Class 12

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 11 Magnetic Materials Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 11 Magnetic Materials

1. Choose the correct option.

i) Intensity of magnetic field of the earth at the point inside a hollow iron box is.
(A) less than that outside
(B) more than that outside
(C) same as that outside
(D) zero
Answer:
(D) zero

ii) Soft iron is used to make the core of the transformer because of its
(A) low coercivity and low retentivity
(B) low coercivity and high retentivity
(C) high coercivity and high retentivity
(D) high coercivity and low retentivity
Answer:
(A) low coercivity and low retentivity

iii) Which of the following statements is correct for diamagnetic materials?
(A) µ_r < 1
(B) χ is negative and low
(C) χ does not depend on temperature
(D) All of above
Answer:
(D) All of above

iv) A rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves ( each having half of the original length) and one piece is made to oscillate freely. Its period of oscillation is T′, the ratio of T′ / T is.
(A) \(\frac{1}2 \sqrt{2}\)
(B) \(\frac{1}{2}\)
(C) 2
(D) \(\frac{1}{4}\)
Answer:
(B) \(\frac{1}{2}\)

v) A magnetising field of 360 Am -1 produces a magnetic flux density (B ) = 0.6 T in a ferromagnetic material. What is its permeability in Tm A^-1 ?
(A) \(\frac{1}{300}\)
(B) 300
(C) \(\frac{1}{600}\)
(D) 600
Answer:
(C) \(\frac{1}{600}\)

2 Answer in brief.

i) Which property of soft iron makes it useful for preparing electromagnet?
Answer:
An electromagnet should become magnetic when a current is passed through its coil but should lose its magnetism once the current is switched off. Hence, the ferromagnetic core (usually iron-based) used for an electromagnet should have high permeability and low retentivity, i.e., it should be magnetically ‘soft’.

ii) What happens to a ferromagnetic material when its temperature increases above curie temperature?
Answer:
A ferromagnetic material is composed of small regions called domains. Within each domain, the atomic magnetic moments of nearest-neighbour atoms interact strongly through exchange interaction, a quantum mechanical phenomenon, and align themselves parallel to each other even in the absence of an external magnetic field. A domain is, therefore, spontaneously magnetized to saturation.

The material retains its domain structure only up to a certain temperature. On heating, the increased thermal agitation works against the spontaneous domain magnetization. Finally, at a certain critical temperature, called the Curie point or Curie temperature, thermal agitation overcomes the exchange forces and keeps the atomic magnetic moments randomly oriented. Thus, above the Curie point, the material becomes paramagnetic. The ferromagnetic to paramagnetic transition is an order to disorder transition. When cooled below the Curie point, the material becomes ferromagnetic again.

iii) What should be retentivity and coercivity of permanent magnet?
Answer:
A permanent magnet should have a large zero-field magnetization and should need a very large reverse field to demagnetize. In other words, it should have a very broad hysteresis loop with high retentivity and very high coercivity.

iv) Discuss the Curie law for paramagnetic material.
Answer:
Curie’s law : The magnetization of a paramagnetic material is directly proportional to the external magnetic field and inversely proportional to the absolute temperature of the material.

If a paramagnetic material at an absolute temperature T is placed in an external magnetic field of induction \(\), the magnitude of its magnetization
M_z ∝ \(\frac{B_{\text {ext }}}{T}\) ∴ M_z = C\(\frac{B_{\text {ext }}}{T}\)
where the proportionality constant C is called the Curie constant.
[Notes : (1) The above law, discovered experimentally in 1895 by Pierre Curie (1859-1906) French physcist, is true only for values of Bext/ T below about 0.5 tesla per kelvin.
(2) [C] = [M_z ∙ T] / [B_ext] = [L^-1I ∙ ] /[MT^-2I^-1]
= [M^-1L^-1T²I²],
where  denotes the dimension of temperature.]

v) Obtain and expression for orbital magnetic moment of an electron rotating about the nucleus in an atom.
Answer:
In the Bohr model of a hydrogen atom, the electron of charge – e performs a uniform circular motion around the positively charged nucleus. Let r, v and T be the orbital radius, speed and period of motion of the electron. Then,
T = \(\frac{2 \pi r}{v}\) …………….. (1)
Therefore, the orbital magnetic moment asso-ciated with this orbital current loop has a magnitude,
I = \(\frac{e}{T}=\frac{e v}{2 \pi r}\) …………… (2)
Therefore, the magnetic dipole moment associated with this electronic current loop has a magnitude
M₀ = current × area of the loop
= I(πr²) = \(\frac{e v}{2 \pi r}\) × πr² = \(\frac{1}{2}\) evr ……………… (3)
Multiplying and dividing the right hand side of the above expression by the electron mass m_e,
M₀ = \(\frac{e}{2 m_{\mathrm{e}}}\) (m_evr) = \(\frac{e}{2 m_{\mathrm{e}}}\) L₀ ……………. (4)
where L₀ = m_evr is the magnitude of the orbital angular momentum of the electron. \(\vec{M}_{0}\) is opposite to \(\vec{L}_{0}\).
∴ \(\vec{M}_{0}=-\frac{e}{2 m_{e}} \overrightarrow{L_{0}}\) ……………. (5)
which is the required expression.

According to Bohr’s second postulate of stationary orbits in his theory of hydrogen atom, the angular momentum of the electron in the nth stationary orbit is equal to n \(\frac{h}{2 \pi}\) , where h is the Planck constant and n is a positive integer. Thus, for an orbital electron,

L₀ = m_evr = \(\frac{nh}{2 \pi}\) …………… (6)
Substituting for L₀ in Eq. (4),
M₀ = \(\frac{e n h}{4 \pi m_{\mathrm{e}}}\)
For n = 1, M₀ = \(\frac{e n h}{4 \pi m_{\mathrm{e}}}\)
The quantity \(\frac{e n h}{4 \pi m_{\mathrm{e}}}\) is a fundamental constant called the Bohr magneton,
µ_B ∙ µ_B = 9.274 × 10^-24 J/T (or A∙m²) = 5.788 × 10^-5 eV/T.
[ Notes : (1) Magnetic dipole moment is conventionally denoted by µ. (2) The magnetic moment of an atom is expressed in terms of Bohr magneton (vµ_B). (3) According to quantum mechanics, an atomic electron also has an intrinsic spin angular momentum and an associated spin magnetic moment of magnitude µ₅. It is this spin magnetic moment that gives rise to magnetism in matter. (4) The total magnetic moment of the atom is the vector sum of its orbital magnetic moment and spin magnetic moment.]

vi) What does the hysteresis loop represents?
Answer:
A magnetic hysteresis loop is a closed curve obtained by plotting the magnetic flux density B of a ferromagnetic material against the corresponding magnetizing field H when the material is taken through a complete magnetizing cycle. The area enclosed by the loop represents the hysteresis loss per unit volume in taking the material through the magnetizing cycle.

vii) Explain one application of electromagnet.
Answer:
Applications of an electromagnet:

1. Electromagnets are used in electric bells, loud speakers and circuit breakers.
2. Large electromagnets are used in junkyard cranes and industrial cranes to lift iron scraps.
3. Superconducting electromagnets are used in MRI and NMR machines, as well as in particle accelerators of cyclotron family.
4. Electromagnets are used in data storage devices such as computer hard disks and magnetic tapes.

Question 3.
When a plate of magnetic material of size 10 cm × 0.5 cm × 0.2 cm (length , breadth and thickness respectively) is located in magnetising field of 0.5 × 10⁴ Am^-1 then a magnetic moment of 0.5 A∙m² is induced in it. Find out magnetic induction in plate.
Answer:
Data : l = 10 cm, b = 0.5 cm, h = 0.2 cm,
H = 0.5 × 10⁴ Am^-1, M = 5 A∙m²
The volume of the plate,
V = 10 × 0.5 × 0.2 = 1 cm² = 10^-6 m²
B = μ₀ (H + M_z) = μ₀ (H + \(\frac{M}{V}\))
The magnetic induction in the plate,
∴ B = 4π × 10^-7 (0.5 × 10⁴ + \(\frac{5}{10^{-6}}\))
= 6.290 T

Question 4.
A rod of magnetic material of cross section 0.25 cm² is located in 4000 Am^-1 magnetising field. Magnetic flux passing through the rod is 25 × 10^-6 Wb. Find out (a) relative permeability (b) magnetic susceptibility and (c) magnetisation of the rod.
Answer:
Data: A = 0.25 cm² = 25 × 10^-6 m²,
H = 4000 A∙m^-1, Φ = 25 × 10^-6 Wb
Magnetic induction is
B = \(\frac{\phi}{A}=\frac{25 \times 10^{-6}}{25 \times 10^{-6}}\) = 1 Wb/m²
(a) B = µ₀µ_rK
∴ The relative permeability of the material,
µ_r = \(\frac{B}{\mu_{0} H}=\frac{1}{4 \times 3.142 \times 10^{-7} \times 4000}\)
= \(\frac{10000}{50.272}\) = 198.91 = 199

(b) µ_r = 1 + χ_m
∴ The magnetic susceptibility of the material,
χ_m = µ_r – 1 = 199 – 1 = 198

(c) χ_m = \(\frac{M_{\mathrm{z}}}{H}\)
The magnetization of the rod,
M_z = χ_mH = 198 × 4000 = 7.92 × 10⁵ A/m

Question 5.
The work done for rotating a magnet with magnetic dipole momentm, through 90° from its magnetic meridian is n times the work done to rotate it through 60°. Find the value of n.
Answer:
Data : θ₀ = 0°, θ₁ = 90°, θ₂ = 60°, W₁ = nW₂
The work done by an external agent to rotate the magnet from θ₀ to θ is
W = MB (cos θ₀ – cos θ)
∴ W1 = MB(cos θ₀ – cosθ₁)
= MB (cos 0° – cos 90°)
= MB (1 – 0)
= MB

∴ W2 = MB (cos 0°- cos 60°)
= MB(1 – \(\frac{1}{2}\))
= 0.5MB
∴ W₁ = 2W₂ = MB
Given W₁ = nW₂. Therefore n = 2.

Question 6.
An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 × 10^-11 m, with a speed of 2 × 10⁶ ms^-1 Find the resultant orbital magnetic moment and angular momentum of electron. (charge on electron e = 1.6 × 10^-19 C, mass of electron m_e = 9.1 × 10^-31 kg.)
Answer:
Data: r = 5.3 × 10^-11 m, v = 2 × 10⁶ m/s,
e = 1.6 × 10^-19 C, m_e = 9.1 × 10^-31 kg
The orbital magnetic moment of the electron is
M₀ = \(\frac{1}{2}\) evr
= \(\frac{1}{2}\) (1.6 × 10^-19) (2 × 10⁶) (5.3 × 10^-11)
= 8.48 × 10^-24 A∙m²
The angular momentum of the electron is
L₀ = m_evr
=(9.1 × 10^-31) (2 × 10⁶) (5.3 × 10^-11)
= 96.46 × 10^-36 = 9.646 × 10^-35 kg∙m²/s

Question 7.
A paramagnetic gas has 2.0 × 10²⁶ atoms/m with atomic magnetic dipole moment of 1.5 × 10^-23 A m² each. The gas is at 27° C. (a) Find the maximum magnetization intensity of this sample. (b) If the gas in this problem is kept in a uniform magnetic field of 3 T, is it possible to achieve saturation magnetization? Why? (k_B = 1.38 × 10^-23 JK^-1)[Answer: 3.0× 103 A m-1, No]
(Hint: Find the ratio of Thermal energy of atom of a gas ( 3/2 kBT) and maximum potential energy of the atom (mB) and draw your conclusion)
Answer:
Data: \(\frac{N}{V}\) = 2.0 × 10²⁶ atoms/m³,
μ = 1.5 × 10^-23 Am2, T = 27 + 273 = 300 K,
B = 3T, k^B = 1.38 × 10^-23 J/K, 1 eV = 1.6 × 10^-19 J
(a) The maximum magnetization of the material,
M_z = \(\frac{N}{V}\)μ =(2.0 × 10²⁶) (1.5 ×10^-23)
= 3 × 10³ A/m

(b) The maximum orientation energy per atom is
U_m = -μB cos 180° = μB
= (1.5 × 10^-23) (3) = \(\frac{4.5 \times 10^{-23}}{1.6 \times 10^{-19}}\)
= 2.8 × 10^-4 eV

The average thermal energy of each atom,
E = \(\frac{3}{2}\) k_BT
where k_B is the Botzmann constant.
∴ E = 1.5(1.38 × 10^-23)(300)
= 6.21 × 10^-21 J = \(\frac{6.21 \times 10^{-21}}{1.6 \times 10^{-19}}\)
= 3.9 × 10^-2 eV
Since the thermal energy of randomization is about two orders of magnitude greater than the magnetic potential energy of orientation, saturation magnetization will not be achieved at 300 K.

Question 8.
A magnetic needle placed in uniform magnetic field has magnetic moment of 2 × 10^-2 A m², and moment of inertia of 7.2 × 10^-7 kg m². It performs 10 complete oscillations in 6 s. What is the magnitude of the magnetic field ?
Answer:
Data: M = 2 × 10^-2 A∙m², I = 7.2 × 10^-7 kg∙m²,
T = \(\frac{6}{10}\) = 0.6 S
T = 2π\(\sqrt{\frac{I}{M B}}\)
The magnitude of the magnetic field is
B = \(\frac{4 \pi^{2} I}{M T^{2}}\)
= \(\frac{(4)(3.14)^{2}\left(7.2 \times 10^{-7}\right)}{\left(2 \times 10^{-2}\right)(0.6)^{2}}\)
= 3.943 × 10^-3 T = 3.943 mT

Question 9.
A short bar magnet is placed in an external magnetic field of 700 guass. When its axis makes an angle of 30° with the external magnetic field, it experiences a torque of 0.014 Nm. Find the magnetic moment of the magnet, and the work done in moving it from its most stable to most unstable position.
Answer:
Data : B = 700 gauss = 0.07 tesla, θ = 30°,
τ = 0.014 N∙m
τ = MB sin θ
The magnetic moment of the magnet is
M = \(\frac{\tau}{B \sin \theta}=\frac{(0.014)}{(0.07)\left(\sin 30^{\circ}\right)}\) = 0.4 A∙m²
The most stable state of the bar magnet is for θ = 0°.
It is in the most unstable state when θ = 180°. Thus, the work done in moving the bar magnet from 0° to 180° is
W = MB(cos θ₀ – cos θ)
= MB (cos 0° – cos 180°)
= MB [1 – (-1)]
= 2 MB = (2) (0.4) (0.07)
= 0.056 J
This the required work done.

Question 10.
A magnetic needle is suspended freely so that it can rotate freely in the magnetic meridian. In order to keep it in the horizontal position, a weight of 0.1 g is kept on one end of the needle. If the pole strength of this needle is 20 Am , find the value of the vertical component of the earth’s magnetic field. (g = 9.8 m s^-2)
Answer:
Data: M = 0.2g = 2 × 10^-4kg, q_m = 20 A∙m, g =9.8 m/s²

Without the added weight at one end, the needle will dip in the direction of the resultant magnetic field inclined with the horizontal. The torque due to the added weight about the vertical axis through the centre balances the torque of the couple due to the vertical component of the Earth’s magnetic field.
∴ (Mg)\(\left(\frac{L}{2}\right)\) = (q_m B_v) L
The vertical component of the Earth’s magnetic field,
B_v = \(\frac{M g}{2 q_{\mathrm{m}}}=\frac{\left(2 \times 10^{-4}\right)(9.8)}{2(20)}\) = 4.9 × 10^-5 T

Question 11.
The susceptibility of a paramagnetic material is χ at 27° C. At what temperature its susceptibility be \(\frac{\chi}{3}\) ?
Answer:
Data: χ_m1 = χ, T₁ = 27°C = 300 K, χ_m2 = \(\frac{\chi}{3}\)
By Curie’s law,
M_z = C\(\frac{B_{0}}{T}\)
Since M_z = χ_mH = B₀ = μ₀H

This gives the required temperature.

12th Physics Digest Chapter 11 Magnetic Materials Intext Questions and Answers

Activity (Textbook Page No. 251)

Question 1.
You have already studied in earlier classes that a short bar magnet suspended freely always aligns in North South direction. Now if you try to forcefully move and bring it in the direction along East West and leave it free, you will observe that the magnet starts turning about the axis of suspension. Do you know from where does the torque which is necessary for rotational motion come from? (as studied in rotational dynamics a torque is necessary for rotational motion).
Answer:
Suspend a short bar magnet such that it can rotate freely in a horizontal plane. Let it come to rest along the magnetic meridian. Rotate the magnet through some angle and release it. You will see that the magnet turns about the vertical axis in trying to return back to its equilibrium position along the magnetic meridian. Where does the torque for the rotational motion come from?

Take another bar magnet and bring it near the suspended magnet resting in the magnetic meridian. Observe the interaction between the like and unlike poles of the two magnets facing each other. Does the suspended magnet rotate continuously or rotate through certain angle and remain stable? Note down your observations and conclusions.

Do you know (Textbook Page No. 255)

Effective magneton numbers for iron group ions (No. of Bohr magnetons)

Ion	Electron configuration	Magnetic moment (in terms of /iB)
Fe3 +	[Ar] 3s23p63d5	5.9
Fe2 +	[Ar] 3s23p63d6	5.4
Co2 +	[Ar] 3s23p63d7	4.8
n2+	[Ar] 3s23p63d8	3.2

(Courtsey: Introduction to solid state physics by Charles Kittel, pg. 306 )
These magnetic moments are calculated from the experimental value of magnetic susceptibility. In several ions the magnetic moment is due to both orbital and spin angular momenta.
Answer:
In terms of Bohr magneton (µ_B), the effective magnetic moments of some iron group ions are as follows. In several cases, the magnetic moment is due to both orbital and spin angular momenta.

Ion	Configuration	Effective magnetic moment in terms of Bohr magneton (B.M) (Expreimental values)
Fe3 +	3d5	5.9
Fe2 +	3d6	5.4
Co2 +	3d7	4.8
n2+	3d8	3.2

Remember this (Textbook Page No. 256)

Question 1.
Permeability and Permittivity:
Magnetic Permeability is a term analogous to permittivity in electrostatics. It basically tells us about the number of magnetic lines of force that are passing through a given substance when it is kept in an external magnetic field. The number is the indicator of the behaviour of the material in magnetic field. For superconductors χ = – 1. If you substitute in the Eq. (11.18), it is observed that permeability of material µ = 0. This means no magnetic lines will pass through the superconductor.

Magnetic Susceptibility (χ) is the indicator of measure of the response of a given material to the external applied magnetic field. In other words it indicates as to how much magnetization will be produced in a given substance when kept in an external magnetic field. Again it is analogous to electrical susceptibility. This means when the substance is kept in a magnetic field, the atomic dipole moments either align or oppose the external magnetic field. If the atomic dipole moments of the substance are opposing the field, χ is observed to be negative, and if the atomic dipole moments align themselves in the direction of field, χ is observed to be positive. The number of atomic dipole moments of getting aligned in the direction of the applied magnetic field is proportional to χ. It is large for soft iron (χ >1000).
Answer:
Magnetic permeability is analogous to electric permittivity, both indicating the extent to which a material permits a field to pass through or permeate into the material. For a superconductor, χ = -1 which makes µ = 0, so that a superconductor does not allow magnetic field lines to pass through it.

Magnetic susceptibility (χ). analogous to electrical susceptibility, is a measure of the response of a given material to an applied magnetic field. That is, it indicates the extent of the magnetization produced in the material when it is placed in an external magnetic field. χ is positive when the atomic dipole moments align themselves in the direction of the applied field; χ is negative when the atomic dipole moments align antiparallel to the field. χ is large for soft iron (χ > 1000).

Use your brain power (Textbook Page No. 259)

Question 1.
Classify the following atoms as diamagnetic or paramagnetic.
H, O, Zn, Fe, F, Ar, He
(Hint : Write down their electronic configurations)
Is it true that all substances with even number of electrons are diamagnetic?
Answer:

Atoms	Electronic configuration	No. of Electrons	Diamagnetic/Paramagnetic
H	1s1„	1	Diamagnetic
0	1s22s22p4	8	Paramagnetic
Zn	1s22s22p63s23p63d104s2	30	Diamagnetic
Fe	1s22s22p63s23p64s23d6	26	Neither diamagnetic nor paramagnetic (ferromagnetic)
F	1s22s22p5	9	Paramagnetic
Ar	1s22s22p63s23p6	18	Diamagnetic
He	Is2	2	Diamagnetic

It can be seen that all substances with an even number of electrons are not necessarily diamagnetic.

Do you know (Textbook Page No. 260)

Question 1.
Exchange Interaction: This exchange interaction in stronger than usual dipole-dipole interaction by an order of magnitude. Due to this exchange interaction, all the atomic dipole moments in a domain get aligned with each other. Find out more about the origin of exchange interaction.
Answer:
Exchange Interaction :
Quantum mechanical exchange interaction be-tween two neighbouring spin magnetic moments in a ferromagnetic material arises as a consequence of the overlap between the magnetic orbitals of two adjacent atoms. The exchange interaction in particular for 3d metals is stronger than the dipole-dipole interaction by an order of magnitude. Due to this, all the atomic dipole moments in a domain get aligned with each other and each domain is spontaneously magnetized to saturation. (Quantum mechanics and exchange interaction are beyond the scope of the syllabus.)

Use your brain power (Textbook Page No. 262)

Question 1.
What does the area inside the curve B – H (hysteresis curve) indicate?
Answer:
A magnetic hysteresis loop is a closed curve obtained by plotting the magnetic flux density B of a ferromagnetic material against the corresponding magnetizing field H when the material is taken through a complete magnetizing cycle. The area enclosed by the loop represents the hysteresis loss per unit volume in taking the material through the magnetizing cycle.

Do you know (Textbook Page No. 262)

Question 1.
What is soft magnetic material?
Soft ferromagntic materials can be easily magnetized and demagnetized.

Hysteresis loop for hard and soft ferramagnetic materials.
Answer:
A soft magnetic material, usually iron-based, has high permeability, low retentivity and low coercivity. In other words, it does not have appreciable hysteresis, i.e., its hysteresis loop is very narrow. Such a material magnetizes and demagnetizes more easily, by small external fields.

Do you know (Textbook Page No. 263)

Question 1.
There are different types of shielding available like electrical and acoustic shielding apart from magnetic shielding discussed above. An electrical insulator functions as an electrical barrier or shield and comes in a wide array of materials. Normally the electrical wires used in our households are also shielded. In the case of audio recording, it is necessary to reduce other stray sounds which may interfere with the sound to be recorded. So the recording studios are sound insulated using acoustic material.
Answer:
There are different types of shielding, such as electrical, electromagnetic. magnetic, RF (radio frequency), and acoustic, to shield a given space or sensitive instrument from unwanted fields of each type.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 10 Magnetic Fields due to Electric Current Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 10 Magnetic Fields due to Electric Current

1. Choose the correct option.

i) A conductor has 3 segments; two straight and of length L each and a semicircular with radius R. It carries a current I. What is the magnetic field B at point P?

Answer:
(C) \(\frac{\mu_{0}}{4} \frac{I}{R}\)

ii) Figure a, b show two Amperian loops associated with the conductors carrying current I in the sense shown. The \(\oint \vec{B} \cdot d \vec{l}\) in the cases a and b will be, respectively,

Answer:
(A) -μ₀ I, 0

iii) A proton enters a perpendicular uniform magnetic field B at origin along the positive x axis with a velocity v as shown in the figure. Then it will follow the following path. [The magnetic field is directed into the paper].

(A) It will continue to move along positive x axis.
(B) It will move along a curved path, bending towards positive x axis.
(C) It will move along a curved path, bending towards negative y axis.
(D) It will move along a sinusoidal path along the positive x axis.
Answer:
(C) It will move along a curved path, bending towards negative y axis.

(iv) A conducting thick copper rod of length 1 m carries a current of 15 A and is located on the Earth’s equator. There the magnetic flux lines of the Earth’s magnetic field are horizontal, with the field of 1.3 × 10^-4 T, south to north. The magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east, are
(A) 14 × 10^-4 N, downward.
(B) 20 × 10^-4 N, downward.
(C) 14 × 10^-4 N, upward.
(D) 20 × 10^-4 N, upward.
Answer:
(D) 20 × 10^-4 N, upward.

v) A charged particle is in motion having initial velocity \(\overrightarrow{\mathrm{V}}\) when it enter into a region of uniform magnetic field perpendicular to \(\overrightarrow{\mathrm{V}}\) . Because of the magnetic force the kinetic energy of the particle will
(A) remain uncharged.
(B) get reduced.
(C) increase.
(D) be reduced to zero.
Answer:
(A) remain uncharged.

Question 2.
A piece of straight wire has mass 20 g and length 1m. It is to be levitated using a current of 1 A flowing through it and a perpendicular magnetic field B in a horizontal direction. What must be the magnetic of B?

Answer:
Data: m = 20 g = 2 × 10^-2 kg, l = 1 m, I = 1 A,
g = 9.8 m/s²
To balance the wire, the upward magnetic force must be equal in magnitude to the downward force due to gravity.
∴ F_m = IlB = mg
Therefore, the magnitude of the magnetic field,
B = \(\frac{m g}{I l}=\frac{\left(2 \times 10^{-2}\right)(9.8)}{(1)(1)}\) = 0.196 T

Question 3.
Calculate the value of magnetic field at a distance of 2 cm from a very long straight wire carrying a current of 5 A (Given: µ₀ = 4π × 10^-7 Wb/Am).
Answer:
Data : I = 5A, a = 0.02 m, \(\frac{\mu_{0}}{4 \pi}\) = 10^-7 T∙m/A
The magnetic induction,
B = \(\frac{\mu_{0} I}{2 \pi a}=\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}\) = 10^-7 × \(\frac{2(5)}{2 \times 10^{-2}}\) = 5 × 10^-5 T

Question 4.
An electron is moving with a speed of 3.2 × 10⁶ m/s in a magnetic field of 6.00 × 10^-4 T perpendicular to its path. What will be the radius of the path? What will be frequency and the kinetic energy in keV ? [Given: mass of electron = 9.1 × 10^-31 kg, charge e = 1.6 × 10^-19 C, 1 eV = 1.6 × 10^-19 J]
Answer:

Question 5.
An alpha particle (the nucleus of helium atom) (with charge +2e) is accelerated and moves in a vacuum tube with kinetic energy = 10.00 MeV.On applying a transverse a uniform magnetic field of 1.851 T, it follows a circular trajectory of radius 24.60 cm. Obtain the mass of the alpha particle. [charge of electron = 1.62 × 10^-19 C]
Answer:
Data: 1 eV = 1.6 × 10^-19 J,
E = 10MeV = 10⁷ × 1.6 × 10^-19 J = 1.6 × 10^-12 J
B = 1.88 T, r = 0.242 m, e = 1.6 × 10^-19 C
Charge of an -partic1e,
q = 2e = 2(1.6 × 10^-19)=3.2 × 10^-19 C

This gives the mass of the α-particle.
[Note : The value of r has been adjusted to match with the answer. The CODATA (Committee on Data for Science and Technology) accepted value of m_x is approximately 6.6446 × 10^-27 kg.]

Question 6.
Two wires shown in the figure are connected in a series circuit and the same amount of current of 10 A passes through both, but in apposite directions. Separation between the two wires is 8 mm. The length AB is S = 22 cm. Obtain the direction and magnitude of the magnetic field due to current in wire 2 on the section AB of wire 1. Also obtain the magnitude and direction of the force on wire 1. [µ₀ = 4π × 10^-7 T.m/A]

Answer:
Data: I₁ = I₂ = 10 A, s = 8 mm = 8 × 10^-3 m, l = 0.22 m
By right hand grip rule, the direction of the magnetic field \(\overrightarrow{B_{2}}\) due to the current in wire 2 at AB is into the page and its magnitude is
B₂ = \(\frac{\mu_{0}}{4 \pi} \frac{2 I}{s}=10^{-7} \times \frac{2(10)}{8 \times 10^{-3}}=\frac{1}{4} \times \mathbf{1 0}^{-3}\) T
The current in segment AB is upwards. Then, by Fleming’s left hand rule, the force on it due to \(\overrightarrow{B_{2}}\) is to the left of the diagram, i.e., away from wire 1, or repulsive. The magnitude of the force is
F_{on 1 by 2} = I₁lB₂ = (10)(0.22) × \(\frac{1}{4}\) × 10^-3
= 5.5 × 10^-4 N

Question 7.
A very long straight wire carries a current 5.2 A. What is the magnitude of the magnetic field at a distance 3.1 cm from the wire? [ µ₀ = 4π × 10^-7 T∙m/A]
Answer:
Data : I = 5.2 A, a = 0.031 m, \(\frac{\mu_{0}}{4 \pi}\) = 10^-7 T∙m/A
The magnetic induction,
B = \(\frac{\mu_{0} I}{2 \pi a}=\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}\) = 10^-7 × \(\frac{2(5.2)}{3.1 \times 10^{-2}}\)
= 3.35 × 10^-5 T

Question 8.
Current of equal magnitude flows through two long parallel wires having separation of 1.35 cm. If the force per unit length on each of the wires in 4.76 × 10^-2 N, what must be I ?
Answer:

Question 9.
Magnetic field at a distance 2.4 cm from a long straight wire is 16 µT. What must be current through the wire?
Answer:
Data: a = 2.4 × 10^-2 m, B = 1.6 × 10^-5 T,
\(\frac{\mu_{0}}{4 \pi}\) = 10^-7 T∙m/A
B = \(\frac{\mu_{0} I}{2 \pi a}=\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}\)
The current through the wire,
I = \(\frac{1}{\mu_{0} / 4 \pi} \frac{a B}{2}=\frac{1}{10^{-7}} \frac{\left(2.4 \times 10^{-2}\right)\left(1.6 \times 10^{-5}\right)}{2}\) = 1.92 A

Question 10.
The magnetic field at the centre of a circular current carrying loop of radius 12.3 cm is 6.4 × 10^-6T. What will be the magnetic moment of the loop?
Answer:
Data: R = 12.3cm = 12.3 × 10^-2 m,
B = 6.4 × 10^-6 T, µ₀ = 4π × 10^-7 T∙m/A

= 5.955 × 10^-2 J/T (or A∙m²)

Question 11.
A circular loop of radius 9.7 cm carries a current 2.3 A. Obtain the magnitude of the magnetic field (a) at the centre of the loop and (b) at a distance of 9.7 cm from the centre of the loop but on the axis.
Answer:
Data: R = z = 9.7 cm = 9.7 × 10^-2 m, I = 2.3A, N = 1
(a) At the centre of the coil :
The magnitude of the magnetic induction,
B = \(\frac{\mu_{0} N I}{2 R}\)

Question 12.
A circular coil of wire is made up of 100 turns, each of radius 8.0 cm. If a current of 0.40 A passes through it, what be the magnetic field at the centre of the coil?
Answer:
Data: N = 100,R = 8 × 10^-2 m, I = 0.4A,
µ₀ = 4π × ^-7 T∙m/A
B = \(\frac{\mu_{0} N I}{2 R}\)
= \(\frac{\left(4 \pi \times 10^{-7}\right)(100)(0.4)}{2\left(8 \times 10^{-2}\right)}\) = 3.142 × 10^-4 T

Question 13.
For proton acceleration, a cyclotron is used in which a magnetic field of 1.4 Wb/m² is applied. Find the time period for reversing the electric field between the two Ds.
Answer:
Data: B = 1.4 Wb/m², m = 1.67 × 10^-27 kg,
q = 1.6 × 10^-19 C
T = \(\frac{2 \pi m}{q B}\)
t = \(\frac{T}{2}=\frac{\pi m}{q B}=\frac{(3.142)\left(1.67 \times 10^{-27}\right)}{\left(1.6 \times 10^{-19}\right)(1.4)}\)
= 2.342 × 10^-8 s
This is the required time interval.

Question 14.
A moving coil galvanometer has been fitted with a rectangular coil having 50 turns and dimensions 5 cm × 3 cm. The radial magnetic field in which the coil is suspended is of 0.05 Wb/m². The torsional constant of the spring is 1.5 × 10^-9 Nm/ degree. Obtain the current required to be passed through the galvanometer so as to produce a deflection of 30°.
Answer:
Data : N = 50, C = 1.5 × 10^-9 Nm/degree,
A = lb = 5 cm × 3 cm = 15 cm² = 15 × 10^-4 m²,
B = 0.05Wb/m², θ = 30°
NIAB = Cθ
∴ The current through the coil, I = \(\frac{C \theta}{N A B}\)
= \(\frac{1.5 \times 10^{-9} \times 30}{50 \times 15 \times 10^{-4} \times 0.05}=\frac{3 \times 10^{-5}}{5 \times 0 \cdot 5}\)
= 1.2 × 10^-5 A

Question 15.
A solenoid of length π m and 5 cm in diameter has winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its centre along the axis.
Answer:
Data: L = 3.142 m, N = 1000, I = 5A,
μ₀ = 4π × 10^-7 T∙m/A
The magnetic induction,
B = μ₀ nI = μ₀(\(\frac{N}{L}\))I
= (4π × 10^-7)(\(\frac{1000}{3.142}\))(5) = \(\frac{20 \times 3.142 \times 10^{-4}}{3.142}\)
= 2 × 10^-3 T

Question 16.
A toroid of narrow radius of 10 cm has 1000 turns of wire. For a magnetic field of 5 × 10^-2 T along its axis, how much current is required to be passed through the wire?
Answer:
Data : CentraI radius, r = 10 cm = 0.1 m, N = 1000,
B = 5 × 10^-2 T, = 4π × 10^-7 T∙m/A
The magnetic induction,
B = \(\frac{\mu_{0} N I}{2 \pi r}=\frac{\mu_{0}}{4 \pi} \frac{2 N I}{r}\)
2irr 4ir r
∴ 5 × 10^-2 = 10^-7 × \(\frac{2(1000) I}{0.1}\)
∴ I = \(\frac{50}{2}\) = 25A
This is the required current.

Question 17.
In a cyclotron protons are to be accelerated. Radius of its D is 60 cm. and its oscillator frequency is 10 MHz. What will be the kinetic energy of the proton thus accelerated?
(Proton mass = 1.67 × 10^-27 kg, e = 1.60 × 10^-19 C, 1eV = 1.6 × 10^-19 J)
Answer:
Data : R = 0.6 m, f = 10⁷ Hz, m_p = 1.67 × 10^-27 kg,
e = 1.6 × 10^-19C, 1 eV = 1.6 × 10^-19 J

Question 18.
A wire loop of the form shown in the figure carries a current I. Obtain the magnitude and direction of the magnetic field at P. Given : \(\frac{\mu_{0} I}{4 \pi R} \sqrt{2}\)

Answer:
The wire loop is in the form of a circular arc AB of radius R and a straight conductor BCA. The arc AB subtends an angle of Φ = 270° = \(\frac{3 \pi}{2}\) rad at the centre of the loop P. Since PA = PB = R and C is the midpoint of AB, AB = \(\sqrt{2} R\) and AC = CB = \(\frac{\sqrt{2} R}{2}\) = \(\frac{R}{\sqrt{2}}\). Therefore, a = PC = \(\frac{R}{\sqrt{2}}\).

The magnetic inductions at P due to the arc AB and the straight conductor BCA are respectively,
B₁ = \(\frac{\mu_{0}}{4 \pi} \frac{I \phi}{R}\) and B₂ = \(\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}\)
Therefore, the net magnetic induction at P is

This is the required expression.

Question 19.
Two long parallel wires going into the plane of the paper are separated by a distance R, and carry a current I each in the same direction. Show that the magnitude of the magnetic field at a point P equidistant from the wires and subtending angle θ from the plane containing the wires, is B = \(\frac{\mu_{0}}{\pi} \frac{I}{R}\) sin 2θ What is the direction of the magnetic field?
Answer:

In above figure, \(\vec{B}_{1}\) and \(\vec{B}_{2}\) are the magnetic fields in the plane of the page due to the currents in wires 1 and 2, respectively. Their directions are given by the right hand grip rule: \(\vec{B}_{1}\) is perpendicular to AP and makes an angle Φ with the horizontal. \(\vec{B}_{2}\) is perpendicular to BP and also makes an angle Φ with the horizontal.
AP = BP = a = \(\frac{R / 2}{\cos \theta}\)
and B₁ = B₂ = \(\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}=\frac{\mu_{0}}{4 \pi} \frac{2 I(2 \cos \theta)}{R}\)
= \(\frac{\mu_{0}}{\pi} \frac{I}{R}\) cos θ
Since the vertical components cancel out, the magnitude of the net magnetic induction at P is
B_net = 2B₁ cos Φ = 2B₁ cos(90° – θ) = 2B₁ sinθ
= 2(\(\frac{\mu_{0}}{\pi} \frac{I}{R}\) cos θ) sin θ = \(\frac{\mu_{0}}{\pi} \frac{I}{R}\) sin 2θ
as required. is in the plane parallel to that of the wires and to the right as shown in the figure.

Question 20.
Figure shows a section of a very long cylindrical wire of diameter a, carrying a current I. The current density which is in the direction of the central axis of the wire varies linearly with radial distance r from the axis according to the relation J = J_or/a. Obtain the magnetic field B inside the wire at a distance r from its centre.
[Answer: B J r

Answer:
Consider an annular differential element of radius r and width dr. The current through the area dA of this element is
dI = JdA = (J_o \(\frac{r}{a}\))2πrdr = \(\frac{2 \pi J_{0} r^{2} d r}{a}\) …………….. (1)

To apply the Ampere’s circuital law to the circular path of integration, we note that the wire has perfect cylindrical symmetry with all the charges moving parallel to the wire. So, the magnetic field must be tangent to circles that are concentric with the wire. The enclosed current is the current within radius r. Thus,

Question 21.
In the above problem, what will be the magnetic field B inside the wire at a distance r from its axis, if the current density J is uniform across the cross section of the wire?
Answer:
Figure shows the cross section of a long straight wire of radius a that carries a current I out of the page. Because the current is uniformly distributed over the cross section of the wire, the magnet ic field \(\vec{B}\) due to the current must be cylindrically symmetrical. Thus, along the Amperian loop of radius r(r < a), symmetry suggests that \(\vec{B}\) is tangent to the loop, as shown in the figure.
\(\oint \vec{B} \cdot \overrightarrow{d l}=B \oint d l\) = B(2πr) ……….. (1)
Because the current is uniformly distributed, the current I_encl enclosed by the loop is proportional to the area encircled by the loop; that is,
I_encl = Jπr²
By right-hand rule, the sign of ‘d is positive. Then, by Ampere’s law,
B (2πr) = µ₀ I_encl = µ₀ Jπr² ……………….. (2)
∴ B = \(\frac{\mu_{0} J}{2} r\) ……………… (3)
OR
I_encl = I\(\frac{\pi r^{2}}{\pi a^{2}}\)
By right-hand rule, the sign of I\frac{\pi r^{2}}{\pi a^{2}} is positive. Then, by Ampere’s law,
\(\oint B d l\) = µ₀ I_encl
∴ B(2πr) = µ₀I \(\frac{r^{2}}{a^{2}}\) …………. (4)
∴ B = (\(\frac{\mu_{0} I}{2 \pi a^{2}}\) )r ………… (5)

[Note: Thus, inside the wire, the magnitude B of the magnetic field is proportional to distance r from the centre. At a distance r outside a straight wire, B = (\(\frac{\mu_{0} I}{2 \pi r}\) i.e., B ∝ \(\frac{1}{r}\).]

Theory Exercise

Question 1.
Distinguish between the forces experienced by a moving charge in a uniform electric field and in a uniform magnetic field.
Answer:
A charge q moving with a velocity \(\vec{v}\) through a magnetic field of induction \(\vec{B}\) experiences a magnetic force perpendicular both to \(\vec{B}\) and \(\vec{v}\) . Experimental observations show that the magnitude of the force is proportional to the magnitude of \(\vec{B}\), the speed of the particle, the charge q and the sine of the angle θ between \(\vec{v}\) and \(\vec{B}\). That is, the magnetic force, F_m = qv B sin θ
∴ \(\vec{F}_{\mathrm{m}}=q(\vec{v} \times \vec{B})\)
Therefore, at every instant \(\vec{F}_{\mathrm{m}}\) acts in a direction perpendicular to the plane of \(\vec{v}\) and \(\vec{B}\) .

If the moving charge is negative, the direction of the force \(\vec{F}_{\mathrm{m}}\) acting on it is opposite to that given by the right-handed screw rule for the cross-product \(\vec{v}\) × \(\vec{B}\).

If the charged particle moves through a region of space where both electric and magnetic fields are present, both fields exert forces on the particle.
The force due to the electric field \(\vec{E}\) is \(\vec{F}_{\mathrm{e}}=q \vec{E}\) .
The total force on a moving charge in electric and magnetic fields is called the Lorentz force :
\(\vec{F}=\vec{F}_{\mathrm{e}}+\vec{F}_{\mathrm{m}}=q(\vec{E}+\vec{v} \times \vec{B})\)
Special cases :
(i) \(\vec{v}\) is parallel or antiparallel to \(\vec{B}\): In this case, F_m = qvB sin 0° = 0. That is, the magnetic force on the charge is zero.
(ii) The charge is stationary {v = 0) : In this case, even if q ≠ 0 and B ≠ 0, F_m = q(0)B sin θ = 0. That is, the magnetic force on a stationary charge is zero.

Question 2.
Under what condition a charge undergoes uniform circular motion in a magnetic field? Describe, with a neat diagram, cyclotron as an application of this principle. Obtain an expression for the frequency of revolution in terms of the specific charge and magnetic field.
Answer:
Suppose a particle of mass m and charge q enters a region of uniform magnetic field of induction \(\vec{B}\). In Fig., \(\vec{B}\) points into the page. The magnetic force \(\vec{F}_{\mathrm{m}}\) on the particle is always perpendicular to the velocity of the particle, \(\vec{v}\) . Assuming the charged particle started moving in a plane perpendicular to \(\vec{B}\), its motion in the magnetic field is a uniform circular motion, with the magnetic force providing the centripetal acceleration.

If the charge moves in a circle of radius R,
F_m = |q|vB = \(\frac{m v^{2}}{R}\)
∴ mv = p = |q|BR …………. (1)
where p = mv is the linear momentum of the particle. Equation (1) is known as the cyclotron formula because it describes the motion of a particle in a cyclotron-the first of the modern particle accelerators.

Question 3.
What is special about a radial magnetic field ? Why is it useful in a moving coil galvanometer ?
Answer:

1. Advantage of radial magnetic field in a moving- coil galvanometer:
   • As the coil rotates, its plane is always parallel to the field. That way, the deflecting torque is always a ximum depending only on the current in the coil, but not on the position of the coil.
   • The restoring torque is proportional to the deflection so that a radial field makes the deflection proportional to the current. The instrument then has a linear scale, i.e., the divisions of the scale are evenly spaced. This makes it particularly straight forward to calibrate and to read.
2. Producing radial magnetic field :
   • The pole pieces of the permanent magnet are made cylindrically concave, concentric with the axis of the coil.
   • A soft iron cylinder is centred between the pole pieces so that it forms a narrow cylindrical gap in which the sides of the coil can move. Together, they produce a radial magnetic, field; that is, the magnetic lines of force in the gap are along radii to the central axis.

Question 4.
State Biot-Savert law. Apply it to
(i) infinitely long current carrying conductor and (ii) a point on the axis of a current carrying circular loop.
Answer:
Consider a very short segment of length dl of a wire carrying a current I. The product I \(\overrightarrow{d l}\) is called a current element; the direction of the vector \(\overrightarrow{d l}\) is along the wire in the direction of the current.

Biot-Savart law (Laplace law) : The magnitude of the incremental magnetic induction \(\overrightarrow{d B}\) produced by a current element I \(\overrightarrow{d l}\) at a distance r from it is directly proportional to the magnitude I \(\overrightarrow{d l}\) of the current element, the sine of the angle between the current element Idl and the unit vector r directed from the current element toward the point in question, and inversely proportional to the square of the distance of the point from the current element; the magnetic induction is directed perpendicular to both I \(\overrightarrow{d l}\) and \(\hat{r}\) as per the cross product rule.

where \(\hat{\mathrm{r}}=\frac{\vec{r}}{r}\) and the constant µ₀ is the permeability of free space. Equations (1) and (2) are called Biot-Savart law.

The incremental magnetic induction \(\overrightarrow{d B}\) is given by the right-handed screw rule of vector crossproduct \(I \overrightarrow{d l} \times \hat{\mathrm{r}}\). In Fig, the current element I\(\overrightarrow{d l}\) and \(\hat{\mathbf{r}}\) are in the plane of the page, so that \(\overrightarrow{d B}\) points out of the page at point P shown by ⊙; at the point Q, \(\overrightarrow{d B}\) points into the page shown by ⊗.

The magnetic induction \(\vec{B}\) at the point due to the entire wire is, by the principle of superposition, the vector sum of the contributions \(\overrightarrow{d B}\) from all the current elements making up the wire.
From Eq. (2),
\(\vec{B}=\int \overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} \int \frac{I \overrightarrow{d l} \times \hat{\mathrm{r}}}{r^{2}}\)
[Notes : (1) The above law is based on experiments by Jean Baptiste Biot (1774-1862) and Felix Savart (1791-1841), French physicists. From their observations Laplace deduced the law mathematically. (2) The Biot- Savart law plays a similar role in magnetostatics as Coulomb’s law does in electrostatics.]

Question 5.
State Ampere’s law. Explain how is it useful in different situations.
Answer:
Ampere’s circuital law : In free space, the line integral of magnetic induction around a closed path in a magnetic field is equal to p0 times the net steady current enclosed by the path.
In mathematical form,
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀ I …………… (1)
where \(\vec{B}\) is the magnetic induction at any point on the path in vacuum, \(\overrightarrow{d l}\) is the length element of the path, I is the net steady current enclosed and μ₀ is the permeability of free space.

Explanation : Figure shows two wires carrying currents I₁ and I₂ in vacuum. The magnetic induction \(\vec{B}\) at any point is the net effect of these currents.

To find the magnitude B of the magnetic induction :
We construct an imaginary closed curve around the conductors, called an Amperian loop, and imagine it divided into small elements of length \(\overrightarrow{d l}\). The direction of dl is the direction along which the loop is traced.

(ii) We assign signs to the currents using the right hand rule : If the fingers of the right hand are curled in the direction in which the loop is traced, then a current in the direction of the outstretched thumb is taken to be positive while a current in the opposite direction is taken to be negative.

For each length element of the Amperian loop, \(\vec{B} \cdot \overrightarrow{d l}\) gives the product of the length dl of the element and the component of \(\vec{B}\) parallel to \(\overrightarrow{d l}\) . If θ r is the angle between \(\overrightarrow{d l}\) and \(\vec{B}\) ,
\(\vec{B} \cdot \overrightarrow{d l}\) = (B cos θ) dl
Then, the line integral,
\(\oint \vec{B} \cdot \overrightarrow{d l}=\oint B \cos \theta d l\) …………. (2)
For the case shown in Fig., the net current I through the surface bounded by the loop is
I = I₂ – I₁
∴ \(\oint B \cos \theta d l\) = μ₀ I
= μ₀(I₂ – I₁) …………… (3)
Equation (3) can be solved only when B is uniform and hence can be taken out of the integral.

[Note : Ampere’s law in magnetostatics plays the part of Gauss’s law of electrostatics. In particular, for currents with appropriate symmetry, Ampere’s law in integral form offers an efficient way of calculating the magnetic field. Like Gauss’s law, Ampere’s law is always true (for steady currents), but it is useful only when the symmetry of the problem enables B to be taken out of the integral \(\oint \vec{B} \cdot \overrightarrow{d l}\). The current configurations that can be handled by Ampere’s law are infinite straight conductor, infinite plane, infinite solenoid and toroid.]

12th Physics Digest Chapter 10 Magnetic Fields due to Electric Current Intext Questions and Answers

Do you know (Textbook Page No. 230)

Question 1.
You must have noticed high tension power transmission lines, the power lines on the big tall steel towers. Strong magnetic fields are created by these lines. Care has to be taken to reduce the exposure levels to less than 0.5 milligauss (mG).
Answer:
With increasing population, many houses are constructed near high voltage overhead power transmission lines, if not right below them. Large transmission lines configurations with high voltage and current levels generate electric and magnetic fields and raises concerns about their effects on humans located at ground surfaces. With conductors typically 20 m above the ground, the electric field 2 m above the ground is about 0.2 kV / m to I kV / m. In comparison, that due to thunderstorms can reach 20 kV/m. For the same conductors, magnetic field 2 m above the ground is less than 6 μT. In comparison, that due to the Earth is about 40 μT.

Do you know (Textbook Page No. 232)

Question 1.
Magnetic Resonance Imaging (MRI) technique used for medical imaging requires a magnetic field with a strength of 1.5 T and even upto 7 T. Nuclear Magnetic Resonance experiments require a magnetic field upto 14 T. Such high magnetic fields can be produced using superconducting coil electromagnet. On the other hand, Earth’s magnetic field on the surface of the Earth is about 3.6 × 10^-5 T = 0.36 gauss.
Answer:
Magnetic Resonance Imaging (MRI) is a non-invasive imaging technology that produces three dimensional detailed anatomical images. Although MRI does not emit the ionizing radiation that is found in X-ray imaging, it does employ a strong magnetic field, e.g., medical MRIs usually have strengths between 1.5 T and 3 T.

The 21.1 T superconducting magnet at Maglab (Florida, US) is the world’s strongest MRI scanner used for Nuclear Magnetic Resonance (NMR) research. Since its inception in 2004, it has been continually conducting electric current of 284 A by itself. Because it is superconducting, the current runs through some 152 km of wire without resistance, so no outside energy source is needed. However, 2400 litres of liquid helium is cycled to keep the magnet at a superconducting temperature of 1.7 K. Even when not in use this magnet is kept cold; if it warms up to room temperature, it takes at least six weeks to cool it back down to operating temperature. The 45 T Hybrid Magnet of the Lab (which combines a superconducting magnet of 11.5 T with a resistive magnet of 33.5 T) is kept at 1.8 K using 2800 L of liquid helium and 15142 L of cold water.

Question 2.
Let us look at a charged particle which is moving in a circle with a constant speed. This is uniform circular motion that you have studied earlier. Thus, there must be a net force acting on the particle, directed towards the centre of the circle. As the speed is constant, the force also must be constant, always perpendicular to the velocity of the particle at any given instant of time. Such a force is provided by the uniform magnetic field \(\vec{B}\) perpendicular to the plane of the circle along which the charged particle moves.
Answer:
When a charged particle moves in uniform circular motion inside a uniform magnetic field \(\vec{B}\) in a plane perpendicular to \(\vec{B}\), the centripetal force is the magnetic force on the particle. As in any UCM, this magnetic force is constant in magnitude and perpendicular to the velocity of the particle.

Remember this (Textbook Page No. 233)

Question 1.
Field penetrating into the paper is represented as ⊗, while that coming out of the paper is shown by ⊙.
Answer:
In a two-dimensional diagram, a vector pointing perpendicularly into the plane of the diagram is shown by a cross ⊗ while that pointing out of the plane is shown by a dot ⊙ .

Do you know (Textbook Page No. 234)

Question 1.
Particle accelerators are important for a variety of research purposes. Large accelerators are used in particle research. There have been several accelerators in India since 1953. The Department of Atomic Energy (DAE), Govt. of India, had taken initiative in setting up accelerators for research. Apart from ion accelerators, the DAE has developed and commissioned a 2 GeV electron accelerator which is a radiation source for research in science. This accelerator, ‘Synchrotron’, is fully functional at Raja Ramanna Centre for Advanced Technology, Indore. An electron accelerator, Microtron with electron energy 8-10 MeV is functioning at Physics Department, Savitribai Phule Pune University, Pune.
Answer:
Particle accelerators are machines that accelerate charged subatomic particles to high energy for research and applications. They play a major role in the field of basic and applied sciences, in our understanding of nature and the universe. The size and cost of particle accelerators increase with the energy of the particles they produce. Medical Cyclotrons across the country are dedicated for medical isotope productions and for medical sciences. There are many existing and upcoming particle accelerators in India in different parts of country.

(https: / / www.researchgate.net/ publication/ 3209480 83_Existing and upcoming_particle_accelerators_in. India). For cutting-edge high energy particle physics, Indian particle physicists collaborate with those at Large Hadron Collider CERN, Geneva.

Can you recall (Textbook Page No. 238)

Question 1.
How does the coil in a motor rotate by a full rotation? In a motor, we require continuous rotation of the current carrying coil. As the plane of the coil tends to become parallel to the magnetic field \(\vec{B}\), the current in the coil is reversed externally. Referring to Fig. the segment ab occupies the position cd. At this position of rotation, the current is reversed. Instead of from b to a, it flows from a to b, force \(\vec{F}_{\mathrm{m}}\) continues to act in the same direction so that the torque continues to rotate the coil. The reversal of the current is achieved by using a commutator which connects the wires of the power supply to the coil via carbon brush contacts.
Answer:
Electric Motor
From Fig. we see that the torque on a current loop rotates the loop to smaller values of 9 until the torque becomes zero, when the plane of the loop is perpendicular to the magnetic field and θ = 0. If the current in the loop remains in the same direction when the loop turns past this position, the torque will reverse direction and turn the loop in the opposite direction, i.e., anticlockwise. To provide continuous rotation in the same sense, the current in the loop must periodically reverse direction, as shown in Fig.

In an electric motor, the current reversal is achieved externally by brushes and a split-ring commutator.

Use your brain power (Textbook Page No. 242)

Question 1.
Currents in two infinitely long, parallel wires exert forces on each other. Is this consistent with Newton’s third law?
Answer:
Yes, they are equal in magnitude and opposite in direction and act on the contrary parts : \(\vec{F}\)_{on 2 by 1} = \(\vec{F}\)_{on 1 by 2}. Thus, they form action-reaction pair.

Do you know (Textbook Page No. 244)

Question 1.
So far we have used the constant µ₀ everywhere. This means in each such case, we have carried out the evaluation in free space (vacuum). µ₀ is the permeability of free space.
Answer:
Permeability of free space or vacuum, µ₀ = 4π × 10^-7 H/m.
Earlier SI (2006) had fixed this value of µ₀ as exact but revised SI fixes the value of e, requiring µ₀ (and ε₀) to be determined experimentally.

Use your brain power (Textbook Page No. 244)

Question 1.
Using electrostatic analogue, obtain the magnetic field \(\vec{B}\)_equator at a distance d on the perpendicular bisector of a magnetic dipole of magnetic length 2l and moment \(\vec{M}\). For far field, verify that
\(\vec{B}_{\text {equator }}=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{-\vec{M}}{\left(d^{2}+l^{2}\right)^{3 / 2}}\)
Answer:
The magnitude of the electric intensity at a point at a distance r from an electric charge q in vacuum is given by
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{|q|}{r^{2}}\)
where ε₀ is the permittivity of free space. This intensity is directed away from the charge, if the charge is positive and towards the charge, if the charge is negative.

A magnetic pole is similar to an electric charge. The N-pole is similar to a positive charge and the S-pole is similar to a negative charge. Like an electric charge, a magnetic pole is assumed to produce a magnetic field in the surrounding region. The magnetic field at any point is denoted by a vector quantity called magnetic induction. Thus, by analogy, the magnitude of the magnetic induction at a point at a distance r from a magnetic pole of strength q_m is given by
B = \(\frac{\mu_{0}}{4 \pi} \frac{q_{\mathrm{m}}}{r^{2}}\)
This induction is directed away from the pole if it is an N-pole (strength + q_m) and towards the pole if it is an S-pole (strength -q_m).

Consider a point P on the equator of a magnetic dipole with pole strengths + q_m and – q_m and of magnetic length 21. Let P be at a distance d from the centre of the dipole,

The magnetic induction of a bar magnet at an equatorial point

The magnetic induction at P due to the N-pole is directed along NP (away from the N-pole) while that due to the S-pole is along PS (towards the S-pole), each having a magnitude
B_N = B_S = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{q_{\mathrm{m}}}{\left(d^{2}+l^{2}\right)}\)
(∵ NP = SP = \(\sqrt{d^{2}+l^{2}}\))
The inductions due to the two poles are equal in magnitude so that the two, oppositely directed equatorial components, B_N sin θ and B_S sin θ, cancel each other.

Therefore, the resultant induction is in a direction’ parallel to the axis of the magnetic dipole and has direction opposite to that of the magnetic moment of the magnetic dipole. The component of the induction due to the two poles along the axis is
\(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{q_{\mathrm{m}}}{\left(d^{2}+l^{2}\right)}\) cos θ
where θ is the angle shown in the diagram. From the diagram,

Thus, for a short dipole the induction varies in-versely as the cube of the distance from it.

Question 2.
What is the fundamental difference between an electric dipole and a magnetic dipole?
Answer:
If a magnet is carefully and repeatedly cut, it would expose two new faces with opposite poles such that each piece would still be a magnet. This suggests that magnetic fields are essentially dipolar in character. The most elementary magnetic structure always behaves as a pair of two magnetic poles of opposite types and of equal strengths. Hence, analogous to an electric dipole, we hypothesize that there are positive and negative magnetic charges (or north and south poles) of equal strengths a finite distance apart within a magnet. Also, they are assumed to act as the source of the magnetic field in exactly the same way that electric charges act as the source of electric field. The magnitude of each ‘magnetic charge’ is referred to as its ‘pole strength’ and is equal to q_m = \(\frac{M}{2 l}\), where \(\vec{M}\) is the magnetic dipole moment, pointing from the negative (or south, S) pole to the positive (or north, N) pole.

However, while two types of electric charges exist in nature and have separate existence, isolated magnetic charges, or magnetic monopoles, are not observed. A magnetic pole is not an experimental fact: there are no real poles. To put it in another way, there are no point sources for \(\vec{B}\), as there are for \(\vec{E}\); there exists no magnetic analog to electric charge. Every experimental effort to demonstrate the existence of magnetic charges has failed. Hence, magnetic poles are called fictitious.

The electric field diverges away from a (positive) charge; the magnetic field line curls around a current. Electric field lines originate on positive charges and terminate on negative ones; magnetic field lines do not begin or end anywhere, they typically form closed loops or extend out to infinity.

Do you know (Textbook Page No. 247)

Question 1.
What is an ideal solenoid?
Answer:
A solenoid is a long wire wound in the form of a helix. An ideal solenoid is tightly wound and infinitely long, i.e., its turns are closely spaced and the solenoid is very long compared to its crosssectional radius.

Each turn of a solenoid acts approximately as a circular loop. Suppose the solenoid carries a steady current I. The net magnetic field due to the current in the solenoid is the vector sum of the fields due to the current in all the turns. In the case of a tightly- wound solenoid of finite length, Fig. 10.39, the magnetic field lines are approximately parallel only near the centre of the solenoid, indicating a nearly uniform field there. However, close to the ends, the field lines diverge from one end and converge at the other end. This field distribution is similar to that of a bar magnet. Thus, one end of the solenoid behaves like the north pole of a magnet and the opposite end behaves like the south pole. The field outside is very weak near the midpoint.

For an ideal solenoid, the magnetic field inside is reasonably uniform over the cross section and parallel to the axis throughout the volume enclosed by the solenoid. The field outside is negligible in this case.

Use your brain power (Textbook Page No. 248)

Question 1.
Choosing different Amperean loops, show that out-side an ideal toroid B = 0.
Answer:
From below figure, the inner Amperean loop does not enclose any current while the outer Amperean loop encloses equal number of I_in and I_out. Hence, by Ampere’s law, B = 0 outside an ideal toroid.

Question 2.
What is an ideal toroid?
Answer:
A toroid is a toroidal solenoid. An ideal toroid consists of a long conducting wire wound tightly around a torus, a doughnut-shaped ring, made of a nonconducting material.

In an ideal toroid carrying a steady current, the magnetic field in the interior of the toroid is tangential to any circle concentric with the axis of the toroid and has the same value on this circle (the dashed line in figure). Also, the magnitude of the magnetic induction external to the toroid is negligible.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 9 Current Electricity Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 9 Current Electricity

1. Choose the correct option.

i) Kirchhoff’s first law, i.e., ΣI = 0 at a junction, deals with the conservation of
(A) charge
(B) energy
(C) momentum
(D) mass
Answer:
(A) charge

ii) When the balance point is obtained in the potentiometer, a current is drawn from
(A) both the cells and auxiliary battery
(B) cell only
(C) auxiliary battery only
(D) neither cell nor auxiliary battery
Answer:
(D) neither cell nor auxiliary battery

iii) In the following circuit diagram, an infinite series of resistances is shown. Equivalent resistance between points A and B is

(A) infinite
(B) zero
(C) 2 Ω
(D) 1.5 Ω
Answer:
(C) 2 Ω

iv) Four resistances 10 Ω, 10 Ω, 10 Ω and 15 Ω form a Wheatstone’s network. What shunt is required across 15 Ω resistor to balance the bridge
(A) 10 Ω
(B) 15 Ω
(C) 20 Ω
(D) 30 Ω
Answer:
(D) 30 Ω

v) A circular loop has a resistance of 40 Ω. Two points P and Q of the loop, which are one quarter of the circumference apart are connected to a 24 V battery, having an internal resistance of 0.5 Ω. What is the current flowing through the battery.
(A) 0.5 A
(B) 1A
(C) 2A
(D) 3A
Answer:
(D) 3A

vi) To find the resistance of a gold bangle, two diametrically opposite points of the bangle are connected to the two terminals of the left gap of a metre bridge. A resistance of 4 Ω is introduced in the right gap. What is the resistance of the bangle if the null point is at 20 cm from the left end?
(A) 2 Ω
(B) 4 Ω
(C) 8 Ω
(D) 16 Ω
Answer:
(A) 2 Ω

2. Answer in brief.

i) Define or describe a Potentiometer.
Answer:
The potentiometer is a device used for accurate measurement of potential difference as well as the emf of a cell. It does not draw any current from the circuit at the null point. Thus, it acts as an ideal voltmeter and it can be used to determine the internal resistance of a cell. It consist of a long uniform wire AB of length L, stretched on a wooden board. A cell of stable emf (E), with a plug key K in series, is connected across AB as shown in the following figure.

ii) Define Potential Gradient.
Answer:
Potential gradient is defined as the potential difference (the fall of potential from the high potential end) per unit length of the wire.

iii) Why should not the jockey be slided along the potentiometer wire?
Answer:
Sliding the jockey on the potentiometer wire decreases the cross sectional area of the wire and thereby affects the fall of potential along the wire. This affects the potentiometer readings. Flence, the jockey should not be slided along the potentiometer wire.

iv) Are Kirchhoff’s laws applicable for both AC and DC currents?
Answer:
Kirchhoff’s laws are applicable to both AC and DC ’ circuits (networks). For AC circuits with different loads, (e.g. a combination of a resistor and a capacitor, the instantaneous values for current and voltage are considered for addition.

[Note : Gustav Robert Kirchhoff (1824-87), German physicist, devised laws of electrical network and, with Robert Wilhelm Bunsen, (1811-99), German chemist, did pioneering work in chemical spectroscopy. He also con-tributed to radiation.]

v) In a Wheatstone’s meter-bridge experiment, the null point is obtained in middle one third portion of wire. Why is it recommended?
Answer:

1. The value of unknown resistance X, may not be accurate due to non-uniformity of the bridge wire and development of contact resistance at the ends of the wire.
2. To minimise these errors, the value of R is so adjusted that the null point is obtained in the middle one-third of the wire (between 34 cm and 66 cm) so that the percentage errors in the measurement of I_X and I_R are minimum and nearly the same.

vi) State any two sources of errors in meterbridge experiment. Explain how they can be minimized.
Answer:
The chief sources of error in the metre bridge experiment are as follows :

1. The bridge wire may not be uniform in cross section. Then the wire will not have a uniform resistance per unit length and hence its resistance will not be proportional to its length.
2. End resistances at the two ends of the wire may be introduced due to
   1. the resistance of the metal strips
   2. the contact resistance of the bridge wire with the metal strips
   3. unmeasured lengths of the wire at the ends because the contact points of the wire with the metal strips do not coincide with the two ends of the metre scale attached.

Such errors are almost unavoidable but can be minimized considerably as follows :

1. Readings must be taken by adjusting the standard known resistance such that the null point is obtained close to the centre of the wire. When several readings are to be taken, the null points should lie in the middle one-third of the wire.
2. The measurements must be repeated with the standard resistance (resistance box) and the unknown resistance interchanged in the gaps of the bridge, obtaining the averages of the two results.

vii) What is potential gradient? How is it measured? Explain.
Answer:
Consider a potentiometer consisting of a long uniform wire AB of length L and resistance R, stretched on a wooden board and connected in series with a cell of stable emf E and internal resistance r and a plug key K as shown in below figure.

Let I be the current flowing through the wire when the circuit is closed.
Current through AB, I = \(\frac{E}{R+r}\)
Potential difference across AB. V_AB = IR
∴ V_AB = \(\frac{E R}{(R+r)}\)
The potential difference (the fall of potential from the high potential end) per unit length of the wire,
\(\frac{V_{\mathrm{AB}}}{L}=\frac{E R}{(R+r) L}\)
As long as E and r remain constant, \(\frac{V_{\mathrm{AB}}}{L}\) will remain constant, \(\frac{V_{\mathrm{AB}}}{L}\) is known as potential gradient along

AB and is denoted by K. Thus the potential gradient is calculated by measuring the potential difference between ends of the potentiometer wire and dividing it by the length of the wire.
Let P be any point on the wire between A and B and AP = l = length of the wire between A and P.
Then V_AP = Kl
∴ V_AP ∝ l as K is constant in a particular case. Thus, the potential difference across any length of the potentiometer wire is directly proportional to that length. This is the principle of the potentiometer.

viii) On what factors does the potential gradient of the wire depend?
Answer:
The potential gradient depends upon the potential difference between the ends of the wire and the length of the wire.

ix) Why is potentiometer preferred over a voltmeter for measuring emf?
Answer:
A voltmeter should ideally have an infinite resistance so that it does not draw any current from the circuit. However a voltmeter cannot be designed to have infinite resistance. A potentiometer does not draw any current from the circuit at the null point. Therefore, it gives more accurate measurement. Thus, it acts as an ideal voltmeter.

x) State the uses of a potentiometer.
Answer:
The applications (uses) of the potentiometer :

1. Voltage divider :
   The potentiometer can be used as a voltage divider to change the output voltage of a voltage supply.
2. Audio control:
   Sliding potentiometers are commonly used in modern low-power audio systems as audio control devices. Both sliding (faders) and rotary potentiometers (knobs) are regularly used for frequency attenuation, loudness control and for controlling different characteristics of audio signals.
3. Potentiometer as a sensor:
   If the slider of the potentiometer is connected to the moving part of a machine, it can work as a motion sensor. A small displacement of the moving part causes a change in potential which is further amplified using an amplifier circuit. The potential difference is calibrated in terms of displacement of the moving part.
4. To measure the emf (for this, the emf of the standard cell and potential gradient must be known).
5. To compare the emf’s of two cells.
6. To determine the internal resistance of a cell.

xi) What are the disadvantages of a potentiometer?
Answer:
Disadvantages of a potentiometer over a voltmeter :

1. The use of a potentiometer is an indirect measurement method while a voltmeter is a direct reading instrument.
2. A potentiometer is unwieldy while a voltmeter is portable.
3. Unlike a voltmeter, the use of a potentiometer in measuring an unknown emf requires a standard . source of emf and calibration.

xii) Distinguish between a potentiometer and a voltmeter.
Answer:

Potentiometer	Voltmeter
1. A potentiometer is used to determine the emf of a cell, potential difference and internal resistance.	1. A voltmeter can be used to measure the potential difference and terminal voltage of a cell. But it cannot be used to measure the emf of a cell.
2. Its accuracy and sensitivity are very high.	2. Its accuracy and sensitivity are less as compared to a potentiometer.
3. It is not a portable instrument.	3. It is a portable instrument.
4. It does not give a direct reading.	4. It gives a direct reading.

xiii) What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is
(i) increased (ii) decreased.
Answer:
(1) On increasing the current through the potentiometer wire, the potential gradient along the wire will increase. Hence, the position of zero deflection will occur at a shorter length.

(2) On decreasing the current through the potentiometer wire, the potential gradient along the wire will decrease. Hence, the position of zero deflection will occur at a longer length.
[Note : In the usual notation,
E₁ = (\(\frac{I R}{L}\)) l₁ = constant
Hence, (i) E, decreases when I is increased (ii) l₁ increases when I is decreased.]

Question 3.
Obtain the balancing condition in case of a Wheatstone’s network.
Answer:
Wheatstone’s network or bridge is a circuit for indirect measurement of resistance by null com-parison method by comparing it with a standard known resistance. It consists of four resistors with resistances P, Q, R and S arranged in the form of a quadrilateral ABCD. A cell (E) with a plug key (K) in series is connected across one diagonal AC and a galvanometer (G) across the other diagonal BD as shown in the following figure.

With the key K dosed, currents pass through the resistors and the galvanometer. One or more of the resistances is adjusted until no deflection in the galvanometer can be detected. The bridge is then said to be balanced.

Let I be the current drawn from the cell. At junction A, it divides into a current I₁ through P and a current I₂ through S.
I = I₁ + I₂ (by Kirchhoff’s first law).
At junction B, current I_g flows through the galvanometer and current I₁ – I_g flows through Q. At junction D, I₂ and I_g combine. Hence, current I₂ + I_g flows through R from D to C. At junction C, I₁ – I_g and I₂ + I_g combine. Hence, current I₁ + I₂(= I) leaves junction C.

Applying Kirchhoff’s voltage law to loop ABDA in a clockwise sense, we get,
– I₁P – I_gG + I₂S = 0 …………… (1)
where G is the resistance of the galvanometer.
Applying Kirchhoff’s voltage law to loop BCDB in a clockwise sense, we get,
– (I₁ – I_g)Q + (I₂ + I_g)R + I_gG = 0 ………….. (2)
When I_g = 0, the bridge (network) is said to be balanced. In that case, from Eqs. (1) and (2), we get,
I₁P = I₂S …………… (3)
and I₁Q = I₂R ………….. (4)
From Eqs. (3) and (4), we get,
\(\frac{P}{Q}=\frac{S}{R}\)
This is the condition of balance.
Alternative Method: When no current flows through the galvanometer, points B and D must be at the same potential.
∴ V_B = V_D
∴ V_A – V_B = V_A – V_D …………. (1)
and V_B – V_C = V_D – V_C ………… (2)
Now, V_A – V_B = I₁P and V_A – V_D = I₂S ………….. (3)
Also, V_B – V_C = I₁Q and V_D – V_C = I₂R …………. (4)
Substituting Eqs. (3) and (4) in Eqs. (1) and (2), we get,
I₁P = I₂S . ………… (5)
and I₁Q = I₂R …(6)
Dividing Eq. (5) by Eq. (6), we get,
\(\frac{P}{Q}=\frac{S}{R}\)
This is the condition of balance.

[ Note : In the determination of an unknown resistance using Wheatstone’s network, the unknown resistance is connected in one arm of the network (say, AB), and a standard known variable resistance is connected in an adjacent arm. Then, the other two arms are called the ratio arms. Also, because the positions of the cell and galvanometer can be interchanged, without a change in the condition of balance, the branches AC and BD in figure are called the conjugate arms. ]

Question 4.
Explain with neat circuit diagram, how you will determine the unknown resistance by using a meter-bridge.
Answer:
A metre bridge consists of a rectangular wooden board with two L-shaped thick metallic strips fixed along its three edges. A single thick metallic strip separates two L-shaped strips. A wire of length one metre and uniform cross section is stretched on a metre scale fixed on the wooden board. The ends of the wire are fixed to the L-shaped metallic strips.

An unknown resistance X is connected in the left gap and a resistance box R is connected in the right gap as shown in above figure. One end of a centre-zero galvanometer (G) is connected to terminal C and the other end is connected to a pencil jockey (J). A cell (E) of emf E, plug key (K) and rheostat (Rh) are connected in series between points A and B.

Working : Keeping a suitable resistance (R) in the resistance box, key K is closed to pass a current through the circuit. The jockey is tapped along the wire to locate the equipotential point D when the galvanometer shows zero deflection. The bridge is then balanced and point D is called the null point and the method is called as null deflection method. The distances l_X and l_R of the null point from the two ends of the wire are measured.

As R, l_X and l_R are known, the unknown resistance X can be calculated.

Question 5.
Describe Kelvin’s method to determine the resistance of a galvanometer by using a meter bridge.
Answer:
Kelvin’s method :
Circuit: The metre bridge circuit for Kelvin’s method of determination of the resistance of a galvanometer is shown in below figure. The galvanometer whose resistance G is to be determined, is connected in one gap of the metre bridge. A resistance box providing a variable known resistance R is connected in the other gap. The junction B of the galvanometer and the resistance box is con-nected directly to a pencil jockey. A cell of emf E, a key (K) and a rheostat (Rh) are connected across AC.

Working : Keeping a suitable resistance R in the resistance box and maximum resistance in the rheostat, key K is closed to pass the current. The rheostat resistance is slowly reduced such that the galvanometer shows about 2 / 3rd of the full-scale deflection.

On tapping the jockey at end-points A and C, the galvanometer deflection should change to opposite sides of the initial deflection. Only then will there be a point D on the wire which is equipotential with point B. The jockey is tapped along the wire to locate the equipotential point D when the galvanometer shows no change in deflection. Point D is now called the balance point and Kelvin’s method is thus an equal deflection method. At this balanced condition,
\(\frac{G}{R}=\frac{\text { resistance of the wire of length } l_{G}}{\text { resistance of the wire of length } l_{R}}\)
where I_G = the length of the wire opposite to the galvanometer, I_R = the length of the wire opposite to the resistance box.
If λ = the resistance per unit length of the wire,
\(\frac{G}{R}=\frac{\lambda l_{G}}{\lambda l_{R}}=\frac{l_{G}}{l_{R}}\)
∴ G = R\(\frac{l_{G}}{l_{R}}\)
The quantities on the right hand side are known, so that G can be calculated.
[Note : The method was devised by William Thom-son (Lord Kelvin, 1824-1907), British physicist.]

Question 6.
Describe how a potentiometer is used to compare the emfs of two cells by connecting the cells individually.
Answer:
A battery of stable emf E is used to set up a potential gradient V/L along a potentiometer wire, where V = potential difference across the length L of the wire. The positive terminals of the cells, whose emf’s (E₁ and E₂) are to be compared, are connected to the high potential terminal A. The negative terminals of the cells are connected to a galvanometer G through a two-way key. The other terminal of the galvanometer is connected to a pencil jockey. The emf E should be greater than both the emf’s E₁ and E₂.

Connecting point P to C, the cell with emf E1 is brought into the circuit. The jockey is tapped along the wire to locate the null point D at a distance l₁ from A. Then,
E₁ = Z₁(V/L)
Now, without changing the potential gradient (i.e., without changing the rheostat setting) point Q (instead of P) is connected to C, bringing the cell with emf E₂ into the circuit. Let its null point D’ be at a distance l₂ from A, so that
E₂ = l₂(V/L)
∴ \(\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}\)
Hence, by measuring the corresponding null lengths l₁ and l₂, E₁/E₂ can be calculated. The experiment is repeated for different potential gradients using the rheostat.
[Note : This method is used when the two emf’s have comparable magnitudes. Then, the errors of measurement of their balancing lengths will also be of comparable magnitudes.]

Question 7.
Describe how a potentiometer is used to compare the emfs of two cells by combination method.
Answer:
A battery of stable emf E is used to set up a potential gradient V/L, along the potentiometer wire, where V = potential difference across length L of the wire. The positive terminal of the cell 1 is connected to the higher potential terminal A of the potentiometer; the negative terminal is connected to the galvanometer G through the reversing key. The other terminal of the galvanometer is connected to a pencil jockey. The cell 2 is connected across the remaining two opposite terminals of the reversing key. The other terminal of the galvanometer is connected to a pencil jockey. The emf E₁ should be greater than the emf E₂; this can be adjusted by trial and error.

Two plugs are inserted in the reversing key in positions 1 – 1. Here, the two cells assist each other so that the net emf is E₁ + E₂. The jockey is tapped along the wire to locate the null point D. If the null point is a distance l₁ from A,
E₁ + E₂ = l₁ (V/L)

For the same potential gradient (without changing the rheostat setting), the plugs are now inserted into position 2 – 2. (instead of 1 – 1). The emf E₂ then opposes E₁ and the net emf is E₁ – E₂. The new null point D’ is, say, a distance l₂ from A and
E₁ – E₂ = l₂ (V/L)
∴ \(\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{l_{1}}{l_{2}}\) ∴ \(\frac{E_{1}}{E_{2}}=\frac{l_{1}-l_{2}}{l_{1}-l_{2}}\)
Here, the emf E should be greater than E₁ + E₂. The experiment is repeated for different potential gradients using the rheostat.
[Note : This method is used when E₁ » E₂, so that E₁ + E₂ and E₁ – E₂ have comparable magnitudes. Then, the errors of measurement of l₁ and l₂ will also be of comparable magnitudes.]

Question 8.
Describe with the help of a neat circuit diagram how you will determine the internal resistance of a cell by using a potentiometer. Derive the necessary formula.
Answer:
Principle : A cell of emf £ and internal resistance r, which is connected to an external resistance R, has its terminal potential difference V less than its emf E. If I is the corresponding current,
\(\frac{E}{V}=\frac{I(R+r)}{I R}=\frac{R+r}{R}\) = 1 + \(\frac{r}{R}\) (when R → ∞, V → E)
∴ r = \(\frac{E-V}{V}\) R
Working : A battery of stable emf E’ is used to set up a potential gradient V/L, along the potentiometer wire, where V = potential difference across the length L of the wire. The negative terminal is connected through a centre-zero galvanometer G to a pencil jockey. A resistance box R with a plug key K in series is connected across the cell.

Firstly, key K is kept open; then, effectively, R = ∞. The jockey is tapped on the potentiometer wire to locate the null point D. Let the null length
AD = l₁, so that
E = (V_AB/L)l₁

With the same potential gradient, and a small resistance R in the resistance box, key K is closed. The new null length AD’ = l2 for the terminal potential difference V is found :

R, l₁, and l₂ being known, r can be calculated. The experiment is repeated with different potential gradients using the rheostat or with different values of R.

Question 9.
On what factors does the internal resistance of a cell depend?
Answer:
The internal resistance of a cell depends on :

1. Nature of the electrolyte :
   The greater the conductivity of the electrolyte, the lower is the internal resistance of the cell.
2. Separation between the electrodes :
   The larger the seperation between the electrodes of the cell, the higher is the internal resistance of the cell. This is because the ions have to cover a greater distance before reaching an electrode.
3. Nature of the electrodes.
4. The internal resistance is inversly proportional to the common area of the electrodes dipping in the electrolyte.

Question 10.
A battery of emf 4 volt and internal resistance 1 Ω is connected in parallel with
another battery of emf 1 V and internal resistance 1 Ω (with their like poles connected together). The combination is used to send current through an external resistance of 2 Ω. Calculate the current through the external resistance.
Answer:
Let I₁ and I₂ be the currents through the two branches as shown in below figure. The current through the 2 Q resistance will be (I₁ + I₂) [Kirchhoff’s current law].
Applying Kirchhoff’s voltage law to loop ABCDEFA, we get,
-2(I₁ + I₂) – 1(I₁) + 4 = 0
∴ 3I₁ + 2I₁ = 4 …………… (1)

Applying Kirchhoff’s voltage law to loop BCDEB, we get,
-2(I₁ + I₂) – 1(I₂) + 1 = 0
2I₁ + 3I₂ = 1 ………… (2)
Multiplying Eq. (1) by 2 and Eq. (2) by 3, we get,
6I₁ + 4I₂ = 8 ………….(3)
and 6I₁ + 9I₂ = 3 ……………. (4)
Subtracting Eq. (4) from Eq. (3), we get,
– 5I₂ =5
∴ I₂ = -1A
The minus sign shows that the direction of current I2 is opposite to that assumed. Substituting this value of 12 in Eq. (1), we get,
3I₁ + 2(-1) = 4
∴3I₁ = 4 + 2 = 6
∴ I₁ = 2A
Current through the 2 0 resistance = I₁ + I₂ = 2 – 1
= 1 A. It is in the direction as shown in the figure.
[Note : We may as well consider loop ABEFA and write the corresponding equation. But it does not provide any additional information.]

Question 11.
Two cells of emf 1.5 Volt and 2 Volt having respective internal resistances of 1 Ω and 2 Ω are connected in parallel so as to send current in same direction through an external resistance of 5 Ω. Find the current through the external resistance.
Answer:
Let I₁ and I₂ be the currents through the two branches as shown in below figure. The current through the 5 Q resistor will be I₁ + I₂ [Kirchhoff’s current law].

Applying Kirchhoff’s voltage law to loop w ABCDEFA, we get,
– 5(I₁ + I₂) – I₁ + 1.5 = 0
∴ 6I₁ + 5I₂ = 1.5 ……………. (1)
Applying Kirchhoff’s voltage law to loop BCDEB, we get,
– 5(I₁ + I₂) – 2I₂ + 2 = 0
∴ 5I₁ + 7I₂ = 2 ……………(2)
Multiplying Eq. (1) by 5 and Eq. (2) by 6, we get,
30I₁ + 25I₂ = 7.5 …………… (3)
and 30I₁ + 42I₂ = 12 …………. (4)
Subtracting Eq. (3) from Eq. (4), we get,
17I₂ = 4.5
∴ I₂ = \(\frac{4.5}{17}\) A
Substituting this value of I2 in Eq. (1), we get,
6I₁ + 5(\(\frac{4.5}{17}\)) = 1.5
∴ 6I₁ + \(\frac{22.5}{17}\) = 1.5
∴ 6I₁ = 1.5 – \(\frac{22.5}{17}\) = \(\frac{28.5-22.5}{17}\)
= \(\frac{3}{17}\)
∴ I₁ = \(\frac{3}{17 \times 6}=\frac{0.5}{17}\) A
Current through the 5 Q resistance (external resistance)
= I₁ + I₂ = \(\frac{0.5}{17}+\frac{4.5}{17}=\frac{5}{17}\) A

Question 12.
A voltmeter has a resistance 30 Ω. What will be its reading, when it is connected across a cell of emf 2 V having internal resistance 10 Ω?
Answer:
Data: E = 2V, r = 10 Ω, R = 30 Ω
The voltmeter reading, V = IR
= (\(\frac{E}{R+r}\)) R
= (\(\frac{2}{30+10}\)) 30
= (\(\frac{2}{40}\)) 30
= 1.5 V

Question 13.
A set of three coils having resistances 10 Ω, 12 Ω and 15 Ω are connected in parallel. This combination is connected in series with series combination of three coils of the same resistances. Calculate the total resistance and current through the circuit, if a battery of emf 4.1 Volt is used for drawing current.
Answer:
Below figure shows the electrical network. For resistances 10 Ω, 12 Ω and 15 Ω connected in parallel the equivalent resistance (R_p) is given by,

For resistance R_p, 10 Ω, 12 Ω and 15 Ω connected in series, the equivalent resistance,
R_s = 4 + 10 + 12 + 15 = 41 Ω
Thus, the total resistance = R_s = 41 Ω
Now, V = IR_s
∴ 4.1 = 1 × 41
∴ I = 0.1A
The total resistance and current through the circuit are 41 Ω and 0.1 A respectively.

Question 14.
A potentiometer wire has a length of 1.5 m and resistance of 10 Ω. It is connected in series with the cell of emf 4 Volt and internal resistance 5 Ω. Calculate the potential drop per centimeter of the wire.
Answer:
Data : L = 1.5 m, R = 10 Ω, E = 4 V, r = 5 Ω

The potential drop per centimeter of the wire is 0.0178 \(\frac{\mathrm{V}}{\mathrm{cm}}\)

Question 15.
When two cells of emfs. E₁ and E₂ are connected in series so as to assist each other, their balancing length on a potentiometer is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emfs of the two cells.
Answer:
Data : l₁ = 2.7m (cells assisting),
l₂ = 0.3 m (cells opposing)
E₁ + E₂ = Kl₁ and E₁ – E₂ = Kl₂
∴\(\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{K l_{1}}{K l_{2}}\)
∴ \(\frac{E_{1}}{E_{2}}=\frac{l_{1}+l_{2}}{l_{1}-l_{2}}=\frac{2.7+0.3}{2.7-0.3}=\frac{3}{2.4}=\frac{30}{24}\) = 1.25
The ratio of the emf’s of the two cells is 1.25.

Question 16.
The emf of a cell is balanced by a length of 120 cm of potentiometer wire. When the cell is shunted by a resistance of 10 Ω, the balancing length is reduced by 20 cm. Find the internal resistance of the cell.
Answer:
Data :R = 10 Ω, l₁ = 120 cm, l₂ = 120 – 20 = 100 cm
r = R(\(\frac{l_{1}-l_{2}}{l_{2}}\))
= 10 (\(\frac{120-100}{100}\))
= 2 Ω
The internal resistance of the cell is 2 Ω.

Question 17.
A potential drop per unit length along a wire is 5 × 10^-3 V/m. If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell.
Answer:
Data: K = 5 × 10^-3 \(\frac{\mathrm{V}}{\mathrm{m}}\), L = 216 cm = 216 × 10^-2 m
E = KL
∴ E = 5 × 10^-3 × 216 × 10^-2
= 1080 × 10^-5
= 0.01080V
The emf of the cell is 0.01080 volt.
(Note: For K = 0.5 V/m, we get, E = 1.08V (reasonable value)]

Question 18.
The resistance of a potentiometer wire is 8 Ω and its length is 8 m. A resistance box and a 2 V battery are connected in series with it. What should be the resistance in the box, if it is desired to have a potential drop of 1 µV/mm?
Answer:
Data: R = 8 Ω, L = 8 m, E = 2 V, K = 1 µV/mm
= 1 × \(\frac{10^{-6} \mathrm{~V}}{10^{-3} \mathrm{~m}}\) = 10^-3 \(\frac{\mathrm{V}}{\mathrm{m}}\)
K = \(\frac{V}{L}=\frac{E R}{\left(R+R_{\mathrm{B}}\right) L^{\prime}}\) where R_B is the resistance in the box.
∴ 10^-3 = \(\frac{2 \times 8}{\left(8+R_{B}\right) 8}\)
∴ 8 + R_B = \(\frac{2}{10^{-3}}\)
= 2 × 10³
∴ R_B = 2000 – 8
= 1992 Ω

Question 19.
Find the equivalent resistance between the terminals F and B in the network shown in the figure below given that the resistance of each resistor is 10 ohm.

Answer:
Applying Kirchhoff’s voltage law to loop FGHF, we get,
– 10I₁ – 10(I₁ – I₂) + 10(I – I₁) + 10(I – I₁) = 0
∴ – 10I₁ – 10I₁ + 10I₂ + 10I – 10I₁ + 10I – 10I₁ = 0
∴ 201 – 40I₁ + 10I₂ = 0
∴ 2I – 4I₁ + I₂ = 0 …………….. (1)
Applying Kirchhoff’s voltage law to loop GABHG, we get,
– 10I₂ – 10I₂ + 10(I – I₂) + 10(I₁ – I₂) = 0
∴ – 20I₂ + 10I – 10I₂ + 10I₁ – 10I₂ = 0
∴ 10I + 10I₁ – 40I₂ = 0 .
∴ I + I₁ – 4I₂ = 0 ……………… (2)
Applying Kirchhoff’s voltage law to loop EFHBCDE, we get,
– 10(I – I₁) – 10(I – I₁) – 10(I – I₂) + E = 0
∴ -10I + 10I₁ – 10I + 10I₁ – 10I + 10I₂ + E = 0
∴E = 30I – 20I₁ – 10I₂ ………….. (3)
From Eq. (1), we get, I₂ = 4I₁ – 2I …………. (4)
From Eqs. (2) and (4), we get,
I + I₁ – 4(4I₁ – 2I) = 0
∴ I + I₁ – 16I₁ + 8I = 0 .
∴ 9I = 15I₁ ∴ I₁ = \(\frac{9}{15}\)I = \(\frac{3}{5}\)I …………. (5)
From Eqs. (4) and (5), we get,
I₂ = 4(\(\frac{3}{5}\)I) – 2I = \(\frac{12}{5}\)I – 2I
= \(\frac{12 I-10 I}{5}\) = \(\frac{2}{5}\) I
From Eqs. (3), (5) and (6), we get
E = 30I – 20(\(\frac{3}{5}\) I) – 10(\(\frac{2}{5}\) I)
= 30I – 12I – 4I = 30I – 16I
∴ E = 14I
If R is the equivalent resistance between E and C,
E = RI
∴ R = 14 Ω

Question 20.
A voltmeter has a resistance of 100 Ω. What will be its reading when it is connected across a cell of emf 2 V and internal resistance 20 Ω?
Answer:
Data: E = 2V, r = 20 Ω, R = 100 Ω
The voltmeter reading, V = IR
V = (\(\frac{2}{100+20}\))100 = \(\frac{200}{120}=\frac{10}{6}\) = 1.667 V.

12th Physics Digest Chapter 9 Current Electricity Intext Questions and Answers

Observe and Discuss (Textbook Page No. 220)

Question 1.
Post Office Box
A post office box (PO Box) is a practical form of Wheatstone bridge as shown in the figure.

It consists of three arms P, Q and R. The resistances in these three arms are adjustable. The two ratio arms P and Q contain resistances of 10 ohm, 100 ohm and 1000 ohm each. The third arm R contains resistances from 1 ohm to 5000 ohm. The unknown resistance X (usually, in the form of a wire) forms the fourth arm of the Wheatstone’s bridge. There are two tap keys K₁ and K₂ .
Answer:
The resistances in the arms P and Q are fixed to a desired ratio. The resistance in the arm R is adjusted so that the galvanometer shows no deflection. Now the bridge is balanced. The unknown resistance X = RQ/P, where P and Q are the fixed resistances in the ratio arms and R is the adjustable known resistance.

If L is the length of the wire used to prepare the resistor with resistance X and r is its radius, then the specific resistance (resistivity) of the material of the wire is given by
ρ = \(\frac{X \pi r^{2}}{L}\)

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 8 Electrostatics Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 8 Electrostatics

1. Choose the correct option

i) A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on a charge, potential, capacitance respectively are
(A) Constant, decreases, decreases
(B) Increases, decreases, decreases
(C) Constant, decreases, increases
(D) Constant, increases, decreases
Answer:
(A) Constant, decreases, decreases

ii) A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has a thickness (3/4d), where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is

Answer:
(D) C = \(\frac{\varepsilon_{0} A}{d}\left(\frac{4 k}{k+3}\right)\)

iii) Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio.
(A) 1 : 1
(B) 1 : 2
(C) 2 : 1
(D) 1 : 3
Answer:
(A) 1 : 1

iv) Charge + q and -q are placed at points A and B respectively which are distance 2L apart. C is the mid point of A and B. The work done in moving a charge +Q along the semicircle CRD as shown in the figure below is

(C) \(\frac{q Q}{6 \pi \varepsilon_{0} L}\)
(D) \(\frac{-q Q}{6 \pi \varepsilon_{0} L}\)
Answer:
(A) \(-\frac{Q q_{1}}{6 \pi \varepsilon_{0} L}\)

v) A parallel plate capacitor has circular plates of radius 8 cm and plate separation 1mm. What will be the charge on the plates if a potential difference of 100 V is applied?
(A) 1.78 × 10^-8 C
(B) 1.78 × 10^-5 C
(C) 4.3 × 10⁴ C
(D) 2 × 10^-9 C
Answer:
(A) 1.78 × 10^-8 C

2. Answer in brief.

i) A charge q is moved from a point A above a dipole of dipole moment p to a point B below the dipole in equitorial plane without acceleration. Find the work done in this process.

Answer:
The equatorial plane of an electric dipole is an equipotential with V = 0. Therefore, the no work is done in moving a charge between two points in the equatorial plane of a dipole.

ii) If the difference between the radii of the two spheres of a spherical capacitor is increased, state whether the capacitance will increase or decrease.
Answer:
The capacitance of a spherical capacitor is C = 4πε₀\(\left(\frac{a b}{b-a}\right)\) where a and b are the radii of the concentric inner and outer conducting shells. Hence, the capacitance decreases if the difference b – a is increased.

iii) A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?
Answer:
Suppose the parallel-plate capacitor has capacitance C₀, plates of area A and separation d. Assume the metal sheet introduced has the same area A.

Case (1) : Finite thickness t. Free electrons in the sheet will migrate towards the positive plate of the capacitor. Then, the metal sheet is attracted towards whichever capacitor plate is closest and gets stuck to it, so that its potential is the same as that of that plate. The gap between the capacitor plates is reduced to d – t, so that the capacitance increases.

Case (2) : Negligible thickness. The thin metal sheet divides the gap into two of thicknesses d₁ and d₁ of capacitances C₁ = ε₀A/d₁ and C₂ = ε₀A/d₂ in series.
Their effective capacitance is
C = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{\varepsilon_{0} A}{d_{1}+d_{2}}=\frac{\varepsilon_{0} A}{d}\) = C₀
i.e., the capacitance remains unchanged.

iv) The safest way to protect yourself from lightening is to be inside a car. Justify.
Answer:
There is danger of lightning strikes during a thunderstorm. Because trees are taller than people and therefore closer to the clouds above, they are more likely to get hit by lightnings. Similarly, a person standing in open ground is the tallest object and more likely to get hit by a lightning. But car with a metal body is an almost ideal Faraday cage. When a car is struck by lightning, the charge flows on the outside surface of the car to the ground but the electric field inside remains zero. This leaves the passengers inside unharmed.

v) A spherical shell of radius b with charge Q is expanded to a radius a. Find the work done by the electrical forces in the process.
Answer:
Consider a spherical conducting shell of radius r placed in a medium of permittivity ε. The mechanical force per unit area on the charged conductor is
f = \(\frac{F}{d S}=\frac{\sigma^{2}}{2 \varepsilon}\)
where a is the surface charge density on the conductor. Given the charge on the spherical shell is Q, (σ = Q/πr². The force acts outward, normal to the surface.

Suppose the force displaces a charged area element adS through a small distance dx, then the work done by the force is
dW = Fdx = (\(\frac{\sigma^{2}}{2 \varepsilon}\) dS) dx
During the displacement, the area element sweeps out a volume dV = dS ∙ dx.

Therefore, the work done by the force in expanding the shell from radius r = b to r = a is

This gives the required expression for the work done.

Question 3.
A dipole with its charges, -q and + q located at the points (0, -b, 0) and (0 +b, 0) is present in a uniform electric field E whose equipotential surfaces are planes parallel to the YZ planes.
(a) What is the direction of the electric field E?
(b) How much torque would the dipole experience in this field?
Answer:
(a) Given, the equipotentials of the external uniform electric field are planes parallel to the yz plane, the electric field \(\vec{E}=\pm E \hat{\mathrm{i}}\) that is, \(\overrightarrow{E}\) is parallel to the x-axis.

(b) From above figure, the dipole moment, \(\vec{p}=q(2 b) \hat{\mathrm{j}}\)
The torque on this dipole,

So that the magnitude of the torque is τ = 2qbE.
If \(\vec{E}\) is in the direction of the + x-axis, the torque \(\vec{\tau}\) is in the direction of – z-axis, while if \(\vec{E}\) is in the direction of the -x-axis, the torque \(\vec{\tau}\) is in the direction of + z-axis.

Question 4.
Three charges – q, + Q and – q are placed at equal distance on straight line. If the potential energy of the system of the three charges is zero, then what is the ratio of Q : q?
Answer:

In the above figure, the line joining the charges is shown as the x-axis with the origin at the + Q charge. Let q₁ = +Q, and q₂ = q₃ = – q. Let the two – q charges be at (- a, 0) and (a, 0), since the charges are given to be equidistant.
∴ r₂₁ = r₃₁ = a and r₃₂ = 2a
The total potential energy of the system of three charges is

This gives the required ratio.

Question 5.
A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, the electric field, charge stored and voltage will increase, decrease or remain constant.
Answer:
Assume a parallel-plate capacitor, of plate area A and plate separation d is filled with a dielectric of relative permittivity (dielectric constant) k. Its capacitance is C = \(\frac{k \varepsilon_{0} A}{d}\) …………. (1)
If it is charged to a voltage (potential) V, the charge on its plates is Q = CV.

Since the battery is disconnected after it is charged, the charge Q on its plates, and consequently the product CV, remain unchanged.

On removing the dielectric completely, its capacitance becomes from Eq. (1),
C’ = \(\frac{\varepsilon_{0} A}{d}=\frac{1}{k} C\) ……………. (2)
that is, its capacitance decreases by the factor k. Since C’V’ = CV, its new voltage is
V’ = \(\frac{C}{C^{\prime}}\) V = kV …………… (3)

so that its voltage increases by the factor k. The stored potential energy, U = \(\frac{1}{2}\) QV, so that Q remaining constant, U increases by the factor k. The electric field, E = V/ d, so that E also increases by a factor k.

Question 6.
Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C₁ and C₂ with their capacitances in the ratio 1 : 2, so that the energy stored in these two cases becomes the same.
Answer:

This gives the required ratio.

Question 7.
Two charges of magnitudes -4Q and +2Q are located at points (2a, 0) and (5a, 0) respectively. What is the electric flux due to these charges through a sphere of radius 4a with its centre at the origin?
Answer:
The sphere of radius 4a encloses only the negative charge Q₁ = -4Q. The positive charge Q₂ = +2Q being located at a distance of 5a from the origin is outside the sphere. Only a part of the electric flux lines originating at Q₂ enters the sphere and exits entirely at other points. Hence, the electric flux through the sphere is only due to Q₁.
Therefore, the net electric flux through the sphere = \(\frac{Q_{1}}{\varepsilon_{0}}=\frac{-4 Q}{\varepsilon_{0}}\) . The minus sign shows that the flux is directed into the sphere, but not radially since the sphere is not centred on Q₁.

Question 8.
A 6 µF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected to another uncharged 3µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?
Answer:
Data: C = 6 µF = 6 × 10^-6 F = C₁, V = 300 V
C₂ = 3 µF
The electrostatic energy in the capacitor
= \(\frac{1}{2}\)Cv² = \(\frac{1}{2}\)(6 × 10^-6)(300)²
= 3 × 10^-6 × 9 × 10⁴ = 0.27J
The charge on this capacitor,
Q = CV = (6 × 10^-6)(300) = 1.8 mC
When two capacitors of capacitances C₁ and C₂ are connected in parallel, the equivalent capacitance C
= C₁ + C₂ = 6 + 3 = 9 µF
= 9 × 10^-6F
By conservation of charge, Q = 1.8 C.
∴ The energy of the system = \(\frac{Q^{2}}{2 C}\)
= \(\frac{\left(1.8 \times 10^{-3}\right)^{2}}{2\left(9 \times 10^{-6}\right)}=\frac{18 \times 10^{-8}}{10^{-6}}\) = 0.18 J
The energy lost = 0.27 – 0.18 = 0.09 J

Question 9.
One hundred twenty five small liquid drops, each carrying a charge of 0.5 µC and each of diameter 0.1 m form a bigger drop. Calculate the potential at the surface of the bigger drop.
Answer:
Data : n = 125, q = 0.5 × 10^-6 C, d = 0.1 m
The radius of each small drop, r = d/2 = 0.05 m
The volume of the larger drop being equal to the volume of the n smaller drops, the radius of the larger drop is
R = \(\sqrt[3]{n} r\) = \(\sqrt[3]{125}\) (0.05) = 5 × 0.05 = 0.25 m
The charge on the larger drop,
Q = nq = 125 × (0.5 × 10^-6) C
∴ The electric potential of the surface of the larger drop,
V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R}\) = (9 × 10⁹) × \(\frac{125 \times\left(0.5 \times 10^{-6}\right)}{0.25}\)
= 9 × 125 × 2 × 10³ = 2.25 × 10⁶ V

Question 10.
The dipole moment of a water molecule is 6.3 × 10^-30 Cm. A sample of water contains 1021 molecules, whose dipole moments are all oriented in an electric field of strength 2.5 × 10⁵ N /C. Calculate the work to be done to rotate the dipoles from their initial orientation θ₁ = 0 to one in which all the dipoles are perpendicular to the field, θ₂ = 90°.
Answer:
Data: p = 6.3 × 10^-30 C∙m, N = 10²¹ molecules,
E = 2.5 × 10⁵ N/C, θ₀ = θ₁ = 0°, θ = θ₂ = 90°
W = pE(cos θ₀ – cos θ)
The total work required to orient N dipoles is
W = NpE(cos θ₁ – cos θ₂)
=(10²¹)(6.3 × 10^-30)(2.5 × 10⁵)
= 15.75 × 10^-4 J = 1.575 mJ

Question 11.
A charge 6 µC is placed at the origin and another charge -5 µC is placed on the y axis at a position A (0, 6.0) m.

(a) Calculate the total electric potential at the point P whose coordinates are (8.0, 0) m
(b) Calculate the work done to bring a proton from infinity to the point P. What is the significance of the sign of the work done ?
Answer:
Data : q₁ = 6 × 10^-6 C, q₂ = -5 × 10^-6 C,
A ≡ (0, 6.0 m), P ≡ (8.0 m, 0), r₁ = OP = 8 m, q = e = 1.6 × 10^-19C, 1/4πε₀ = 9 × 10⁹ N∙m²/C²
r₂ = AP = \(\sqrt{(8-0)^{2}+(0-6)^{2}}\) = \(\sqrt{64+36}\) = 10 m

(a) The net electric potential at P due to the system of two charges is

(b) The electric potential V at the point P is the negative of the work done per unit charge, by the electric field of the system of the charges q₁ and q₂, in bringing a test charge from infinity to that point.
V = \(-\frac{W}{q_{0}}\)
∴ W = -qV= -(1.6 × 10^-19)(2.25 × 10³)
= -3.6 × 10^-16 J= -2.25 keV
That is, in bringing the positively charged proton from a point of lower potential to a point of higher potential, the work done by the electric field on it is negative, which means that an external agent must bring the proton against the electric field of the system of the two source charges.

[Note : The potential V at a point is the work done per unit charge (W_ext) by an external agent in bringing a test charge from infinity to that point. In the above case, the work done by an external agent will be positive. The question does not specify this.]

Question 12.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^-3 m² and the separation between the plates is 2 mm.
i) Calculate the capacitance of the capacitor.
ii) If this capacitor is connected to 100 V supply, what would be the charge on each plate?
iii) How would charge on the plates be affected if a 2 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected ?
Answer:
Data: k = 1(air), A = 6 × 10^-3 m², d = 2
mm = 2 × 10^-3 m,V = 100V, t = 2 mm = d, k₁ = 6,
ε₀ = 8.85 × 10^-12 F/m
(i) The capacitance of the air capacitor, C₀ = \(\frac{\varepsilon_{0} A}{d}\)
= \(\frac{\left(8.85 \times 10^{-12}\right)\left(6 \times 10^{-3}\right)}{2 \times 10^{-3}}\)
= 26.55 × 10^-12 F = 26.55 pF

(ii) Q₀ = C₀V = (26.55 × 10^-12)(100)
= 26.55 × 10^-10 C = 2.655 nC

(iii) The dielectric of relative permittivity k₁ completely fills the space between the plates (∵t = d), so that the new capacitance is C = k₁C₀.
With the supply still connected, V remains the same.
∴ Q = CV = kC₀V = kQ₀ =6(2.655 nF) = 15.93 nC
Therefore, the charge on the plates increases.
[Note: Ck₁C₀ = 6(26.55 pF)= 159.3 pF.]

Question 13.
Find the equivalent capacitance between P and Q. Given, area of each plate = A and separation between plates = d.

Answer:
(i) The capacitor in figure is a series combination of three capacitors of plate separations d/3 and plate areas A, with C₁ filled with air (k₁ = 1), C₂ filled with dielectric of k₂ = 3 and C₃ filled with dielectric of k₃ = 6

(ii) In figure, a series combination of two capacitors C₂(k₂ = 3) and C₃(k₃ = 6), of plate areas A/2 and plate separations d/2, is in parallel with a capacitor C₁ (k₁ = 4) of plate area A/2 and plate separation d.

12th Physics Digest Chapter 8 Electrostatics Intext Questions and Answers

Can you recall (Textbook Page No. 188)

Question 1.
What is gravitational Potential ?
Answer:
We measure the gravitational potential energy U of a body (1) by assigning U = 0 for a reference configuration (such as the body at a reference level) (2) then equating U to the work W done by an external force to move the body up or down from that level to a point. We then define gravitational potential of the point as gravitational potential energy per unit mass of the body.

We follow the same procedure with the electric force, which is also a conservative force with the only difference that while the gravitational force is always attractive, electric force can be attractive (for unlike charges) or repulsive (for like charges).

Remember this (Textbook Page No. 191)

Question 1.
Due to a single charge at a distance r, Force (F ) α 1/r², Electric field (E ) α 1/r² but Potential (V) α 1/ r.
Answer:
At a point a distance r from an isolated point charge, the force F on a point charge and the electric field E both vary as 1/r², while the potential energy U of a point charge and the electric potential V at the point both vary as 1/r.

Use your brain power (Textbook Page No.194)

Question 1.
Is electrostatic potential necessarily zero at a point where electric field strength is zero? Justify.
Answer:
Electric potential is a scalar quantity while electric field intensity is a vector quantity. When we add potentials at a point due two or more point charges, the operation is simple scalar addition along with the sign of V, determined by the sign of the q that produces V. At a point, the net field is the vector sum of the fields due to the individual charges. Midway between the two charges of an electric dipole, the potentials due to the two charges are equal in magnitude but opposite in sign, and thus add up to zero. But the electric fields due to the charges are equal in magnitude and direction-towards the negative charge-so that the net field there is not zero. But midway between two like charges of equal magnitudes, the potentials are equal in magnitude and have the same sign, so that the net potential is nonzero. However, the fields due to the two equal like charges are equal in magnitude but opposite in direction, and thus vectorially add up to zero.

Do you know (Textbook Page No. 203)

Question 1.
If we apply a large enough electric field, we can ionize the atoms and create a condition for electric charge to flow like a conductor. The fields required for the breakdown of dielectric is called dielectric strength.
Answer:
In a sufficiently strong electric field, the molecules of a dielectric material become ionized, allowing flow of charge. The insulating properties of the dielectric breaks down, permanently or temporarily, and the phenomenon is called dielectric breakdown. During dielectric breakdown, electrical discharge through the dielectric follows random-path patterns like tree branches, called a Lichtenberg figure. Dielectric strength is the voltage that an insulating material can withstand before breakdown occurs. It usually depends on the thickness of the material. It is expressed in kV/mm. For example, the dielectric strengths of air, polystyrene and mica in kV/mm are 3, 20 and 118. Higher dielectric strength corresponds to better insulation properties.

Remember this (Textbook Page No. 205)

Question 1.
Series combination is used when a high voltage is to be divided on several capacitors. Capacitor with minimum capacitance has the maximum potential difference between the plates.
Answer:
Series combination of capacitors

1. Equivalent capacitance is less than the smallest capacitance in series. For several capacitors of given capacitances, the equivalent capacitance of their series combination is minimum.
2. All capacitors in the combination have the same charge but their potential differences are in the inverse ratio of their capacitances.
3. Series combination of capacitors is sometimes used when a high voltage, which exceeds the breakdown voltage of a single capacitor, is to be divided on more than one capacitors. Capacitive voltage dividers are only useful in AC circuits, since capacitors do not pass DC signals.

Parallel combination of capacitors

1. For several capacitors of given capacitances, the equivalent capacitance of their parallel combination is maximum.
2. The same voltage is applied to all capacitors in the combination, but the charge stored in the combination is distributed in proportion to their capacitances. The maximum rated voltage of a parallel combination is only as high as the lowest voltage rating of all the capacitors used. That is, if several capacitors rated at 500 V are connected in parallel to a capacitor rated at 100 V, the maximum voltage rating of the capacitor bank is only 100 V.
3. Parallel combination of capacitors is used when a large capacitance is required, i.e., a large charge is to be stored, at a small potential difference.

Remember this (Textbook Page No. 207)

Question 1.
(1) If there are n parallel plates then there will be (n-1) capacitors, hence
C = (n – 1) \(\frac{A \varepsilon_{0}}{d}\)
(2) For a spherical capacitor, consisting of two concentric spherical conducting shells with inner and outer radii as a and b respectively, the capacitance C is given by
C = 4πε₀\(\left(\frac{a b}{b-a}\right)\)
(3) For a cylindrical capacitor, consisting of two coaxial cylindrical shells with radii of the inner and outer cylinders as a and b, and length l, the capacitance C is given by
C = \(\frac{2 \pi \varepsilon_{0} \ell}{\log _{e} \frac{b}{a}}\)
Answer:
1. Stacking together n identical conducting plates equally spaced and with alternate plates connected to two points P and Q, forms a parallel combination of n -1 identical capacitors between P and Q. Then, the capacitance between the points is (n – 1) times the capacitance between any two adjacent plates.

2. A cylindrical capacitor consists of a solid cylindrical conductor of radius a is surrounded by coaxial cylindrical shell of inner radius b. The length of both cylinders is L, such that L is much larger than b – a, the separation of the cylinders, so that edge effects can be ignored. The capacitance of the capacitor is C = \(\frac{2 \pi \varepsilon_{0} L}{\log _{e}(b / a)}\).

The capacitance depends only on the geometrical factors, L, a and b, as for a parallel-plate capacitor.

3. A spherical capacitor which consists of two concentric spherical shells of radii a and b. The capacitance of the capacitor is C = 4πε₀\(\left(\frac{a b}{b-a}\right)\)

Again, the capacitance depends only on the geometrical factors, a and b.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 7 Wave Optics Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 7 Wave Optics

1. Choose the correct option.

i) Which of the following phenomenon proves that light is a transverse wave?
(A) reflection
(B) interference
(C) diffraction
(D) polarization
Answer:
(D) polarization

ii) Which property of light does not change when it travels from one medium to another?
(A) velocity
(B) wavelength
(C) amplitude
(D) frequency
Answer:
(D) frequency

iii) When unpolarized light is passed through a polarizer, its intensity
(A) increases
(B) decreases
(C) remains unchanged
(D) depends on the orientation of the polarizer
Answer:
(B) decreases

iv) In Young’s double slit experiment, the two coherent sources have different intensities. If the ratio of maximum intensity to the minimum intensity in the interference pattern produced is 25:1. What was the ratio of intensities of the two sources?
(A) 5:1
(B) 25:1
(C) 3:2
(D) 9:4
Answer:
(D) 9:4

v) In Young’s double slit experiment, a thin uniform sheet of glass is kept in front of the two slits, parallel to the screen having the slits. The resulting interference pattern will satisfy
(A) The interference pattern will remain unchanged
(B) The fringe width will decrease
(C) The fringe width will increase
(D) The fringes will shift.
Answer:
(A) The interference pattern will remain unchanged

2. Answer in brief.

i) What are primary and secondary sources of light?
Answer:
(1) Primary sources of light: The sources that emit light on their own are called primary sources. This emission of light may be due to
(a) the high temperature of the source, e.g., the Sun, the stars, objects heated to high temperature, a flame, etc.
(b) the effect of current being passed through the source, e.g., tubelight, TV, etc.
(c) chemical or nuclear reactions taking place in the source, e.g., firecrackers, nuclear energy generators, etc.

(2) Secondary sources of light: Some sources are not self luminous, i.e., they do not emit light on their own, but reflect or scatter the light incident on them. Such sources of light are called secondary sources, e.g. the moon, the planets, objects such as humans, animals, plants, etc. These objects are visible due to reflected light.
Many of the sources that we see around are secondary sources and most of them are extended sources.

ii) What is a wavefront? How is it related to rays of light? What is the shape of the wavefront at a point far away from the source of light?
Answer:
Wavefront or wave surface : The locus of all points where waves starting simultaneously from a source reach at the same instant of time and hence the particles at the points oscillate with the same phase, is called a wavefront or wave surface.

Consider a point source of light O in a homogeneous isotropic medium in which the speed of light is v. The source emits light in all directions. In time t, the disturbance (light energy) from the source, covers a distance vt in all directions, i.e., it reaches out to all points which are at a distance vt from the point source. The locus of these points which are in the same phase is the surface of a sphere with the centre O and radius vt. It is a spherical wavefront.

In a given medium, a set of straight lines can be drawn which are perpendicular to the wavefront. According to Huygens, these straight lines are the rays of light. Thus, rays are always normal to the wavefront. In the case of a spherical wavefront, the rays are radial.

If a wavefront has travelled a large distance away from the source, a small portion of this wavefront appears to be plane. This part is a plane wavefront.

iii) Why are multiple colours observed over a thin film of oil floating on water? Explain with the help of a diagram.
Answer:
Several phenomena that we come across in our day to day life are caused by interference and diffraction of light. These are the vigorous colours of soap bubbles as well as those seen in a thin oil film on the surface of water, the bright colours of butterflies and peacocks etc. Most of these colours are not due to pigments which absorb specific colours but are due to interference of light waves that are reflected by different layers.

Interference due to a thin film:

The brilliant colours of soap bubbles and thin films on the surface of water are due to the interference of light waves reflected from the upper and lower surfaces of the film. The two rays have a path difference which depends on the point on the film that is being viewed. This is shown in above figure.

The incident wave gets partially reflected from upper surface as shown by ray AE. The rest of the light gets refracted and travels along AB. At B it again gets partially reflected and travels along BC. At C it refracts into air and travels along CF. The parallel rays AE and CF have a phase difference due to their different path lengths in different media. As can be seen from the figure, the phase difference depends on the angle of incidence θ₁, i.e., the angle of incidence at the top surface which is the angle of viewing, and also on the wavelength of the light as the refractive index of the material of the thin film depends on it. The two waves propagating along AE and CF interfere producing maxima and minima for different colours at different angles of viewing. One sees different colours when the film is viewed at different angles.

As the reflection is from the denser boundary, there is an additional phase difference of π radians (or an
additional path difference λ). This should be taken into account for mathematical analysis.

iv) In Young’s double slit experiment what will we observe on the screen when white light is incident on the slits but one slit is covered with a red filter and the other with a violet filter? Give reasons for your answer.
Answer:
In Young’s double-slit experiment, when white light is incident on the slits and one of the slit is covered with a red filter, the light passing through this slit will emerge as the light having red colour. The other slit which is covered with a violet filter, will give light having violet colour as emergent light. The interference fringes will involve mixing of red and violet colours. At some points, fringes will be red if constructive interference occurs for red colour and destructive interference occurs for violet colour. At some points, fringes will be violet if constructive interference occurs for violet colour and destructive interference occurs for red colour. The central fringe will be bright with mixing of red and violet colours.

v) Explain what is optical path length. How is it different from actual path length?
Answer:
Consider, a light wave of angular frequency ω and wave vector k travelling through vacuum along the x-direction. The phase of this wave is (kx-ωt). The speed of light in vacuum is c and that in medium is v.
k = \(\frac{2 \pi}{\lambda}\) = \(\frac{2 \pi v}{v \lambda}\) = \(\frac{\omega}{v}\) as ω = 2πv and v = vλ, where v is the frequency of light.

If the wave travels a distance ∆x, its phase changes by ∆φ = k∆x = ω∆x/v.
Similarly, if the wave is travelling in vacuum, k = ω/c and ∆φ = ω∆x/c
Now, consider a wave travelling a distance ∆x in the medium, the phase difference generated is,
∆φ’ = k’∆x = ωn∆x/c = ω∆x’/c … (1)
where ∆x’ = n∆x … (2)
The distance n∆x is called the optical path length of the light in the medium; it is the distance the light would have travelled in the same time t in vacuum (with the speed c).

The optical path length in a medium is the corresonding path in vacuum that the light travels in the same time as it takes in the given medium.

Thus, a distance d travelled in a medium of refractive index n introduces a path difference = nd – d = d(n – 1) over a ray travelling equal distance through vacuum.

Question 3.
Derive the laws of reflection of light using Huygens’ principle.
Answer:
Consider a plane wavefront AB of monochromatic light propagating in the direction A’A incident obliquely at an angle i on a plane refracting surface MN. This plane refracting surface MN separates two uniform and optically transparent mediums.
Let v₁ and v₂ be the speeds of light in medium 1 (say, a rarer medium) and medium 2 (a denser medium) respectively.

When the wavefront reaches MN at point A at t = 0, A becomes a secondary source and emits secondary waves in the second medium, while ray B’B reaches the surface MN at C at time t = T. Thus, BC = v₁T. During the time T, the secondary wavelet originating at A covers a distance AE in the denser medium with radius v₂T.

As all the points on CE are in the same phase of wave motion, CE represents the refracted wavefront in the denser medium. CE is the tangent to the secondary wavelet starting from A. It is also a common tangent to all the secondary wavelets emitted by points between A and C. PP’ is the normal to the boundary at A.
∠A’AP = ∠BAC = the angle of incidence (i) and ∠P’AE = ∠ACE = the angle of refraction (r).
From ∆ABC and ∆AEC,

By definition, the refractive index of medium 2 with respect to medium 1,

Here, n₁ and n₂ are the absolute refractive indices of medium 1 and medium 2 respectively. Eq. (1) is Snell’s law of refraction. Also, it can be seen from the figure, that the incident ray and the refracted ray lie on the opposite sides of the normal and all three of them lie in the same plane.

Thus, the laws of refraction of light can be deduced by Huygens’ construction of a plane wavefront.
If v₁ > v₂, i.e. n₁ < n₂, then r < i (bending of the refracted ray towards the normal).
[Notes :
(1) Quite often, the terms vacuum and free space are used in the same sense. Absolute vacuum or a perfect vacuum-a region of space devoid of material, particles – does not exist. The term vacuum is also used to mean a region of space occupied by a gas at very low pressure. Free space means a region of space devoid of matter and fields. Its refractive index is 1 (by definition). Its temperature is 0 K. ε₀ and μ₀ are defined for free space. The refractive index of a medium with respect to air is very close to the absolute refractive index of the medium as the speed of light in air is very close to that in free space.
(2) There is no lateral inversion in refraction.
(3) There is no bending of light when the angle of incidence is zero (normal incidence), r = 0 for i = 0.]

Question 4.
Derive the laws of refraction of light using Huygens’ principle.
Answer:
Consider a plane wavefront AB of monochromatic light propagating in the direction A’A incident obliquely at an angle i on a plane refracting surface MN. This plane refracting surface MN separates two uniform and optically transparent mediums.
Let v₁ and v₂ be the speeds of light in medium 1 (say, a rarer medium) and medium 2 (a denser
medium) respectively.

When the wavefront reaches MN at point A at t = 0, A becomes a secondary source and emits secondary waves in the second medium, while ray B’B reaches the surface MN at C at time t = T. Thus, BC = v₁T. During the time T, the secondary wavelet originating at A covers a distance AE in the denser medium with radius v₂T.

As all the points on CE are in the same phase of wave motion, CE represents the refracted wavefront in the denser medium. CE is the tangent to the secondary wavelet starting from A. It is also a common tangent to all the secondary wavelets emitted by points between A and C. PP’ is the normal to the boundary at A.
∠A’AP = ∠BAC = the angle of incidence (i) and ∠P’AE = ∠ACE = the angle of refraction (r).
From ∆ABC and ∆AEC,

By definition, the refractive index of medium 2 with respect to medium 1,

Here, n₁ and n₂ are the absolute refractive indices of medium 1 and medium 2 respectively. Eq. (1) is Snell’s law of refraction. Also, it can be seen from the figure, that the incident ray and the refracted ray lie on the opposite sides of the normal and all three of them lie in the same plane.

Thus, the laws of refraction of light can be deduced by Huygens’ construction of a plane wavefront.
If v₁ > v₂, i.e. n₁ < n₂, then r < i (bending of the refracted ray towards the normal).
[Notes :

1. Quite often, the terms vacuum and free space are used in the same sense. Absolute vacuum or a perfect vacuum-a region of space devoid of material, particles – does not exist. The term vacuum is also used to mean a region of space occupied by a gas at very low pressure. Free space means a region of space devoid of matter and fields. Its refractive index is 1 (by definition). Its temperature is 0 K. ε₀ and μ₀ are defined for free space. The refractive index of a medium with respect to air is very close to the absolute refractive index of the medium as the speed of light in air is very close to that in free space.
2. There is no lateral inversion in refraction.
3. There is no bending of light when the angle of incidence is zero (normal incidence), r = 0 for i = 0.]

Question 5.
Explain what is meant by polarization and derive Malus’ law.
Answer:
According to the electromagnetic theory of light, a light wave consists of electric and magnetic fields vibrating at right angles to each other and to the direction of propagation of the wave. If the vibrations of \(\vec{E}\) in a light wave are in all directions perpendicular to the direction of propagation of light, the wave is said to be unpolarized.

If the vibrations of the electric field \(\vec{E}\) in a light wave are confined to a single plane containing the direction of propagation of the wave so that its electric field is restricted along one particular direction at right angles to the direction of propagation of the wave, the wave is said to be plane-polarized or linearly polarized.

This phenomenon of restricting the vibrations of light, i.e., of the electric field vector in a particular direction, which is perpendicular to the direction of the propagation of the wave is called polarization of light.

Consider an unpolarized light wave travelling along the x-direction. Let c, v and λ be the speed, frequency and wavelength, respectively, of the wave. The magnitude of its electric field (\(\vec{E}\)) is,
E = E₀ sin (kx – ωt), where E₀ = E_max = amplitude of the wave, ω = 2πv = angular frequency of the wave and k = \(\frac{2 \pi}{\lambda}\) = magnitude of the wave vector or propagation vector.

The intensity of the wave is proportional to \(\left|E_{0}\right|^{2}\). The direction of the electric field can be anywhere in the y-z plane. This wave is passed through two identical polarizers as shown in below figure.

When a wave with its electric field inclined at an angle φ to the axis of the first polarizer is passed through the polarizer, the component E₀ cos φ will pass through it. The other component E₀ sin φ which is perpendicular to it will be blocked.

Now, after passing through this polarizer, the intensity of this wave will be proportional to the square of its amplitude, i.e., proportional to |E₀ cos φ|².

The intensity of the plane-polarized wave emerging from the first polarizer can be obtained by averaging | E₀ cos φ|² over all values of φ between 0 and 180°. The intensity of the wave will be
proportional to \(\frac{1}{2}\)|E₀|² as the average value of cos² φ over this range is \(\frac{1}{2}\). Thus the intensity of an unpolarized wave reduces by half after passing through a polarizer.

When the plane-polarized wave emerges from the first polarizer, let us assume that its electric field (\(\overrightarrow{E_{1}}\)) is along the y-direction. Thus, this electric field is,
\(\overrightarrow{E_{1}}\) = \(\hat{\mathrm{j}}\) E₁₀ sin (kx – ωt) …. (1)
where, E₁₀ is the amplitude of this polarized wave. The intensity of the polarized wave,
I₁ ∝ |E₁₀|² …(2)

Now this wave passes through the second polarizer whose polarization axis (transmission axis) makes an angle θ with the y-direction. This allows only the component E₁₀ cos θ to pass through it. Thus, the amplitude of the wave which passes through the second polarizer is E₂₀ = E₁₀ cos θ and its intensity,
I₂ ∝ | E₂₀|²
∴ I₂ ∝ | E₁₀|² cos²θ
∴ I₂ = I₁ cos₂θ … (3)

Thus, when plane-polarized light of intensity I₁ is incident on the second identical polarizer, the intensity of light transmitted by the second polarizer varies as cos²θ, i.e., I₂ = I₁ cos²θ, where θ is the angle between the transmission axes of the two polarizers. This is known as Malus’ law

[Note : Etienne Louis Malus (1775-1812), French military engineer and physicist, discovered in 1809 that light can be polarized by reflection. He was the first to use the word polarization, but his arguments were based on Newton’s corpuscular theory.]

Question 6.
What is Brewster’s law? Derive the formula for Brewster angle.
Answer:
Brewster’s law : The tangent of the polarizing angle is equal to the refractive index of the reflecting medium with respect to the surrounding (₁n₂). If θ_B is the polarizing angle,
tan θ_B = ₁n₂ = \(\frac{n_{2}}{n_{1}}\)
Here n₁ is the absolute refractive index of the surrounding and n₂ is that of the reflecting medium.
The angle θ_B is called the Brewster angle.

Consider a ray of unpolarized monochromatic light incident at an angle θ_B on a boundary between two transparent media as shown in below figure. Medium 1 is a rarer medium with refractive index n₁ and medium 2 is a denser medium with refractive index n₂. Part of incident light gets refracted and the rest

gets reflected. The degree of polarization of the reflected ray varies with the angle of incidence.
The electric field of the incident wave is in the plane perpendicular to the direction of propagation of incident light. This electric field can be resolved into a component parallel to the plane of the paper, shown by double arrows, and a component perpendicular to the plane of the paper shown by dots, both having equal magnitude. Generally, the reflected and refracted rays are partially polarized, i.e., the two components do not have equal magnitude.

In 1812, Sir David Brewster discovered that for a particular angle of incidence. θ_B, the reflected wave is completely plane-polarized with its electric field perpendicular to the plane of the paper while the refracted wave is partially polarized. This particular angle of incidence (θ_B) is called the Brewster angle.

For this angle of incidence, the refracted and reflected rays are perpendicular to each other.
For angle of refraction θ_r,
θ_B + θ_r = 90° …… (1)
From Snell’s law of refraction,
∴ n₁ sin θ_B = n₂ sin θ_r … (2)
From Eqs. (1) and (2), we have,
n₁ sin θ_B = n₂ sin (90° – θ_B) = n₂ cos θ_B

This is called Brewster’s law.

Question 7.
Describe Young’s double slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.
Answer:
Description of Young’s double-slit interference experiment:

1. A plane wavefront is obtained by placing a linear source S of monochromatic light at the focus of a convex lens. It is then made to pass through an opaque screen AB having two narrow and similar slits S₁ and S₂. S₁ and S₂ are equidistant from S so that the wavefronts starting simultaneously from S and reaching S₁ and S₂ at the same time are in phase. A screen PQ is placed at some distance from screen AB as shown in below figure
   
2. S₁ and S₂ act as secondary sources. The crests/-troughs of the secondary wavelets superpose and interfere constructively along straight lines joining the black dots shown in above figure. The point where these lines meet the screen have high intensity and are bright.
3. Similarly, there are points shown with red dots where the crest of one wave coincides with the trough of the other. The corresponding points on the screen are dark due to destructive interference.
4. These dark and bright regions are called fringes or bands and the whole pattern is called interference pattern.

Conditions for occurence of dark and bright fringes on the screen :

Consider Young’s double-slit experimental set up. Two narrow coherent light sources are obtained by wavefront splitting as monochromatic light of wavelength λ emerges out of two narrow and closely spaced, parallel slits S₁ and S₂ of equal widths. The separation S₁S₂ = d is very small. The interference pattern is observed on a screen placed parallel to the plane of and at considerable distance D (D » d) from the slits. OO’ is the perpendicular bisector of segment S₁S₂.

Consider, a point P on the screen at a distance y from O’ (y « D). The two light waves from S₁ and S₂ reach P along paths S₁P and S₂P, respectively. If the path difference (∆l) between S₁P and S₂P is an integral multiple of λ, the two waves arriving there will interfere constructively producing a bright fringe at P. On the contrary, if the path difference between S₁P and S₂P is half integral multiple of λ, there will be destructive interference and a dark fringe will be produced at P.
From above figure,
(S₂P)² = (S₂S₂)² + (PS₂‘)²
= (S₂S₂‘)² + (PO’ + O’S₂‘)²

Expression for the fringe width (or band width) : The distance between consecutive bright (or dark) fringes is called the fringe width (or band width) W. Point P will be bright (maximum intensity), if the path difference, ∆l = y_n\(\frac{d}{D}\) = nλ where n = 0,1, 2, 3…, Point P will be dark (minimum intensity equal to zero), if y_m\(\frac{d}{D}\) = (2m – 1)\(\frac{\lambda}{2}\), where, m = 1, 2, 3…,
Thus, for bright fringes (or bands),

These conditions show that the bright and dark fringes (or bands) occur alternately and are equally – spaced. For Point O’, the path difference (S₂O’ – S₁O’) = 0. Hence, point O’ will be bright. It corresponds to the centre of the central bright fringe (or band). On both sides of O’, the interference pattern consists of alternate dark and bright fringes (or band) parallel to the slit.

Let y_n and y_{n + 1}, be the distances of the nth and (n + 1)^th bright fringes from the central bright fringe.

Alternately, let y_m and y_{m + 1} be the distances of the m th and (m + 1)^th dark fringes respectively from the central bright fringe.

Eqs. (7) and (11) show that the fringe width is the same for bright and dark fringes.

[Note : In the first approximation, the path difference is d sin θ.]

Question 8.
What are the conditions for obtaining good interference pattern? Give reasons.
Answer:
The conditions necessary for obtaining well defined and steady interference pattern :

1. The two sources of light should be coherent:
   The two sources must maintain their phase relation during the time required for observation. If the phases and phase difference vary with time, the positions of maxima and minima will also change with time and consequently the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed. For coherence, the two secondary sources must be derived from a single original source.
2. The light should be monochromatic :
   Otherwise, interference will result in complex coloured bands (fringes) because the separation of successive bright bands (fringes) is different for different colours. It also may produce overlapping bands.
3. The two light sources should be of equal brightness, i.e., the waves must have the same amplitude.
   The interfering light waves should have the same amplitude. Then, the points where the waves meet in opposite phase will be completely dark (zero intensity). This will increase the contrast of the interference pattern and make it more distinct.
4. The two light sources should be narrow :
   If the source apertures are wide in comparison with the light wavelength, each source will be equivalent to multiple narrow sources and the superimposed pattern will consist of bright and less bright fringes. That is, the interference pattern will not be well defined.
5. The interfering light waves should be in the same state of polarization :
   Otherwise, the points where the waves meet in opposite phase will not be completely dark and the interference pattern will not be distinct.
6. The two light sources should be closely spaced and the distance between the screen and the sources should be large : Both these conditions are desirable for appreciable fringe separation. The separation of successive bright or dark fringes is inversely proportional to the closeness of the slits and directly proportional to the screen distance.

Question 9.
What is meant by coherent sources? What are the two methods for obtaining coherent sources in the laboratory?
Answer:
Coherent sources : Two sources of light are said to be coherent if the phase difference between the emitted waves remains constant.

It is not possible to observe interference pattern with light from any two different sources. This is because, no observable interference phenomenon occurs by superposing light from two different sources. This happens due to the fact that different sources emit waves of different frequencies. Even if the two sources emit light of the same frequency, the phase difference between the wave trains from them fluctuates randomly and rapidly, i.e., they are not coherent.
Consequently, the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed.

In the laboratory, coherent sources can be obtained by using
(1) Lloyd’s mirror and
(2) Fresnel’s biprism.

(1) Lloyd’s mirror : A plane polished mirror is kept at some distance from the source of monochromatic light and light is made incident on the mirror at a grazing angle.

Some light falls directly on the screen as shown by the black lines in above figure, while some light falls on the screen after reflection from the mirror as shown by red lines. The reflected light appears to come from a virtual source and thus two sources can be obtained. These two sources are coherent as they are derived from a single source. Superposition of the waves coming from these coherent sources, under appropriate conditions, gives rise to interference pattern consisting of alternate bright and dark bands on the screen as shown in the figure.

(2) Fresnel’s biprism : It is a single prism having an obtuse angle of about 178° and the other two angles of about 1° each. The biprism can be considered as made of two thin prisms of very small refracting angle of about 1°. The source, in the form of an illuminated narrow slit, is aligned parallel to the refracting edge of the biprism. Monochromatic light from the source is made to pass through that narrow slit and fall on the biprism.

Two virtual images S₁ and S₂ are formed by the two halves of the biprism. These are coherent sources which are obtained from a single secondary source S. The two waves coming from S₁ and S₂ interfere under appropriate conditions and form interference fringes, like those obtained in Young’s double-slit experiment, as shown in the figure in the shaded region. The formula for y is the same as in Young’s experiment.

Question 10.
What is diffraction of light? How does it differ from interference? What are Fraunhoffer and Fresnel diffractions?
Answer:
1. Phenomenon of diffraction of light: When light passes by the edge of an obstacle or through a small opening or a narrow slit and falls on a screen, the principle of rectilinear propagation of light from geometrical optics predicts a sharp shadow. However, it is found that some of the light deviates from its rectilinear path and penetrates into the region of the geometrical shadow. This is a general characteristic of wave phenomena, which occurs whenever a portion of the wavefront is obstructed in some way. This bending of light waves at an edge into the region of geometrical shadow is called diffraction of light.

2. Differences between interference and diffraction :

1. The term interference is used to characterise the superposition of a few coherent waves (say, two). But when the superposition at a point involves a large number of waves coming from different parts of the same wavefront, the effect is referred to as diffraction.
2. Double-slit interference fringes are all of equal width. In single-slit diffraction pattern, only the non-central maxima are of equal width which is half of that of the central maximum.
3. In double-slit interference, the bright and dark fringes are equally spaced. In diffraction, only the non-central maxima lie approximately halfway between the minima.
4. In double-slit interference, bright fringes are of equal intensity. In diffraction, successive non-central maxima decrease rapidly in intensity.

[Note : Interference and diffraction both have their origin in the principle of superposition of waves. There is no physical difference between them. It is just a question of usage. When there are only a few sources, say two, the phenomenon is usually called interference. But, if there is a large number of sources the word diffraction is used.]

3. Diffraction can be classified into two types depending on the distances involved in the experimental setup :

(A) Fraunhofer diffraction : In this class of diffraction, both the source and the screen are at infinite distances from the aperture. This is achieved by placing the source at the focus of a convex lens and the screen at the focal plane of another convex lens.

(B) Fresnel diffraction : In this class of diffraction, either the source of light or the screen or both are at finite distances from the diffracting aperture. The incident wavefront is either cylindrical or spherical depending on the source. A lens is not needed to observe the diffraction pattern on the screen.

Question 11.
Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.
Answer:
When a parallel beam of monochromatic light of wavelength λ illuminates a single slit of finite width a, we observe on a screen some distance from the slit, a broad pattern of alternate dark and bright fringes. The pattern consists of a central bright fringe, with successive dark and bright fringes of diminishing intensity on both sides. This is called ‘ the diffraction pattern of a single slit.

Consider a single slit illuminated with a parallel beam of monochromatic light perpendicular to the plane of the slit. The diffraction pattern is obtained on a screen at a distance D (» a) from the slit and at the focal plane of the convex lens, Fig. 7.33.

We can imagine the single slit as being made up of a large number of Huygens’ sources evenly distributed over the width of the slit. Then the maxima and minima of the pattern arise from the interference of the various Huygens’ wavelets.

Now, imagine the single slit as made up of two adjacent slits, each of width a/2. Since, the incident plane wavefronts are parallel to the plane of the slit, all the Huygens sources at the slit will be in phase. They will therefore also in phase at the point P₀ on the screen, where P₀ is equidistant from all the Huygens sources. At P₀, then, we get the central maximum.

For the first minimum of intensity on the screen, the path difference between the waves from the Huygens sources A and O (or O and B) is λ/2, which is the condition for destructive interference. Suppose, the nodal line OP for the first minimum subtends an angle θ at the slit; θ is very small. With P as the centre and PA as radius, strike an arc intersecting PB at C. Since, D » a, the arc AC can be considered a straight line at right angles to PB. Then, ∆ABC is a right-angled triangle similar to ∆OP₀P.
This means that, ∠BAC = θ
∴ BC = a sinθ
∴ Difference in path length,
BC = PB – PA = (PB – PO) + (PO – PA)

(∵ θ is very small and in radian)
The other nodal lines of intensity minima can be understood in a similar way. In general, then, for the with minimum (m = ±1, ±2, ±3, …).
θ_m = \(\frac{m \lambda}{a}\) (mth minimum) … (2)
as θ_m is very small and in radian.
Between the successive minima, the intensity rises to secondary maxima when the path difference is an odd-integral multiple of \(\frac{\lambda}{2}\) :
a sin θ_m = (2m + 1)\(\frac{\lambda}{2}\) = (m + \(\frac{1}{2}\))λ
i.e., at angles given by,
θ_m \(\simeq\) sin θ_m = (m + \(\frac{1}{2}\))\(\frac{\lambda}{a}\)
(with secondary maximum) … (3)

Question 12.
Describe what is Rayleigh’s criterion for resolution. Explain it for a telescope and a microscope.
Answer:
Rayleigh’s criterion for minimum resolution : Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.

The ‘sharpness’ of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the first minimum. It gives the limit of resolution.

Two overlapping diffraction patterns due to two point sources are not resolved if the angular separation between the central peaks is less than the limit of resolution. They are said to be just separate, or resolved, if the angular separation between the central peaks is equal to the limit of resolution. They are said to be well resolved if the angular separation between the central peaks is more than the limit of resolution.

Resolving power of an optical instrument:
The primary aim of using an optical instrument is to see fine details, whether observing a star system through a telescope or a living cell through a microscope. After passing through an optical system, light from two adjacent parts of the object should produce sharp, distinct (separate) images of those parts. The objective lens or mirror of a telescope or microscope acts like a circular aperture. The diffraction pattern of a circular aperture consists of a central bright spot (called the Airy disc and corresponds to the central maximum) and concentric dark and bright rings.

Light from two close objects or parts of an object after passing through the aperture of an optical system produces overlapping diffraction patterns that tend to obscure the image. If these diffraction patterns are so broad that their central maxima overlap substantially, it is difficult to decide if the intensity distribution is produced by two separate objects or by one.

The resolving power of an optical instrument, e.g., a telescope or microscope, is a measure of its ability to produce detectably separate images of objects that are close together.

Definition : The smallest linear or angular separation between two point objects which appear just resolved when viewed through an optical instrument is called the limit of resolution of the instrument and its reciprocal is called the resolving power of the instrument.

Rayleigh’s criterion for minimum resolution : Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.

The ‘sharpness’ of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the first minimum. It gives the limit of resolution.

Two overlapping diffraction patterns due to two point sources are not resolved if the angular separation between the central peaks is less than the limit of resolution. They are said to be just separate, or resolved, if the angular separation between the central peaks is equal to the limit of resolution. They are said to be well resolved if the angular separation between the central peaks is more than the limit of resolution.

The resolving power of a telescope is defined as the reciprocal of the angular limit of resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope.

Consider two stars seen through a telescope. The diameter (D) of the objective lens or mirror corresponds to the diffracting aperture. For a distant point source, the first diffraction minimum is at an angle θ away from the centre such that

Dsinθ = 1.22 λ
where λ is the wavelength of light. The angle θ is usually so small that we can substitute sin θ \(\approx\) θ (θ in radian). Thus, the Airy disc for each star will be spread out over an angular half-width θ = 1.22 λ/D about its geometrical image point. The radius of the Airy disc at the focal plane of the objective lens is r = fθ = 1.22fλ/D, where / is the focal length of the objective.

When observing two closely-spaced stars, the Rayleigh criterion for just resolving the images as that of two point sources (instead of one) is met when the centre of one Airy disc falls on the first minimum of the other pattern. Thus, the angular limit (or angular separation) of resolution is
θ = \(\frac{1.22 \lambda}{D}\) …. (1)
and the linear separation between the images at the focal plane of the objective lens is
y = fθ …(2)
∴ Resolving power of a telescope,
R = \(\frac{1}{\theta}\) = \(\frac{D}{1.22 \lambda}\) … (3)
It depends

1. directly on the diameter of the objective lens or mirror,
2. inversely on the wavelength of the radiation.

Question 13.
Whitelight consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55? [Ans: 258.1 – 451.6 nm]
Answer:
Let λ₁ and λ₂ be the wavelengths of light in water for 400 nm and 700 nm (wavelengths in vacuum) respectively. Let λ_a be the wavelength of light in vacuum.

The wavelength range seen when white light is passed through the glass would be 258.06 nm to 451.61 nm.

Question 14.
The optical path of a ray of light of a given wavelength travelling a distance of 3 cm in flint glass having refractive index 1.6 is same as that on travelling a distance x cm through a medium having refractive index 1.25. Determine the value of x.
[Ans: 3.84 cm]
Answer:
Let d_fg and d_m be the distances by the ray of light in the flint glass and the medium respectively. Also, let n_fg and n_m be the refractive indices of the flint glass and the medium respectively.
Data : d_fg = 3 cm, n_fg = 1.6, nm = 1.25,
Optical path = n_m × d_m = n_fg × d_fg

Thus, x cm = 3.84 cm
∴ x = 3.84
This is the value of x.

Question 15.
A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.20° apart. What
is the angular fringe separation if the entire arrangement is immersed in water (n = 1.33)? [Ans: 0.15°]
Answer:
Data : θ₁ = 0.20°, n_w = 1.33
In the first approximation,
D sin θ₁ = y₁ and D sin θ₂ = y₂

∴ θ₂ = sin^-10.026
= 9′ = 0.15°
This is the required angular fringe separation.
OR
In the first approximation,

Question 16.
In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits.
(a) What is the angular separation in radians between the central maximum and an adjacent maximum?
(b) What is the distance between these maxima on a screen 50.0 cm from the slits?
[Ans: 0.01 rad, 0.5 cm]
Answer:
Data : d = 100λ, D = 50.0 cm
(a) The condition for maximum intensity in Young’s experiment is, d sin θ = nλ, n = 0, 1, 2 …,
The angle between the central maximum and its adjacent maximum can be determined by setting n equal to 1,
∴ d sin θ = λ

(b) The distance between these maxima on the screen is D sin θ = D\(\left(\frac{\lambda}{d}\right)\)
= (50.0 cm)\(\left(\frac{\lambda}{100 \lambda}\right)\)
= 0.50 cm

Question 17.
Unpolarized light with intensity I₀ is incident on two polaroids. The axis of the first polaroid makes an angle of 50° with the vertical, and the axis of the second polaroid is horizontal. What is the intensity of the light after it has passed through the second polaroid? [Ans: I₀/2 × (cos 40°)²]
Answer:
According to Malus’ law, when the unpolarized light with intensity I₀ is incident on the first polarizer, the polarizer polarizes this incident light. The intensity of light becomes I₁ = I₀/2.
Now, I₂ = I₁ cos²θ
∴ I₂ = \(\left(\frac{I_{0}}{2}\right)\) cos²θ
Also, the angle θ between the axes of the two polarizers is θ₂ – θ₁.

The intensity of light after it has passed through the second polaroid = \(\left(\frac{\boldsymbol{I}_{0}}{\mathbf{2}}\right)\)cos²40° = \(\frac{I_{0}}{2}\)(0.7660)²
= 0.2934 I₀

Question 18.
In a biprism experiment, the fringes are observed in the focal plane of the eyepiece at a distance of 1.2 m from the slits. The distance between the central bright band and the 20^th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eyepiece, 90 cm from the eyepiece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used. [Ans: 5000 Å]
Answer:
Data : D = 1.2 m
The distance between the central bright band and the 20^th bright band is 0.4 cm.
∴ y₂₀ = 0.4 cm = 0.4 × 10^-2 m
W = \(\frac{y_{20}}{20}\) = \(\frac{0.4}{20}\) × 10^-2 m = 2 × 10^-4 m,
d₁ = 0.9 cm = 0.9 × 10^-2m, v₁ = 90 cm = 0.9 m
∴ u₁ = D – v₁ = 1.2m – 0.9m = 0.3 m

Question 19.
In Fraunhoffer diffraction by a narrow slit, a screen is placed at a distance of 2 m from the lens to obtain the diffraction pattern. If the slit width is 0.2 mm and the first minimum is 5 mm on either side of the central maximum, find the wavelength of light. [Ans: 5000 Å]
Answer:
Data : D = 2 m, y_1d = 5 mm = 5 × 10^-3 m, a = 0.2 mm = 0.2 × 10^-3 m = 2 × 10^-4 m

This is the wavelength of light.

Question 20.
The intensity of the light coming from one of the slits in Young’s experiment is twice the intensity of the light coming from the other slit. What will be the approximate ratio of the intensities of the bright and dark fringes in the resulting interference pattern? [Ans: 34]
Answer:
Data : I₁ : I₂ = 2 : 1
If E₁₀ and E₂₀ are the amplitudes of the interfering waves, the ratio of the maximum intensity to the minimum intensity in the fringe system is


The ratio of the intensities of the bright and dark fringes in the resulting interference pattern is 34 : 1.

Question 21.
A parallel beam of green light of wavelength 550 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen which is 40 cm away. What is the distance between the two first order minima? [Ans: 1.1 mm]
Answer:
Data : λ = 550 nm = 546 × 10^-9 m, a = 0.4 mm = 4 × 10^-4 m, D = 40 cm = 40 × 10^-2 m
y_md = m\(\frac{\lambda D}{a}\)
∴ y_1d = 1\(\frac{\lambda D}{a}\) and
2y_1d = \(\frac{2 \lambda D}{a}\)
= \(\frac{2 \times 550 \times 10^{-9} \times 40 \times 10^{-2}}{4 \times 10^{-4}}\)
= 2 × 550 × 10^-6 = 1092 × 10^-6
= 1.100 × 10^-3m = 1.100 mm
This is the distance between the two first order minima.

Question 22.
What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 45.0°? [Ans: 1.274]
Answer:
Data : θ = 45°, m = 1
a sin θ = mλ for (m = 1, 2, 3… minima)
Here, m = 1 (First minimum)
∴ a sin 45° = (1) λ
∴ \(\frac{a}{\lambda}\) = \(\frac{1}{\sin 45^{\circ}}\) = 1.414
This is the required ratio.

Question 23.
Monochromatic electromagnetic radiation from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width if the wavelength is
(a) 500 nm (visible light);
(b) 50 µm (infrared radiation);
(c) 0.500 nm (X-rays)?
[Ans: 0.4167 mm, 41.67 mm, 4.167 × 10^-4 mm]
Answer:
Data:2W = 6mm ∴ W= 3 mm = 3 × 10^-3 m, y = 2.5 m,
(a) λ₁ = 500 nm = 5 × 10^-7 m
(b) λ₂ = 50 μm = 5 × 10^-5 m
(c) λ₃ = 0.500 nm = 5 × 10^-10 m
Let a be the slit width.

Question 24.
A star is emitting light at the wavelength of 5000 Å. Determine the limit of resolution of a telescope having an objective of diameter of 200 inch. [Ans: 1.2 × 10^-7 rad]
Answer:
Data : λ = 5000 Å = 5 × 10^-7 m,
D = 200 × 2.54 cm = 5.08 m
θ = \(\frac{1.22 \lambda}{D}\)
= \(\frac{1.22 \times 5 \times 10^{-7}}{5.08}\)
= 1.2 × 10^-7 rad
This is the required quantity.

Question 25.
The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm by how much will the distance change if light of wavelength 4800 Å is used? [Ans: 0.064 mm]
Answer:
Data : λ₁ = 6000 Å = 6 × 10^-7m, λ₂ = 4800 Å = 4.8 × 10^-7m, W₁ = 0.32 mm = 3.2 × 10^-4m
Distance between consecutive bright fringes,

∴ ∆W = W₁ – W₂
= 3.2 × 10^-4m – 2.56 × 10^-4m
= 0.64 × 10^-4m
= 6.4 × 10^-5
= 0.064 mm
This is the required change in distance.

12th Physics Digest Chapter 7 Wave Optics Intext Questions and Answers

Use your brain power (Textbook Page No. 167)

What will you observe if

Question 1.
you look at a source of unpolarized light through a polarizer ?
Answer:
When a source of unpolarized light is viewed through a polarizer, the intensity of the light transmitted by the polarizer is reduced and hence the source appears less bright.

Question 2.
you look at the source through two polarizers and rotate one of them around the path of light for one full rotation?
Answer:
When a source of unpolarized light is viewed through two polarizers and the second polarizer is rotated gradually, the intensity of the light transmitted by the second polarizer goes on decreasing. When the axes of polarization of the two polarizers are at 90° to each other, light almost disappears depending on the quality of the polarizers. (Ideally the intensity of the transmitted light should be zero.) The light reappears, i.e., its intensity increases, when the second polarizer is rotated further, and the intensity of the light becomes maximum when the axes of polarization are parallel again.

Question 3.
instead of rotating only one of the polaroids, you rotate both polaroids simultaneously in the same direction?
Answer:
If both the polaroids are rotated simultaneously in the same direction with the same angular velocity, then there would be no change in the intensity of the transmitted light observed.

Can you tell? (Textbook Page No. 168)

Question 1.
If you look at the sky in a particular direction through a polaroid and rotate the polaroid around that direction what will you see ?
Answer:
As the scattered light is polarized, the sky appears bright and dim alternately.

Question 2.
Why does the sky appear to be blue while the clouds appear white ?
Answer:
The blue colour of the sky is because of the scattering of light by air molecules and dust particles in the atmosphere. As the wavelength of blue light is less than that of red light, the blue light is preferentially scattered than the light corresponding to other colours in the visible region. Clouds are seen due to scattering of light from lower parts of the atmosphere. The clouds contain the dust particles and water droplets which are sometimes large enough to scatter light of all the wavelengths such that the combined effect makes the clouds appear white.

Remember this (Textbook Page No. 171)

Question 1.
For the interference pattern to be clearly visible on the screen, the distance (D) between the slits and the screen should be much larger than the distance (d) between the two slits (S₁ and S₂), i.e., D » d.
Answer:
The condition for constructive interference at P is,
∆l = y_n\(\frac{d}{D}\) = nλ …. (1)
y_n being the position (y-coordinate) of nth bright fringe (n = 0, ±1, ±2, …).
∴ y_n = nλ\(\frac{D}{d}\) ….. (2)
Similarly, the position of mth (m = +1, ±2,…) dark fringe (destructive interference) is given by,
∆l = y_m\(\frac{d}{D}\) = (2m – 1)λ giving
y_m = (2m – 1)λ\(\frac{D}{d}\) …(3)
The distance between any two consecutive bright or dark fringes, i.e., the fringe width
= W = ∆y = y_{n + 1} – y_n = λ\(\frac{D}{d}\) …(4)
Conditions given by Eqs. (1) to (4) and hence the location of the fringes are derived assuming that the two sources S₁ and S₂ are in phase. If there is a non-zero phase difference between them it should be added appropriately. This will shift the entire fringe pattern but will not change the fringe width.

Do you know (Textbook Page No. 172 & 173)

Several phenomena that we come across in our day to day life are caused by interference and diffraction of light. These are the vigorous colours of soap bubbles as well as those seen in a thin oil film on the surface of water, the bright colours of butterflies and peacocks etc. Most of these colours are not due to pigments which absorb specific colours but are due to interference of light waves that are reflected by different layers.

Interference due to a thin film :
The brilliant colours of soap bubbles and thin films on the surface of water are due to the interference of light waves reflected from the upper and lower surfaces of the film. The two rays have a path difference which depends on the point on the film that is being viewed. This is shown in above figure.

The incident wave gets partially reflected from upper surface as shown by ray AE. The rest of the light gets refracted and travels along AB. At B it again gets partially reflected and travels along BC. At C it refracts into air and travels along CF. The parallel rays AE and CF have a phase difference due to their different path lengths in different media. As can be seen from the figure, the phase difference depends on the angle of incidence θ₁ i.e., the angle of incidence at the top surface which is the angle of viewing, and also on the wavelength of the light as the refractive index of the material of the thin film depends on it. The two waves propagating along AE and CF interfere producing maxima and minima for different colours at different angles of viewing. One sees different colours when the film is viewed at different angles.

As the reflection is from the denser boundary, there is an additional phase difference of π radians (or an additional path difference λ). This should be taken into account for mathematical analysis.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 6 Superposition of Waves Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 6 Superposition of Waves

1. Choose the correct option.

i) When an air column in a pipe closed at one end vibrates such that three nodes are formed in it, the frequency of its
vibrations are…….times the fundamental frequency.
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(D) 5

ii) If two open organ pipes of length 50 cm and 51 cm sounded together to produce 7 beats per second, the speed of sound is.
(A) 307 m/s
(B) 327m/s
(C) 350m/s
(D) 357m/s
Answer:
(D) 357m/s

iii) The tension in a piano wire is increased by 25%. Its frequency becomes ….. times the original frequency.
(A) 0.8
(B) 1.12
(C) 1.25
(D) 1.56
Answer:
(B) 1.12

iv) Which of the following equations represents a wave travelling along the y-axis?
(A) x = A sin(ky – ωt)
(B) y = A sin(kx – ωt)
(C) y = A sin(ky) cos(ωt)
(D) y = A cos(ky)sin(ωt)
Answer:
(A) x = A sin(ky – ωt)

v) A standing wave is produced on a string fixed at one end with the other end free. The length of the string
(A) must be an odd integral multiple of λ/4.
(B) must be an odd integral multiple of λ/2.
(C) must be an odd integral multiple of λ.
(D) must be an even integral multiple ofλ.
Answer:
(A) must be an odd integral multiple of λ/4.

2. Answer in brief.

i) A wave is represented by an equation y = A sin (Bx + Ct). Given that the constants A, B, and C are positive, can you tell in which direction the wave is moving?
Answer:
The wave is travelling along the negative x-direction.

ii) A string is fixed at the two ends and is vibrating in its fundamental mode. It is known that the two ends will be at rest. Apart from these, is there any position on the string which can be touched so as not to disturb the motion of the string? What will be the answer to this question if the string is vibrating in its first and second overtones?
Answer:
Nodes are the points where the vibrating string can be touched without disturbing its motion.
When the string vibrates in its fundamental mode, the string vibrates in one loop. There are no nodes formed between the fixed ends. Hence, there is no point on the string which can be touched without disturbing its motion.

When the string vibrates in its first overtone (second harmonic), there are two loops of the stationary wave on the string. Apart from the two nodes at the two ends, there is now a third node at its centre. Hence, the string can be touched at its centre without disturbing the stationary wave pattern.

When the string vibrates in its second overtone (third harmonic), there are three loops of the stationary wave on the string. So, apart from the two end nodes, there are two additional nodes in between, at distances one-third of the string length from each end. Thus, now the string can be touched at these two nodes.

iii) What are harmonics and overtones?
Answer:
A stationary wave is set up in a bounded medium in which the boundary could be a rigid support (i.e., a fixed end, as for instance a string stretched between two rigid supports) or a free end (as for instance an air column in a cylindrical tube with one or both ends open). The boundary conditions limit the possible stationary waves and only a discrete set of frequencies is allowed.

The lowest allowed frequency, n₁, is called the fundamental frequency of vibration. Integral multiples of the fundamental frequency are called the harmonics, the fundamental frequency being the fundamental or 2n₁, the third harmonic is 3n₁, and so on.

The higher allowed frequencies are called the overtones. Above the fundamental, the first allowed frequency is called the first overtone, the next higher frequency is the second overtone, and ‘so on. The relation between overtones and allowed harmonics depends on the system under consideration.

iv) For a stationary wave set up in a string having both ends fixed, what is the ratio of the fundamental frequency to the
second harmonic?
Answer:
The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (n₁) is 1 : 2.

v) The amplitude of a wave is represented by y = 0.2 sin 4π[\(\frac{t}{0.08}\) – \(\frac{x}{0.8}\)] in SI units. Find
(a) wavelength,
(b) frequency and
(c) amplitude of the wave. [(a) 0.4 m (b) 25 Hz (c) 0.2 m]
Answer:

y = 0.2 sin 2π[\(\frac{t}{0.04}\) – \(\frac{x}{0.4}\)]
Let us compare above equation with the equation of a simple harmonic progressive wave:
y = A sin 2π[\(\frac{t}{T}\) – \(\frac{x}{\lambda}\)] = 0.2 sin 2π[\(\frac{t}{0.04}\) – \(\frac{x}{0.4}\)]
Comparing the quantities on both sides, we get,
A = 0.2 m, T = 0.04 s, λ = 0.4 m
∴ (a) Wavelength (λ) = 0.4 m
(b) Frequency (n) = \(\frac{1}{T}\) = \(\frac{1}{0.04}\) = 25 Hz
(c) Amplitude (A) = 0.2 m

Question 3.
State the characteristics of progressive waves.
Answer:
Characteristics of a progressive wave :

1. Energy is transmitted from particle to particle without the physical transfer of matter.
2. The particles of the medium vibrate periodically about their equilibrium positions.
3. In the absence of dissipative forces, every particle vibrates with the same amplitude and frequency, but differs in phase from its adjacent particles. Every particle lags behind in its state of motion compared to the one before it.
4. A wave motion is doubly periodic, i.e., it is periodic in time and periodic in space.
5. The velocity of propagation through a medium depends upon the properties of the medium.
6. Progressive waves are of two types : transverse and longitudinal. In a transverse mechanical wave, the individual particles of the medium vibrate perpendicular to the direction of propagation of the wave. The progressively changing phase of the successive particles results in the formation of alternate crests and troughs that are periodic in space and time. In an em wave, the electric and magnetic fields oscillate in mutually perpendicular directions, perpendicular to the direction of propagation.
   In a longitudinal mechanical wave, the individual particles of the medium vibrate along the line of propagation of the wave. The progressively changing phase of the successive particles results in the formation of typical alternate regions of compressions and rarefactions that are periodic in space and time. Periodic compressions and rarefactions result in periodic pressure and density variations in the medium. There are no longitudinal em wave.
7. A transverse wave can propagate only through solids, but not through liquids and gases while a longitudinal wave can propagate through any material medium.

Question 4.
State the characteristics of stationary waves.
Answer:
Characteristics of stationary waves :

1. Stationary waves are produced by the interference of two identical progressive waves travelling in opposite directions, under certain conditions.
2. The overall appearance of a standing wave is of alternate intensity maximum (displacement antinode) and minimum (displacement node).
3. The distance between adjacent nodes (or antinodes) is λ/2.
4. The distance between successive node and antinode is λ/4.
5. There is no progressive change of phase from particle to particle. All the particles in one loop, between two adjacent nodes, vibrate in the same phase, while the particles in adjacent loops are in opposite phase.
6. A stationary wave does not propagate in any direction and hence does not transport energy through the medium.
7. In a region where a stationary wave is formed, the particles of the medium (except at the nodes) perform SHM of the same period, but the amplitudes of the vibrations vary periodically in space from particle to particle.

[Note : Since the nodes are points where the particles are always at rest, energy cannot be transmitted across a node. The energy of the particles within a loop remains localized, but alternates twice between kinetic and potential energy during each complete vibration. When all the particles are in the mean position, the energy is entirely kinetic. When they are in their extreme positions, the energy is entirely potential.]

Question 5.
Derive an expression for equation of stationary wave on a stretched string.
Answer:
When two progressive waves having the same amplitude, wavelength and speed propagate in opposite directions through the same region of a medium, their superposition under certain conditions creates a stationary interference pattern called a stationary wave.

Consider two simple harmonic progressive waves, of the same amplitude A, wavelength A and frequency n = ω/2π, travelling on a string stretched along the x-axis in opposite directions. They may be represented by
y₁ = A sin (ωt – kx) (along the + x-axis) and … (1)
y₂ = A sin (ωt + kx) (along the – x-axis) …. (2)
where k = 2π/λ is the propagation constant.

By the superposition principle, the resultant displacement of the particle of the medium at the point at which the two waves arrive simultaneously is the algebraic sum
y = y₁ + y₂ = A [sin (ωt – kx) + sin (ωt + kx)]
Using the trigonometrical identity,
sin C + sin D = 2 sin \(\left(\frac{C+D}{2}\right)\) cos \(\left(\frac{C-D}{2}\right)\),
y = 2A sin ωt cos (- kx)
= 2A sin ωt cos kx [∵ cos(- kx) = cos(kx)]
= 2A cos kx sin ωt … (3)
∴ y = R sin ωt, … (4)
where R = 2A cos kx. … (5)
Equation (4) is the equation of a stationary wave.

Question 6.
Find the amplitude of the resultant wave produced due to interference of two waves given as y₁ = A₁ sin ωt y₂ = A₂ sin (ωt + φ)
Answer:
The amplitude of the resultant wave produced due to the interference of the two waves is
A = \(\sqrt{A_{1}^{2}+2 A_{1} A_{2} \cos \varphi+A_{2}^{2}}\).

Question 7.
State the laws of vibrating strings and explain how they can be verified using a sonometer.
Answer:
The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (n₁) is 1 : 2.

Here, L = 3\(\frac{\lambda}{2}\)
∴ Wavelength, λ = \(\frac{2 L}{3}\) = \(\frac{2 \times 30}{3}\) = 20 cm.

Question 8.
Show that only odd harmonics are present in the vibrations of air column in a pipe closed at one end.
Answer:
Consider a narrow cylindrical pipe of length l closed at one end. When sound waves are sent down the air column in a cylindrical pipe closed at one end, they are reflected at the closed end with a phase reversal and at the open end without phase reversal. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.

The stationary waves in the air column in this case are subject to two boundary conditions that there must be a node at the closed end and an antinode at the open end.

Taking into account the end correction e at the open end, the resonating length of the air column is L = l + e.

Let v be the speed of sound in air. In the simplest mode of vibration, there is a node at the closed end and an antinode at the open end. The distance between a node and a consecutive anti-node is \(\frac{\lambda}{4}\), where λ is the wavelength of sound. The corresponding wavelength λ and frequency n are
λ = 4L and n = \(\frac{v}{\lambda}\) = \(\frac{v}{4 L}\) = \(\frac{v}{4(l+e)}\) …… (1)
This gives the fundamental frequency of vibration and the mode of vibration is called the fundamental mode or first harmonic.

In the next higher mode of vibration, the first overtone, two nodes and two antinodes are formed. The corresponding wavelength λ₁ and frequency n₁ are
λ₁ = \(\frac{4 L}{3}\) and n₁ = \(\frac{v}{\lambda_{1}}\) = \(\frac{3 v}{4 L}\) = \(\frac{3 v}{4(l+e)}\) = 3n … (2)
Therefore, the frequency in the first overtone is three times the fundamental frequency, i.e., the first overtone is the third harmonic.

In the second overtone, three nodes and three antinodes are formed. The corresponding wavelength λ₂ and frequency n₂ are

which is the fifth harmonic.
Therefore, in general, the frequency of the pth overtone (p = 1, 2, 3, ,..) is
n_p = (2p + 1)n … (4)
i.e., the pth overtone is the (2p + 1)th harmonic.

Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe closed at one end are n, 3n, 5n, …. That is, only odd harmonics are present as overtones.

Question 9.
Prove that all harmonics are present in the vibrations of the air column in a pipe open at both ends.
Answer:
Consider a cylindrical pipe of length l open at both the ends. When sound waves are sent down the air column in a cylindrical open pipe, they are reflected at the open ends without a change of phase. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.

The stationary waves in the air column in this case are subject to the two boundary conditions that there must be an antinode at each open end.

Taking into account the end correction e at each of the open ends, the resonating length of the air column is L = l + 2e.

Let v be the speed of sound in air. In the simplest mode of vibration, the fundamental mode or first harmonic, there is a node midway between the two antinodes at the open ends. The distance between two consecutive antinodes is λ/2, where λ is the wavelength of sound. The corresponding wavelength λ and the fundamental frequency n are
λ = 2L and n = \(\frac{v}{\lambda}\) = \(\frac{v}{2 L}\) = \(\frac{v}{2(l+2 e)}\) …. (1)

In the next higher mode, the first overtone, there are two nodes and three antinodes. The corresponding wavelength λ₁ and frequency n₁
λ₁ = L and n₁ = \(\frac{v}{\lambda_{1}}\) = \(\frac{v}{L}\) = \(\frac{v}{(l+2 e)}\) = 2n …. (2)
i.e., twice the fundamental. Therefore, the first overtone is the second harmonic.

In the second overtone, there are three nodes and four antinodes. The corresponding wavelength λ₂ and frequency n₂ are
λ₂ = \(\frac{2 L}{3}\) and n₂ = \(\frac{v}{\lambda_{2}}\) = \(\frac{3v}{2L}\) = \(\frac{3 v}{2(l+2 e)}\) = 3n …. (3)
or thrice the fundamental. Therefore, the second overtone is the third harmonic.

Therefore, in general, the frequency of the pth overtone (p = 1, 2, 3, …) is
n_p = (p + 1)n … (4)
i.e., the pth overtone is the (p + 1)th harmonic.
Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe open at both ends are n, 2n, 3n, …. That is, all the harmonics are present as overtones.

Question 10.
A wave of frequency 500 Hz is travelling with a speed of 350 m/s.
(a) What is the phase difference between two displacements at a certain point at times 1.0 ms apart?
(b) what will be the smallest distance between two points which are 45º out of phase at an instant of time?
[Ans : π, 8.75 cm ]
Answer:
Data : n = 500 Hz, v = 350 m/s
D = n × λ
∴ λ = \(\frac{350}{500}\) = 0.7 m
(a) in t = 1.0 ms = 0.001 s, the path difference is the distance covered v × t = 350 × 0.001 = 0.35 m
∴ Phase difference = \(\frac{2 \pi}{\lambda}\) × Path difference
= \(\frac{2 \pi}{0.7}\) × 0.35 = π rad

(b) Phase difference = 45° = \(\frac{\pi}{4}\) rad
∴ Path difference = \(\frac{\lambda}{2 \pi}\) × Phase difference
= \(\frac{0.7}{2 \pi}\) × \(\frac{\pi}{4}\) = 0.0875 m

Question 11.
A sound wave in a certain fluid medium is reflected at an obstacle to form a standing wave. The distance between two successive nodes is 3.75 cm. If the velocity of sound is 1500 m/s, find the frequency. [Ans : 20 kHz]
Answer:
Data : Distance between two successive nodes =
\(\frac{\lambda}{2}\) = 3.75 × 10^-2 m, v = 1500 m/s
∴ λ = 7.5 × 10^-2m
v = n × λ
∴ n = \(\frac{1500}{7.5 \times 10^{-2}}\) = 20 kHz

Question 12.
Two sources of sound are separated by a distance 4 m. They both emit sound with the same amplitude and frequency (330 Hz), but they are 180º out of phase. At what points between the two sources, will the sound intensity be maximum? (Take velocity of sound to be 330 m/s) [Ans: ± 0.25, ± 0.75, ± 1.25 and ± 1.75 m from the point at the center]
Answer:
∴ λ = \(\frac{v}{n}\) = \(\frac{330}{330}\) = 1 m
Directly at the cenre of two sources of sound, path difference is zero. But since the waves are 180° out of phase, two maxima on either sides should be at a distance of \(\frac{\lambda}{4}\) from the point at the centre. Other
maxima will be located each \(\frac{\lambda}{2}\) further along.
Thus, the sound intensity will be maximum at ± 0.25, ± 0.75, ± 1.25, ± 1.75 m from the point at the centre.

Question 13.
Two sound waves travel at a speed of 330 m/s. If their frequencies are also identical and are equal to 540 Hz, what will be the phase difference between the waves at points 3.5 m from one source and 3 m from the other if the sources are in phase? [Ans : 1.636 π]
Answer:
Data : v = 330 m/s, n₁ = n₂ = 540 Hz
v = n × λ
∴ λ = \(\frac{330}{540}\) = 0.61 m
Here, the path difference = 3.5 – 3 m = 0.5 m
Phase difference = \(\frac{2 \pi}{\lambda}\) × Path difference
= \(\frac{2 \pi}{0.61}\) × 0.5 = 1.64π rad

Question 14.
Two wires of the same material and same cross-section are stretched on a sonometer. One wire is loaded with 1.5 kg and another is loaded with 6 kg. The vibrating length of first wire is 60 cm and its fundamental frequency of vibration is the same as that of the second wire. Calculate vibrating length of the other wire. [Ans: 1.2 m]
Answer:
Data : m₁ = m₂ = m, L₁ = 60 cm = 0.6 m, T₁ = 1.5 kg = 14.7 N, T₂ = 6 kg = 58.8 N

The vibrating length of the second wire is 1.2 m.

Question 15.
A pipe closed at one end can produce overtones at frequencies 640 Hz, 896 Hz and 1152 Hz. Calculate the fundamental
frequency. [Ans: 128 Hz]
Answer:
The difference between the given frequencies of the overtones is 256 Hz. This implies that they are consecutive overtones. Let n_C be the fundamental frequency of the closed pipe and n_q, n_q-1, n_q-1 = the frequencies of the q^th, (q + 1)^th and (q + 2)^th consecutive overtones, where q is an integer.

Data : n_q = 640 Hz, n_q-1 = 896 Hz, n_q+2 = 1152 Hz
Since only odd harmonics are present as overtones, n_q = (2q +1) n_C
and n_q+1 = [2(q + 1) + 1] n_C = (2q + 3) n_C

∴ 14q + 7 = 10q + 15 ∴ 4q = 8 ∴ q = 2
Therefore, the three given frequencies correspond to the second, third and fourth overtones, i.e., the fifth, seventh and ninth harmonics, respectively.
∴ 5n_C = 640 ∴ b_C = 128Hz

Question 16.
A standing wave is produced in a tube open at both ends. The fundamental frequency is 300 Hz. What is the length
of tube in the fundamental mode? (speed of the sound = 340 m s^-1). [Ans: 0.5666 m]
Answer:
Data : For the tube open at both the ends, n = 300 Hz and v = 340 m / s Igonoring end correction, the fundamental frequency of the tube is
n = \(\frac{v}{2 L}\) ∴ L = \(\frac{v}{2 n}\) = \(\frac{340}{2 \times 300}\) = 0.566m
The length of the tube open at both the ends is 0.5667 m.

Question 17.
Find the fundamental, first overtone and second overtone frequencies of a pipe, open at both the ends, of length 25 cm if the speed of sound in air is 330 m/s. [Ans: 660 Hz, 1320 Hz, 1980 Hz]
Answer:
Data : Open pipe, ∠25 cm = 0.25 m, v = 330 m / s
The fundamental frequency of an open pipe ignoring end correction,

Since all harmonics are present as overtones, the first overtone is, n₁ = 2n₀ = 2 × 660 = 1320 Hz
The second overtone is n₂ = 3n = 3 × 660 = 1980 Hz

Question 18.
A pipe open at both the ends has a fundamental frequency of 600 Hz. The first overtone of a pipe closed at one end has the same frequency as the first overtone of the open pipe. How long are the two pipes? (Take velocity of sound to be 330 m/s) [Ans : 27.5 cm, 20.625 cm]
Answer:
Data : Open pipe, n_o = 600 Hz, n_{C, 1} = n_{o, 1} (first overtones)
For an open pipe, the fundamental frequency,
n_o = \(\frac{v}{2 L_{O}}\)
∴ The length of the open pipe is
L₀ = \(\frac{v}{2 n_{O}}\) = \(\frac{330}{2 \times 600}\) = 0.275 m
For the open pipe, the frequency of the first overtone is
2n₀ = 2 × 600 = 1200 Hz
For the pipe closed at one end, the frequency of the first overtone is \(\frac{3 v}{L_{O}}\).
By the data, \(\frac{3 v}{4 L}\) = 1200
∴ L_C = \(\frac{3 \times 330}{4 \times 1200}\) = 0.206 m
The length of the pipe open at both ends is 27.5 cm

Question 19.
A string 1m long is fixed at one end. Transverse vibrations of frequency 15 Hz are imposed at the free end. Due to this, a stationary wave with four complete loops, is produced on the string. Find the speed of the progressive wave which produces the stationary wave.[Hint: Remember that the free end is an antinode.] [Ans: 6.67 m s^-1]
Answer:
Data : L = 1 m, n = 15 Hz.
The string is fixed only at one end. Hence, an antinode will be formed at the free end. Thus, with four and half loops on the string, the length of the string is
L = \(\frac{\lambda}{4}\) + 4\(\left(\frac{\lambda}{2}\right)\) = \(\frac{9}{4}\)λ
∴ λ = \(\frac{4 L}{9}\) = \(\frac{4}{9}\) × 1 = \(\frac{4}{9}\) m
v = nλ
∴ Speed of the progressive wave
v = 15 × \(\frac{4}{9}\) = \(\frac{60}{9}\) =6.667m/s

Question 20.
A violin string vibrates with fundamental frequency of 440Hz. What are the frequencies of first and second overtones? [Ans: 880 Hz, 1320 Hz]
Answer:
Data: n =440Hz
The first overtone, n₁ = 2n =2 × 400 = 880 Hz
The second overtone, n₁ = 3n = 3 × 400 = 1320 Hz

Question 21.
A set of 8 tuning forks is arranged in a series of increasing order of frequencies. Each fork gives 4 beats per second with the next one and the frequency of last for k is twice that of the first. Calculate the frequencies of the first and the last for k. [Ans: 28 Hz, 56 Hz]
Answer:
Data : n₈ = 2n₁, beat frequency = 4 Hz
The set of tuning fork is arranged in the increasing order of their frequencies.
∴ n₂ = n₁ + 4
n₃ = n₂ + 4 = n₁ + 2 × 4
n₄ = n₃ + 4 = n₁ + 3 × 4
∴ n₈ = n₇ + 4 = n₁ + 7 × 4 = n₁ + 28
Since n₈ = 2n₁,
2n₁ = n₁ + 28
∴ The frequency of the first fork, n₁ = 28 Hz
∴ The frequency of the last fork,
n₈ = n₁ + 28 = 28 + 28 = 56 Hz

Question 22.
A sonometer wire is stretched by tension of 40 N. It vibrates in unison with a tuning fork of frequency 384 Hz. How
many numbers of beats get produced in two seconds if the tension in the wire is decreased by 1.24 N? [Ans: 12 beats]
Answer:
Data : T₁ =40N, n₁ = 384 Hz, T₂ = 40 – 1.24 = 38.76 N

∴ The number of beats produced in two seconds = 2 × 6 = 12

Question 23.
A sonometer wire of length 0.5 m is stretched by a weight of 5 kg. The fundamental frequency of vibration is 100 Hz. Calculate linear density of wire. [Ans: 4.9 × 10^-3 kg/m]
Answer:
Data : L = 0.5 m, T = 5 kg = 5 × 9.8 = 49 N, n = 100 Hz
n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
∴ Linear density, m = \(\frac{T}{4 L^{2} n^{2}}\)
= \(\frac{49}{4(0.5)^{2}(100)^{2}}\)
= 4.9 × 10^-3 kg/m

Question 24.
The string of a guitar is 80 cm long and has a fundamental frequency of 112 Hz. If a guitarist wishes to produce a frequency
of 160 Hz, where should the person press the string? [Ans : 56 cm from one end]
Answer:
Data : L₁ = 80 cm n₁ = 112 Hz, n₂ = 160 Hz
According to the law of length, n₁L₁ = n₂L₂.
∴ The vibrating length to produce the fundamental frequency of 160 Hz,
L₂ = \(\frac{n_{1} L_{1}}{n_{2}}\) = \(\frac{112(80)}{160}\) = 56 cm

12th Physics Digest Chapter 6 Superposition of Waves Intext Questions and Answers

Can you tell? (Textbook Page No. 132)

Question 1.
What is the minimum distance between any two particles of a medium which always have the same speed when a sinusoidal wave travels through the medium ?
Answer:
When a sinusoidal wave travels through a medium the minimum distance between any two particles of the medium which always have the same speed is \(\frac{\lambda}{2}\).
Such particles are opposite in phase, i.e., their instantaneous velocities are opposite in direction.
[Note : The minimum distance between any two particles which have the same velocity is λ]

Do you know? (Textbook Page No. 140)

Question 1.
What happens if a simple pendulum is pulled aside and released ?
Answer:
If a simple pendulum is pulled aside and released, it oscillates freely about its equilibrium position at its natural frequency which is inversely proportional to the square root of its length and directly proportional to the square root of the acceleration of gravity at the place. These oscillations, called as free oscillations, are periodic and tautochronous if the displacement of its bob is small and the dissipative forces can be ignored.

Question 2.
What happens when a guitar string is plucked ?
Answer:
When a guitar string is plucked, two wave pulses of the same amplitude, frequency and phase move out from that point towards the fixed ends of the string where they get reflected. For certain ratios of wavelength to length of the string, these reflected pulses moving towards each other will meet in phase to form standing waves on the string. The vibrations of the string cause the air molecules to oscillate, forming sound waves that radiate away from the string. The frequency of the sound waves is equal to the frequency of the vibrating string. In general, the wavelengths of the sound waves and the waves on the string are different because their speeds in the two mediums are not the same.

Question 3.
Have you noticed vibrations in a drill machine or in a washing machine ? How do they differ from vibrations in the above two cases ?
Answer:
Vibrations in the body of a drill machine or that of a washing machine are forced vibrations induced by the vibrations of the motors of these machines. On the other hand, the oscillations of a simple pendulum or a guitar string are free oscillations, produced when they are disturbed from their equilibrium position and released.

Question 4.
A vibrating tuning fork of certain frequency is held in contact with a tabletop and its vibrations are noticed and then another vibrating tuning fork of different frequency is held on the tabletop. Are the vibrations produced in the tabletop the same for both the tuning forks ? Why ?
Answer:
No. Because the tuning forks have different frequencies, the forced vibrations in the tabletop differ both in frequency and amplitude. The tuning fork whose frequency is closer to a natural frequency of the tabletop induces forced vibrations of a larger amplitude.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 5 Oscillations Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 5 Oscillations

1. Choose the correct option.

i) A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and its time period is T. At the instance when its speed is half the maximum speed, its displacement x is
(A) \(\frac{\sqrt{3}}{2}\)A
(B) \(\frac{2}{\sqrt{3}}\)
(C) A/2
(D) \(\frac{1}{\sqrt{2}}\)
Answer:
(A) \(\frac{\sqrt{3}}{2}\)A

ii) A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + π/4). Maximum kinetic energy of the body is
(A) 36 J
(B) 9 J
(C) 27 J
(D) 18 J
Answer:
(D) 18 J

iii) The length of the second’s pendulum on the surface of the earth is nearly 1 m. Its length on the surface of the moon should be [Given: acceleration due to gravity (g) on the moon is 1/6 th of that on the earth’s surface]
(A) \(\frac{1}{6}\) m
(B) 6 m
(C) \(\frac{1}{36}\) m
(D) \(\frac{1}{\sqrt{6}}\) m.
Answer:
(A) \(\frac{1}{6}\) m

iv) Two identical springs of constant k are connected, first in series and then in parallel. A metal block of mass m is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio
(A) 1:4
(B) 1:2
(C) 2:1
(D) 4:1
Answer:
(B) 1:2

v) The graph shows the variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph?
(A) The acceleration is maximum at time T.
(B) The force is maximum at time 3T/4.
(C) The velocity is zero at time T/2.
(D) The kinetic energy is equal to total energy at time T/4.

Answer:
(B) The force is maximum at time 3T/4.

2. Answer in brief.

i) Define linear simple harmonic motion.
Answer:
Definition: Linear simple harmonic motion (SHM) is defined as the linear periodic motion of a body, in which the force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.
OR
A particle is said to execute linear SHM if the particle undergoes oscillations about a point of stable equilibrium, subject to a linear restoring force always directed towards that point and whose magnitude is proportional to the magnitude of the displacement of the particle from that point.
Examples: The vibrations of the tines (prongs) of a tuning fork, the oscillations of the needle of a sewing machine.

ii) Using the differential equation of linear S.H.M, obtain the expression for
(a) velocity in S.H.M.,
(b) acceleration in S.H.M.
Answer:
The general expression for the displacement of a particle in SHM at time t is x = A sin (ωt + α) … (1)
where A is the amplitude, ω is a constant in a particular case and α is the initial phase.
The velocity of the particle is

Equation (2) gives the velocity as a function of x.
The acceleration of the particle is
a = \(\frac{d v}{d t}\) = \(\frac{d}{d t}\) [Aω cos (ωt + α) J at at
∴ a = – ω² A sin (ωt + α)
But from Eq. (1), A sin (ωt + α) = x
∴ a = -ω²x … (3)
Equation (3) gives the acceleration as a function of x. The minus sign shows that the direction of the acceleration is opposite to that of the displacement.

iii) Obtain the expression for the period of a simple pendulum performing S.H.M.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.

In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.

Consider a simple pendulum of length L₁ – suspended from a rigid support O. When displaced from its initial position of rest through a small angle θ in a vertical plane and released, it performs oscillations between two extremes, B and C, as shown in below figure. At B, the forces on the bob are its weight \(m \vec{g}\) and the tension \(\overrightarrow{F_{1}}\) in the string. Resolve \(m \vec{g}\) into two components : mg cos θ in the direction opposite to that of the tension and mg sin θ perpendicular to the string.

mg cos θ balanced by the tension in the string. mg sin θ restores the bob to the equilibrium position.
Restoring force, F = – mg sin θ
If θ is small and expressed in radian,
sin θ \(\approx\) θ = \(\frac{\text { arc }}{\text { radius }}\) = \(\frac{\mathrm{AB}}{\mathrm{OB}}\) = \(\frac{x}{L}\)
∴ F = – mgθ = -mg\(\frac{x}{L}\) …. (1)
Since m, g and L are constant,
F ∝ (-x) …. (2)

Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, a = \(\frac{F}{m}\) = –\(\frac{g}{L}\)x … (3)
Therefore, acceleration per unit displacement
= |\(\frac{a}{x}\)| = \(\frac{g}{L}\) ….. (4)
Period of SHM,

This gives the expression for the period of a simple pendulum.

iv) State the laws of simple pendulum.
Answer:
The period of a simple pendulum at a given place is
T = 2π\(\sqrt{\frac{L}{g}}\)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

(1) Law of length : The period of a simple pendulum at a given place (g constant) is directly proportional to the square root of its length.
∴ T ∝\(\sqrt{L}\)
(2) Law of acceleration due to gravity : The period of a simple pendulum of a given length (L constant) is inversely proportional to the square root of the acceleration due to gravity.
∴ T ∝ \(\frac{1}{\sqrt{g}}\)
(3) Law of mass : The period of a simple pendulum does not depend on the mass or material of the bob of the pendulum.
(4) Law of isochronism : The period of a simple pendulum does not depend on the amplitude of oscillations, provided that the amplitude is small.

v) Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M.
Answer:
Consider a bar magnet of magnetic moment μ, suspended horizontally by a light twistless fibre in a region where the horizontal component of the Earth’s magnetic field is B_h. The bar magnet is free to rotate in a horizontal plane. It comes to rest in approximately the North-South direction, along B_h. If it is rotated in the horizontal plane by a small

displacement θ from its rest position (θ = 0), the suspension fibre is twisted. When the magnet is released, it oscillates about the rest position in angular or torsional oscillation.

The bar magnet experiences a torque \(\tau\) due to the field B_h. Which tends to restore it to its original orientation parallel to B_h. For small θ, this restoring torque is
\(\tau\) = – μB_h sin θ = – μB_hμ …. (1)

where the minus sign indicates that the torque is opposite in direction to the angular displacement θ. Equation (1) shows that the torque (and hence the angular acceleration) is directly proportional in magnitude of the angular displacement but opposite in direction. Hence, for small angular displacement, the oscillations of the bar magnet in a uniform magnetic field is simple harmonic.

Question 3.
Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.
Answer:
Definition : Angular SHM is defined as the oscillatory motion of a body in which the restoring torque responsible for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement.
The differential equation of angular SHM is
I\(\frac{d^{2} \theta}{d t^{2}}\) + cθ = 0 ….. (1)
where I = moment of inertia of the oscillating body, \(\frac{d^{2} \theta}{d t^{2}}\) = angular acceleration of the body when its angular displacement is θ, and c = torsion constant of the suspension wire,
∴ \(\frac{d^{2} \theta}{d t^{2}}\) + \(\frac{c}{I}\)θ = 0
Let \(\frac{c}{I}\) = ω², a constant. Therefore, the angular frequency, ω = \(\sqrt{c / I}\) and the angular acceleration,
a = \(\frac{d^{2} \theta}{d t^{2}}\) = -ω²θ … (2)
The minus sign shows that the α and θ have opposite directions. The period T of angular SHM is

This is the expression for the period in terms of torque constant. Also, from Eq. (2),

Question 4.
Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.
Answer:
Consider a particle which moves anticlockwise around a circular path of radius A with a constant angular speed ω. Let the path lie in the x-y plane with the centre at the origin O. The instantaneous position P of the particle is called the reference point and the circle in which the particle moves as the reference circle.

The perpendicular projection of P onto the y-axis is Q. Then, as the particle travels around the circle, Q moves to-and-fro along the y-axis. Line OP makes an angle α with the x-axis at t = 0. At time t, this angle becomes θ = ωt + α.
The projection Q of the reference point is described by the y-coordinate,
y = OQ = OP sin ∠OPQ, Since ∠OPQ = ωt + α, y = A sin(ωt + α)
which is the equation of a linear SHM of amplitude A. The angular frequency w of a linear SHM can thus be understood as the angular velocity of the reference particle.

The tangential velocity of the reference particle is v = ωA. Its y-component at time t is v_y = ωA sin (90° – θ) = ωA cos θ
∴ v_y = ωA cos (ωt + α)
The centripetal acceleration of the reference particle is a_r = ω²A, so that its y-component at time t is a_x = a_r sin ∠OPQ
∴ a_x = – ω²A sin (ωt + α)

Question 5.
Draw graphs of displacement, velocity and acceleration against phase angle, for a particle performing linear S.H.M. from (a) the mean position
(b) the positive extreme position. Deduce your conclusions from the graph.
Answer:
Consider a particle performing SHM, with amplitude A and period T = 2π/ω starting from the mean position towards the positive extreme position where ω is the angular frequency. Its displacement from the mean position (x), velocity (v) and acceleration (a) at any instant are
x = A sin ωt = A sin\(\left(\frac{2 \pi}{T} t\right)\) (∵ω = \(\frac{2 \pi}{T}\))
v = \(\frac{d x}{d t}\) = ωA cos ωt = ωA cos\(\left(\frac{2 \pi}{T} t\right)\)
a = \(\frac{d v}{d t}\) = -ω² A sin ωt = – ω²A sin\(\left(\frac{2 \pi}{T} t\right)\) as the initial phase α = 0.
Using these expressions, the values of x, v and a at the end of every quarter of a period, starting from t = 0, are tabulated below.

Using the values in the table we can plot graphs of displacement, velocity and acceleration with time.

Conclusions :

1. The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting at the mean position, the x-t and a-t graphs are sine curves. The v-t graph is a cosine curve.
2. There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
3. There is a phase difference of π radians between x and a.

Consider a particle performing linear SHM with amplitude A and period T = 2π/ω, starting from the positive extreme position, where ω is the angular frequency. Its displacement from the mean position (x), velocity (y) and acceleration (a) at any instant (t) are
x = A cos ωt = A cos \(\left(\frac{2 \pi}{T} t\right)\) (∵ω = \(\frac{2 \pi}{T}\))
v = -ωA sin ωt = – ωA sin \(\left(\frac{2 \pi}{T} t\right)\)
a = -ω²A cos ωt = -ω²A cos \(\left(\frac{2 \pi}{T} t\right)\)
Using these expressions, the values of x, y and a at the end of every quarter of a period, starting from t = 0, are tabulated below.

Using these values, we can plot graphs showing the variation of displacement, velocity and acceleration with time.

Conclusions :

1. The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting from an extreme position, the x-t and a-t graphs are cosine curves; the v-t graph is a sine curve.
2. There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
3. There is a phase difference of n radians between x and a.

Explanations :
(1) v-t graph : It is a sine curve, i.e., the velocity is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π/2 rad between a and v.
v is minimum (equal to zero) at the extreme positions (i.e., at x = ± A) and v is maximum ( = ± ωA) at the mean position (x = 0).

(2) a-t graph : It is a cosine curve, i.e., the acceleration is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π rad between v and a. a is minimum (equal to zero) at the mean position (x = 0) and a is maximum ( = \(\mp\)ω²A) at the extreme positions (x = ±A).

Question 6.
Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.
Answer:
Consider a particle of mass m performing linear SHM with amplitude A. The restoring force acting r on the particle is F = – kx, where k is the force constant and x is the displacement of the particle from its mean position.
(1) Kinetic energy : At distance x from the mean position, the velocity is
v = ω\(\sqrt{A^{2}-x^{2}}\)
where ω = \(\sqrt{k / m} .\) The kinetic energy (KE) of the particle is
KE = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) mω² (A² – x²)
= \(\frac{1}{2}\)k(A² – x²) … (1)
If the phase of the particle at an instant t is θ = ωt + α, where α is initial phase, its velocity at that instant is
v = ωA cos (ωt + α)
and its KE at that instant is
KE = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\)mω²A² cos²(ωt + α) ….. (2)
Therefore, the KE varies with time as cos² θ.

(2) Potential energy : The potential energy of a particle in linear SHM is defined as the work done by an external agent, against the restoring force, in taking the particle from its mean position to a given point in the path, keeping the particle in equilibirum.

Suppose the particle in below figure is displaced from P₁ to P₂, through an infinitesimal distance dx against the restoring force F as shown.

The corresponding work done by the external agent will be dW = ( – F)dx = kx dx. This work done is stored in the form of potential energy. The potential energy (PE) of the particle when its displacement from the mean position is x can be found by integrating the above expression from 0 to x.
∴ PE = \(\int\)dW = \(\int_{0}^{x}\) kx dx = \(\frac{1}{2}\) kx² … (3)
The displacement of the particle at an instant t being
x = A sin (wt + α)
its PE at that instant is
PE = \(\frac{1}{2}\)kx² = \(\frac{1}{2}\)kA² sin²(ωt + α) … (4)
Therefore, the PE varies with time as sin²θ.

(3) Total energy : The total energy of the particle is equal to the sum of its potential energy and kinetic energy.
From Eqs. (1) and (2), total energy is E = PE + KE
= \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)k(A² – x²)
= \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)kA² – \(\frac{1}{2}\)kx²
∴ E = \(\frac{1}{2}\)kA² = \(\frac{1}{2}\)mω²A² … (5)
As m is constant, ω and A are constants of the motion, the total energy of the particle remains constant (or its conserved).

Question 7.
Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.

In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.

Consider a simple pendulum of length L₁ – suspended from a rigid support O. When displaced from its initial position of rest through a small angle θ in a vertical plane and released, it performs oscillations between two extremes, B and C, as shown in below figure. At B, the forces on the bob are its weight \(m \vec{g}\) and the tension \(\overrightarrow{F_{1}}\) in the string. Resolve \(m \vec{g}\) into two components : mg cos θ in the direction opposite to that of the tension and mg sin θ perpendicular to the string.

mg cos θ balanced by the tension in the string. mg sin θ restores the bob to the equilibrium position.
Restoring force, F = – mg sin θ
If θ is small and expressed in radian,
sin θ \(\approx\) θ = \(\frac{\text { arc }}{\text { radius }}\) = \(\frac{\mathrm{AB}}{\mathrm{OB}}\) = \(\frac{x}{L}\)
∴ F = – mgθ = -mg\(\frac{x}{L}\) …. (1)
Since m, g and L are constant,
F ∝ (-x) …. (2)

Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, a = \(\frac{F}{m}\) = –\(\frac{g}{L}\)x … (3)
Therefore, acceleration per unit displacement
= |\(\frac{a}{x}\)| = \(\frac{g}{L}\) ….. (4)
Period of SHM,

This gives the expression for the period of a simple pendulum.

The period of a simple pendulum at a given place is
T = 2π\(\sqrt{\frac{L}{g}}\)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

(1) Law of length : The period of a simple pendulum at a given place (g constant) is directly proportional to the square root of its length.
∴ T ∝\(\sqrt{L}\)
(2) Law of acceleration due to gravity : The period of a simple pendulum of a given length (L constant) is inversely proportional to the square root of the acceleration due to gravity.
∴ T ∝ \(\frac{1}{\sqrt{g}}\)

Question 8.
At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given path length of S.H.M. = 10 cm. [Ans: 4.33 cm]
Answer:
Data : v = \(\frac{1}{2}\)v_max, 2A = 10 cm
∴ A = 5 cm
v = ω\(\sqrt{A^{2}-x^{2}}\) and v_max = ωA
Since v = \(\frac{1}{2}\)v_max,

This gives the required displacement.

Question 9.
In SI units, the differential equation of an S.H.M. is \(\frac{d^{2} x}{d t^{2}}\) = -36x. Find its frequency and period. Find its frequency and period.
[Ans: 0.9548 Hz, 1.047 s]
Answer:
\(\frac{d^{2} x}{d t^{2}}\) = -36x
Comparing this equation with the general equation,
\(\frac{d^{2} x}{d t^{2}}\) = -ω²x
We get, ω² = 36 ∴ ω = 6 rad/s
ω = 2πf
∴ The frequency, f = \(\frac{\omega}{2 \pi}\) = \(\frac{6}{2(3.142)}\) = \(\frac{6}{6.284}\) = 0.9548 Hz
and the period, T = \(\frac{1}{f}\) = \(\frac{1}{0.9548}\) = 1.047 s

Question 10.
A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration \(\frac{1}{30}\)s after it has crossed the mean position. [Ans: 34.2 m/s²]
Answer:
Data : A = 4 cm = 4 × 10^-2 m, f = 5Hz, t = \(\frac{1}{30}\)s
ω = 2πf = 2π (5) = 10π rad/s
Therefore, the magnitude of the acceleration,
|a| = ω²x = ω²A sin ωt
= (10π)² (4 × 10²)
= 10π² sin \(\frac{\pi}{3}\) = 10 (9.872)(0.866) = 34.20 m/s²

Question 11.
Potential energy of a particle performing linear S.H.M is 0.1 π² x² joule. If mass of the particle is 20 g, find the frequency of S.H.M. [Ans: 1.581 Hz]
Answer:
Data : PE = 0.1 π² x² J, m = 20 g = 2 × 10^-2 kg
PE = \(\frac{1}{2}\)mω²x² = \(\frac{1}{2}\)m (4π²f²)x²
∴ \(\frac{1}{2}\)m(4π²f²)x² = 0.1 π² x²
∴ 2mf² = 0.1 ∴ f² = \(\frac{1}{20\left(2 \times 10^{-2}\right)}\) = 2.5
∴ The frequency of SHM is f = \(\sqrt{2.5}\) = 1.581 Hz

Question 12.
The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. [Ans: 6.324 cm/s]
Answer:
Data : m = 2 kg, E = 40 J
The speed of the body while crossing the centre of the path (mean position) is v_max and the total energy is entirely kinetic energy.

Question 13.
A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position, will the displacement of the bob be one third of its amplitude. [Ans: 0.2163 s]
Answer:
Data : T = 4 s, x = A/3
The displacement of a particle starting into SHM from the mean position is x = A sin ωt = A sin \(\frac{2 \pi}{T}\) t

∴ the displacement of the bob will be one-third of its amplitude 0.2163 s after crossing the mean position.

Question 14.
A simple pendulum of length 100 cm performs S.H.M. Find the restoring force acting on its bob of mass 50 g when the displacement from the mean position is 3 cm. [Ans: 1.48 × 10^-2 N]
Answer:
Data : L = 100 cm, m = 50 g = 5 × 10^-2 kg, x = 3 cm, g = 9.8 m/s²
Restoring force, F = mg sin θ = mgθ
= (5 × 10^-2)(9.8)\(\left(\frac{3}{100}\right)\)
= 1.47 × 10^-2 N

Question 15.
Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75
m/s² to 9.80 m/s². [Ans: Decreases by 5.07 mm]
Answer:
Data : g₁ = 9.75 m/s², g₂ = 9.8 m/s²
Length of a seconds pendulum, L = \(\frac{g}{\pi^{2}}\)

∴ The length of the seconds pendulum must be increased from 0.9876 m to 0.9927 m, i.e., by 0.0051 m.

Question 16.
At what distance from the mean position is the kinetic energy of a particle performing S.H.M. of amplitude 8 cm, three times its potential energy? [Ans: 4 cm]
Answer:
Data : A = 8 cm, KE = 3 PE
KE = \(\frac{1}{2}\) (A² – x²) and PE = \(\frac{1}{2}\)kx²
Given, KE = 3PE.
∴ \(\frac{1}{2}\)k(A² – x²) = 3\(\left(\frac{1}{2} k x^{2}\right)\)
∴ A² – x² = 3x² ∴ 4x² = A²
∴ the required displacement is
x = ±\(\frac{A}{2}\) = ±\(\frac{8}{2}\) = ± 4 cm

Question 17.
A particle performing linear S.H.M. of period 2π seconds about the mean position O is observed to have a speed of \(b \sqrt{3}\) m/s, when at a distance b (metre) from O. If the particle is moving away from O at that instant, find the
time required by the particle, to travel a further distance b. [Ans: π/3 s]
Answer:
Data : T = 2πs, v = b\(\sqrt{3}\) m/s at x = b

∴ Assuming the particle starts from the mean position, its displacement is given by
x = A sin ωt = 2b sin t
If the particle is at x = b at t = t₁,
b = 2b sint₁ ∴ t₁ = sin^-1 \(\frac{1}{2}\) = \(\frac{\pi}{6}\)s
Also, with period T = 2πs, on travelling a further distance b the particle will reach the positive extremity at time t₂ = \(\frac{\pi}{2}\)s.
∴ The time taken to travel a further distance b from x = b is t₂ – t₁ = \(\frac{\pi}{2}\) – \(\frac{\pi}{6}\) = \(\frac{\pi}{3}\)s.

Question 18.
The period of oscillation of a body of mass m₁ suspended from a light spring is T. When a body of mass m₂ is tied to the first body and the system is made to oscillate, the period is 2T. Compare the masses m₁ and m₂ [Ans: 1/3]
Answer:

This gives the required ratio of the masses.

Question 19.
The displacement of an oscillating particle is given by x = asinωt + bcosωt where a, b and ω are constants. Prove that the particle performs a linear S.H.M. with amplitude A = \(\sqrt{a^{2}+b^{2}}\)
Answer:
x = asinωt + bcosωt
Let a = A cos φ and b = A sin φ, so that
A² = a² + b² and tan φ = \(\frac{b}{a}\)
∴ x = A cos φ sin ωt + A sin φ cos ωt
∴ x = A sin (ωt + φ)
which is the equation of a linear SHM with amplitude A = \(\sqrt{a^{2}+b^{2}}\) and phase constant φ = tan^-1 \(\frac{b}{a}\), as required.

Question 20.
Two parallel S.H.M.s represented by x₁ = 5sin (4πt + \(\frac{\pi}{3}\)) cm and x₂ = 3sin(4πt + π/4) cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M. [Ans: 7.936 cm, 54° 23′]
Answer:
Data: x₁ = 5 sin (4πt + \(\frac{\pi}{3}\)) = A₁ sin(ωt + α),
x₂ = 3 sin (4πt + \(\frac{\pi}{4}\)) = A₂ sin(ωt + β)
∴ A₁ = 5 cm, A₂ = 3 cm, α = \(\frac{\pi}{3}\) rad, β = \(\frac{\pi}{4}\) rad
(i) Resultant amplitude,

(ii) Epoch of the resultant SHM,

Question 21.
A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60° and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped conditions. (π³ ≈ 31)
[Ans: 0.04133 N m]
Answer:
Data: R = 10cm = 0.1 m, M = 0.2 kg, θ_m = 60° = \(\frac{\pi}{3}\) rad, T = 1 s, π³ ≈ 31
The Ml of the disc about the rotation axis (perperdicular through its centre) is
I = \(\frac{1}{2}\)MR² = (0.2)(0.1)² = 10^-3 kg.m²
The period of torsional oscillation, T = 2π\(\sqrt{\frac{I}{c}}\)
∴ The torsion constant, c = 4πr²\(\frac{I}{T^{2}}\)
The magnitude of the maximum restoring torque,

Question 22.
Find the number of oscillations performed per minute by a magnet is vibrating in the plane of a uniform field of 1.6 × 10^-5 Wb/m². The magnet has moment of inertia 3 × 10^-6 kgm² and magnetic moment 3 A m². [Ans:38.19 osc/min.]
Answer:
Data : B = 1.6 × 10^-5 T, I = 3 × 10^-6kg/m²,
µ = 3 A.m²
The period of oscillation, T = 2π \(\sqrt{\frac{I}{\mu B_{\mathrm{h}}}}\)
∴ The frequency of oscillation is
f = \(\frac{1}{2 \pi}\)\(\sqrt{\frac{\mu B}{I}}\)
∴ The number of oscillations per minute

= 38.19 per minute

Question 23.
A wooden block of mass m is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don’t want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25 cm? In this case, how much is the energy per unit mass of the block? (g ≈ π² ≈ 10 m s^-2)
[Ans: n_max = 1/s, E/m = 1.25 J/kg]
Answer:
Data : A = 0.25 m, g = π² = 10 m/s²
During vertical oscillations, the acceleration is maximum at the turning points at the top and bottom. The block will just lose contact with the piston when its apparent weight is zero at the top, i. e., when its acceleration is a_max = g, downwards.

This gives the required frequency of the piston.

12th Physics Digest Chapter 5 Oscillations of Waves Intext Questions and Answers

Can you tell? (Textbook Page No. 112)

Question 1.
Why is the term angular frequency (ω) used here for a linear motion?
Answer:
A linear SHM is the projection of a UCM on the diameter of the circle. The angular speed co of a particle moving along this reference circle is called the angular frequency of the particle executing linear SHM.

Can you tell? (Textbook Page No. 114)

Question 1.
State at which point during an oscillation the oscillator has zero velocity but positive acceleration?
Answer:
At the left extreme, i.e., x = – A, so that a = – ω²x = – ω²(- A) = ω²A = a_max

Question 2.
During which part of the simple harmonic motion velocity is positive but the displacement is negative, and vice versa?
Answer:
Velocity v is positive (to the right) while displacement x is negative when the particle in SHM is moving from the left extreme towards the mean position. Velocity v is negative (to the left) while displacement x is positive when the particle in SHM is moving from the right extreme towards the mean position.

Can you tell? (Textbook page 76)

Question 1.
To start a pendulum swinging, usually you pull it slightly to one side and release it. What kind of energy is transferred to the mass in doing this?
Answer:
On pulling the bob of a simple pendulum slightly to one side, it is raised to a slightly higher position. Thus, it gains gravitational potential energy.

Question 2.
Describe the energy changes that occur when the mass is released.
Answer:
When released, the bob oscillates in SHM in a vertical plane and the energy oscillates back and forth between kinetic and potential, going completely from one form of energy to the other as the pendulum oscillates. In the case of undamped SHM, the motion starts with all of the energy as gravitational potential energy. As the object starts to move, the gravitational potential energy is converted into kinetic energy, becoming entirely kinetic energy at the equilibrium position. The velocity becomes zero at the other extreme as the kinetic energy is completely converted back into gravitational potential energy,
and this cycle then repeats.

Question 3.
Is/are there any other way/ways to start the oscillations of a pendulum? Which energy is supplied in this case/cases?
Answer:
The bob can be given a kinetic energy at its equilibrium position or at any other position of its path. In the first case, the motion starts with all of the energy as kinetic energy. In the second case, the motion starts with partly gravitational potential energy and partly kinetic energy.

Can you tell? (Textbook Page No. 109)

Question 1.
Is the motion of a leaf of a tree blowing in the wind periodic ?
Answer:
The leaf of a tree blowing in the wind oscillates, but the motion is not periodic. Also, its displacement from the equilibrium position is not a regular function of time.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 4 Thermodynamics Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 4 Thermodynamics

1. Choose the correct option.

i) A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is the total change in the internal energy of the system?
(A) 10J
(B) 4J
(C) -10J
(D) – 4J
Answer:
(B) 4J

ii) Which of the following is an example of the first law of thermodynamics?
(A) The specific heat of an object explains how easily it changes temperatures.
(B) While melting, an ice cube remains at the same temperature.
(C) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(D) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.
Answer:
(B) While melting, an ice cube remains at the same temperature. [Here, ∆u = 0, W = Q]
(C) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(D) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.

iii) Efficiency of a Carnot engine is large when
(A) T_H is large
(B) T_C is low
(C) T_H – T_C is large
(D) T_H – T_C is small
Answer:
(A) T_H is large
(B) T_C is low
(C) T_H – T_C is large
[η = \(\frac{T_{\mathrm{H}}-T_{\mathrm{c}}}{T_{\mathrm{H}}}\) = 1 – \(\frac{T_{\mathrm{c}}}{T_{\mathrm{H}}}\)]

iv) The second law of thermodynamics deals with transfer of:
(A) work done
(B) energy
(C) momentum
(D) mass
Answer:
(B) energy

v) During refrigeration cycle, heat is rejected by the refrigerant in the :
(A) condenser
(B) cold chamber
(C) evaporator
(D) hot chamber
Answer:
closed tube[See the textbook]

2. Answer in brief.

i) A gas contained in a cylinder surrounded by a thick layer of insulating material is quickly compressed.
(a) Has there been a transfer of heat?
(b) Has work been done?
Answer:
(a) There is no transfer of heat.
(b) The work is done on the gas.

ii) Give an example of some familiar process in which no heat is added to or removed form a system, but the temperature of the system changes.
Answer:
Hot water in a container cools after sometime. Its temperature goes on decreasing with time and after sometime it attains room temperature.
[Note : Here, we do not provide heat to the water or remove heat from the water. The water cools on exchange of heat with the surroundings. Recall the portion covered in chapter 3.]

iii) Give an example of some familiar process in which heat is added to an object, without changing its temperature.
Answer:

1. Melting of ice
2. Boiling of water.

iv) What sets the limits on efficiency of a heat engine?
Answer:
The temperature of the cold reservoir sets the limit on the efficiency of a heat engine.
[Notes : (1) η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\)
This formula shows that for maximum efficiency, T_C should be as low as possible and T_H should be as high as possible.
(2) For a Carnot engine, efficiency
η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\).η → 1 T_C → 0.]

v) Why should a Carnot cycle have two isothermal two adiabatic processes?
Answer:
With two isothermal and two adiabatic processes, all reversible, the efficiency of the Carnot engine depends only on the temperatures of the hot and cold reservoirs.
[Note : This is not so in the Otto cycle and Diesel cycle.]

3. Answer the following questions.

i) A mixture of hydrogen and oxygen is enclosed in a rigid insulting cylinder. It is ignited by a spark. The temperature and the pressure both increase considerably. Assume that the energy supplied by the spark is negligible, what conclusions may be drawn by application of the first law of thermodynamics?
Answer:
The internal energy of a system is the sum of potential energy and kinetic energy of all the constituents of the system. In the example stated above, conversion of potential energy into kinetic energy is responsible for a considerable rise in pressure and temperature of the mixture of hydrogen and oxygen ignited by the spark.

ii) A resistor held in running water carries electric current. Treat the resistor as the system
(a) Does heat flow into the resistor?
(b) Is there a flow of heat into the water?
(c) Is any work done?
(d) Assuming the state of resistance to remain unchanged, apply the first law of thermodynamics to this process.
Answer:
(a) Heat is generated into the resistor due to the passage of electric current. In the usual notation, heat generated = I2Rt.
(b) Yes. Water receives heat from the resistor.

Here, I = current through the resistor, R = resistance of the resistor, t = time for which the current is passed through the resistor, M = mass of the water, S = specific heat of water, T = rise in the temperature of water, P = pressure against which the work is done by the water, ∆u= increase in the volume of the water.

iii) A mixture of fuel and oxygen is burned in a constant-volume chamber surrounded by a water bath. It was noticed that the temperature of water is increased during the process. Treating the mixture of fuel and oxygen as the system,
(a) Has heat been transferred ?
(b) Has work been done?
(c) What is the sign of ∆u ?
Answer:
(a) Heat has been transferred from the chamber to the water bath.
(b) No work is done by the system (the mixture of fuel and oxygen) as there is no change in its volume.
(c) There is increase in the temperature of water. Therefore, ∆u is positive for water.
For the system (the mixture of fuel and oxygen), ∆u is negative.

iv) Draw a p-V diagram and explain the concept of positive and negative work. Give one example each.
Answer:
Consider some quantity of an ideal gas enclosed in a cylinder fitted with a movable, massless and frictionless piston.

Suppose the gas is allowed to expand by moving the piston outward extremely slowly. There is decrease in pressure of the gas as the volume of the gas increases. Below figure shows the corresponding P-V diagram.

In this case, the work done by the gas on its surroundings,
W = \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\) (= area under the curve) is positive as the volume of the gas has increased from V_i to V_f.
Let us now suppose that starting from the same

initial condition, the piston is moved inward extremely slowly so that the gas is compressed. There is increase in pressure of the gas as the volume of the gas decreases. Figure shows the corresponding P-V diagram.
In this case, the work done by the gas on its surroundings, W = \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\) (= area under the curve) is negative as the volume of the gas has decreased from V_i to V_f.

v) A solar cooker and a pressure cooker both are used to cook food. Treating them as thermodynamic systems, discuss the similarities and differences between them.
Answer:
Similarities :

1. Heat is added to the system.
2. There is increase in the internal energy of the system.
3. Work is done by the system on its environment.

Differences : In a solar cooker, heat is supplied in the form of solar radiation. The rate of supply of heat is relatively low.

In a pressure cooker, usually LPG is used (burned) to provide heat. The rate of supply of heat v is relatively high.

As a result, it takes very long time for cooking when a solar cooker is used. With a pressure cooker, it does not take very long time for cooking.
[Note : A solar cooker can be used only when enough solar radiation is available.]

Question 4.
A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system. [Ans: -106.5 J]
Answer:
Data : P = 1 atm = 1.013 × 10⁵ Pa, V₁ = 5 litres = 5 × 10^-3 m³ V₂ = 10 litres = 10 × 10^-3 m³, Q = 400J.
The work done by the system (gas in this case) on its surroundings,
W = P(V₂ – V₁)
= (1.013 × 10⁵ Pa) (10 × 10^-3 m³ – 5 × 10^-3 m³)
= 1.013 (5 × 10²)J = 5.065 × 10²J
The change in the internal energy of the system, ∆u = Q – W = 400J – 506.5J = -106.5J
The minus sign shows that there is a decrease in the internal energy of the system.

Question 5.
A system releases 130 kJ of heat while 109 kJ of work is done on the system. Calculate the change in internal energy.
[Ans: ∆U = 21 kJ]
Answer:
Data : Q = -130kj, W= – 109kJ
∆u = Q – W = – 130kJ – ( – 109kJ)
= (-130 + 104) kJ = – 26 kj.
This is the change (decrease) in the internal energy.

Question 6.
Efficiency of a Carnot cycle is 75%. If temperature of the hot reservoir is 727ºC, calculate the temperature of the cold reservoir. [Ans: 23ºC]
Data : η = 75% = 0.75, T_H = (273 + 727) K = 1000 K
η = 1 – \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\) ∴ \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\) = 1 – η
∴ T_C = T_C(1 – η) = 1000 K (1 – 0.75)
= 250K = (250 – 273)°C
= -23 °C
This is the temperature of the cold reservoir.

Question 7.
A Carnot refrigerator operates between 250K and 300K. Calculate its coefficient of performance. [Ans: 5]
Answer:
Data : T_C = 250 K, T_H = 300 K
K = \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}\) = \(\frac{250 \mathrm{~K}}{300 \mathrm{~K}-250 \mathrm{~K}}\) = \(\frac{250}{50}\) = 5
This is the coefficient of performance of the refrigerator.

Question 8.
An ideal gas is taken through an isothermal process. If it does 2000 J of work on its environment, how much heat is added to it? [Ans: 2000J]
Answer:
Data : W = 2000 J, isothermal process
In this case, the change in the internal energy of the gas, ∆u, is zero as the gas is taken through an isothermal process.
Hence, the heat added to it,
Q = ∆ u + W = 0 + W = 200J

Question 9.
An ideal monatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. What is the ratio of the final pressure to its initial pressure? [Ans: 5.656]
Answer:
Data : T_f = 2T_i, monatomic gas ∴ γ = 5/3


This is the ratio of the final pressure (P_f) to the initial pressure (P_i).

Question 10.
A hypothetical thermodynamic cycle is shown in the figure. Calculate the work done in 25 cycles.

[Ans: 7.855 × 10⁴ J]
Answer:


25 cycles
The work done in one cycle, \(\oint\)PdV
= πab = (3.142) (2 × 10^-3 m³) (5 × 10⁵ Pa)
= 3.142 × 10³J
Hence, the work done in 25 cycles
= (25) (3.142 × 10³ J) = 7.855 × 10⁴J

Question 11.
The figure shows the V-T diagram for one cycle of a hypothetical heat engine which uses the ideal gas. Draw the p-V diagram and P-V diagram of the system. [Ans: (a)]

[Ans: (b)]
Answer:
(a) P-V diagram (Schematic)

ab: isobaric process,
bc : isothermal process,
cd : isobaric process,
da : isothermal process
\(\frac{P_{\mathrm{a}} V_{\mathrm{a}}}{T_{\mathrm{a}}}\) = \(\frac{P_{\mathrm{b}} V_{\mathrm{b}}}{T_{\mathrm{b}}}\) = \(\frac{P_{\mathrm{c}} V_{\mathrm{c}}}{T_{\mathrm{c}}}\) = \(\frac{P_{\mathrm{d}} V_{\mathrm{d}}}{T_{\mathrm{d}}}\) = nR

(b) P—T diagram (Schematic)

Question 12.
A system is taken to its final state from initial state in hypothetical paths as shown figure. Calculate the work done in each case.

[Ans: AB = 2.4 × 10⁶ J, CD = -8 × 10⁵ J, BC and DA zero, because constant volume change]
Answer:

Data: P_A = P_B = 6 × 10⁵ Pa, P_C = P_D = 2 × 10⁵ Pa V_A = V_D = 2 L, V_B = V_C6 L, 1 L = 10^-3m³
(i) The work done along the path A → B (isobaric process), W_AB = P_A (V_B – V_A) = (6 × 10⁵ Pa)(6 – 2)(10^-3 m³) = 2.4 × 10³ J
(ii) W_BC = zero as the process is isochoric (V = constant).
(iii) The work done along the path C → D (isobaric process), W_CD = P_C (V_D – V_C)
= (2 × 10⁵ Pa) (2 – 6) (10^-3m³) = -8 × 10²J
(iv) W_DA = zero as V = constant.

12th Physics Digest Chapter 4 Thermodynamics Intext Questions and Answers

Can you tell? (Textbook Page No. 76)

Question 1.
Why is it that different objects kept on a table at room temperature do not exchange heat with the table ?
Answer:
The objects do exchange heat with the table but there is no net transfer of energy (heat) as the objects and the table are at the same temperature.

Can you tell? (Textbook Page No. 77)

Question 1.
Why is it necessary to make a physical contact between a thermocouple and the object for measuring its temperature ?
Answer:
For heat transfer to develop thermoemf.

Can you tell? (Textbook Page No. 81)

Question 1.
Can you explain the thermodynamics involved in cooking food using a pressure cooker ?
Answer:
Basically, heat is supplied by the burning fuel causing increase in the internal energy of the food (system), and the system does some work on its surroundings. In the absence of any data about the components of food and their thermal and chemical properties, we cannot evaluate changes in internal energy and work done.

Use your brain power (Textbook Page No. 85)

Question 1.
Verify that the area under the P-V curve has dimensions of work.
Answer:
Area under the P-V curve is \(\int_{V_{\mathrm{i}}}^{V_{\mathrm{f}}} P d V\), where P is the pressure and V is the volume.

Use your brain power (Textbook Page No. 91)

Question 1.
Show that the isothermal work may also be expressed as W = nRT ln\(\left(\frac{\boldsymbol{P}_{\mathrm{i}}}{\boldsymbol{P}_{\mathrm{f}}}\right)\),
Answer:
In the usual notation,
W = nRT In \(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\) and P_i V_i = P_f V_f = nRT in an isothermal process
∴ \(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\) = \(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\) and W = nRT ln \(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\)

Use your brain power (Textbook Page No. 94)

Question 1.
Why is the P-V curve for an adiabatic process steeper than that for an isothermal process ?
Answer:
Adiabatic process : PV^γ = constant
∴ V^γdP + γPV^γ-1 dV = 0
∴ \(\frac{d P}{d V}\) = – \(\frac{\gamma P}{V}\)
Isothermal process : PV = constant
∴ pdV + VdP = 0 ∴ \(\frac{d P}{d V}\) = \(\frac{P}{-V}\)
Now, γ > 1

∴ \(\frac{d P}{d V}\) is the slope of the P – V curve.
∴The P – V curve for an adiabatic process is steeper than that for an isothermal process.

Question 2.
Explain formation of clouds at high altitude.
Answer:
As the temperature of the earth increases due to absorption of solar radiation, water from rivers, lakes, oceans, etc. evaporates and rises to high altitude. Water vapour forms clouds as water molecules come together under appropriate conditions. Clouds are condensed water vapour and are of various type, names as cumulus clouds, nimbostratus clouds, stratus clouds and high-flying cirrus clouds.

Can you tell? (Textbook Page No. 95)

Question 1.
How would you interpret Eq. 4.21 (Q = W) for a cyclic process ?
Answer:
It means ∆ u = 0 for a cyclic process as the system returns to its initial state.

Question 2.
An engine works at 5000 rpm, and it performs 1000 J of work in one cycle. If the engine runs for 10 min, how much total work is done by the engine ?
Answer:
The total work done by the engine = (1000 J/cycle) (5000 cycles/min) (10 min) = 5 × 10⁷ J.

Do you know? (Textbook Page No. 101)

Question 1.
The capacity of an air conditioner is expressed in a tonne. Do you know why?
Answer:
Before refrigerators and AC were invented, cooling was done by using blocks of ice. When cooling machines were invented, their capacity was expressed in terms of the equivalent amount of ice melted in a day (24 hours). The same term is used even today.
(Note : 1 tonne = 1000 kg = 2204.6 pounds, 1 ton (British) = 2240 pounds = 1016.046909 kg, 1 ton (US) = 2000 pounds = 907.184 kg.]

Use your brainpower (Textbook Page No. 105)

Question 1.
Suggest a practical way to increase the efficiency of a heat engine.
Answer:
The efficiency of a heat engine can be increased by choosing the hot reservoir at very high temperature and cold reservoir at very low temperature.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 3 Kinetic Theory of Gases and Radiation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 3 Kinetic Theory of Gases and Radiation

1. Choose the correct option.

i) In an ideal gas, the molecules possess
(A) only kinetic energy
(B) both kinetic energy and potential energy
(C) only potential energy
(D) neither kinetic energy nor potential energy
Answer:
(A) only kinetic energy

ii) The mean free path λ of molecules is given by
(A) \(\sqrt{\frac{2}{\pi n d^{2}}}\)
(B) \(\frac{1}{\pi n d^{2}}\)
(C) \(\frac{1}{\sqrt{2} \pi n d^{2}}\)
(D) \(\frac{1}{\sqrt{2 \pi n d^{1}}}\)
where n is the number of molecules per unit volume and d is the diameter of the molecules.
Answer:
(C) \(\frac{1}{\sqrt{2} \pi n d^{2}}\)

iii) If pressure of an ideal gas is decreased by 10% isothermally, then its volume will
(A) decrease by 9%
(B) increase by 9%
(C) decrease by 10%
(D) increase by 11.11%
Answer:
(D) increase by 11.11% [Use the formula P₁V₁ = P₂V₂. It gives \(\frac{V_{2}}{V_{1}}\) = \(\frac{1}{0.9}\) = 1.111 ∴ \(\frac{V_{2}-V_{1}}{V_{1}}\) = 0.1111, i.e., 11.11%

iv) If a = 0.72 and r = 0.24, then the value of t_r is
(A) 0.02
(B) 0.04
(C) 0.4
(D) 0.2
Answer:
(B) 0.04

v) The ratio of emissive power of a perfect blackbody at 1327°C and 527°C is
(A) 4 : 1
(B) 16 : 1
(C) 2 : 1
(D) 8 : 1
Answer:
(B) 16 : 1

2. Answer in brief.

i) What will happen to the mean square speed of the molecules of a gas if the temperature of the gas increases?
Answer:
If the temperature of a gas increases, the mean square speed of the molecules of the gas will increase in the same proportion.
[Note: \(\overline{v^{2}}\) = \(\frac{3 n R T}{N m}\) ∴ \(\overline{v^{2}}\) ∝ T for a fixed mass of gas.]

ii) On what factors do the degrees of freedom depend?
Answer:
The degrees of freedom depend upon
(i) the number of atoms forming a molecule
(ii) the structure of the molecule
(iii) the temperature of the gas.

iii) Write ideal gas equation for a mass of 7 g of nitrogen gas.
Answer:
In the usual notation, PV = nRT.

Therefore, the corresponding ideal gas equation is
PV = \(\frac{1}{4}\)RT.

iv) What is an ideal gas ? Does an ideal gas exist in practice ?.
Answer:
An ideal or perfect gas is a gas which obeys the gas laws (Boyle’s law, Charles’ law and Gay-Lussac’s law) at all pressures and temperatures. An ideal gas cannot be liquefied by application of pressure or lowering the temperature.

A molecule of an ideal gas is an ideal particle having only mass and velocity. Its structure and size are ignored. Also, intermolecular forces are zero except during collisions

v) Define athermanous substances and diathermanous substances.
Answer:

1. A substance which is largely opaque to thermal radiations, i.e., a substance which does not transmit heat radiations incident on it, is known as an athermanous substance.
2. A substance through which heat radiations can pass is known as a diathermanous substance.

Question 3.
When a gas is heated its temperature increases. Explain this phenomenon based on kinetic theory of gases.
Answer:
Molecules of a gas are in a state of continuous random motion. They possess kinetic energy. When a gas is heated, there is increase in the average kinetic energy per molecule of the gas. Hence, its temperature increases (the average kinetic energy per molecule being proportional to the absolute temperature of the gas).

Question 4.
Explain, on the basis of kinetic theory, how the pressure of gas changes if its volume is reduced at constant temperature.
Answer:
The average kinetic energy per molecule of a gas is constant at constant temperature. When the volume of a gas is reduced at constant temperature, the number of collisions of gas molecules per unit time with the walls of the container increases. This increases the momentum transferred per unit time per unit area, i.e., the force exerted by the gas on the walls. Hence, the pressure of the gas increases.

Question 5.
Mention the conditions under which a real gas obeys ideal gas equation.
Answer:
A real gas obeys ideal gas equation when temperature is vey high and pressure is very low.
[ Note : Under these conditions, the density of a gas is very low. Hence, the molecules, on an average, are far away from each other. The intermolecular forces are then not of much consequence. ]

Question 6.
State the law of equipartition of energy and hence calculate molar specific heat of mono-and di-atomic gases at constant volume and constant pressure.
Answer:
Law of equipartition of energy : For a gas in thermal equilibrium at absolute temperature T, the average energy for a molecule, associated with each quadratic term (each degree of freedom), is \(\frac{1}{2}\)k_BT,
where k_B is the Boltzmann constant. OR

The energy of the molecules of a gas, in thermal equilibrium at a thermodynamic temperature T and containing large number of molecules, is equally divided among their available degrees of freedom, with the energy per molecule for each degree of freedom equal to \(\frac{1}{2}\)k_BT, where k_B is the Boltzmann constant.
(a) Monatomic gas : For a monatomic gas, each atom has only three degrees of freedom as there can be only translational motion. Hence, the average energy per atom is \(\frac{3}{2}\)k_BT. The total internal energy per mole of the gas is E = \(\frac{3}{2}\)N_Ak_BT, where N_A is the Avogadro
number.
Therefore, the molar specific heat of the gas at constant volume is
C_V = \(\frac{d E}{d T}\) = \(\frac{3}{2}\)N_Ak_B = \(\frac{3}{2}\)R,
where R is the universal gas constant.
Now, by Mayer’s relation, C_p — C_v = R, where C_p is the specific heat of the gas at constant pressure.
∴ C_P = C_V + R = \(\frac{3}{2}\)R + R = \(\frac{5}{2}\)R

(b) Diatomic gas : Treating the molecules of a diatomic gas as rigid rotators, each molecule has three translational degrees of freedom and two rotational degrees of freedom. Hence, the average energy per molecule is
3(\(\frac{1}{2}\)k_BT) + 2(\(\frac{1}{2}\)k_BT) = \(\frac{5}{2}\)k_BT
The total internal energy per mole of the gas is
E = \(\frac{5}{2}\)N_Ak_BT.
∴ C_V = \(\frac{d E}{d T}\) = \(\frac{5}{2}\)N_Ak_B = \(\frac{5}{2}\)R and
C_P = C_V + R = \(\frac{5}{2}\)R + R = \(\frac{7}{2}\)R
A soft or non-rigid diatomic molecule has, in addition, one frequency of vibration which contributes two quadratic terms to the energy.
Hence, the energy per molecule of a soft diatomic molecule is

Therefore, the energy per mole of a soft diatomic molecule is
E = \(\frac{7}{2}\)k_BT × N_A = \(\frac{7}{2}\) RT
In this case, C_V = \(\frac{d E}{d T}\) = \(\frac{7}{2}\)R and
C_P = C_V + R = \(\frac{7}{2}\)R + R = \(\frac{9}{2}\)R
[Note : For a monatomic gas, adiabatic constant,
γ = \(\frac{C_{p}}{C_{V}}\) = \(\frac{5}{3}\). For a diatomic gas, γ =\(\frac{7}{5}\) or \(\frac{9}{7}\).]

Question 7.
What is a perfect blackbody ? How can it be realized in practice?
Answer:
A perfect blackbody or simply a blackbody is defined as a body which absorbs all the radiant energy incident on it.

Fery designed a spherical blackbody which consists of a hollow double-walled, metallic sphere provided with a tiny hole or aperture on one side, in below figure. The inside wall of the sphere is blackened with lampblack while the outside is silver-plated. The space between the two walls is evacuated to minimize heat loss by conduction and convection.

Any radiation entering the sphere through the aperture suffers multiple reflections where about 97% of it is absorbed at each incidence by the coating of lampblack. The radiation is almost completely absorbed after a number of internal reflections. A conical projection on the inside wall opposite the hole minimizes probability of incident radiation escaping out.

When the sphere is placed in a bath of suitable fused salts, so as to maintain it at the desired temperature, the hole serves as a source of black-body radiation. The intensity and the nature of the radiation depend only on the temperature of the walls.

A blackbody, by definition, has coefficient of absorption equal to 1. Hence, its coefficient of reflection and coefficient of transmission are both zero.

The radiation from a blackbody, called blackbody radiation, covers the entire range of the electromagnetic spectrum. Hence, a blackbody is called a full radiator.

Question 8.
State (i) Stefan-Boltmann law and
(ii) Wein’s displacement law.
Answer:
(i) The Stefan-Boltzmann law : The rate of emission of radiant energy per unit area or the power radiated per unit area of a perfect blackbody is directly proportional to the fourth power of its absolute temperature. OR
The quantity of radiant energy emitted by a perfect blackbody per unit time per unit surface area of the body is directly proportional to the fourth power of its absolute temperature.

(ii) Wien’s displacement law : The wavelength for which the emissive power of a blackbody is maximum, is inversely proportional to the absolute temperature of the blackbody.
OR
For a blackbody at an absolute temperature T, the product of T and the wavelength λ_m corresponding to the maximum radiation of energy is a constant.
λ_mT = b, a constant.
[Notes: (1) The law stated above was stated by Wilhelm Wien (1864-1928) German Physicist. (2) The value of the constant b in Wien’s displacement law is 2.898 × 10^-3 m.K.]

Question 9.
Explain spectral distribution of blackbody radiation.
Answer:
Blackbody radiation is the electromagnetic radiation emitted by a blackbody by virtue of its temperature. It extends over the whole range of wavelengths of electromagnetic waves. The distribution of energy over this entire range as a function of wavelength or frequency is known as the spectral distribution of blackbody radiation or blackbody radiation spectrum.

If R_λ is the emissive power of a blackbody in the wavelength range λ and λ + dλ, the energy it emits per unit area per unit time in this wavelength range depends on its absolute temperature T, the wavelength λ and the size of the interval dλ.

Question 10.
State and prove Kirchoff’s law of heat radiation.
Answer:
Kirchhoff’s law of heat radiation : At a given temperature, the ratio of the emissive power to the coefficient of absorption of a body is equal to the emissive power of a perfect blackbody at the same temperature for all wavelengths.
OR
For a body emitting and absorbing thermal radiation in thermal equilibrium, the emissivity is equal to its absorptivity.

Theoretical proof: Consider the following thought experiment: An ordinary body A and a perfect black body B are enclosed in an athermanous enclosure as shown in below figure.

According to Prevost’s theory of heat exchanges, there will be a continuous exchange of radiant energy between each body and its surroundings. Hence, the two bodies, after some time, will attain the same temperature as that of the enclosure.

Let a and e be the coefficients of absorption and emission respectively, of body A. Let R and R_b be the emissive powers of bodies A and B, respectively. ;

Suppose that Q is the quantity of radiant energy incident on each body per unit time per unit surface area of the body.

Body A will absorb the quantity aQ per unit time per unit surface area and radiate the quantity R per unit time per unit surface area. Since there is no change in its temperature, we must have,
aQ = R … (1)
As body B is a perfect blackbody, it will absorb the quantity Q per unit time per unit surface area and radiate the quantity R_b per unit time per unit surface area.

Since there is no change in its temperature, we must have,
Q = R_b ….. (2)
From Eqs. (1) and (2), we get,
a = \(\frac{R}{Q}\) = \(\frac{R}{R_{\mathrm{b}}}\) ….. (3)
From Eq. (3), we get, \(\frac{R}{a}\) = R_b OR

By definition of coefficient of emission,
\(\frac{R}{R_{\mathrm{b}}}\) …(4)
From Eqs. (3) and (4), we get, a = e.
Hence, the proof of Kirchhoff ‘s law of radiation.

Question 11.
Calculate the ratio of mean square speeds of molecules of a gas at 30 K and 120 K. [Ans: 1:4]
Answer:
Data : T₁ = 30 K, T₂ = 120 K

This is the required ratio.

Question 12.
Two vessels A and B are filled with same gas where volume, temperature and pressure in vessel A is twice the volume, temperature and pressure in vessel B. Calculate the ratio of number of molecules of gas in vessel A to that in vessel B.
[Ans: 2:1]
Answer:
Data : V_A = 2V_B, T_A = 2T_B, P_A = 2P_B PV = Nk_BT

This is the required ratio.

Question 13.
A gas in a cylinder is at pressure P. If the masses of all the molecules are made one third of their original value and their speeds are doubled, then find the resultant pressure. [Ans: 4/3 P]
Answer:
Data : m₂ = m₁/3, v_{rms 2} = 2v_{rms 1} as the speeds of all molecules are doubled

This is the resultant pressure.

Question 14.
Show that rms velocity of an oxygen molecule is \(\sqrt{2}\) times that of a sulfur dioxide molecule at S.T.P.
Answer:

Question 15.
At what temperature will oxygen molecules have same rms speed as helium molecules at S.T.P.? (Molecular masses of oxygen and helium are 32 and 4 respectively)
[Ans: 2184 K]
Answer:
Data : T₂ = 273 K, M₀₁ (oxygen) = 32 × 10^-3 kg/mol, M₀₂ (hydrogen) = 4 × 10^{– 3} kg/mol
The rms speed of oxygen molecules, v₁ = \(\sqrt{\frac{3 R T_{1}}{M_{01}}}\) and that of helium molecules, v₂ = \(\sqrt{\frac{3 R T_{2}}{M_{02}}}\)
When v₁ = v₂,

Question 16.
Compare the rms speed of hydrogen molecules at 127 ºC with rms speed of oxygen molecules at 27 ºC given that molecular masses of hydrogen and oxygen are 2 and 32 respectively. [Ans: 8: 3]
Answer:
Data : M₀₁ (hydrogen) = 2 g/mol,
M₀₂ (oxygen) = 32 g/mol,
T₁ (hydrogen) = 273 + 127 = 400 K,
T₂ (oxygen) = 273 + 27 = 300 K
The rms speed, v_rms = \(\sqrt{\frac{3 R T}{M_{0}}}\),
where M₀ denotes the molar mass

Question 17.
Find kinetic energy of 5000 cc of a gas at S.T.P. given standard pressure is 1.013 × 10⁵ N/m².
[Ans: 7.598 × 10² J]
Answer:
Data : P = 1.013 × 10⁵ N/m², V = 5 litres
= 5 × 10^-3 m³
E = \(\frac{3}{2}\)PV
= \(\frac{3}{2}\)(1.013 × 10⁵ N/m²) (5 × 10^-3 m³)
= 7.5 × 1.013 × 10² J = 7.597 × 10² J
This is the required energy.

Question 18.
Calculate the average molecular kinetic energy
(i) per kmol
(ii) per kg
(iii) per molecule of oxygen at 127 ºC, given that molecular weight of oxygen is 32, R is 8.31 J mol^-1 K^-1 and Avogadro’s number N_A is 6.02 × 10²³ molecules mol^-1.
[Ans: 4.986 × 10⁶J, 1.558 × 10²J 8.282 × 10^-21 J]
Answer:
Data : T = 273 +127 = 400 K, molecular weight = 32 ∴ molar mass = 32 kg/kmol, R = 8.31 Jmol^-1 K^-1, N_A = 6.02 × 10²³ molecules mol^-1

(i) The average molecular kinetic energy per kmol of oxygen = the average kinetic energy per mol of oxygen × 1000
= \(\frac{3}{2}\)RT × 1000 = \(\frac{3}{2}\) (8.31) (400) (10³)\(\frac{\mathrm{J}}{\mathrm{kmol}}\)
= (600)(8.31)(10³) = 4.986 × 10⁶ J/kmol

(ii) The average molecular kinetic energy per kg of

(iii) The average molecular kinetic energy per molecule of oxygen

Question 19.
Calculate the energy radiated in one minute by a blackbody of surface area 100 cm² when it is maintained at 227ºC. (Take Stefen’s constant σ = 5.67 × 10^-8J m^-2s^-1K^-4)
[Ans: 2126.25 J]
Answer:
Data : t = one minute = 60 s, A = 100 cm²
= 100 × 10^-4 m² = 10^-2 m², T = 273 + 227 = 500 K,
σ = 5.67 × 10^-8 W/m².K⁴
The energy radiated, Q = σAT⁴t
= (5.67 × 10^-8)(10^-2)(500)⁴(60) J
= (5.67)(625)(60)(10^-2)J = 2126 J

Question 20.
Energy is emitted from a hole in an electric furnace at the rate of 20 W, when the temperature of the furnace is 727 ºC. What is the area of the hole? (Take Stefan’s constant σ to be 5.7 × 10^-8 J s^-1 m^-2 K^-4) [Ans: 3.509 × 10^-4m²]
Answer:
Data : \(\frac{Q}{t}\) = 20 W, T = 273 + 727 = 1000 K

Question 21.
The emissive power of a sphere of area 0.02 m² is 0.5 kcal s^-1 m^-2. What is the
amount of heat radiated by the spherical surface in 20 second? [Ans: 0.2 kcal]
Answer:
Data : R = 0.5 kcal s^-1m^-2, A = 0.02 m², t = 20 s Q = RAt = (0.5) (0.02) (20) = 0.2 kcal
This is the required quantity.

Question 22.
Compare the rates of emission of heat by a blackbody maintained at 727ºC and at 227ºC, if the black bodies are surrounded
by an enclosure (black) at 27ºC. What would be the ratio of their rates of loss of heat ?
[Ans: 18.23:1]
Answer:
Data : T₁ = 273 + 727 = 1000 K, T₂ = 273 + 227 = 500 K, T₀ = 273 + 27 = 300 K.
(i) The rate of emission of heat, \(\frac{d Q}{d t}\) = σAT⁴.
We assume that the surface area A is the same for the two bodies.

Question 23.
Earth’s mean temperature can be assumed to be 280 K. How will the curve of blackbody radiation look like for this temperature? Find out λ_max. In which part of the electromagnetic spectrum, does this value lie? (Take Wien’s constant b = 2.897 × 10^-3 m K) [Ans: 1.035 × 10^-5m, infrared region]
Answer:
Data : T = 280 K, Wien’s constant b = 2.897 × 10^-3 m.K
λ_maxT = b

This value lies in the infrared region of the electromagnetic spectrum.
The nature of the curve of blackbody radiation will be the same as shown in above, but the maximum will occur at 1.035 × 10^-5 m.

Question 24.
A small-blackened solid copper sphere of radius 2.5 cm is placed in an evacuated chamber. The temperature of the chamber is maintained at 100 ºC. At what rate energy must be supplied to the copper sphere to maintain its temperature at 110 ºC? (Take Stefan’s constant σ to be 5.670 × 10^-8 J s^-1 m^-2 K^-4 , π = 3.1416 and treat the sphere as a blackbody.)
[Ans: 0.9624 W]
Answer:
Data : r = 2.5 cm = 2.5 × 10^-2 m, T₀ = 273 + 100 = 373 K, T = 273 + 110 = 383 K,
σ = 5.67 × 10^-8 J s^-1 m^-2 k^-4
The rate at which energy must be supplied
σA(T⁴ — T₀⁴) = σ 4πr²(T⁴ – T₀⁴)
= (5.67 × 10^-8) (4) (3.142) (2.5 × 10^-2)² (383⁴ – 373⁴)
= (5.67) (4) (3.142) (6.25) (3.83⁴ – 3.73⁴) × 10^-4
= 0.9624W

Question 25.
Find the temperature of a blackbody if its spectrum has a peak at (a) λ_max = 700 nm (visible), (b) λ_max = 3 cm (microwave region) and (c) λ_max = 3 m (short radio waves) (Take Wien’s constant b = 2.897 × 10^-3 m K). [Ans: (a) 4138 K, (b) 0.09657 K, (c) 0.9657 × 10^-3 K]
Answer:
Data:(a) λ_max = 700nm=700 × 10^-9m,
(b) λ_max = 3cm = 3 × 10^-2 m, (c) λ_max = 3 m, b = 2.897 × 10^-3 m.K
λ_maxT = b

12th Physics Digest Chapter 3 Kinetic Theory of Gases and Radiation Intext Questions and Answers

Remember This (Textbook Page No. 60)

Question 1.
Distribution of speeds of molecules of a gas.
Answer:
Maxwell-Boltzmann distribution of molecular speeds is a relation that describes the distribution of speeds among the molecules of a gas at a given temperature.

The root-mean-square speed v_rms gives us a general idea of molecular speeds in a gas at a given temperature. However, not all molecules have the same speed. At any instant, some molecules move slowly and some very rapidly. In classical physics, molecular speeds may be considered to cover the range from 0 to ∞. The molecules constantly collide with each other and with the walls of the container and their speeds change on collisions. Also the number of molecules under consideration is very large statistically. Hence, there is an equilibrium distribution of speeds.

If dN_v represents the number of molecules with speeds between v and v + dv, dN_v remains fairly constant at equilibrium. We consider a gas of total N molecules. Let r\vdv be the probability that a molecule has its speed between v and v + dv. Then, dN_v = Nη_vdv
so that the fraction, i.e., the relative number of molecules with speeds between v and v + dv is dN_v/N = η_vdv
Below figure shows the graph of η_v against v. The area of the strip with height η_v and width dv gives the fraction dN_v/N.

Remember This (Textbook Page No. 64)

Question 1.
If a hot body and a cold body are kept in vacuum, separated from each other, can they exchange heat ? If yes, which mode of transfer of heat causes change in their temperatures ? If not, give reasons.
Answer:
Yes. Radiation.

Remember This (Textbook Page No. 66)

Question 1.
Can a perfect blackbody be realized in practice ?
Answer:
For almost all practical purposes, Fery’s blackbody is very close to a perfect blackbody.

Question 2.
Are good absorbers also good emitters ?
Answer:
Yes.

Use your brain power (Textbook Page No. 68)

Question 1.
Why are the bottom of cooking utensils blackened and tops polished ?
Answer:
The bottoms of cooking utensils are blackened to increase the rate of absorption of radiant energy and tops are polished to increase the reflection of radiation.

Question 2.
A car is left in sunlight with all its windows closed on a hot day. After some time it is observed that the inside of the car is warmer than the outside air. Why?
Answer:
The air inside the car is trapped and hence is a bad conductor of heat.

Question 3.
If surfaces of all bodies are continuously emitting radiant energy, why do they not cool down to 0 K?
Answer:
Bodies absorb radiant energy from their surroundings.

Can you tell? (Textbook Page No. 71)

Question 1.
λ_max the wavelength corresponding to maximum intensity for the Sun is in the blue-green region of the visible spectrum. Why does the Sun then appear yellow to us?
Answer:
The colour that we perceive depends upon a number of factors such as absorption and scattering by the atmosphere (which in turn depends upon the composition of air) and the spectral response of the human eye. The colour maybe yellow/orange/ red/white.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 2 Mechanical Properties of Fluids Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 2 Mechanical Properties of Fluids

1) Multiple Choice Questions

i) A hydraulic lift is designed to lift heavy objects of a maximum mass of 2000 kg. The area of cross-section of piston carrying
the load is 2.25 × 10^-2 m². What is the maximum pressure the piston would have to bear?
(A) 0.8711 × 10⁶ N/m²
(B) 0.5862 × 10⁷ N/m²
(C) 0.4869 × 10⁵ N/m²
(D) 0.3271 × 10⁴ N/m²
Answer:
(A) 0.8711 × 10⁶ N/m²

ii) Two capillary tubes of radii 0.3 cm and 0.6 cm are dipped in the same liquid. The ratio of heights through which the liquid will rise in the tubes is
(A) 1:2
(B) 2:1
(C) 1:4
(D) 4:1
Answer:
(B) 2:1

iii) The energy stored in a soap bubble of diameter 6 cm and T = 0.04 N/m is nearly
(A) 0.9 × 10^-3 J
(B) 0.4 × 10^-3 J
(C) 0.7 × 10^-3 J
(D) 0.5 × 10^-3 J
Answer:
(A) 0.9 × 10^-3 J

iv) Two hail stones with radii in the ratio of 1:4 fall from a great height through the atmosphere. Then the ratio of their terminal velocities is
(A) 1:2
(B) 1:12
(C) 1:16
(D) 1:8
Answer:
(C) 1:16

v) In Bernoulli’s theorem, which of the following is conserved?
(A) linear momentum
(B) angular momentum
(C) mass
(D) energy
Answer:
(D) energy

2) Answer in brief.

i) Why is the surface tension of paints and lubricating oils kept low?
Answer:
For better wettability (surface coverage), the surface tension and angle of contact of paints and lubricating oils must below.

ii) How much amount of work is done in forming a soap bubble of radius r?
Answer:
Let T be the surface tension of a soap solution. The initial surface area of soap bubble = 0
The final surface area of soap bubble = 2 × 4πr²
∴ The increase in surface area = 2 × 4πr^2-
The work done in blowing the soap bubble is W = surface tension × increase in surface area = T × 2 × 4πr² = 8πr²T

iii) What is the basis of the Bernoulli’s principle?
Answer:
Conservation of energy.

iv) Why is a low density liquid used as a manometric liquid in a physics laboratory?
Answer:
An open tube manometer measures the gauge pressure, p — p₀ = hpg, where p₀ is the pressure being measured, p₀ is the atmospheric pressure, h is the difference in height between the manometric liquid of density p in the two arms. For a given pressure p, the product hp is constant. That is, p should be small for h to be large. Therefore, for noticeably large h, laboratory manometer uses a low density liquid.

v) What is an incompressible fluid?
Answer:
An incompressible fluid is one which does not undergo change in volume for a large range of pressures. Thus, its density has a constant value throughout the fluid. In most cases, all liquids are incompressible.

Question 3.
Why two or more mercury drops form a single drop when brought in contact with each other?
Answer:
A spherical shape has the minimum surface area- to-volume ratio of all geometric forms. When two drops of a liquid are brought in contact, the cohesive forces between their molecules coalesces the drops into a single larger drop. This is because, the volume of the liquid remaining the same, the surface area of the resulting single drop is less than the combined surface area of the smaller drops. The resulting decrease in surface energy is released into the environment as heat.

Proof : Let n droplets each of radius r coalesce to form a single drop of radius R. As the volume of the liquid remains constant, volume of the drop = volume of n droplets
∴ \(\frac{4}{3}\)πR³ = n × \(\frac{4}{3}\)πr³
∴ R³ = nr³ ∴ R = \(\sqrt[3]{n}\)r
Surface area of n droplets = n × πR²
Surface area of the drop = 4πR² = n^2/3 × πR²
∴ The change in the surface area = surface area of drop – surface area of n droplets
= πR²(n^2/3 – n)
Since the bracketed term is negative, there is a decrease in surface area and a decrease in surface energy.

Question 4.
Why does velocity increase when water flowing in broader pipe enters a narrow pipe?
Answer:
When a tube narrows, the same volume occupies a greater length, as schematically shown in below figure. A₁ is the cross section of the broader pipe and that of narrower pipe is A₂. By the equation of continuity, V₂ = (A₁/A₂)V₁

Since A₁/A₂ > v₂ > v₁. For the same volume to pass points 1 and 2 in a given time, the speed must be greater at point 2.
The process is exactly reversible. If the fluid flows in the opposite direction, its speed decreases when the tube widens.

Question 5.
Why does the speed of a liquid increase and its pressure decrease when a liquid passes through constriction in a horizontal pipe?
Answer:

Consider a horizontal constricted tube.
Let A₁ and A₂ be the cross-sectional areas at points 1 and 2, respectively. Let v₁ and v₂ be the corresponding flow speeds, ρ is the density of the fluid in the pipeline. By the equation of continuity,
v₁A₁ = v₂A₂ …… (1)
∴ \(\frac{v_{2}}{v_{1}}\) = \(\frac{A_{1}}{A_{2}}\) > 1 (∵ A₁ > A₂)
Therefore, the speed of the liquid increases as it passes through the constriction. Since the meter is assumed to be horizontal, from Bernoulli’s equation we get,

Again, since A₁ > A₂, the bracketed term is positive so that p₁ > p₂. Thus, as the fluid passes through the constriction or throat, the higher speed results in lower pressure at the throat.

Question 6.
Derive an expression of excess pressure inside a liquid drop.
Answer:
Consider a small spherical liquid drop with a radius R. It has a convex surface, so that the pressure p on the concave side (inside the liquid) is greater than the pressure p0 on the convex side (outside the liquid). The surface area of the drop is
A = 4πR² … (1)
Imagine an increase in radius by an infinitesimal amount dR from the equilibrium value R. Then, the differential increase in surface area would be dA = 8πR ∙ dR …(2)
The increase in surface energy would be equal to the work required to increase the surface area :
dW = T∙dA = 8πTRdR …..(3)

We assume that dR is so small that the pressure inside remains the same, equal to p. All parts of the surface of the drop experience an outward force per unit area equal to ρ — ρ₀. Therefore, the work done by this outward pressure-developed force against the surface tension force during the increase in radius dR is
dW = (excess pressure × surface area) ∙ dR
= (ρ – ρ₀) × 4πnR² ∙ dR …..(4)
From Eqs. (3) and (4),
(ρ — ρ₀) × 4πR² ∙ dR = 8πTRdR
∴ ρ – ρ₀ = \(\frac{2 T}{R}\) …… (5)
which is called Laplace’s law for a spherical membrane (or Young-Laplace equation in spherical form).
[Notes : (1) The above method is called the principle of virtual work. (2) Equation (5) also applies to a gas bubble within a liquid, and the excess pressure in this case is also called the gauge pressure. An air or gas bubble within a liquid is technically called a cavity because it has only one gas-liquid interface. A bubble, on the other hand, such as a soap bubble, has two gas-liquid interfaces.]

Question 7.
Obtain an expression for conservation of mass starting from the equation of continuity.
Answer:

Consider a fluid in steady or streamline flow, that is its density is constant. The velocity of the fluid within a flow tube, while everywhere parallel to the tube, may change its magnitude. Suppose the velocity is \(\vec{v}_{1}\), at point P and \(\vec{v}_{2}\) at point. Q. If A₁ and A₂ are the cross-sectional areas of the tube at these two points, the volume flux across A₁, \(\frac{d}{d t}\)(V₂) = A₁v₁ and that across A₂, \(\frac{d}{d t}\)(V₂) = A₂v₂
By the equation of continuity of flow for a fluid, A₁v₁ = A₂V₂
i.e., \(\frac{d}{d t}\)(V₁) = \(\frac{d}{d t}\)(V₂)
If ρ₁ and ρ₁ are the densities of the fluid at P and Q, respectively, the mass flux across A₁, \(\frac{d}{d t}\)(m₁) = \(\frac{d}{d t}\)(ρ₁ v₁) = A₁ρ₁v₁
and that across A₂, \(\frac{d}{d t}\)(m₂) = \(\frac{d}{d t}\)(ρ₂V₂) = A₂ρ₂v₂
Since no fluid can enter or leave through the boundary of the tube, the conservation of mass requires the mass fluxes to be equal, i.e.,
\(\frac{d}{d t}\)(m₁) = \(\frac{d}{d t}\)(m₂)
i.e., A₁ρ₁v₁ = A₂ρ₂v₂
i. e., Apv = constant
which is the required expression.

Question 8.
Explain the capillary action.
Answer:
(1) When a capillary tube is partially immersed in a wetting liquid, there is capillary rise and the liquid meniscus inside the tube is concave, as shown in below figure.

Consider four points A, B, C, D, of which point A is just above the concave meniscus inside the capillary and point B is just below it. Points C and D are just above and below the free liquid surface outside.

Let P_A, P_B, P_C and P_D be the pressures at points A, B, C and D, respectively.
Now, P_A = P_C = atmospheric pressure
The pressure is the same on both sides of the free surface of a liquid, so that

The pressure on the concave side of a meniscus is always greater than that on the convex side, so that
P_A > P_B
∴ P_D > P_B (∵ P_A = P_D)

The excess pressure outside presses the liquid up the capillary until the pressures at B and D (at the same horizontal level) equalize, i.e., P_B becomes equal to P_D. Thus, there is a capillary rise.

(2) For a non-wetting liquid, there is capillary depression and the liquid meniscus in the capillary tube is convex, as shown in above figure.

Consider again four points A, B, C and D when the meniscus in the capillary tube is at the same level as the free surface of the liquid. Points A and B are just above and below the convex meniscus. Points C and D are just above and below the free liquid surface outside.

The pressure at B (P_B) is greater than that at A (P_A). The pressure at A is the atmospheric pressure H and at D, P_D \(\simeq\) H = P_A. Hence, the hydrostatic pressure at the same levels at B and D are not equal, P_B > P_D. Hence, the liquid flows from B to D and the level of the liquid in the capillary falls. This continues till the pressure at B’ is the same as that D’, that is till the pressures at the same level are equal.

Question 9.
Derive an expression for capillary rise for a liquid having a concave meniscus.
Answer:
Consider a capillary tube of radius r partially immersed into a wetting liquid of density p. Let the capillary rise be h and θ be the angle of contact at the edge of contact of the concave meniscus and glass. If R is the radius of curvature of the meniscus then from the figure, r = R cos θ.

Surface tension T is the tangential force per unit length acting along the contact line. It is directed into the liquid making an angle θ with the capillary wall. We ignore the small volume of the liquid in the meniscus. The gauge pressure within the liquid at a depth h, i.e., at the level of the free liquid surface open to the atmosphere, is
ρ – ρ_o = ρgh …. (1)
By Laplace’s law for a spherical membrane, this gauge pressure is
ρ – ρ_o = \(\frac{2 T}{R}\) ….. (2)
∴ hρg = \(\frac{2 T}{R}\) = \(\frac{2 T \cos \theta}{r}\)
∴ h = \(\frac{2 T \cos \theta}{r \rho g}\) …. (3)
Thus, narrower the capillary tube, the greater is the capillary rise.
From Eq. (3),
T = \(\frac{h \rho r g}{2 T \cos \theta}\) … (4)
Equations (3) and (4) are also valid for capillary depression h of a non-wetting liquid. In this case, the meniscus is convex and θ is obtuse. Then, cos θ is negative but so is h, indicating a fall or depression of the liquid in the capillary. T is positive in both cases.
[Note : The capillary rise h is called Jurin height, after James Jurin who studied the effect in 1718. For capillary rise, Eq. (3) is also called the ascent formula.]

Question 10.
Find the pressure 200 m below the surface of the ocean if pressure on the free surface of liquid is one atmosphere. (Density of sea water = 1060 kg/m³) [Ans. 21.789 × 10⁵ N/m²]
Answer:
Data : h = 200 m, p = 1060 kg/m³,
p₀ = 1.013 × 10⁵ Pa, g = 9.8 m/s²
Absolute pressure,
p = p₀ + hρg
= (1.013 × 10³) + (200)(1060)(9.8)
= (1.013 × 10⁵) + (20.776 × 10⁵)
= 21.789 × 10⁵ = 2.1789 MPa

Question 11.
In a hydraulic lift, the input piston had surface area 30 cm² and the output piston has surface area of 1500 cm². If a force of 25 N is applied to the input piston, calculate weight on output piston. [Ans. 1250 N]
Answer:
Data : A₁ = 30 cm² = 3 × 10^-3 m²,
A₂ = 1500 cm² = 0.15 m², F₁ = 25 N
By Pascal’s law,
\(\frac{F_{1}}{A_{1}}\) = \(\frac{F_{2}}{A_{2}}\)
∴ The force on the output piston,
F₂ = F₁\(\frac{A_{2}}{A_{1}}\) = (25)\(\frac{0.15}{3 \times 10^{-3}}\) = 25 × 50 = 1250 N

Question 12.
Calculate the viscous force acting on a rain drop of diameter 1 mm, falling with a uniform velocity 2 m/s through air. The coefficient of viscosity of air is 1.8 × 10^-5 Ns/m².
[Ans. 3.393 × 10^-7 N]
Answer:
Data : d = 1 mm, v₀ = 2 m / s,
η = 1.8 × 10^-5 N.s/m²
r = \(\frac{d}{2}\) = 0.5 mm = 5 × 10^-4 m
By Stokes’ law, the viscous force on the raindrop is f = 6πηrv₀
= 6 × 3.142 (1.8 × 10^-5 N.s/m² × 5 × 10^-4 m)(2 m/s)
= 3.394 × 10^-7 N

Question 13.
A horizontal force of 1 N is required to move a metal plate of area 10^-2 m² with a velocity of 2 × 10^-2 m/s, when it rests on a layer of oil 1.5 × 10^-3 m thick. Find the coefficient of viscosity of oil. [Ans. 7.5 Ns/m²]
Answer:
Data : F = 1 N, A = 10^-2m², v₀ = 2 × 10^-2 y = 1.5 × 10^-3 m
Velocity gradient, \(\frac{d v}{d y}\) = \(\frac{2 \times 10^{-2}}{1.5 \times 10^{-3}}\) = \(\frac{40}{3}\)s^-1

Question 14.
With what terminal velocity will an air bubble 0.4 mm in diameter rise in a liquid of viscosity 0.1 Ns/m² and specific gravity 0.9? Density of air is 1.29 kg/m³. [Ans. – 0.782 × 10^-3 m/s, The negative sign indicates that the bubble rises up]
Answer:
Data : d = 0.4 mm, η = 0.1 Pa.s, ρ_L = 0.9 × 10³ kg/m³ = 900 kg/m³, ρ_air = 1.29 kg/m³, g = 9.8 m/s².
Since the density of air is less than that of oil, the air bubble will rise up through the liquid. Hence, the viscous force is downward. At terminal velocity, this downward viscous force is equal in magnitude to the net upward force.
Viscous force = buoyant force – gravitational force

Question 15.
The speed of water is 2m/s through a pipe of internal diameter 10 cm. What should be the internal diameter of nozzle of the pipe if the speed of water at nozzle is 4 m/s?
[Ans. 7.07 × 10^-2m]
Answer:
Data : d₁ = 10 cm = 0.1 m, v₁ = 2 m/s, v₂ = 4 m/s
By the equation of continuity, the ratio of the speed is

Question 16.
With what velocity does water flow out of an orifice in a tank with gauge pressure 4 × 10⁵ N/m² before the flow starts? Density of water = 1000 kg/m³. [Ans. 28.28 m/s]
Answer:
Data : ρ — ρ₀ = 4 × 10⁵ Pa, ρ = 10³ kg/m³
If the orifice is at a depth h from the water surface in a tank, the gauge pressure there is
ρ – ρ₀ = hρg … (1)
By Toricelli’s law of efflux, the velocity of efflux,
v = \(\sqrt{2 g h}\) …(2)
Substituting for h from Eq. (1),

Question 17.
The pressure of water inside the closed pipe is 3 × 10⁵ N/m². This pressure reduces to 2 × 10⁵ N/m² on opening the value of the pipe. Calculate the speed of water flowing through the pipe. (Density of water = 1000 kg/m³). [Ans. 14.14 m/s]
Answer:
Data : p₁ = 3 × 10⁵ Pa, v₁ = 0, p₂ = 2 × 10⁵ Pa, ρ = 10³ kg/m³
Assuming the potential head to be zero, i.e., the pipe to be horizontal, the Bernoulli equation is

Question 18.
Calculate the rise of water inside a clean glass capillary tube of radius 0.1 mm, when immersed in water of surface tension 7 × 10^-2 N/m. The angle of contact between water and glass is zero, density of water = 1000 kg/m³, g = 9.8 m/s².
[Ans. 0.1429 m]
Answer:
Data : r = 0.1 mm = 1 × 10^-4m, θ = 0°,
T = 7 × 10^-2 N/m, r = 10³ kg/m³, g = 9.8 m/s²
cos θ = cos 0° = 1

= 0.143 m

Question 19.
An air bubble of radius 0.2 mm is situated just below the water surface. Calculate the gauge pressure. Surface tension of water = 7.2 × 10^-2 N/m.
[Ans. 720 N/m²]
Answer:
Data : R = 2 × 10^-4m, T = 7.2 × 10^-2N/m, p = 10³ kg/m³
The gauge pressure inside the bubble = \(\frac{2 T}{R}\)
= \(\frac{2\left(7.2 \times 10^{-2}\right)}{2 \times 10^{-4}}\) = 7.2 × 10² = 720 Pa

Question 20.
Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m. [Ans. 1.628 × 10^-7 J = 1.628 erg]
Answer:
Data : r = 1 mm = 1 × 10^-3 m, T = 0.472 J/m²
Let R be the radius of the single drop formed due to the coalescence of 8 droplets of mercury.
Volume of 8 droplets = volume of the single drop as the volume of the liquid remains constant.
∴ 8 × \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\)πR³
∴ 8r³ = R³
∴ 2r = R
Surface area of 8 droplets = 8 × 4πr²
Surface area of single drop = 4πR²
∴ Decrease in surface area = 8 × 4πr² – 4πR²
= 4π(8r² – R²)
= 4π[8r² – (2r)²]
= 4π × 4r²
∴ The energy released = surface tension × decrease in surface area = T × 4π × 4r²
= 0.472 × 4 × 3.142 × 4 × (1 × 10^-3)²
= 2.373 × 10^-5 J
The decrease in surface energy = 0.072 × 4 × 3.142 × 18 × (1 × 10^-4)²
= 1.628 × 10^-7 J

Question 21.
A drop of mercury of radius 0.2 cm is broken into 8 identical droplets. Find the work done if the surface tension of mercury is 435.5 dyne/cm. [Ans. 2.189 × 10^-5J]
Answer:
Let R be the radius of the drop and r be the radius of each droplet.
Data : R = 0.2 cm, n = 8, T = 435.5 dyn/cm
As the volume of the liquid remains constant, volume of n droplets = volume of the drop
∴ n × \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\)πR³

Surface area of the drop = 4πR²
Surface area of n droplets = n × 4πR²
∴ The increase in the surface area = surface area of n droplets-surface area of drop
= 4π(nr² – R²) = 4π(8 × \(\frac{R^{2}}{4}\) – R²)
= 4π(2 — 1)R² = 4πR²
∴ The work done
= surface tension × increase in surface area
= T × 4πR² = 435.5 × 4 × 3.142 × (0.2)²
= 2.19 × 10² ergs = 2.19 × 10^-5 J

Question 22.
How much work is required to form a bubble of 2 cm radius from the soap solution having surface tension 0.07 N/m.
[Ans. 0.7038 × 10^-3 J]
Answer:
Data : r = 4 cm = 4 × 10^-2 m, T = 25 × 10^-3 N/m
Initial surface area of soap bubble = 0
Final surface area of soap bubble = 2 × 4πr²
∴ Increase in surface area = 2 × 4πr²
The work done
= surface tension × increase in surface area
= T × 2 × 4πr²
= 25 × 10^-3 × 2 × 4 × 3.142 × (4 × 10^-2)²
= 1.005 × 10^-3 J
The work done = 0.07 × 8 × 3.142 × (2 × 10^-2)²
= 7.038 × 10^-4 J

Question 23.
A rectangular wire frame of size 2 cm × 2 cm, is dipped in a soap solution and taken out. A soap film is formed, if the size of the film is changed to 3 cm × 3 cm, calculate the work done in the process. The surface tension of soap film is 3 × 10^-2 N/m. [Ans. 3 × 10^-5 J]
Answer:
Data : A₁ = 2 × 2 cm² = 4 × 10^-4 m²,
A₂ = 3 × 3 cm² =9 × 10^-4 m², T = 3 × 10^-2 N/m
As the film has two surfaces, the work done is W = 2T(A₂ – A₁)
= 2(3 × 10^-2)(9 × 10^-4 × 10^-4)
= 3.0 × 10^-5 J = 30 µJ

12th Physics Digest Chapter 2 Mechanical Properties of Fluids Intext Questions and Answers

Can you tell? (Textbook Page No. 27)

Question 1.
Why does a knife have a sharp edge or a needle has a sharp tip ?
Answer:
For a given force, the pressure over which the force is exerted depends inversely on the area of contact; smaller the area, greater the pressure. For instance, a force applied to an area of 1 mm² applies a pressure that is 100 times as great as the same force applied to an area of 1 cm². The edge of a knife or the tip of a needle has a small area of contact. That is why a sharp needle is able to puncture the skin when a small force is exerted, but applying the same force with a finger does not.

Use your brain power

Question 1.
A student of mass 50 kg is standing on both feet. Estimate the pressure exerted by the student on the Earth. Assume reasonable value to any quantity you need; justify your assumption. You may use g = 10 m/s², By what factor will it change if the student lies on back ?
Answer:
Assume area of each foot = area of a 6 cm × 25 cm rectangle.
∴ Area of both feet = 0.03 m²
∴ The pressure due to the student’s weight
= \(\frac{m g}{A}\) = \(\frac{50 \times 10}{0.03}\) = 16.7 kPa

According to the most widely used Du Bois formula for body surface area (BSA), the student’s BSA = 1.5 m², so that the area of his back is less than half his BSA, i.e., < 0.75 m². When the student lies on his back, his area of contact is much smaller than this. So, estimating the area of contact to be 0.3 m², i.e., 10 times more than the area of his feet, the pressure will be less by a factor of 10 or more, [Du Bois formula : BSA = 0.2025 × W^0.425 × H^0.725, where W is weight in kilogram and H is height in metre.]

Can you tell? (Textbook Page No. 30)

Question 1.
The figures show three containers filled with the same oil. How will the pressures at the reference compare ?

Answer:
Filled to the same level, the pressure is the same at the bottom of each vessel.

Use Your Brain Power (Textbook Page 35)

Question 1.
Prove that equivalent SI unit of surface tension is J/m².
Answer:
The SI unit of surface tension =

Try This (Textbook Page No. 36)

Question 1.
Take a ring of about 5 cm in diameter. Tie a thread slightly loose at two diametrically opposite points on the ring. Dip the ring into a soap solution and take it out. Break the film on any one side of the thread. Discuss what happens.
Answer:
On taking the ring out, there is a soap film stretched over the ring, in which the thread moves about quite freely. Now, if the film is punctured with a pin on one side-side A in below figure-then immediately the thread is pulled taut by the film on the other side as far as it can go. The thread is now part of a perfect circle, because the surface tension on the side F of the film acts everywhere perpendicular to the thread, and minimizes the surface area of the film to as small as possible.

Can You Tell ? (Textbook Page No. 38)

Question 1.
How does a waterproofing agent work ?
Answer:
Wettability of a surface, and thus its propensity for penetration of water, depends upon the affinity between the water and the surface. A liquid wets a surface when its contact angle with the surface is acute. A waterproofing coating has angle of contact obtuse and thus makes the surface hydrophobic.

Brain Teaser (Textbook Page No. 41)

Question 1.
Can you suggest any method to measure the surface tension of a soap solution? Will this method have any commercial application?
Answer:
There are more than 40 methods for determining equilibrium surface tension at the liquid-fluid and solid-fluid boundaries. Measuring the capillary rise (see Unit 2.4.7) is the laboratory method to determine surface tension.

Among the various techniques, equilibrium surface tension is most frequently measured with force tensiometers or optical (or the drop profile analysis) tensiometers in customized measurement setups.
[See https: / / www.biolinscientific.com /measurements /surface-tension]

Question 2.
What happens to surface tension under different gravity (e.g., aboard the International Space Station or on the lunar surface)?
Answer:
Surface tension does not depend on gravity.
[Note : The behaviour of liquids on board an orbiting spacecraft is mainly driven by surface tension phenomena. These make predicting their behaviour more difficult than under normal gravity conditions (i.e., on the Earth’s surface). New challenges appear when handling liquids on board a spacecraft, which are not usually present in terrestrial environments. The reason is that under the weightlessness (or almost weightlessness) conditions in an orbiting spacecraft, the different inertial forces acting on the bulk of the liquid are almost zero, causing the surface tension forces to be the dominant ones.

In this ‘micro-gravity’ environment, the surface tension forms liquid drops into spheres to minimize surface area, causes liquid columns in a capillary rise up to its rim (without over flowing). Also, when a liquid drop impacts on a dry smooth surface on the Earth, a splash can be observed as the drop disintegrates into thousands of droplets. But no splash is observed as the drop hits dry smooth surface on the Moon. The difference is the atmosphere. As the Moon has no atmosphere, and therefore no gas surrounding a falling drop, the drop on the Moon does not splash.
(See http://mafija.fmf.uni-Ij.si/]

Can you tell? (Textbook Page No. 45)

Question 1.
What would happen if two streamlines intersect?
Answer:
The velocity of a fluid molecule is always tangential to the streamline. If two streamlines intersect, the velocity at that point will not be constant or unique.

Activity

Question 1.
Identify some examples of streamline flow and turbulent flow in everyday life. How would you explain them ? When would you prefer a streamline flow?
Answer:
Smoke rising from an incense stick inside a wind-less room, air flow around a car or aeroplane in motion are some examples of streamline flow, Fish, dolphins, and even massive whales are streamlined in shape to reduce drag. Migratory birds species that fly long distances often have particular features such as long necks, and flocks of birds fly in the shape of a spearhead as that forms a streamlined pattern.

Turbulence results in wasted energy. Cars and aeroplanes are painstakingly streamlined to reduce fluid friction, and thus the fuel consumption. (See ‘Disadvantages of turbulence’ in the following box.) Turbulence is commonly seen in washing machines and kitchen mixers. Turbulence in these devices is desirable because it causes mixing. (Also see ‘Advantages of turbulence’ in the following box.) Recent developments in high-speed videography and computational tools for modelling is rapidly advancing our understanding of the aerodynamics of bird and insect flights which fascinate both physicists and biologists.

Use your Brain power (Textbook Page No. 46)

Question 1.
The CGS unit of viscosity is the poise. Find the relation between the poise and the SI unit of viscosity.
Answer:
By Newton’s law of viscosity,
\(\frac{F}{A}\) = η\(\frac{d v}{d y}\)
where \(\frac{F}{A}\) is the viscous drag per unit area, \(\frac{d v}{d y}\) is the velocity gradient and η is the coefficient of viscosity of the fluid. Rewriting the above equation as

SI unit : the pascal second (abbreviated Pa.s), 1 Pa.s = 1 N.m^-2.s
CGS unit: dyne.cm^-2.s, called the poise [symbol P, named after Jean Louis Marie Poiseuille (1799 -1869), French physician].
[Note : Thè most commonly used submultiples are the millipascalsecond (mPa.s) and the centipoise (cP). 1 mPa.s = 1 cP.]

Use your Brain power (Textbook Page No. 49)

Question 1.
A water pipe with a diameter of 5.0 cm is connected to another pipe of diameter 2.5 cm. How would the speeds of the water flow compare ?
Answer:
Water is an incompressible fluid (almost). Then, by the equation of continuity, the ratio of the speeds, is

Do you know? (Textbook Page No. 50)

Question 1.
How does an aeroplane take off?
Answer:
A Venturi meter is a horizontal constricted tube that is used to measure the flow speed through a pipeline. The constricted part of the tube is called the throat. Although a Venturi meter can be used for a gas, they are most commonly used for liquids. As the fluid passes through the throat, the higher speed results in lower pressure at point 2 than at point 1. This pressure difference is measured from the difference in height h of the liquid levels in the U-tube manometer containing a liquid of density ρ_m. The following treatment is limited to an incompressible fluid.

Let A₁ and A₂ be the cross-sectional areas at points 1 and 2, respectively. Let v₁ and v₂ be the corresponding flow speeds. ρ is the density of the fluid in the pipeline. By the equation of continuity,
v₁A₁ = v₂A₂ … (1)
Since the meter is assumed to be horizontal, from Bernoufli’s equation we get,

The pressure difference is equal to ρ_mgh, where h is the differences in liquid levels in the manometer.
Then,

Equation (3) gives the flow speed of an incompressible fluid in the pipeline. The flow rates of practical interest are the mass and volume flow rates through the meter.
Volume flow rate =A₁v₁ and mass flow rate = density × volume flow
rate = ρA₁v₁
[Note When a Venturi meter is used in a liquid pipeline, the pressure difference is measured from the difference in height h of the levels of the same liquid in the two vertical tubes, as shown in the figure. Then, the pressure difference is equal to ρgh.


The flow meter is named after Giovanni Battista Venturi (1746—1822), Italian physicist.]

Question 2.
Why do racer cars and birds have typical shape ?
Answer:
The streamline shape of cars and birds reduce drag.

Question 3.
Have you experienced a sideways jerk while driving a two wheeler when a heavy vehicle overtakes you ?
Answer:
Suppose a truck passes a two-wheeler or car on a highway. Air passing between the vehicles flows in a narrower channel and must increase its speed according to Bernoulli’s principle causing the pressure between them to drop. Due to greater pressure on the outside, the two-wheeler or car veers towards the truck.

When two ships sail parallel side-by-side within a distance considerably less than their lengths, since ships are widest toward their middle, water moves faster in the narrow gap between them. As water velocity increases, the pressure in between the ships decreases due to the Bernoulli effect and draws the ships together. Several ships have collided and suffered damage in the early twentieth century. Ships performing At-sea refueling or cargo transfers performed by ships is very risky for the same reason.

Question 4.
Why does dust get deposited only on one side of the blades of a fan ?
Answer:
Blades of a ceiling/table fan have uniform thickness (unlike that of an aerofoil) but are angled (cambered) at 8° to 12° (optimally, 10°) from their plane. When they are set rotating, this camber causes the streamlines above/behind a fan blade to detach away from the surface of the blade creating a very low pressure on that side. The lower/front streamlines however follow the blade surface. Dust particles stick to a blade when it is at rest as well as when in motion both by intermolecular force of adhesion and due to static charges. However, they are not dislodged from the top/behind surface because of complete detachment of the streamlines.

The lower/front surface retains some of the dust because during motion, a thin layer of air remains stationary relative to the blade.

Question 5.
Why helmets have specific shape?
Answer:
Air drag plays a large role in slowing bike riders (especially, bicycle) down. Hence, a helmet is aerodynamically shaped so that it does not cause too much drag.

Use your Brainpower (Textbook Page No. 52)

Question 1.
Does Bernoulli’s equation change when the fluid is at rest? How?
Answer:
Bernoulli’s principle is for fluids in motion. Hence, it is pointless to apply it to a fluid at rest. Nevertheless, for a fluid is at rest, the Bernoulli equation gives the pressure difference due to a liquid column.

For a static fluid, v₁ = v₂ = 0. Bernoulli’s equation in that case is p₁ + ρgh₁ = ρ₂ + ρgh₂

Further, taking h₂ as the reference height of zero, i.e., by setting h₂ = 0, we get p₂ = p₁ + ρgh₁

This equation tells us that in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by h₁ and consequently, p₂ is greater than p₁ by an amount ρgh₁.

In the case, p₁ = p₀, the atmospheric pressure at the top of the fluid, we get the familiar gauge pressure at a depth h₁ = ρgh₁. Thus, Bernoulli’s equation confirms the fact that the pressure change due to the weight of a fluid column of length h is ρgh.