Maharashtra Board Class 12 Physics Sample Paper Set 3 with Solutions

Maharashtra State Board Class 12th Physics Sample Paper Set 3 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Physics Model Paper Set 3 with Solutions

General instructions:

The question paper is divided into four sections.

1. Section A: Q. No. 1 contains Ten multiple choice type of questions carrying One mark each.
   Q. No. 2 contains Eight very short answer type of questions carrying One mark each.
2. Section B: Q. No. 3 to Q. No. 14 are Twelve short answer type of questions carrying Two marks each. (Attempt any Eight).
3. Section C: Q. No. 15 to Q. No. 26 are Twelve short answer type of questions carrying Three marks each. (Attempt any Eight).
4. Section D: Q. No. 27 to Q. No. 31 are Five long answer type of questions carrying Four marks each. (Attempt any Three).
5. Use of log table is allowed. Use of calculator is not allowed.
6. Figures to the right indicate full marks.
7. For each MCQ, correct answer must be written along with its alphabet.
   e.g., (a) /(b)…./ (c)…J (d) Only first attempt will be considered for evaluation.
8. Physical constants:
   1. Universal gas constant, R = 8.3 J/mol -K.
   2. Mass of neutron, m_n = 1.67 × 10^-27 kg
   3. Acceleration due to gravity, g = 9.8 m/s²

Section – A

Question 1.
Select and write the correct answer for the following multiple choice type of questions: [10 Marks]

i. Soft iron is used to make the core of 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

ii. A sonometer wire vibrates with three nodes and two antinodes, the corresponding mode of vibration is ______
(A) First overtone
(B) Second overtone
(C) Third overtone
(D) Fourth overtone
Answer:
(A) First overtone

iii. In hydrogen atom, electron jumps from the 3^rd orbit to the 1^st orbit. The change in angular momentum is ____.
(A) 1.05 × 10^-34 Js
(B) 2.11 × 10^-34 Js
(C) 3.16 × 10^-34 Js
(D) 4.22 × 10^-34 Js
Answer:
(B) 2.11 × 10^-34 Js

Explanation:

iv. In a series C-R circuit, impedance of the circuit is
(A) \(\sqrt{\mathrm{R}^2+(\omega \mathrm{C})^2}\)
(B) \(\sqrt{\mathrm{R}^2+(1 / \omega \mathrm{C})^2}\)
(C) R + ωC
(D) R + (I/ωC)
Answer:
(B) \(\sqrt{R^2+(1 / \omega C)^2}\)

v. Which gate can act as building block for the digital circuits?
(A) OR
(B) AND
(C) NOT
(D) NAND
Answer:
(D) NAND

vi. In a Young’s double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fourth minimum has path difference
(A) 11\(\frac{\lambda}{2}\)
(B) 5\(\frac{\lambda}{2}\)
(C) 10\(\frac{\lambda}{2}\)
(D) 7\(\frac{\lambda}{2}\)
Answer:
(D) 7\(\frac{\lambda}{2}\) [1 Mark]

Explanation:

In Young’s double slit experiment, for n^th minimum,
∆x_n = \(\frac{(2 n+1) \lambda}{2}\) for n = o, 1, 2, 3,…
∴ ∆x₄ = \(\frac{[(2 \times 3)+1] \lambda}{2}\) = \(\frac{7 \lambda}{2}\)

vii. The _______ dielectrics develop net dipole moment in presence of an electric field.
(A) polar
(B) non-polar
(C) polar and non-polar
(D) solid
Answer:
(C) polar and non-polar

viii. If the frequency’ of incident light falling on a photosensitive material is doubled, then the kinetic energy of the emitted photoelectron will be
(A) same as its initial value.
(B) two times its initial value.
(C) more than two times its initial value.
(D) less than two times its initial value.
Answer:
(C) more than two times its initial value.

ix. A body performing linear S. H. M. experiences a force of 0.2 N when displaced through 4 cm from the mean position. Its force constant will be ______.
(A) 2N/m
(B) 2.5 N/m
(C) 5 N/m
(D) 8 N/m
Answer:
(C) 5 N/m

Explanation:
F = – kx
∴ k = \(\left|\frac{F}{x}\right|\) = \(\frac{0.2}{0.04}\) = 5 N/m

x. 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) 30Ω
Answer:
(D) 30 Ω

Explanation:

According to Wheatstone balancing condition,
∴ \(\frac{1}{15}+\frac{1}{X}\) = \(\frac{1}{10}\)
∴ x = 30 Ω

Question 2.
Answer the following questions: [8 Marks]

i. State relation for magnetic force in vector form.
Answer:
\(\vec{F}\) = q \(|\vec{v} \times \vec{B}| \hat{n}\) = qvB sinθ\(\hat{\mathrm{n}}\)A
Where, θ is the angle between \(\vec{v}\) and \(\vec{B}\) and \(\hat{n}\) is unit vector in the direction of force.

ii. If the frequency of the input voltage 100 Hz is applied to a
(a) half wave rectifier and
(b) full wave rectifier, what is the output frequency in both cases?
Answer:
a. In case of a half wave rectifier, for one pulsating AC input we get one cycle of DC. Thus, the output frequency will be 100 Hz.
b. In case of full wave rectifier, for one pulsating AC input, we get two cycles of DC. Thus, the output frequency will be 200 Hz.

iii. Write the differential equation for angular S.H.M.
Answer:
I\(\frac{d^2 \theta}{d t^2}\) + cθ = 0

iv. If gyromagnetic ratio of the electron revolving in a circular orbit of hydrogen atom is
8.8 × 10^1o C kg^-1, then what would be the mass of the electron? (Given: charge of the electron = 1.6 × 10^-9C.)
Answer:
Gyromagnetic ratio = \(\frac{c}{2 m_e}\)
∴ m_e = \(\frac{1.6 \times 10^{-19}}{2 \times 8.8 \times 10^{10}}\) = \(\frac{1}{11}\) × 10^-29 = 9.09 × 10^-31 kg

v. The optical path of a ray of light of a given wavelength travelling a distance of 2.5 cm in flint glass having refractive index 1.5 is same as that on travelling a distance x cm through a medium having refractive index 1.25. Determine the value of x.
Answer:
The optical paths in two mediums of refractive indices n₁ and n₂ are related as, n₁d₁ = n₂d₂
∴ d₂ = x = \(\frac{n_1 d_1}{n_2}\) = \(\frac{1.5 \times 2.5}{1.25}\) = 3 cm

vi. Categorize the following into polar and non-polar dielectrics:
a. H₂O
b. CO₂
Answer:
a. H₂O – polar dielectric
b. CO₂ – non-polar dielectric

vii. What is photoelectric effect?
Answer:
The phenomenon of emission of electrons from a metal surface, when radiation of appropriate frequency is incident on it, is known as photoelectric effect.

viii. Define the SI unit of mutual inductance.
Answer:
If corresponding to 1 A/s rate of change of current in the primary circuit, the induced emf produced in the secondary circuit is 1 volt, then the mutual inductance (M) of the two circuits is 1 H.

Section – B

Attempt any EIGHT of the following questions: [16 Marks]

Question 3.
Discuss the Curie law for paramagnetic material.
Answer:
Curie’s law:
Magnetization of a paramagnetic sample is directly proportional to the external magnetic field and inversely proportional to the absolute temperature.
Mathematically,
M_Z ∝ B_ext and M_Z ∝ \(\frac{1}{\mathrm{~T}}\)
∴ M_Z ∝ \(\frac{B_{e x}}{T}\)
∴ M_Z = C × \(\frac{B_{e x t}}{T}\)
where, C is called Curie constant.
Above equation represents Curie’s law for magnetization.

Question 4.
The primary and secondary coil of a transformer each have an inductance of 500 × 10^-6 H. The mutual inductance (M) between the windings is 2.5 × 10^-6 H. What percentage of the flux from one coil reaches the other?
Answer:
Solution:
Self inductance = 500 × 10^-6 H
Mutual inductance = 2.5 × 10^-6 H
Percentage of flux transfer = \(\frac{M}{L}\) × 100 = \(\frac{2.5 \times 10^{-6}}{500 \times 10^{-6}}\) × 100 = 0.5 %
Ans: 0.5 % of the flux from one coil reaches the other.

Question 5.
Derive an expression for effective spring constant when two springs are connected in series.
Answer:
i. Consider an arrangement of two such springs of spring constants k₁ and k₂ as shown in figure.

ii. If the springs are massless, each will have the same stretching force as f.
iii. For vertical arrangement, it will be the weight (mg). If e₁ and e₂ are the respective extensions, then, f = k₁e₁ = k₂e₂
∴ e₁ = \(\frac{f}{k_1}\) and e₂ = \(\frac{f}{k_2}\)
iv. The total extension is given by, e = e₁ + e₂ = f\(\left(\frac{1}{k_1}+\frac{1}{k_2}\right)\)
v. If k_s is the effective spring constant, then
e = \(\frac{f}{k_s}\) = f\(\left(\frac{1}{k_1}+\frac{1}{k_2}\right)\)
∴ \(\frac{1}{\mathrm{k}_{\mathrm{s}}}\) = \(\frac{1}{k_1}+\frac{1}{k_2}\)
vi. For two massless sprinqs of sprinq constant k₁ and k₂ in series, k_s = \(\frac{k_1 k_2}{k_1+k_2}\) \(=\frac{\text { Product }}{\text { Sum }}\)

Question 6.
A capacitor of capacitance 2 pF is connected to a source of alternating e.m.f. of frequency 100 Hz. What is the capacitive reactance? (π = 3.142)
Answer:
Given: C = 2 μF = 2 × 10^-6F, f = 100 Hz
To find: Capacitive reactance (X_C)
Formula: X_C = \(\frac{1}{2 \pi f C}\)
Calculation: From formula,
X_C = \(\frac{1}{2 \times 3.142 \times 100 \times 2 \times 10^{-6}}\)
= 0.795 × 10³Ω
≈ 0.8 kΩ
Ans: The capacitive reactance of a capacitor is 0.8 kΩ.

Question 7.
Distinguish between interference and diffraction of light.
Answer:

Question 8.
Distinguish between progressive waves and stationary waves.
Answer:

Progressive waves	Stationary waves
i. The disturbance travels from one region to the other with definite velocity.	Disturbance remains in the region where it is produced, velocity of the wave is zero.
ii. Amplitudes of all particles are same.	Amplitudes of particles are different.
iii. Particles do not cross each other.	All the particles cross their mean positions simultaneously.
iv. All the particles are moving.	Particles at the position of nodes are alwayš at rest.
v. Energy is transmitted from one region to another.	There is no transfer of energy.
vi. Phases of adjacent particles are different.	All particles between two consecutive nodes are moving in the same direction and are in phase while those in adjacent loops are moving in opposite directions and differ in phase by 180°.

Question 9.
Why is potentiometer preferred over a voltmeter for measuring emf?
Answer:
i. Potentiometer is more sensitive than a voltmeter.

ii. A potentiometer can be used to measure a potential difference as well as an emf of a cell. A voltmeter always measures terminal potential difference, and as it draws some current, it cannot be used to measure an emf of a cell.

iii. Measurement of potential difference or emf is very accurate in the case of a potentiometer. A very small potential difference of the order 10‘6 volt can be measured with it. Least count of a potentiometer is much better compared to that of a voltmeter. Due to all these reasons potentiometer is preferred over a voltmeter for measuring emf.

Question 10.
Derive an expression for root mean square speed of a molecule of a gas in terms of its absolute temperature.
Answer:
i. Average value of the pressure of the gas is, P = \(\frac{1}{3} \frac{\mathrm{~N}}{\mathrm{~V}} \mathrm{~m} \overline{v^2}\)
ii. Thus, the mean square velocity of molecule will be, \(\overline{v^2}\) = \(\frac{3 P V}{N m}\)
iii.

Question 11.
Three capacitors of capacities 8 µF, 8 µF and 4 µF are connected in a series and potential difference of 120 volt is maintained across the combination. Calculate the charge on capacitor of capacity 4 µF.
Answer:
Given: C₁ = 8 μF, C₂ = 8 μF, C₃ = 4 μF, V = 120 volt
To find: Charge on the capacitor C₃
Formulae:
i. \(\frac{1}{C_{\mathrm{s}}}\) = \(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)
ii. C = \(\frac{\mathrm{Q}}{\mathrm{~V}}\)
Calculation: Using formula (i),
\(\frac{1}{C_s}\) = \(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\) = \(\frac{1}{8}+\frac{1}{8}+\frac{1}{4}\)
∴ C_s = 2µF = 2 × 10^-6F
In series combination,
Q₁ = Q₂ = Q₃
Using formula (ii),
∴ Q = C_s V
∴ Q = 2 × 10^-6 × 120
= 240 × 10^-6 C
Ans: The charge on the capacitor C₃ is 240 × 10^-6 C.

Question 12.
Explain zeroth law of thermodynamics using a schematic representation.
Answer:
Statement: If two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other.

1. Figure shows a schematic representation of the zeroth law of thermodynamics.
2. The double arrow represents thermal equilibrium between systems.
3. If system A and C are in thermal equilibrium, and systems A and B are in thermal equilibrium, then systems B and C must be in thermal equilibrium.
4. Then it can be inferred that, systems A, B and C are at the same temperature.

Question 13.
A rectangular coil in a moving coil galvanometer has 100 turns, each of length 5 cm and breadth 6 cm, which is suspended in a radial magnetic field of 0.050 Wb/m². The twist constant of suspension is 1.5 × 10^-9 Nm/degree. Calculate the current through the coil which will deflect it through 60°.
Answer:
Given: N = 100, K = 1.5 × 10^-9 Nm/degree, B = 0.05 Wb/m², φ = 60°,
A = l × b = 5 × 6 = 30 cm² = 30 × 10^-4 m²
To find: Current (I)
Formula: I = \(\frac{K \phi}{N A B}\)
Calculation: From formula,
I = \(\frac{1.5 \times 10^{-9} \times 60}{100 \times 30 \times 10^{-4} \times 0.05}\) = 6 × 10^-6 A
Ans: The current through the coil which will deflect it through 30° is 6 × 10^-6 A.

Question 14.
Thorium ₉₀Th²³² is disintegrated into lead ₈₂Pb²⁰⁰. Find the number of α and β particles emitted in disintegration.
Answer:
When an a – particle is emitted by an atom its atomic number decreases by 2 and mass number decreases by 4.
Whereas, when α β – particle is emitted by an atom its atomic number increases by 1 and mass number does not change.
When ₉₀Th²³² is disintegrated to ₈₂Pb²⁰⁰ its mass number is changed due to emission of α – particles.
∴ Number of α – particles emitted = \(\frac{232-200}{4}\) = 8.
Due to emission of a – particles atomic number of _9oTh²³² has decreased to 90 – (2 × 8) = 74.
But atomic number of Th is 82. The difference is due to emission β – particles.
∴ Number of p – particles emitted = 82 – 74 = 8

Section – C

Attempt any EIGHT of the following questions: [24 Marks]

Question 15.
With a neat labelled diagram, explain Ferry’s perfectly blackbody.
Answer:

1. Ferry’s perfectly blackbody consists of a double walled hollow sphere having tiny hole or aperture, through which radiant heat can enter.
2. The space between the walls is evacuated and outer surface of the sphere is silvered.
3. The inner surface of sphere is coated with lampblack.
4. There is a conical projection on the inner surface of sphere opposite the aperture. The projection ensures that a ray travelling along the axis of the aperture is not incident normally on the surface and is therefore not reflected back along the same path.
5. A heat ray entering the sphere through the aperture suffers multiple reflections and is almost completely absorbed inside.
6. Thus, the aperture behaves like a perfect blackbody.
7. The effective area of perfectly blackbody is equal to the area of the aperture.

Question 16.
In a biprism experiment, light of wavelength 5200 A is used to obtain an interference pattern on the screen. The fringe width changes by 1.3 mm when the screen is moved towards the biprism by 50 cm. Calculate the distance between the two virtual images of the slit.
Answer:
Given: λ, = 5200 Å, X₂ – X₁ = 1.3 mm, D₂ – D₁ = 50 cm = 0.5 m
To find: distance between two virtual images (d)
Formula: d = \(\frac{5200 \times 10^{-10} \times 0.5}{1.3 \times 10^{-3}}\)
= 2 × 10^-4 m
Ans: The distance between the two virtual images of the slit is 2 × 10^-4 m.

Question 17.
A motor cyclist at a speed of 5 m/s is describing a circle of radius 25 m. Find his inclination with vertical. What is the value of coefficient of friction between tyre and ground?
Answer:
Given: v = 5 m/s, r = 25 m, g = 9.8 m/s²

To find:
i. Inclination with vertical (θ)
ii. Coefficient of friction (μ)

Formulae:
i. tan θ = \(\frac{v^2}{r g}\)
ii. \(\frac{v^2}{r}\) = μg
Calculation: From formula (i),
tanθ = \(\frac{(5)^2}{25 \times 9.8}\) = \(\frac{1}{9.8}\) = 0.1021
∴ θ = tan^-1 (0.1021) = 5°50’
From formula (ii),
μ = \(\frac{5^2}{25 \times 9.8}\)
= 0.1021

Ans:
i. The inclination of the motor cyclist with vertical is 5°50’.
ii. The value of coefficient of friction between tyre and ground is 0.1021.

Question 18.
What is post office box? How does it work?
Answer:

1. A post office box (PO Box) s a practical form of Wh&itstone bridgeas shown in the figure.
2. It consists of three arms P, Q and R. The resistances in these three arms are adjustable.
3. The two arms P and Q contain resistances 10 ohm, 100 ohm and 1000 ohm each.
4. The third arm R contains resistances from 1 ohm to 5000 ohm. The unknown resistance X forms the fourth resistance. There are two tap keys K₁ and K₂.
5. The resistances in the arms P and Q are fixed to desired ratio. The resistance in the arm R is adjusted so that the galvanometer shows no deflection.
6. Now the calculated as, bridge is balanced. The unknown resistance is calculated as X = \(\frac{R Q}{P}\).
   where P and Q are the fixed resistances in the ratio arms and R is an adjustable known resistance.
7. If L is the length of the wire and r is its radius then the specific resistance of the material of the wire is given by
   ∴ ρ = \(\frac{X_A}{L}\) = \(\frac{X_{\pi r^2}}{L}\) …… (∵ Resistance = \(\frac{\rho L}{A}\))

Question 19.
A heat engine operates between a cold reservoir at temperature T_C = 300 K and a hot reservoir at v temperature T_H. It takes 200 J of heat from the hot reservoir and delivers 120 J of heat to the cold reservoir in a cycle. What could be the minimum temperature of the hot reservoir?
Answer:
The work done by the engine in a cycle is,
W = |Q_H| – |Q_C|
∴ W = 200 – 120 = 80 J
The efficiency of the engine is,
η = \(\frac{W}{Q_H}\) = \(\frac{80}{200}\) = 0.40
From Carnot’s theorem, no engine can have an efficiency greater than that of a Carnot engine.
Thus, 0.40 ≤ 1 – \(\frac{T_C}{T_H}\) = 1 – \(\frac{300}{T_H}\)
∴ \(\frac{300}{T_H}\) ≤ 1 – 0.40 = 0.60
∴ T_H ≥ \(\frac{300}{0.60}\)
∴ T_H ≥ 500 K
Ans: The minimum temperature of the hot reservoir may be 500 K.

Question 20.
Show that the deflection produced in a moving coil galvanometer is’ directly proportional to the current flowing through its coil or vice-versa.
Answer:
Working:
i. The coil rotates due to a torque acting on it as the current flows through it.
Torque acting on current carrying coil is τ = NIAB Sinθ.
Here θ = 90° as the field is radial.
∴ τ = NIAB
where A is the area of the coil B the strength of the magnetic field N the number of turns of the coil and I the current in the coil.

ii. This torque is counter balanced by a torque due to a spring fitted at the bottom so that a fixed steady current I in the coil produces a steady angular deflection φ.

iii. Larger the current is, larger is the deflection and larger is the torque due to the sprinq. If the deflection is φ, the restorinq torque due to the sprinq is equal to Kφ where K is the torsional constant of the spring.
Thus, Kφ = NIAB, and the deflection φ = \(\left(\frac{N A B}{K}\right) I\)
This means, the deflection 4 is proportional to the current I i.e., φ ∝ I.

Question 21.
Discuss hydrogen spectrum in detail.
Answer:

1. When a hydrogen gas is heated inside a glass tube to high temperature, it emits radiation of few selected wavelengths (410, 434, 486 and 656 nm) in visible region.
2. When these wavelengths are allowed to pass through prism, the lines are seen in the spectrum called as emission lines.
3. It also emits radiation of specific wavelengths in ultraviolet (UV), infrared (IR) and of longer wavelengths.
4. The spectral lines can be divided into groups known as series with names of the scientists who studied them.
5. The series, starting from shorter wavelengths and going to larger wavelengths are called Lyman series, Balmer series, Paschen series, Brackett series, Pfund series, etc.
6. In each series, the separation between successive lines decreases as we go towards shorter wavelength and they reach a limiting value.
7. The observed wavelengths of the emission lines are found to obey the relation.
   \(\frac{1}{\lambda}\) = \(R\left[\frac{1}{n^2}-\frac{1}{m^2}\right]\)
   where, λ = wavelength of a line,
   R = constant
   n and m = integers such that
   n = 1, 2, 3, …. respectively, for Lyman, Balmer, Paschen…. series, such that m > n.
8. The wavelength decreases with increase in m.
9. The successive lines in a given series come closer and closer and ultimately reach the values of λ = \(\frac{n^2}{R}\) in the limit m → ∞, for different values of n.

Question 22.
i. A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?
ii. Define magnetization.
Answer:
i.
a. Let ± Q be the charges on the capacitor plates.
When a conducting slab of area A and thickness t < d is introduced between the plates, the free charges in the conductor flow as shown in figure below.

b. The charges ± Q appear on the two faces of the slab as shown reducing the electric field in the interior of slab to zero. The original uniform electric field E_o now exists over a distance (d – t), so that potential difference between the plates

This implies C > C₀ i.e., capacitance increases on introducing metal slab (conducting plate) in between the plates. (2 Marks)

ii. The ratio of magnetic moment to the volume of the material is called magnetization. It is denoted by \(\vec{M}\).

Question 23.
Explain working of a transistor as an amplifier with the help of proper diagram.
Answer:
Working of an amplifier:
i. The circuit of an amplifier using a n-p-n transistor in CE configuration is shown in the figure.

ii. When the input voltage V_in is not applied, applying the Kirchhoffs law to the output loop, we can write,
V_CC = V_CE + I_CR_L

iii. Similarly, for input loop,
V_BB = V_BE + I_BR_B

iv. When input AC signal is applied, V is not zero. Thus, the voltage drop across the input loop will now be,
V_BB + V_in = V_BE + I_BR_B + ∆I_B (R_B + r_i) …….. (I)

v. The AC signal applied adds the current of AIB to the original current flowing through the circuit. Therefore, the additional voltage drop in the input loop will be across resistor R_B (= ∆I_BR_B) and across the input dynamic resistance of the transistor (= ∆I_Br_i).

vi. From equation (1),
V_in = ∆I_B (R_B + r_i)
As, R_B is very small, we can consider,
V_in = ∆I_Rr_i

vii. The changes in the base current ∆V_CE cause changes in the collector current I_C. This changes the voltage drop across the load resistance because V_CC is constant. We can write,
∆V_CC = ∆V_CE + R_LI_C = 0
∴ ∆V_CE = – R_LI_C

viii. The change in the output voltage ∆V_CE is the output voltage V₀ hence we can write, V₀ = ∆V_CE = β_ACR_L∆I_B

Question 24.
Define terminal velocity. Obtain an expression for the terminal velocity of a small sphere falling under gravity through a viscous fluid.
Answer:
Terminal velocity: The constant maximum velocity acquired by a body falling through a viscous liquid is called as terminal velocity.
Expression for terminal velocity:

i. Consider a sphere of radius (r) and density (p) falling under gravity through a liquid of density (σ) and coefficient of viscosity (η) as shown in figure.

ii. Forces acting on the sphere during downward motion are,
a. Viscous force = F_v = 6πηrv (directed upwards)
b. Weight of the sphere, (F_g)
mg = \(\frac{4}{3}\)πr³ ρg (directed downwards)
c. Upward thrust as Buoyant force (F_u)
F_u = \(\frac{4}{3} \pi r^3 \sigma g\) (directed upwards)

iii. As the downward velocity increases, the viscous force increases. A stage is reached, when sphere attains terminal velocity.

iv. When the sphere attains the terminal velocity, the total downward force acting on the sphere is balanced by the total upward force acting on the sphere.
∴ Total downward force = Total upward force

Question 25.
Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the
i. same speed,
ii. same energy and
iii. same momentum?
iv. State which of the two will have longer wavelength in each case?
Answer:
Ratio of de Broglie wavelengths for electron and proton is,

i. For proton and electron moving with the same speed,
\(\frac{\Lambda_e}{\Lambda_p}\) = \(\frac{m_p}{m_e}\) = 1836
Thus, de Broglie wavelength of electron is longer than that of proton.

ii. For proton and electron having same energy,
\(\frac{\Lambda_e}{\Lambda_p}\) = \(\sqrt{\frac{m_p}{m_z}}\) = \(\sqrt{1836}\) = 42.85
Thus, de Broglie wavelength of electron is longer than that of proton.

iii. For proton and electron having same momentum,
\(\frac{\Lambda_e}{\Lambda_p}\) = \(\frac{\mathrm{P}_{\mathrm{p}}}{\mathrm{P}_{\mathrm{e}}}\) = 1
Thus, both electron and proton will have equal de Broglie wavelength.

Question 26.
A coil of resistance 5 Ω and self inductance 0.2 H is connected in series with a variable capacitor across 30 volt, 50 Hz supply. At what value of capacitor will resonance occur? Find the corresponding value of current.
Answer:

∴ C = 50.65 μF
At series resonance, the current is maximum and it is given by,
i_rms = \(\frac{e_{\mathrm{rms}}}{R}\) = \(\frac{30}{5}\) = 6
∴ i_rms = 6 A

Ans:
i. The value of capacitor at which the resonance will occur is 50.65 μF.
ii. The corresponding value of current is 6 A.

Section – D

Attempt any THREE of the following questions: [12 Marks]

Question 27.
i. Derive an expression for the self-inductance of a toroid of circular cross-section of radius r and major radius R.
ii. Calculate the self inductance (L) of toroid for major radius (R) = 30 cm, cross-section of toroid having radius (r) = 1.5 cm and the number of turns (n) =1200.
Answer:
i. The magnetic field inside a toroid, B = \(\frac{\mu_0}{2 \pi} \frac{N i}{R}\)
where N = number of turns and r the distance from the toroid axis.
As r << R, magnetic field (B) in the cavity of toroid is uniform and can be written as,
B = \(\frac{\mu_0}{2 \pi} \frac{N i}{R}\) …………(1)
The magnetic flux (4) passing through cavity that links each turn is,
φ = BA = \(\frac{\mu_0}{2 \pi} \frac{\mathrm{Ni}}{\mathrm{R}}\)(πr²) = \(\frac{\mu_0 \mathrm{Nir}^2}{2 R}\)
When the current i varies with time, the induced emf e across the terminals of toroid is given by Faraday’s law.

ii.
Given: N = 1200, r = 1.5 cm = 1.5 × 10^-2 m, R = 30 cm: 30 × 10^-2 m, μo = 4π × 10^-7 Tm/A
To find: Self inductance (L)
Formula: L = \(\frac{\mu_0}{2} \frac{N^2 r^2}{R}\)
Calculation: From formula,
L = \(\frac{4 \times 3.142 \times 10^{-7} \times(1200)^2 \times\left(1.5 \times 10^{-2}\right)^2}{2 \times 30 \times 10^{-2}}\) = 6.79 × 10^-4 H
Ans: The self inductance of the toroid is 6.79 × 10^-4 H.

Question 28.
i. Define ideal simple pendulum.
ii. Derive an expression for equation of stationary wave on a stretched string.
Answer:
i. Ideal simple pendulum: An ideal simple pendulum is a heavy particle suspended by a massless, inextensible, flexible string from a rigid support.

ii.

a. Consider two simple harmonic progressive waves of equal amplitudes (a) and wavelength (λ) propagating on a long uniform string in opposite directions.

b. The equation of wave travelling along the X-axis in the negative direction is given by,
y₁ = asin[2π(nt – \(\frac{x}{\lambda}\))]
The equation of wave travelling along the X-axis in the negative direction is given by,
y₂ = asin[2π(nt + \(\frac{x}{\lambda}\))]

c. When these waves interfere, the resultant displacement of particles of string is given by the principle of superposition of waves as
y = y₁ + y₂
∴ y = asin[2π(nt – \(\frac{x}{\lambda}\))] + asin[2π(nt + \(\frac{x}{\lambda}\))]
d. By using trigonometry formula,
sin C + sin D = 2 sin\(\left(\frac{C+D}{2}\right)\) cos \(\left(\frac{C-D}{2}\right)\)
∴ y = 2a sin (2πnt) cos \(\frac{2 \pi x}{\lambda}\)
y = 2a cos\(\frac{2 \pi x}{\lambda}\) sin (2πnt) ….(1)
e. Substituting 2a cos\(\frac{2 \pi x}{\lambda}\) = A in equation (1),
y = A sin(2πnt)
∴ y = A sinωt ………(∴ ω = 2πn)
This is the equation of a stationary wave which gives resultant displacement due to two simple harmonic progressive waves.

Question 29.
Explain the phenomenon of surface tension on the basis of molecular theory.
Answer:
Molecular theory of surface tension:

i. Let PQRS = Surface film of liquid in a container containing liquid.
PS is the free surface of liquid and QR is the inner layer parallel to PS at distance equal to the range of molecular force.

ii. Now consider three molecules A, B and C in a liquid in a vessel such that the molecule
A is well inside the liquid, the molecule B within surface film and molecule C is on the surface of liquid as shown in the figure.

iii. The sphere of influence of the molecule A is entirely inside the liquid. As a result, molecule A is acted upon by equal cohesive forces in all directions. Thus, the net cohesive force acting on molecule A is zero.

iv. For molecule B, a large part of its sphere of influence is inside the liquid and a smaller part is outside the surface (in air). The adhesive force acting on molecule B due to air molecules above it and within its sphere of influence is weak compared to the strong downward cohesive force acting on the molecule. As a result, the molecule B gets attracted inside the liquid.

v. For molecule C, half of the sphere of influence is in air and half is in liquid. As density of air is much less than that of liquid, number of air molecules within the sphere of influence of the molecule C above the free surface of the liquid is much less than the numbers of liquid molecules within the sphere of influence that lies within the liquid. Thus, the adhesive force due to the air molecules acting on molecule C is weak compared to the cohesive force acting on the molecule. As a result, the molecule C also gets attracted inside the liquid.

vi. Thus, all molecules in the surface film are acted upon by an unbalanced net cohesive force directed into the liquid. Therefore, the molecules in the surface film are pulled inside the liquid. This minimizes the total number of molecules in the surface film. As a result, the surface film remains under tension. The surface film of a liquid behaves like a stretched elastic membrane. This tension is known as surface tension and the force due to it acts tangential to the free surface of a liquid.

Question 30.
i. Deduce Boyle’s law using the expression for pressure exerted by the gas.
ii. A system releases 235 kJ of heat while 25 kJ of work is done on the system. Calculate the
change in internal energy.
Answer:
i. Suppose, P = pressure exerted by the gas
V = volume of the gas
N = number of molecules of the gas
m = mass of each molecule of the gas
∴ Total mass of the gas, M = mN

ii. From kinetic theory of gases, P = \(\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{~V}} \overline{\mathrm{v}^2}\)
∴ Pressure exerted by a gas in an enclosed vessel, P = \(\frac{2}{3} \frac{N}{V}\left(\frac{1}{2} m \overline{v^2}\right)\)

iii. But, \(\frac{1}{2} m \overline{v^2}\) K.E = constant for all the gases at a given temperature.
N = number of molecules which is constant for a given mass of the gas.
∴ P = \(\frac{\text { constant }}{V}\) ⇒ P ∝ \(\frac{1}{V}\)
Hence, at constant temperature, pressure of the gas is increased if its volume is reduced.

ii. Solution:
Given: W = -25 kJ, Q = -235 kJ
To find: Change in internal energy (∆U)
Formula: ∆U = |Q| – |W|
Calculation: From formula,
∆U = |Q| – |W| = (235 – 25) = 210 kJ
Ans: Change in internal energy is 210 kJ.

Question 31.
i. State an expression for moment of inertia of a thin uniform disc about
a. an axis passing through its centre and perpendicular to its plane.
b. its diameter.
ii. A uniform solid sphere has radius 0.2 m and density 8 × 10³ kg/m³. Find the moment of inertia about the tangent to its surface.
(π = 3.142)
Answer:
Expression of M.I of uniform disc about an axis passing through its centre and perpendicular to its plane:

a. Let, M = mass of disc, R = radius of disc ZZ’ = axis passinq through the centre of disc and perpendicular to the plane.
b. The M. I of a thin uniform disc about an axis passing through its centre and perpendicular to its plane is given by, I_c = \(\frac{1}{2}\)MR²

i. Expression of M.I of uniform disc about its diameter:

a. Let, I_X = M.I of disc about diameter XX’,
I_Y = M.I of disc about yy’⊥^ar to XX’
I_Z = M.I of disc about ⊥^ar to plane

b. Since the disc is symmetrical about any of its diameter,

∴ I_X = I_Y = I_d
where I_d is M.I of disc about any of its diameter.
c. Applying perpendicular axis theorem,
I_Z = I_X + I_Y …..[∵ I_Z = I]
∴ I = I_d + I_d = 2 I_d
∴ I_d = \(\frac{I}{2}\) = \(\frac{1}{2} \times \frac{1}{2} M R^2\)
∴ I_d = \(\frac{M R^2}{4}\)

ii.
Solution:
Given: R = 0.2m, ρ = 8000 Kg/m³
To find: Moment of inertia (I)

Formulae.

i. I₀ = \(\frac{7}{5} M R^2\)
ii. Mass (M) = volume × density
Calculation: From formula (ii),

Ans: M.I. of the uniform solid sphere about a tangent to its surface is 15.02 kg m².

Maharashtra Board Class 12 Physics Previous Year Question Papers