Maharashtra Board Class 12 Physics Sample Paper Set 5 with Solutions

Maharashtra State Board Class 12th Physics Sample Paper Set 5 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Physics Model Paper Set 5 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)…/ (d) Only first attempt will be considered for evaluation.
8. Physical constants:
   1. Latent heat of vaporisation, L_vap = 2256 kJ/kg
   2. 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. If the incident energy is 200 cal, absorptive power is 0.31 and reflection coefficient is 0.41, then the amount of energy transmitted will be
(A) 48 cal
(B) 56 cal
(C) 58 cal
(D) 54 cal
Answer:
(B) 56 cal

ii. A parallel plate capacitor is charged. If the plates are pulled apart,
(A) the potential difference increases.
(B) the capacitance increases.
(C) the total-charge increases.
(D) the charge and the potential difference remain the same.
Answer:
(A) the potential difference increases.

iii. If A.C. voltage is applied to a pure capacitor, then voltage across the capacitor
(A) leads the current by phase angle \(\left(\frac{\pi}{2}\right)\) rad.
(B) leads the current by phase angle (π) rad.
(C) lags behind the current by phase angle \(\left(\frac{\pi}{2}\right)\) rad.
(D) lags behind the current by phase angle (π) rad.
Answer:
(C) lags behind the current by phase angle \(\left(\frac{\pi}{2}\right)\) rad.

iv. In the hydrogen atom spectrum, the series which lies in ultraviolet region is
(A) Lyman series
(B) Balmer series
(C) Paschen series
(D) Brackett series
Answer:
(A) Lyman series

v. In a photocell, increasing the intensity of incident light increases ____.
(A) the stopping potential
(B) the photoelectric current
(C) the energy of the incident photons
(D) maximum kinetic energy of the photoelectron
Answer:
(B) the photoelectric current

vi. An ideal ammeter has ____.
(A) low resistance
(B) high resistance
(C) infinite resistance
(D) zero resistance
Answer:
(D) zero resistance

vii. The minimum velocity (in m s^-1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is (g = 10 m/s²)
(A) 60
(B) 30
(C) 15
(D) 25
Answer:
(B) 30

viii. A circular coil of 100 turns with a cross-sectional area (A) of 1 m² is kept with its plane perpendicular to the magnetic field (B) of 1 T. What is the magnetic flux linkage with the coil?
(A) 1 Wb
(B) 100 Wb
(C) 50 Wb
(D) 200 Wb
Answer:
(B) 100 Wb
Explanation: Nφ = NBA
= 100 × 1 × 1
= 100 Wb [1 Mark]

ix. In a moving coil galvanometer, we use a radial magnetic field so that the galvanometer scale is
(A) linear
(B) algebraic
(C) logarithmic
(D) exponential
Answer:
(A) linear [1 Mark]

x. The forward voltage drop of an LED is
(A) much larger than the ordinary diode.
(B) much less than that of the ordinary diode.
(C) almost 1.5 to 3.5 volts.
(D) Both (A) and (C)
Answer:
(D) Both (A) and (C)

Question 2.
Answer the following questions:

i. In a common-base connection, a certain transistor has an emitter current of 11 mA and collector current of 10.7 mA. Calculate the value of the base current.
Answer:
For a common base connection, I_E = I_B + I_C
∴ I_B = I_E – I_C = 11 – 10.7 = 0.3 mA

ii. State principle of moving coil galvanometer (M.C.G.)
Answer:
Principle of M.C.G:
When a coil carrying an electric current is suspended in a uniform magnetic field, a torque acts on it. This torque tends to rotate the coil about the axis of suspension so that the magnetic flux passing through the coil is maximum.

iii. What will happen to the mean square speed of the molecules of a gas if the temperature of the gas decreases?
Answer:
The mean square speed of the molecules of the gas is, \(\overline{v^2}\) = \(\frac{3 R T}{M_0}\)
∴ \(\overline{v^2}\) ∝ T
Hence, the mean square speed of molecules of the gas decreases in same proportion with the decrease in temperature.

iv. Define Potential Gradient.
Answer:
Potential gradient is defined as potential difference per unit length of wire.

v. What is electromagnetic induction?
Answer:
Electromagnetic induction:
The phenomenon of producing an induced e.m.f in a conductor or conducting coil due to changing magnetic flux is called electromagnetic induction.

vi. 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 moving in the direction of positive X-axis.

vii. The half life of radium is 1600 years. Find the fraction of a sample of radium that would remain after 6400 years.
Answer:
The fraction of sample that is left after disintegration is,

viii. What will be the direction of angular displacement and angular velocity, if angular acceleration is constant and is along the axis of rotation?
Answer:
If the angular acceleration a is constant and is along the axis of rotation, then all \(\vec{\theta}\), \(\vec{\omega}\), and \(\vec{a}\) will be directed along the same axis.

Section – B

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

Question 3.
Plane wavefront of light of wavelength 5500 A is incident on two slits in a screen perpendicular to the direction of light rays. If the total separation of 10 bright fringes on a screen 2 m away is 2 cm, find the distance between the slits.
Answer:
Given: λ = 5500 A, D = 2 m
Distance between 10 fringes = 2 cm = 0.02 m.
Fringe width W = 0.02/10 = 0.002 m = 2 × 10^-3 m
To find: Distance between slits (d)
Formula: W = \(\frac{\Lambda D}{d}\)
Calculation: From formula,
2 × 10^-3 = \(\frac{5500 \times 10^{-10} \times 2}{d}\)
∴ d = \(\frac{5.5 \times 10^{-7} \times 2}{2 \times 10^{-3}}\)
= 5.5 × 10^-4 m
Ans: The distance between two slits is 5.5 × 10^-4 m.

Question 4.
Derive an expression for the energy stored in a magnetic field.
Answer:
i. Changing magnetic flux in a coil causes an induced emf.
ii. The induced emf so produced opposes the change and hence the energy has to be spent to overcome it to build up the magnetic field.
iii. This energy may be recovered as heat in a resistance of the circuit.
iv. The induced emf is given as, e = -L \(\frac{d I}{d t}\)
v. The work done in moving a charge dg against this emf is,

vi. Equation (1) gives the energy stored (U_B) in magnetic field and is analogous to the energy stored (U_E) in the electric field in a capacitor.

Question 5.
Derive an expression for ‘Half Life Time’ of a radioactive material using the ‘Law of Radioactive Decay’.
Answer:
Expression for half life period:
From law of radioactive decay,

This is the required expression for half life period of radioactive substance.

Question 6.
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 4.5 cm. If the velocity of sound is 1629 m/s, find the frequency.
Answer:
Given: \(\left(\frac{\lambda}{2}\right)\) = 4.5 cm = 4.5 × 10 m,
velocity of sound (v) = 16294 m/S
To find: Frequency of sound wave (n)
Formula: n = \(\frac{v}{\lambda}\)
Calculation:
As \(\frac{\lambda}{2}\) = 4.5 × 10^-2 m
∴ λ = 9 × 10^-2 m
From formula,
n = \(\frac{1629}{7 \times 10^{-2}}\) = 18.1 × 10³ = 18.1 kHz
Ans: The frequency of the sound wave is 18.1 kHz.

Question 7.
Define magnetization. State its formula and S.I. unit.
Answer:

1. Definition: The ratio of magnetic moment to the volume of the material is called magnetization. It is denoted by M.
2. If magnetic specimen of volume V acquires net magnetic dipole moment ‘mnef due to the magnetising field, then \(\vec{M}\) = \(\frac{m_{\text {net }}}{\mathrm{V}}\)
3. It is a vector quantity.
4. Unit: Am^-1 in SI system.
5. Dimensions: [M⁰L^-1T⁰I¹]

Question 8.
The common-base DC current gain of a transistor is 0.982. If the emitter current is 15 mA, what is the value of base current?
Answer:
Given: α_DC = 0.982, I_E = 15 mA
To find: Base current (IB)

Formulae:

i. α_DC = \(\frac{I_c}{I_E}\)

ii. I_E = I_B + I_C
Calculation: From formula (i),
I_C = α_DC I_E = 0.982 × 15 = 14.73 mA
From formula (ii),
I_B = I_E – I_C = 15 – 14.73 = 0.27 mA
Ans: The value of base current is 0.27 mA.

Question 9.
A particle performing linear S.H.M. has maximum velocity 25 cm/s and maximum acceleration 100 cm/s². Find period of oscillations.
Answer:
Given: v_max = 25 cm/s, a_max = 100 cm/s2
To find: Amplitude (A), Period (T)

Formulae:

1. v_max = Aω
2. a_max = Aω²
3. ω = \(\frac{2 \pi}{T}\)

Calculation:
From formula (i) and (ii),
\(\frac{a_{\max }}{v_{\max }}\) = ω = \(\frac{2 \pi}{T}\)
∴ T = \(\frac{2 \pi \times v_{\max }}{a_{\max }}\) = \(\frac{2 \times 3.142 \times 25}{100}\) = 1.571 s
Ans: Time period of oscillation is 1.571 s.

Question 10.
Explain relation between electric field and electric potential.
Answer:
i. Consider the electric field produced by a charge +q kept at point O.

ii. A unit positive charge (+q_o) is present in vicinity is moved towards charge +q through small distance dx.

iii. As direction of electric field of charge +q is outward, displacement dx is in direction opposite to field as shown in figure below.

iv. As electrostatic force is along \(\vec{E}\), work done is
dW = \(\vec{F} \cdot \overrightarrow{d x}\) = -Fdx = -Edx
Negative sign indicates that displacement of charge is in direction opposite to field.

v. But dW = dV × q_o = dV ….(∵ q_o is a unit charge)
∴ Potential difference between M and N,
dV = -Edx
∴ E = \(\frac{-d V}{d x}\)
Thus the electric field at a point in an electric field is the negative of the potential gradient at that point.

Question 11.
State the properties of an ideal fluid.
Answer:
An ideal fluid has the following properties:

1. It is incompressible i.e.( its density is constant.
2. Its flow is irratational i.e., its flow is smooth with no turbulences in the flow.
3. It is non-viscous i.e., there is no internal friction in the flow and hence the fluid has no viscosity.
4. Its flow is steady i.e., its velocity at each point is constant in time.

Question 12.
Calculate molar specific heat of mono-atomic gases at constant volume and constant pressure.
Answer:
i. For a monatomic gas enclosed in a container, held at a constant temperature T and containing N_A atoms, each atom has only 3 translational degree of freedom.

ii. Therefore, average energy per atom is \(\frac{3}{2}\)k_BT and the total internal energy per mole is, E = \(\frac{3}{2}\)N_Ak_BT

iii. Molar specific heat at constant volume, C_V = \(\frac{d E}{d T}\) = \(\frac{3}{2}\)N_AK_B = \(\frac{3}{2}\)R

iv. Using Mayer’s relation, C_P = R + C_M
∴ C_P = \(\frac{5}{2}\)R

Question 13.
Find the time required for a 25 Hz alternating current to change its value from zero to the r.m.s. value.
Answer:
Given: f = 25 Hz; i_rms = \(\frac{i_0}{\sqrt{2}}\)
To find: Time required (t)
Formula: i = i_o sin ωt
Calculation:
From formula,
\(\frac{i_0}{\sqrt{2}}\) = i_osin (2π × 25 × t) ……(∵ ω = 2πf)
∴ \(\frac{1}{\sqrt{2}}\) = sin(50πt)
∴ sin \(\left(\frac{\pi}{4}\right)\) = sin(50π t)
On comparing, we get
∴ 50 π t = \(\left(\frac{\pi}{4}\right)\)
∴ t = \(\frac{1}{200}\) = 5 × 10^-3 s
Ans: The time required for the alternating current to change its value from zero to rms is 5 × 10^{-33 s.}

Question 14.
On what factors, does the frequency of a conical pendulum depend? Is it independent of some factors?
Answer:
i. Frequency of conical pendulum depends on following factors:

a. Length of pendulum (L): Frequency of conical pendulum increases with decrease in length of pendulum.
i.e., n ∝ \(\frac{1}{\sqrt{L}}\)
b. Acceleration due to gravity (g): Frequency of conical pendulum increases with increase in g. i.e., n ∝ \(\sqrt{g}\)
c. Angle of inclination (θ): As θ increases, cos θ decreases, hence, frequency of conical pendulum increases with increase in θ.
i.e., n ∝ \(\frac{1}{\sqrt{\cos \theta}}\) (For 0 < θ < π)

ii. Frequency at conical pendulum is independent of mass of the bob.

Section – C

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

Question 15.
Derive Malus’ law.
Answer:
i. Consider the linearly polarized wave (electric field vectors are confined to one plane and in unique direction) emerging from first polarizer.

ii. If the polarized wave has its electric field \(\left(\vec{E}_1\right)\) along the Y-direction. Then the electric field is,
\(\vec{E}_1\) = \(\hat{j} E_{10}\) sin (kx – ωt) ……. (1)

iii. The intensity of the polarized wave is given by
I₁ ∝ \(\left|E_{10}\right|^2\) ….(2)

iv. Now if this wave passes through second polarizer whose polarization axis makes an angle θ with the Y-direction, only the component E₁₀ cosθ will pass through.

v. Thus, the amplitude of the wave which passes through (say E₂₀) is now E_1o cosθ and its intensity I₂ will be
I₂ ∝ \(\left|E_{20}\right|^2\)
∴ I₂ ∝ \(\left|E_{10}\right|^2\) cos²θ
∴ I₂ = I₁ cos²θ ……(3)
This is known as Malus’ law which gives the intensity of a linearly polarized wave after it passes through a polarizer.

Question 16.
2.0 kg of liquid water is boiled at 100 °C and all of it is converted to steam. If the change of state, takes place at the atmospheric pressure (1.01 × 10⁵ Pa), calculate
i. the energy transferred to the system,
ii. the work done by the system during this change, and
iii. the change in the internal energy of the system. Given, the volume of 2 kg of water changes from 2.0 × 10^-3 m³ in liquid form to 3.342 m³ when in the form of steam.
Answer:
Given: m = 2.0 kg, p = 1.01 × 10⁵ Pa, T = 100 °C = 373 K,
V_steam = 3.342 m³, V_liq = 2.0 × 10^-3 m³
∴ dv = (V_steam – V_liq) = (3.342 – 0.002) = 3.340 m³
We know that, L_vap = 2256 kJ/kg

To find:
i. Energy absorbed by system (Q)
ii. Work done by the system (W)
iii. Change in internal energy (∆U)

Formulae:
i. Q = mL
ii. W = p∆V
iii. ∆U = Q – W

Calculation:
From formula (i),
Q = 2.0 × 2256 = 4512 kJ
From formula (ii),
W = (1.01 × 10⁵) × 3.340 = 3.373 × 10⁵ J ≈ 337 kJ
From formula (iii),
∆U = 4512 – 337 = 4175 kJ

Ans:

i. Energy absorbed by system is 4512 kJ.
ii. Work done by the system is 337 kJ.
iii. Change in the internal energy is 4175 kJ.

Question 17.
Define the following terms:
i. Excitation energy of an electron in an atom
ii. Binding energy of an electron in an atom
iii. Dark resistance of a photodiode
Answer:
i. Excitation energy of an electron: The energy required to take an electron from the ground state to an excited state is called the excitation energy of the electron in that state.

ii. Binding energy of an electron: Binding energy of an electron is the minimum energy required to make it free from the nucleus.

iii. Dark resistance of a photodiode: Dark resistance of a photodiode (R_d) is defined as the ratio of the maximum reverse voltage and its dark current. It is the resistance of a photodiode when it is not illuminated.
R_d \(=\frac{\text { Maximum reverse voltage }}{\text { Dark current }}\)

Question 18.
An electric dipole consists of two opposite charges each of magnitude 1 µC separated by 2.5 cm.
The dipole is placed in an external electric field of 1.2 × 10⁵ NC^-1. Find:
i. The maximum torque exerted by the field on the dipole ’
ii. The work the external agent will have to do in turning the dipole through 180° starting from the position θ = 0°
Answer:
Given: q = 1 μC = 10^-6 C, 2l = 2.5 cm = 2.5 × 10^-2 m, E = 1.2 × 10⁵ NC^-1

To find:
i. Maximum torque
ii. Work done in rotating dipole

Formulae:
i. p = q × 2l
ii. \(\tau\) = pE sinθ
iii. W = pE (cosθ₁ – cosθ₂)

Calculation:
From formula (i)
p = 10^-6 × 2.5 × 10^-2 = 2.5 × 10^-8 Cm
For maximum torque, θ = 90°
\(\tau_{\max }\) = p E sin 90°
= 2.5 × 10^-8 × 1.2 × 10⁵ × 1 = 3 × 10^-3 Nm
As θ₁ = 0° and θ₂ = 180°
W = pE (cos θ₁ – cos θ₂)
= 2.5 × 10^-8 × 1.2 × 10⁵ × (cos 0° – cos180°)
= 3 × 10^-3 (1 + 1) = 6 × 10^-3 J.

Ans:

i. The maximum torque exerted by field on dipole is 3 × 10^-3 Nm.
ii. Work done in rotating dipole is 6 × 10^-3 J.

Question 19.
State Lenz’s law and explain how it is incorporated in Faraday’s law.
Answer:
Statement: The direction of induced current in a circuit is such that the magnetic field produced by the induced current opposes the change in the magnetic flux that induces the current. The direction of induced emf is same as that of induced current.

Explanation:

i. According to Faraday’s laws of electromagnetic induction,
e = \(\frac{-d \phi}{d t}\) ….(1)

ii. Consider that area vector \(\vec{A}\) of the loop perpendicular to the plane of the loop is fixed and oriented parallel (θ = 0) to magnetic field \(\vec{B}\). The magnetic field \(\vec{B}\) increases with time.

iii. By definition of flux, φ = \((\vec{B} \cdot \vec{A})\)
∴ e = \(-\frac{d}{d t}(\vec{B} \cdot \vec{A})\) = \(-|\vec{A}| \frac{d|\vec{B}|}{d t}\) ……. (1)
iv. Here, \(-|\vec{A}| \frac{d|\vec{B}|}{d t}\) is positive and \(\frac{d B}{d t}\) is positive as B increases with time whereas e.m.f is a negative quantity,

v. The screw driver rule fixes the positive sense of circulation around the loop as the clockwise direction.

vi. As the sense of the induced current in the loop is counter clockwise (negative), the sense of induced emf also is negative (-ve). That is, the LHS of equation (1) is indeed a negative (-ve) quantity in order to be equal to the RHS.

vii. Thus the negative (-ve) sign in the equation e = \(-\frac{d \phi}{d t}\) incorporates Lenz’s law into Faraday’s law.

Question 20.
The maximum value of permeability of a metal (77% Ni, 16% Fe, 5% Cu, 2% Cr) is 0.126 TmA^-1 Find the maximum relative permeability and susceptibility.
Answer:
Given: μ = 0.126 T mA^-1
To find: Relative permeability (pr), Susceptibility (x)

Formulae:
i. μ_r = \(\frac{\mu}{\mu_0}\)
ii. μ_r = 1 + χ

Calculation:
From formula (i),
μ_r = \(\frac{0.126}{4 \pi \times 10^{-7}}\)
= 1.0 × 10⁵
From formula (ii),
χ = μ_r – 1
χ = 1.0 × 10⁵ – 1
∴ χ = 99.99 × 10³

Ans:
i. The relative permeability is 1.0 × 10⁵.
ii. The susceptibility is 99.99 × 10³.

Question 21.
i. What is a perfect blackbody? How can it be realized in practice?
ii. State Prevost’s theory of exchange of heat.
Answer:
i.

a. A body, which absorbs the entire radiant energy incident on it, is called an ideal or perfect blackbody.
b. Perfectly blackbody does not exist in nature. However, for practical purposes, lamp black can be treated as perfectly blackbody.
c. A blackbody is most nearly realized in practice by the use of a small hole in the wall of a uniform temperature enclosure.
d. Such an enclosure has its walls maintained at the same temperature. Thus, the radiation coming out of the hole is approximately blackbody radiation.
e. In practice, Ferry’s absorber behaves as a perfectly blackbody. [2 Marks]

ii. All bodies at all temperatures above 0 K (absolute zero temperature) radiate thermal energy and at the same time, they absorb radiation received from the surroundings.

Question 22.
A wire of length 0.5 m is stretched by 2 kg wt. If mass of the wire is 0.98 × 10^-3 kg, find the velocity of transverse wave along the wire and its fundamental frequency.
Answer:
Given: M = 0.98 × 10^-3 kg, I = 0.5 m,
T = 2 kg wt = 2 × 9.8 N = 19.6 N,
∴ m = \(\frac{M}{I}\) = \(\frac{0.98 \times 10^{-3}}{0.5}\)
m = 1.96 × 10^-3 kg/m
To find: Velocity (v), Fundamental frequency (n)
Formulae:
i. v = \(\sqrt{\frac{T}{m}}\)
ii. n = \(\frac{1}{21} \sqrt{\frac{T}{m}}\)
Calculation:
From formula (i),
v = \(\sqrt{\frac{19.6}{1.96 \times 10^{-3}}}\)
∴ v = 100 ms^-1
From formula (ii).
n = \(\frac{1}{2 \times 0.5} \sqrt{\frac{19.6}{1.96 \times 10^{-3}}}\) = \(\frac{100}{2 \times 0.5}\)
∴ n = 100 Hz
Ans: The velocity of the transverse wave along the wire is 100 ms^-1 and its fundamental frequency is 100 Hz.

Question 23.
A rigid object is rolling down an inclined plane. Derive expressions for the acceleration along the track and the speed after falling through a certain vertical distance.
Answer:
i. Consider a rigid object of mass m and radius R, rolling down an inclined plane, without slipping. Inclination of the plane with the horizontal is θ.

ii. As the objects starts rollinq down, its gravitational P.E. is converted into K.E. of rolling.

iii. Starting from rest, let v be the speed of the centre of mass as the object comes down through a vertical distance h.

iv. Total kinetic energy,

v. Linear distance travelled along the plane is
s = \(\frac{h}{\sin \theta}\)

vi. During this distance, the linear velocity has increased from zero to v.

vii. If a is the linear acceleration along the plane,
v² = u² + 2as
∴ 2as = v² – u²
∴ 2a\(\left(\frac{h}{\sin \theta}\right)\) = \(\frac{2 g h}{\left(1+\frac{K^2}{R^2}\right)}\) – 0
∴ a = \(\frac{g \sin \theta}{\left(1+\frac{K^2}{R^2}\right)}\)

viii. For pure sliding, without friction, the acceleration is g sinθ and final velocity is \(\sqrt{2 g h}\).
Thus, during pure rolling, the factor (1 + \(\frac{\mathrm{K}^2}{\mathrm{R}^2}\)) is effective for both the expressions.

Question 24.
What is series LCR resonant circuit? State conditions for series resonance. Obtain an expression for resonant frequency.
Answer:
i. A circuit in which inductance L, capacitance C and resistance R are connected in series and the circuit admits maximum current corresponding to a given frequency of AC, is called a series resonance circuit.

ii. The impedance (Z) of an LCR circuit is given by, Figure(a)
Z = \(\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)

iii. At very low frequencies, inductive reactance X_L = ωL is negligible but capacitive reactance X_C = \(\frac{1}{\omega C}\) is very high.

iv. As we increase the applied frequency then X_L increases and X_C decreases.

v. At some angular frequency (ω_r), X_L = X_C

Where f_r is called the resonant frequency.

vi. At this particular frequency f_r, since X_L: X_C we get Z = \(\sqrt{R^2+0}\) = R. This is the least value of Z.

vii. Thus, when the impedance of an LCR circuit is minimum, circuit is said to be purely
resistive, current and voltage are in phase and hence the current i₀ = \(\frac{e_0}{Z}\) = \(\frac{e_0}{R}\) is maximum. This condition of the LCR circuit is called resonance condition and this frequency iS called series resonant frequency.

Question 25.
Derive the relation between surface tension and surface energy per unit area.
Answer:
Relation between surface tension and surface energy:

i. Let ABCD be a rectangular frame of wire, fitted with a movable arm PQ.

ii. The frame held in horizontal position is dipped into soap solution and taken out so that a soap film APQB is formed. Due to surface tension of soap solution, a force ‘F’. will act on each arm of the frame. Under the action of this force, the movable arm PQ moves towards AB.

iii. Magnitude of force due to surface tension is,
F = 2Tl …… (∵ T = F/l)
(A factor of 2 appears because soap film has two surfaces which are in contact with wire.)

iv. Let the wire PQ be pulled outwards through a small distance ‘dx’ to the position P’Q’, by applying an external force F’ isothermally, which is equal and opposite to F. Work done by this force, dW = F’dx = 2T/dx.

v. But, 2ldx = dA = increase in area of two surfaces of film.
∴ dW = T dA

vi. This work done in stretching the film is stored in the area dA in the form of potential energy (surface energy).
∴ Surface energy, E = T dA
∴ \(\frac{E}{d A}\) = T
Hence,
surface tension = surface energy per unit area.

vii. Thus, surface tension is equal to the mechanical work done per unit surface area of the liquid, which is also called as surface energy.

Question 26.
A body of mass 1 kg is made to oscillate on a spring of force constant 16 N/m. Calculate:
i. Angular frequency,
ii. Frequency of vibration
Answer:
Given: m = 1 kg, k = 16 N/m
To find: Angular frequency (ω), Frequency of vibration (n)

Formulae:
i. ω = \(\sqrt{\frac{k}{m}}\)
ii. ω = 2πn

Calculation:
From formula (i)
We have, for S.H.M.
ω = \(\sqrt{\frac{\mathrm{k}}{\mathrm{~m}}}\) = \(\sqrt{\frac{16}{1}}\)
∴ ω = 4 rad/s
From formula (ii),
n = \(\frac{\omega}{2 \pi}\) = \(\frac{4}{2 \pi}\) = \(\frac{2}{\pi}\)Hz
∴ n = \(\frac{2}{3.142}\) = 0.6365 Hz.

Ans: The angular frequency of the body is 4 rad/s and the frequency of vibration is 0.6365 Hz.

Section – D

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

Question 27.
i. State and Explain Newton’s law of viscosity.
ii. What is the minimum distance between two objects which can be resolved by a microscope having the visual angle of 30° when light of wavelength 500 nm is used?
Answer:
i. a. Statement: For streamline flow, the viscous force acting on any layers is directly proportional to

1. area of the layer (A)
2. velocity gradient (dv/dx)

b. Explanation:
Let A be the area of layer parallel to the direction of flow and \(\left|\frac{d v}{d x}\right|\) be the velocity gradient, then the viscous force F is given by,
F ∝ A …… (1)
F ∝ \(\frac{d v}{d x}\) ……. (2)
Combining equations (1) and (2) we have, F ∝ A\(\left(\frac{d v}{d x}\right)\)
∴ F = ηA\(\left(\frac{d v}{d x}\right)\)
where η constant called coefficient of viscosity of the liquid which depends upon the nature of the liquid.

ii. Solution;
Given: a = 30°, λ = 500 nm = 500 × 10^-9 m
To find: Minimum distance of resolution (a)
Formula: a_min = \(\frac{1.22 \lambda}{2 \sin a}\) = \(\frac{0.61 \lambda}{\sin a}\)
Calculation:
From formula.
a_min = \(\frac{0.61 \times 500 \times 10^{-9}}{\sin \left(30^{\circ}\right)}\)
= 6.1 × 10^-7 m
Ans: The minimum distance of resolution is 6.1 × 10^-7 m.

Question 28.
i. Derive the relation for magnetic force acting on an arbitrarily shaped wire assuming relation for a straight wire.
ii. A straight conductor 1 m long carrying a current of 12 A is kept at right angles to a uniform magnetic field of induction 5.5 × 10^-3 Wb/m². What is the force acting upon it?
Answer:
i. a. Consider a segment of infinitesimal length dl along the wire as shown in figure below:

b. If I is the current flowing, the magnetic force due to perpendicular magnetic field \(\vec{B}\) (coming out of the plane of the paper) is given by
\(d \vec{F}_m\) = \(I \vec{d} \times \vec{B}\)
c. The force on the total length of wire is
\(\vec{F}_m\) = \(\int d \vec{F}_m\) = \(I \int \vec{d} \times \vec{B}\)
d. If \(\vec{B}\) is uniform over the whole wire then,
\(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) = \(\mathrm{I}\left[\int \overrightarrow{\mathrm{~d} l}\right] \times \overrightarrow{\mathrm{B}}\)
This is the required relation. [2 Marks]

ii.
Given: I = 12 A, l = 1 m, B = 5.5 × 10^-3 Wb/m², θ = 90°
To find: Force (F)
Formula: F = I/B Sinθ
Calculation:
From formula,
F = 12 × 1 × 5.5 × 10^-3 × sin 90°
= 12 × 1 × 5.5 × 10^-3 × 1
∴ F = 0.066 N
Ans: The force acting on the conductor is 0.066 N.

Question 29.
With the help of p-V diagram explain the work done by the system while expanding when
i. System undergoes first volume change at constant pressure and then the pressure changes at constant volume.
ii. System undergoes first pressure change at constant volume and then the volume change at constant pressure.
iii. In which case will the work done be more?
Answer:
The state of a system can be changed from initial to final in different ways.

i. a. Consider the system changes its state from A to B as shown in figure (a).
b. In this case, the volume increases to V_i from the point A up to the point C at the constant pressure p_i.

c. After point C, the pressure of the system decreases to p_f at constant volume as shown in the figure (a).
d. The system thus, reaches its final state B with co-ordinates (V_f, p_f). Work done in this process is represented by the shaded area under the curve in figure (a).

ii. a. Consider the system changes its state from A to B as shown in figure (b).

b. In this case, the pressure decreases from p_i to p_f at constant volume V_i along the path AD.
c. After point b, the volume of the system increases to V_f at constant pressure p_f as shown in figure (b).
d. Work done in this process is represented by the shaded area under the curve in figure (b).

iii. From figures (a) and (b) we can conclude that the work done is more when the system follows path ACB than the work done by the system along the path ADB.

Question 30.
Explain with neat circuit diagram, how will you determine the unknown resistance ‘X’ by using a metrebridge experiment.
Answer:
Construction:

1. Metrebridge consists of a one metre long wire of uniform cross section, stretched on a metre scale which iS fixed on a wooden table.
2. The ends of the wire are fixed below two L shaped metallic strips. A single metallic
   strip separates the two L shaped strips leaving two gaps, left gap and right gap.
3. Usually, an unknown resistance X is connected in the left gap and a resistance box is connected in the other gap.
4. One terminal of a galvanometer is connected to the point C on the central strip, while
   the other terminal of the galvanometer carries the jockey (J). Temporary contact with the wire AB can be established with the help of the jockey.
5. A cell of emf E along with a key and a rheostat are connected between the points A and B.

Working:

I. A suitable resistance R is selected from resistance box.

ii. The jockey is brought in contact with AB at various points on the wire AB and the balance point (null point), D is obtained. The galvanometer shows no deflection when the jockey is at the balance point (point D).

iii. Let the respective lengths of the wire between A and D, and that between D and B be I_X and I_R.

iv. Then using the balancing conditions,
\(\frac{X}{R}\) = \(\frac{R_{A D}}{R_{D B}}\) …..(1)
where R_AD and R_DB are resistance of the parts AD and DB of the wire respectively.

v. If I_X and I_R are lengths of the parts AD and DB of the wire AB, p is its specific resistance of the wire, and A is its area of cross section of wire AB then

Thus, knowing R, I_X and I_R, the value of the unknown resistance X can be determined.

Question 31.
Describe the experimental set-up for a photoelectric effect with the help of neat and labelled schematic diagram.
Answer:
i. A laboratory experimental set-up for the photoelectric effect consists of an evacuated glass tube with a quartz window.

ii. The glass tube contains photosensitive metal plates. One is the emitter E and another plate is the collector C.

iii. The emitter and collector are connected to a voltage source whose voltage can be changed and to an ammeter to measure the current in the circuit.

iv. A potential difference of V. as measured by the voltmeter, is maintained between the
emitter E and collector C. Generally, C (the anode) is at a positive potential with respect to the emitter E (the cathode). This potential difference can be varied and C can even be at negative potential with respect to E.

v. When the anode potential (V) is positive, it accelerates the electrons. This potential
is called accelerating potential. When the anode potential (V) is negative, it retards the flow of electrons. This potential is known as retarding potential.

vi. A source S of monochromatic light of sufficiently high frequency (short wavelength
≤ 10^-7 m) is used.

Maharashtra Board Class 12 Physics Previous Year Question Papers