Maharashtra State Board Class 12th Physics Question Paper 2024 with Solutions Answers Pdf Download.

## Class 12 Physics Question Paper 2024 Maharashtra State Board with Solutions

Time: 3 Hrs.

Max. Marks: 70

General instructions: The question paper is divided into four sections.

- Section A: Q. No’.l 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. - Section B: Q.No.3 to Q.No.14 contains Twelve short answer type of questions carrying Two marks each. (Attempt any Eight)
- Section C: Q.No.15 to Q.No.26 contain Twelve short answer type of questions carrying Three marks each. (Attempt any Eight)
- Section D: Q.No.27 to Q.No.31 contain Five long answer type questions carrying Four marks each. (Attempt any Three)
- Use of the log table is allowed. Use of calculator is not allowed.
- Figures to the right indicate full marks.
- For multiple choice type question, only the first attempt will be considered for evaluation.
- Physical Constants:
- Mass of electron m = 9.1 × 10
^{-31}kg - ∈
_{0}= 8.85 × 10^{-12}C^{2}/Nm^{2} - ⚿ = 3.142
- Charge on electron e = 1.6 × 10
^{-19}C - µ
_{0}= 4⚿ × 10^{-7}Wb/Am - Planck’s constant h = 6.63 × 10
^{-34}J.s. - Speed of light c = 3 × 10
^{8}m/s - g = 9.8 m/s
^{2} - Rydberg’s constant RH = 1.097 × 10
^{7}m^{-1} - Stefan’s constant σ = 5.67 × 10
^{-8}J m^{-2}s^{-1}K^{-4}

- Mass of electron m = 9.1 × 10

**Section – A**

Question 1.

Select and write the correct answer for the following multiple choice questions: [10]

(i) The moment of inertia (MI) of a disc of radius Rand mass M about its central axis is

(a) \(\frac{M R^2}{4}\)

(b) \(\frac{M R^2}{2}\)

(c) MR^{2}

(d) \(\frac{3 M R^2}{2}\)

Answer:

(b) \(\frac{1}{2}\)MR^{2}

Explanation :

Total moment of intertia of the disc

(ii) The dimensional formula of surface tension is

(a) [L^{-1}M^{1}T^{-2}]

(b) [L^{1}M^{1}T^{-1}]

(c) [L^{2}M^{1}T^{-2}]

(d) [L^{0}M^{1}T^{-2}]

Answer:

(d) [L^{0}M^{1}T^{-2}]

Explanation :

Surface tension T = \(\frac{\mathrm{F}}{\mathrm{~L}}\)

= \(\left[\frac{M L T^{-2}}{\mathrm{~L}}\right]\) = L^{0}M^{1}T^{-2}

(iii) Phase difference between a node and an adjacent antinode in a stationary wave is ………

(a) \(\frac{\pi}{4}\)rad

(b) \(\frac{\pi}{2}\)rad

(c) \(\frac{3 \pi}{2}\)rad

(d) π rad

Answer:

(b) \(\frac{\pi}{2}\) rad

Explanation :

The phase difference between a node and an adjacent antinode in a stationary wave is \(\frac{\pi}{2}\) radians or 90 degrees, due to the nature of their constructive and destructive interference.

(iv) The work done/in bringing a unit positive charge from infinity to a given point against the direction of electric field is known as ……..

(a) electric flux

(b) magnetic potential

(c) electric potential

(d) gravitational potential

Answer:

(c) electric potential

(v) To convert a moving coil galvanometer into an ammeter we need to connect a …..

(a) small resistance in parallel with it

(b) large resistance in series with it

(c) small resistance in series with it

(d) large resistance in parallel with it

Answer:

(a) small resistance in parallel with it

(vi) If the frequency of incident light falling on a photosensitive material is doubled, then kinetic energy of the emitted photoelectron will be …….

(a) the 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

Explanation :

We know that

E = ω_{0} + K_{max}

K_{max} = E – ω_{0}

but E = hv

(v frequency of incident Light)

Then

K_{1} = hv – ω_{0}

⇒ hv = K_{1} + ω_{0}

According to question

K_{2} = 2hv – ω_{0}

Put the value hv in equation (i)

K_{2} = 2K_{2} + 2ω_{0} – ω_{0}

K_{2} = 2K_{1} + ω_{0}

Hence,

K_{2} > 2K_{1}

(vii) In a cyclic process, if ∆U= internal energy, W = work done, Q = Heat supplied then ……..

(a) ∆U = Q

(b) Q = O

(c) W = O

(d) W = Q

Answer:

(d) W = Q

Explanation:

Heat supplied ∆Q = ∆V + W

In a cyclic process

∆U = O

Then,

∆Q = W

(viii) The current in a coil changes from 50A to 10A in 0.1 second. The self-inductance of the coil is 20H. The induced e.m.f in the coil is ……

(a) 800 V

(b) 6000 V

(c) 7000 V

(d) 8000 V

Answer:

(d) 8000 V

Explanation :

Self Inductance

e = -L\(\frac{d l}{d t}\) = -L\(\frac{\Delta l}{\Delta t}\)

Given,

L = 20 H

Initial, current I_{1} = 50A

Final current I_{2} = 10A

So, ∆I = I_{2} – I_{1} = -40

∆t = 0.1 sec

Then,

e = -20 × \(\frac{-40}{0.1}\) = 8000V

(ix) The velocity of bob of a second’s pendulum when it is 6 cm from its mean position and amplitude of 10 cm, is …….

(a) 8π cm/s

(b) 6π cm/s

(c) 4π cm/s

(d) 2π cm/s

Answer:

(a) 8π cm/s

Explanation:

bob at any position x from the mean position to its amplitude (A) and angular frequency (ω)

V = ω\(\sqrt{\mathrm{A}^2-x^2}\)

Here, ω = \(\frac{2 \pi}{T}\)

Given,

A = 10 cm, x = 6 cm, T = 2 second

Then ω = \(\frac{2 \pi}{2}\) = π rad/second

Then

V = \(\pi \sqrt{10^2-6^2}\)

= 8π cm/s

(x) In biprism experiment, the distance of 20^{th} bright band from the central bright band is 1.2 cm. Without changing the experimental set-up, the distance of 30^{th} bright band from the central bright band will be …….

(a) 0.6 cm

(b) 0.8 cm

(c) 1.2 cm

(d) 1.8 cm

Answer:

(d) 1.8 cm

Explanation:

y_{n} = K.n

n = 20

Then y_{20} = k.20

Given y_{20} = 1.2

Then k = \(\frac{1.2}{20}\)

y_{30} = k × 30

= \(\frac{1.2}{20}\) × 30

= 1.8 cm

Question 2.

Answer the following questions: [8]

(i) Define centripetal force.

Answer:

Centripetal. force is the force acting on an object performing circular motion which is along the radius of the circle directed towards the centre of circle.

(ii) Why a detergent powder is mixed with water to wash clothes?

Answer:

When a detergent powder is mixed with water then the surface tension of water decreases due to which water make good contact with the fabric and is able to remove tough stains.

(iii) What is the resistance of an ideal voltmeter?

Answer:

Resistance of an ideal voltmeter is infinite.

(iv) Write the formula for torque acting on rotating current carrying coil in terms of magnetic dipole moment, in vector form.

Answer:

\(\vec{\tau}\) = \(\vec{M} \times \vec{B}\)

Here,

\(\vec{\tau}\) = torque

\(\vec{M}\) = Magnetic dipole moment

\(\vec{B}\) = Magnetic field

(v) What is binding energy of a hydrogen atom?

Answer:

13.6eV

(vi) What is surroundings in thermodynamics?

Answer:

In thermodynamics, Surrounding refers to everything outside of the system being studied.

OR

It contains everything other than the system.

(vii) In a photoelectric experiment, the stopping potential is 1.5V. What is the maximum kinetic energy of a photoelectron?

Answer:

Given: V_{0} = 1.5 V, KE_{max} = ?

Formula: KE_{max} = eV_{0}

∴ KE_{max} = e × 1.5V

∴ KE_{max} = 1.5 eV

(viii) Two capacitors of capacities 5 μF and 10 μF respectively are connected in series. Calculate the resultant capacity of the combination.

Answer:

Given : C_{1} = 5µF, C_{2} = 10µF, C_{s} = ?

Formula: C_{s} = \(\frac{C_1 \times C_2}{C_1+C_2}\)

C_{s} = \(\frac{5 \times 10}{5+10}\)

= \(\frac{50}{15}\)

∴ C_{s} = 3.33 µF

**Section – B**

Attempt any EIGHT questions of the following: [16]

Question 3.

Explain the change in internal energy of a thermodynamic system (the gas) by heating it.

Answer:

(1) When a thermodynamic system, such as a gas, is heated, its internal energy typically increases. Internal energy refers to the total kinetic energy and potential energy of the particles within the system.

(2) Heating a gas increases its internal energy primarily by transferring/energy to the gas particles in the form of heat. This additional energy causes the gas molecules to move faster, leading to an increase in their kinetic energy. Additionally, heating can also lead to an increase in the potential energy of the gas molecules as the intermolecular forces between them weaken, allowing them to move farther apart.

(3) From a thermodynamic perspective, the change in internal energy (∆U) of the gas due to heating can be quantified by the first law of thermodynamics, which states:

∆U = Q – W

Where, ∆U = change in the internal energy of the system.

Q = Heat added to the system

W = Work done by the system

Question 4.

Explain the construction of a spherical wavefront by using Huygens’ principle.

Answer:

Construction of a spherical wave front by using Huygens’ principle:

(1) Spherical wave front is formed when source of light is at a finite distance from point of observation.

(2) Let S be the point source of light in air. PQR represents spherical wave front at any instant. The wavefront PQR acts as a primary wave which is propagated through air.

PQR: Primary wave front

P_{1}Q_{1}R_{1}: Secondary wave front after time

SPN_{1}, SQN_{2}, SRN_{3}: Wave normals at P, Q, R.

(3) According to Huygens’ principle, all the points on PQR will act as secondary sources of light and send secondary wavelets with same velocity ‘c’ in air.

(4) To find out new wavefront at a later instant ‘t’, draw hemispheres with P, Q, R,…….. as centres and ‘ct’ as radius in the forward direction.

The secondary waves moving in a backward direction does not exist.

Question 5.

Define magnetisation. State its SI unit and dimensions.

Answer:

The ratio of net magnetic moment to the volume of the material is called magnetisation.

Magnetisation \(=\frac{\text { Net Magnetic Moment }}{\text { Volume }}\)

S.I. unit: Am^{-1}

Dimensions: [L^{-1}M^{0}T^{0}]^{1}]

Question 6.

Obtain the differential equation of linear simple harmonic motion.

Answer:

(1) In a Linear S.H.M.. the force is directed towards the mean position and its magnitude is directly proportional to the displacement of the body from mean position.

Restoring force, f = -kx …..(i)

Where k is force constant and x is displacement from mean position.

(2) According to Newtons second law of motion,

f = Ma

Put the value f in equation (i)

∴ Ma = -kx …….. (ii)

(3) The velocity of the particle is, v = \(\frac{d x}{d t}\) and its acceleration, a = \(\frac{d v}{d t}\) = \(\frac{d^2 x}{d t^2}\)

(4) Substituting value of a n equation (ii), we get

M\(\frac{d^2 x}{d t^2}\) = -kx

∴ \(\frac{d^2 x}{d t^2}+\frac{k}{M} x\) = 0

Substituting \(\frac{\mathrm{K}}{\mathrm{M}}\) = ω^{2}, where ω is the angular frequency

∴ \(\frac{d^2 x}{d t^2}\) + ω^{2}.x = 0

Which is the required equation.

Question 7.

A galvanometer has a resistance of 30Ω and its full scale deflection current is 20 microampere (μA). What resistance should be added to it to have a range 0-10 volt?

Answer:

Given: G = 30Ω, I_{g} = 20µA = 20 × 10^{-6}A, R = ?

Deflection of voltage needed V = 10 volt

∴ V = I_{g}(R + G)

∴ 10 = 20 × 10^{-6} (R + 30)

∴ \(\frac{10^6}{2}\) = R + 30

R = 500000 – 30

∴ R = 499970Ω

Hence, 499970Ω resistance should be added to it to have a range 0 – 10 volt.

Question 8.

Explain Biot-Savart law.

Answer:

(1) In the above figure, there is a current carrying wire of arbitrary shape, carrying a current I. The current in the differential length element dl produces differential magnetic field d\(\vec{B}\) at a point Pat a distance \(\vec{r}\) from the element dl. The \(\otimes\) indicates that d\(\vec{B}\) directed into the plane of the paper.

(2) According to Biot-Savart’s law, the magnitude of the magnetic field induction dB at a point P due to current carrying conductor:

(i) dB ∝ I

Here, I = current

(ii) dB ∝ dl

Here, dl = differential length element

(iii) dB ∝ sin θ

Here, θ = angle between dl and r

(iv) d ∝ \(\frac{1}{r^2}\)

Here, r = distance of point P.

Combining these factors, we get

∴ dB = \(\frac{\mu_0}{4 \pi} \frac{\mid d l \sin \theta}{r^2}\)

Above equation is known as Biot-Savart’s law.

Question 9.

What is a Light Emitting Diode? Draw its circuit symbol.

Answer:

The light Emitting Diode (LED) is a diode which emits light when large forward current passes through it.

Question 10.

An aircraft of wing span of 60 m flies horizontally in earth’s magnetic field of 6 × 10^{-5} T at a speed of 500 m/s. Calculate the e.m.f induced between the tips of wings of aircraft.

Answer:

Given: B = 6 × 10^{-5}T, l = 60m, V = 500m/s, e = ? Formula: Induced emf, e = Blv

∴ e = 6 × 10^{-5} × 60 × 500

∴ e = 1.8 V

Question 11.

Derive an expression for maximum speed of a vehicle moving along a horizontal circular track.

Answer:

(1) Forces acting on the cor (considered to be a particle) are:

- Weight Mg, vertically downwards.
- Normal reaction N, vertically upwards that balances the weight Mg and
- Force of static friction f
_{s}between rood the tyres to prevents the vehicle from outward slipping.

(2) f_{s} balances the centrifugal force

∴ Mg = N and f_{s} = \(\frac{m v^2}{r}\)

∴ \(\frac{f_s}{N}\) = \(\frac{m v^2}{r m g}\) = \(\frac{v^2}{r g}\)

(3) As the speed v increases, then force of static friction f_{s} also increases.

(4) But maximum force of static friction (f_{s})_{max} = μ_{s} • N

This is the required expression for maximum speed of a vehicle moving along a horizontal circular track.

Question 12.

A horizontal force of 0.5N is required to move a metal plate of area 10^{-2} m^{2} with a velocity of 3 × 10^{-2} m/s, when it rests on 0.5 × 10^{-3} m thick layer of glycerin. Find the coefficient of viscosity of glycerin.

Answer:

Given : F = 0.5 N, A = 10^{-2} m^{2}, dv = 3 × 10^{-2} m/s, dx = 0.5 × 10^{-3} m

To find: n = ?

Formula: n = \(\frac{F \cdot d x}{A \cdot d v}\)

∴ n = \(\frac{0.5 \times 0.5 \times 10^{-3}}{10^{-2} \times 3 \times 10^{-2}}\)

∴ n = 0.833 Ns/m^{2}

Question 13.

Two tuning forks having frequencies 320 Hz and 340 Hz are sounded together to produce sound waves. The velocity of sound in air is 340 m/s. Find the difference in wavelength of these waves.

Answer:

Given : n_{1} = 320 Hz, n_{2} = 340 Hz, v = 340 m/s

Question 14.

Calculate the change in angular momentum of electron when it jumps from third orbit to first orbit in hydrogen atom.

Answer:

To find : L_{3} – L_{1} =?

Formula : L = \(\frac{n h}{2 \pi}\)

∴ L_{3} – L_{1} = \(\frac{3 h}{2 \pi}-\frac{h}{2 \pi}\) = \(\frac{2 h}{2 \pi}\) = \(\frac{6.63 \times 10^{-34}}{3.142}\)

∴ L_{3} – L_{1} = 2.11 × 10^{-34} Nm^{-s}

**Section – C**

Attempt any EIGHT questions of the following: [24]

Question 15.

A circular coil of wire is made up of 200 turns, each of radius. 10 cm. If a current of 0.5 A passes through it, what will be the magnetic field at the centre of the coil?

Answer:

Given : N = 200 turns, R = 10 cm = 0.1 m, I = 0.5A, B = ?

Formula: B ‘= ?

Formula: B = \(\frac{\mu_0 N I}{2 R}\)

∴ B = \(\frac{4 \pi \times 10^{-7} \times 200 \times 0.5}{2 \times 0.1}\)

B = 6.28 × 10^{-4}T

Question 16.

Define photoelectric effect and explain the experimental set-up of 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. Experimental set-up of photoelectric effect:

1. A Laboratory experimental set-up for the photoelectric effect consists of an evacuated glass tube with a quartz window.

2. The glass tube contains photosensitive metaL plate. One is the emitter E and another plate is the collector C.

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

4. 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 a negative potential with respect to E.

5. 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.

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

7. When light is incident on the emitter plate the photoelectrons are emitted. The photoelectrons are attracted by positive plate A. The emission of electrons causes flow of electric current in the circuit. The current flowing in the circuit is measured by the ammeter connected in the circuit.

Question 17.

Define the current gain α_{DC} and β_{DC} for a transistor. Obtain the relation between them.

Answer:

α_{DC} : It is defined as the ratio of collector current to emitter current.

∴ α_{DC} = \(\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{E}}}\) ……… (i)

β_{DC}: It is defined as the ratio of collector current to base current

∴β_{DC} = \(\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}\) …… (ii)

Relation between α_{DC} and β_{DC} : We know that,

I_{E} = I_{B} + I_{C}

Dividing both sides by I_{C}

Above expression represents the relationship between α_{DC} and β_{DC}

Question 18.

Define surface energy of the liquid. Obtain the relation between the surface energy and surface tension.

Answer:

Surface Energy : The extra energy of the molecules on the surface layer of a liquid is called surface energy of the liquid.

Fig.: Surface energy of a liquid

Consider a rectangular frame P’ PSS’ having a movable wire QR. Let QR = PS = L If soap film is formed on the frame PQRS then surface tension will pull the wire inward by a force F.

∴ Surface tension \(=\frac{\text { Force }}{\text { Free surface }}\)

∴ T = \(\frac{F}{2 L}\)

∴ F = T(2L)

If the wire is pulled out to Q’R’ through distance ‘dx’ then work is given by

dW = F.dx

∴ dW = T • 2L • dx

But increase in area = d∆ = 2L • dx

∴ dW = T • dA

This work done is stretching the film is stored in the area dA of the film as its potential energy. This energy is called surface energy.

∴ Surface energy = T • dA

Above expression represents. The relationship between surface energy and surface tension.

Question 19.

What is an isothermal process? Obtain an expression for work done by a gas in an isothermal process.

Answer:

A process in which change in pressure and volume takes place at a constant temperature is called an isothermal process.

Expression of the work-done during an isothermal process:

Consider the isothermal expansion of an ideal gas. Let its initial volume be V_{i} and the final volume be V_{f}. The work done in an infinitesimally small isothermal expansion is given by

dW = PdV

The total work done in bringing out the expansion from the initial volume V_{i} to the final volume V_{f} is given by:

This is the required expression for the work done during an isothermal process.

Question 20.

Derive an expression for equation of stationary wave on a stretched string. Show that the distance between two successive nodes or antinodes is λ/2.

Answer:

(1) 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 condition creates a stationary interference pattern called a stationary waves.

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

(3) The equation of wave travelling along the x-axis in the positive direction is

y_{1} = a sin{2π(nt – \(\frac{x}{\lambda}\))} …….. (i)

The equation of wave travelling along the x-axis in the negative direction is

y_{2} = a sin{2π(nt + \(\frac{x}{\lambda}\))} …… (ii)

When these waves interfere, the resultant displacement of particles of string is given by the principle of superposition of wave as

Using the trigonometrical identify,

This is the equation of a stationary wave which gives resultant displacement due to two simple harmonic progressive waves.

(4) Nodes are the points of minimum displacement. This possible if the amplitude is minimum (zero), i.e.

A = 0

Distance between two successive nodes

= \(\frac{3 \lambda}{4}-\frac{\lambda}{4}\) = \(\frac{\lambda}{2}\) …… (iii)

(5) Antinodes are the points of maximum displacement, i.e., A = ± 2a

Distance between two successive antinodes

= \(\frac{\lambda}{2}\) – 0 = \(\frac{\lambda}{2}\) ……. (iv)

(6) From (iii) and (iv), distance between two successive nodes or antinodes is \(\frac{\lambda}{2}\).

Question 21.

Derive an expression for the impedance of an LCR circuit connected to an AC power supply. Draw

Answer:

(1) The following figure shown an inductor of inductance (L) capacitor of capacitance (C) a resistor of resistance (R) are connected to an alternating emf (e) to form a closed series circuit

(2) As the inductor, capacitor, and resistor are connected in series, they all carry the same current at any given time.

i = i_{0} sin ωt

(3) The current is in phase with the voltage across the resistor, e_{R} = Ri. The voltage across the inductor, e_{L} = X_{L}i is \(\frac{\pi}{2}\) rad. ahead of the current, while the voltage across the capacitor, e_{c} = X_{c}i, is \(\frac{\pi}{2}\) rad. behind the current.

(4) Phasor diagram:

(5) From the figure,

This is the effective resistance of the circuit. It is called the impedance (Z).

Question 22.

Calculate the wavelength of the first two lines in Balmer series of hydrogen atom.

Answer:

Given: R = 1.097 × 10^{7} m^{-1}, n = 2

To find : λ_{1} = ?, λ_{2} = ?

Question 23.

A current carrying toroid winding is internally filled with lithium having susceptibility χ = 2.1 × 10^{-5}. What is the percentage increase in the magnetic field in the presence of lithium over that without it?

Answer:

The magnetic field inside the solenoid without lithium,

B_{0} = μ_{0}H

The magnetic field inside the solenoid with lithium,

B = μH

Question 24.

The radius of a circular track is 200 m. Find the angle of banking of the track, if the maximum speed at which a car can be driven safely along it is 25 m/sec.

Answer:

Given : V = 25 m/s, r = 200 m, g = 9.8 m/s^{2}, θ =?

Formula: θ = tan^{-1}\(\left(\frac{\mathrm{V}^2}{r g}\right)\)

θ = tan^{-1}\(\left(\frac{25^2}{200 \times 9.8}\right)\)

θ = tan^{-1}(0.3188)

θ = 17.68°

Therefore, the angle of banking of the track should be approximately 17.68° to allow a car to be driven safely along it at a maximum speed of 25 m/s.

Question 25.

Prove the Mayer’s relation: C_{P} – C_{V} = \(\frac{R}{J}\)

Answer:

(1) Consider on mole of an ideal gas. Let the gas be heated at a constant value so that its temperature increase by dT.

(2) If Q_{1} = heat supplied to 1 mole of gas at constant volume, then Q_{1} = C_{V} • dT

When gas is heated at constant volume, it will not perform external work. According to the first law of thermodynamics, the heat supplied will just increase the internal energy of the gas.

Q_{1} = dU = C_{V} • dT …….(i)

Where, C_{V} is the molar specific heat of the gas at constant volume.

(3) Now, let the gas be heated at a constant pressure until its temperature rises by cfT.

If Q_{2} = Heat supplied to 1 mole of gas constant pressure, then

Q_{2} = C_{P} • dT ……. (ii)

Where C_{P} is the molar specific heat of the gas at constant pressure.

(4) When heat is given under constant pressure, it increases energy and allows the gas to perform work (dW). As the volume of the gas increases (dV), the work done by the gas (dW) is equal to the PdV.

∴ Q_{2} = dU + dW

∴ C_{P} • dT = C_{V} • dT + PdV

[from equations (i) to (ii)]

∴ (C_{P} – C_{V}) • dT = PdV

But, for one mole of an ideal gas,

PdV = RdT

∴ (C_{P} – C_{V}) • dT = RdT

∴ C_{P} – C_{V} = R

This is known as Mayer’s relation between C_{P} and C_{V}.

But if heat supplied is measured in calories and work done is measured in joules. Then the above relation is modified to,

C_{P} – C_{V }= \(\frac{R}{J}\), where J is the mechanical equivalent of heat.

Question 26.

An alternating voltage is given by e = 8 sin 628.4t. Find

(i) peak value of e.m.f

(ii) frequency of e.m.f

(iii) instantaneous value of e.m.f. at time t = 10 ms.

Answer:

Given:

e = 8 sin 628.4 t

Comparing e = 8 sin 628.4 t with e = e_{0} sin ωt, we get e_{0} = 8 volts, ω = 628.4 rad/s.

(i) Peak value of e.m.f. is 8 volts

(ii) f = \(\frac{\omega}{2 \pi}\) = \(\frac{628.4}{2 \times 3.142}\) = 100 Hz

(iii) To find instantaneous value of e.m.f. at time t = 10 ms, we need to put the value oft in the given equation.

∴ e(t) = 8 sin 628.4 × 10 × 10^{-3}

e(t) = 8 × 0.1094 = 0.8752 volts.

**Section – D**

Attempt any THREE questions of the following: [12]

Question 27.

What is a transformer? Explain construction and working of a transformer. Derive the equation for a transformer.

Answer:

Transformer:

(1) A transformer is a device used for changing the voltage of alternating current from low value to high value or vice versa.

(2) The transformer which changes the voltage of ac current from low value to high value is called step-up transformer. While the transformer which changes the voltage of ac current from high value to low value is called step-down transformer.

Construction and Working of a Transformer:

(1) The transformer work on the principle of Faraday’s law of electromagnetic induction and mutual induction.

(2) There are usually two coils primary coil and secondary coil on the transformer core.

(3) The core laminations are jointed in the form of strips.

(4) The two coils have high mutual inductance.

(5) When an alternating current pass through the primary coil it. Creates a varying magnetic flux.

(6) As per Faraday’s law of electromagnetic induction this change in magnetic flux induces an emf (Electromotive Force) in the secondary coil which is linked to the core having a primary coil. This is called mutual inductance.

(7) Let φ be the magnetic flux Linked per turn with both the coils at an instant t. N_{P} and N_{S} be the number of turns in the primary and secondary coil respectively.

Then induced emf in primary and secondary coil will be,

Above equation is known as equation for transformer.

(8) The ratio \(\frac{N_S}{N_p}\) is called turn ratio (transformer ratio) of the transformer.

Question 28.

Using the geometry of the double slit experiment, derive the expression for fringe width of interference bands.

Answer:

(1) Let the distance between the two slits S_{1} and S_{2} be d and that between O and O’ be D.

(2) The point O’ on the screen is equidistant from S_{1} and S_{2}. Thus, the distance travelled by the wavelets starting from S_{1} and S_{2} to reach O’ will be equal. The two waves will be in phase at O’, resulting in constructive interference. Thus, there will a bright spot at O’ and a bright fringe at the centre of the screen.

(3) Now, consider any point P on the screen. The two wavelets from S_{1} and S_{2} travel different distance to reach point P and so the phases of the waves reaching P will not be the same. If the path difference (∆L) between S_{1}P and S_{2}P is a integral multiple of λ, the two waves arriving there will interfere. Constructively producing a bright fringe at P. If the path difference between S_{1}P, S_{2}P is half integral multiple of λ, there will be destructive interference and a dark fringe will be located at P.

(4) Considering triangle S_{1}S_{1} ‘P and S_{2}S_{2}‘P, we can write

(5) Thus, the condition for constructive interference at P can be written as

∆L = y_{n}\(\frac{d}{D}\) = nλ

(6) y_{n} being the position (y-coordinate) of n^{th}

Bring fringe (n = 0, ± 1, ± 2,……..). It is given by

nλ = y_{n}\(\frac{d}{D}\)

y_{n}\(\frac{n \lambda D}{d}\)

(7) Similarly, the position of n^{th} (n = ± 1, ± 2,….)

Dark-fringe (destructive interference) is given by

Put n = n + 1

y_{n+1} = \(\frac{(n+1) \lambda D}{d}\)

(8) The distance between any two successive dark or any two successive bright fringes (y_{n+1} – y_{n}) is equal. This is called the fringe width and given by,

Fringe width = ω = ∆y = y_{n+1} – y_{n}

= (n+1)\(\frac{\lambda D}{d}\) – n\(\frac{\lambda D}{d}\) = \(\frac{\lambda D}{d}\)

This is the required expression for the fringe width.

Question 29.

Distinguish between an ammeter and a voltmeter. (Two points each).

The displacement of a particle performing simple harmonic motion is \(\frac{1}{3}\)rd of its amplitude. What fraction of total energy will be its kinetic energy?

Answer:

Ammeter | Voltmeter |

1. It measure current | 1. It measure the potential difference. |

2. It is connected in series. | 2. It is connected in paralleL |

3. It’s resistance is negligible | 3. It’s resistance is infinite |

Question 30.

Draw a neat labelled diagram of Ferry’s perfectly black body. Compare the rms speed of hydrogen molecules at 227°C with rms speed of oxygen molecule at 127°C. Given that molecular masses of hydrogen and oxygen are 2 and 32 respectively.

Answer:

Question 31.

Derive an expression for energy stored in a charged capacitor. A spherical metal ball of radius 15 cm carries a charge of 2μC. Calculate the electric field at a distance of 20 cm from the center of the sphere.

Answer:

Consider a capacitor of capacitance C connected to an emf source of voltage (V).

Let ‘q’ be the magnitude of charge on any plate of a capacitor at any instant Then