ = r sin \(\frac{2 \pi}{\lambda}(v t+x)\) —– (i)
travelling along + direction of x-axis is super imposed by the reflected wave
y = y₁ + y₂ = -2rsin\(\frac{2 \pi}{\lambda}\) xcos \(\frac{2 \pi}{\lambda}\) vt is formed. ——- (ii)
Comparing (i) & (ii) we find that
\(\frac{2 \pi}{\lambda}\) = \(\frac{2 \pi}{3}\) ⇒ λ = 3m.
Also \(\frac{2 \pi}{\lambda} v\) = 120π (Or)
V = 60λ = 60 × 3 = 180m/s.
frequency, v = \(\frac{v}{\lambda}\) = \(\frac{180}{3}\) = 60 hertz.
Note that both the waves have same wave length, same frequency and same speed.

c) Velocity of transverse waves is
υ = \(\sqrt{\frac{T}{\mu}}\) (or) υ² = T/μ
T = V² × μ where μ = \(\frac{3 \times 10^{-2}}{1.5}\)
= 2 × 10^-2 kg/m
T = (180)² × 2 × 10^-2 = 648 N.

Question 12.
i) For the wave on a string described in previous problem do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers.
(ii) What is the amplitude of a point 0.375 m away from one end?
Answer:
i) All the points on the string
a) have the same frequency except at the nodes (where frequency is cos θ),
b) have the same phase every where in one loop except at the nodes,
c) however, the amplitude of vibration at different points is different.

ii) From y(x, t) = 0.06 sin\(\left(\frac{2 \pi}{3} x\right)\) cos (120 πt)
The amplitude at x = 0.375 m is 0.06
sin \(\frac{2 \pi}{3} x \times 1\) = 0.06 × sin \(\frac{2 \pi}{3} \times 0.375\)
= 0.06sin\(\frac{\pi}{4}\) = \(\frac{0.06}{\sqrt{2}}\) = 0.042 M

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave State which of these represent
(i) a travelling wave,
(ii) a stationary wave or
(iii) none at all:
a) y = 2 cos (3x) sin (10t)
b) y = \(2 \sqrt{x-v t}\)
c) y = 3 sin(5x – 0.5t) + 4 cos(5x – 0.5t)
d) y = cos x sin t + cos 2x sin 2t
Answer:
a) It represents a stationary wave as harmonic functions of x & t are contained separetely in the equation.
b) It cannot represent any type of wave.
c) It represents a progressive / travelling harmonic wave.
d) This equation is sum of two functions each representing a stationary wave. Therefore it represents superposition of two stationary waves.

Question 14.
A wire stretched between two right supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10^-2 kg and its linear mass density is 4.0 × 10^-2 kg m^-1. What is (a) the speed of transverse wave on the string, and (b) the tension in the string?
Answer:
Here, v = 45Hz, μ = 3.5 × 10^-2 kg
Mass/length = μ = 4.0 × 10^-2 kg/m
l = \(\frac{\mu}{\mu}\) = \(\frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}}\) = \(\frac{7}{8}\)
As \(\frac{\lambda}{2}\) = l = \(\frac{7}{8}\) ∴ λ = \(\frac{7}{4}\)m = 1.75m
a) The speed of transverse wave
υ = vλ = 45 × 1.75 = 78.75 m/s.

b) As υ = \(\sqrt{\frac{\mathrm{T}}{\mu}}\)
∴ T = υ² × μ = (78.75)² × 4.0 × 10^-2
= 248.06 N.

Question 15.
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length Is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.
Answer:
As there is a piston at one end of the tube, it behaves as a closed organ pipe, which produces odd harmonics only. Therefore the pipe is in resonance with the fundamental note at the third harmonic (79.3 cm is about 3 times 25.5 cm)
In the fundamental note = \(\frac{\lambda}{4}\) = l₁ = 25.5
λ = 4 × 25.5 = 102 cm = 1.02m
Speed of sound in air.
υ = vλ = 340 × 1.02
= 346.0 m/s

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
Answer:
Here, l = 100 cm = 1 m, y = 2.53 KHz
= 2.53 × 10³ Hz
When the rod is clamped at the middle, then in the fundamental mode of vibration of the rod, anode is formed at the middle and antinode is formed at each end.
Therefore, it is clear from fig

l = \(\frac{\lambda}{4}\) + \(\frac{\lambda}{4}\) + \(\frac{\lambda}{2}\)
λ = 2l = 2m
As v = λl
v = 2.53 × 10³ × 2
= 5.06 × 10³ ms^-1

Question 17.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s^-1).
Answer:
Here l = 20 cm = 0.2m, v_n = 430 Hz,
υ = 340 m/s
The frequency of n^th normal mode of vibration of closed pipe is
v_n = (2n – 1)\(\frac{v}{4l}\)
∴ 430 = (2n – 1)\(\frac{340}{4 \times 0.2}\)
2n – 1 = \(\frac{430 \times 4 \times 0.2}{340}\) = 1.02
2n = 2.02, n = 1.01
Hence it will be the 1^st normal mode of vibration. In a pipe, open at both ends we have
v_n = n × \(\frac{\mathrm{v}}{2l}\) = \(\frac{\mathrm{n} \times 340}{2 \times 0.2}\) = 430.
∴ n = \(\frac{430 \times 2 \times 0.2}{340}\) = 0.5
As n has to be an integer, therefore open organ pipe cannot be in resonance with the source.

Question 18.
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324Hz. What is the frequency of B?
Answer:
Let original frequency of sitar string A be n_a & original frequeny of sitar string B be n_b.
As number of beats / sec = 6
∴ n_b = n_a ± 6 = 330 (or) 318Hz.
When tension in A is reduced, its frequency reduces (∴ n ∝ \(\sqrt{T}\))
As number of beats/sec decreases to 3 therefore, frequency of B = 324 – 6
= 318Hz.

Question 19.
Explain why (or how):
a) in a sound wave, a displacement node is a pressure antinode and vice versa,
b) bats can ascertain distances, directions, nature and sizes of the obstacles without any ‘eyes”.
c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
d) Soils can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases and
e) The shape of a pulse gets distorted during propagation In a dispersive medium.
Answer:
a) Node (N) is a point where the amplitude of oscillation is 0. (and pressure is maximum)
Antinode (A) is a point where the amplitude of oscillation is maximum (and pressure is min).
These nodes & antinodes do not coincide with pressure nodes & antinodes.
Infact, N coincides with pressure antinode and A coincides with pressure node, as is clear from the definitions.

b) Bats emit ultrasonic wave of large frequencies, when these waveš are reflected from the obstacles in their path,
they give them the idea about the distance, direction, size & nature of the obstacle.