Maharashtra State Board Class 12th Physics Question Paper 2023 with Solutions Answers Pdf Download.

## Class 12 Physics Question Paper 2023 Maharashtra State Board with Solutions

Time: 3 Hours

Max. Marks: 70

**Section – A**

Select and write the correct answers for the following multiple choice type of questions: (10)

(i) If’n’ is the number of molecules per unit volume and ‘d is the diameter of the molecules, the mean free path ‘λ’ of molecules is

(a) \(\sqrt{\frac{2}{\pi n d}}\)

(b) \(\frac{1}{2 \pi n d^2}\)

(c) \(\frac{1}{\sqrt{2} \pi n d^2}\)

(d) \(\frac{1}{\sqrt{2 \pi n d}}\)

Answer:

\(\frac{1}{\sqrt{2} \pi n d^2}\)

(ii) The first law of thermodynamics is consistent with the law of conservation of …….

(a) momentum

(b) energy

(c) mass

(d) velocity

Answer:

(b) energy

(iii) Y = \(\overline{A+B}\) is the Boolean expression for

(a) OR – gate

(c) NOR – gate

(b) AND – gate

(d) NAND – gate

Answer:

(c) NOR-gate

(iv) The property of light which remains unchanged when it travels from one medium to another is ……

(a) velocity

(b) wavelength

(c) amplitude

(d) frequency

Answer:

(d) frequency

(v) If a circular coil of 100 turns with a cross-sectional area of 1 m^{2} is kept with its plane perpendicular to the magnetic field of 1 T, the magnetic flux linked with the coil will be

(a) 1 Wb

(b) 50 Wb

(c) 100 Wb

(d) 200 Wb

Answer:

(c) 100 Wb

(vi) If ‘θ’ represents the angle of contact made by a liquid which completely wets the surface of the container then ………

(a) θ = 0

(b) 0 < θ < \(\frac{\pi}{2}\)

(c) θ = \(\frac{\pi}{2}\)

(d) \(\frac{\pi}{2}\) < θ < π

Answer:

(a) 0 = 0

(vii) The LED emits visible light when its…….

(a) junction is reverse biased

(b) depletion region widens

(c) holes and electrons recombine

(d) junction becomes hot

Answer:

(c) holes and electrons recombine

(viii) 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:

(d) high coercivity and low retentivity

(ix) If the maximum kinetic energy of emitted electrons in photoelectric effect is 2eV, the stopping potential will be ………

(a) 0.5V

(b) 1.0 V

(c) 1.5 V

(d) 2.0 V

Answer:

(d) 2.0 V

(x) The radius of eighth orbit of electron in H-atom will be more than that of fourth orbit by a factor of ……..

(a) 2

(b) 4

(c) 8

(d) 16

Answer:

(b) 4

Question 2.

Answer the following questions: (8)

(i) What is the value of resistance for an ideal voltmeter?

Answer:

Infinite.

(ii) What is the value of force on a closed circuit in a magnetic field?

Answer:

zero

(iii) What is the average value of alternating current over a complete cycle?

Answer:

zero

(iv) An electron is accelerated through a potential difference of 100 volt. Calculate de-Broglie wavelength in nm.

Answer:

λ = \(\frac{12.27}{\sqrt{v}}\)Å = \(\frac{12.27}{\sqrt{100}}\)Å = 1.228 Å

(v) If friction is made zero for a road, can a vehicle move safely on this road?

Answer:

No

(vi) State the formula giving relation between electric field intensity and potential gradient?

Answer:

E = –\(\frac{d v}{d x}\)

(vii) Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by x = 5 sin \(\left(\frac{\pi t}{3}\right)\)m.

Answer:

t = 1s, x = 5 sin 60

Á = 5.x = 2.5\(\sqrt{3}\) (given)

v = w\(\sqrt{A^2-x^2}\) Formula

v = \(\frac{\pi}{3} \sqrt{25-18.74}\)

v = 1.04 × 2.5

v = 2.61 m/s

(viii) Write the mathematicalformula for Bohr magneton for an electron revolving in n^{th} orbit.

Answer:

μ_{B} = \(\frac{n e h}{4 \pi m_e}\)

**Section – B**

Attempt any EIGHT questions of the following: (16)

Question 3.

Define coefficient of viscosity. State its formula and S.I. units.

Answer:

The coefficient of viscosity of a liquid is defined as the viscous force acting tangentially per unit area of a liquid layer having a unit velocity gradient in direction perpendicular to the direction of flow of the liquid.

Formula: \(\eta\) = \(\frac{F}{A\left(\frac{d v}{d x}\right)}\)

S.I. unit: Ns/m^{2}

Question 4.

Obtain an expression for magnetic induction of a toroid of ‘N’ turns about an axis passing through its centre and perpendicular to its plane.

Answer:

The toroid is a solenoid bent into a shape of the hollow doughnut

According to Ampere’s circuital law,

\(\oint \vec{B} \cdot \overrightarrow{d L}\) = μ_{0}I

Here current ‘I’ flow through the ring as many times as there are the N no. of turns

Question 5.

State and prove principle of conservation of angular momentum.

Answer:

The angular momentum of a body remains constant if resultant external torque acting on the body is zero.

Consider a particle of mass m, relating about an axis torque \((\tau)\).

\(\vec{P}\) be linear momentum and \(\vec{r}\) be position vector

By definition of angular momentum,

If resultant external torque \((\tau)\) acting on the particle is zero, then \(\frac{\overrightarrow{d \mathrm{~L}}}{d t}\) = 0

∴ L = constant

Question 6.

Obtain an expression for equivalent capacitance of two capacitors C_{1} and C_{2} connected in series.

Answer:

All two capacitors C_{1} and C_{2} are connected in series. The P.D. across them,

V_{1} = \(\frac{\mathrm{Q}}{\mathrm{C}_1}\), V_{2} = \(\frac{\mathrm{Q}}{\mathrm{C}_2}\)

Question 7.

Explain, why the equivalent inductance of two coils connected in parallel is less than the inductance of either of the coils.

Answer:

- For parallel combination of two coils, the current through each parallel inductor is a fraction of the total current of the voltage across each parallel inductor is the same.
- As a result, a change in total current will result in less voltage dropped across the parallel array than for any one of the individual inductors
- There will be less voltage drop across parallel inductors for a given rate of charge in current than for any of the individual inductors.
- Less voltage for the same rate of charge in current result in less inductance.

Question 8.

How will you convert a moving coil galvanometer into an ammeter?

Answer:

A moving cell galvanometer can be converted into an ammeter by connecting a law resistance (called shunt) in parallel to galvanometer.

Question 9.

A 100 Ω resistor is connected to a 220V, 50 Hz supply. Calculate:

(a) r.m.s. value of current and

(b) net power consumed over the full cycle

Answer:

Given,

R = 100 Ω

V = 220 V

n = 50 Hz

(a) The rms value of current in circuit is

I_{rms} = \(\frac{V_{r m s}}{R}\) = \(\frac{220}{100}\) = 2.2 A

(b) The net power consumed over a full cycle

P = V_{rms} I_{rms}

= 220 × 2.2

= 484 W

Question 10.

A bar magnet of mass 120 g in the form of a rectangular parallelopiped, has dimensions l = 40 mm, b = 10 mm and h = 80 mm, with its dimension ‘h’ vertical, the magnet performs angular oscillations in the plane of the magnetic field with period, π seconds. If the magnetic moment is 3.4 Am^{2}, determine the influencing magnetic field.

Answer:

Given, m = 120g = 0.12 kg

l = 4 × 10^{-2} m

b = 10^{-2} m

h = 80 m = 8 × 10^{-2} m

T = π sec

μ = 3.4 Am^{2}

To find: B

Question 11.

Distinguish between free vibrations and forced vibrations (Two points).

Answer:

Free vibrations | Forced vibrations |

1. Free vibration are produced when body is disturbed from its equilibrium position. | 1. It is produced by an external periodic force of any frequency. |

2. To start free vibrations only, the force is required initially. | 2. Continuous external periodic force is required. |

Question 12.

Compare the rate of loss of heat from a metal sphere at 827°C with rate of loss of heat from the same at 427°C, if the temperature of surrounding is 27°C.

Answer:

Temp. of metal sphere, T_{1} = 827 °C = 1100 K

Temp. of surrounding, T = 27°C = 300 K

Rate of loss of heat

Question 13.

An ideal mono-atomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. Calculate the ratio of final pressure to its initial pressure.

Answer:

Question 14.

Disintegration rate of a radio-active sample is 10^{10} per hour at 20 hours from the start. It reduces to 5 × 10^{9} per hour after 30 hours. Calculate the decay constant.

Answer:

A(t_{1}) = 10^{10} per hour, t_{1} = 20 h

A(t_{2}) = 5 × 10 per hour, t_{2} = 30 h

A(t) = A_{0}e^{-λt}

A(t_{2}) = \(A_0 e^{-\lambda t_2}\)

∴ A(t_{1}) = A_{0}\(e^{-\lambda t_1}\)

∴ \(\frac{\mathrm{A}\left(t_1\right)}{\mathrm{A}\left(t_2\right)}\) = \(\left[\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}\right]\) = \(e^{\lambda\left(t_2-t_1\right)}\)

∴ \(\frac{10^{10}}{5 \times 10^9}\) = e^{λ(30-20)} = e^{10λ}

2 = 1.e^{10 λ}

10λ = 2.303 log_{10} [2]

10λ = 2.303 × 0.3010

10λ = 0.6932

λ = 0.06932 per hours

**Section – C**

Attempt any EIGHT questions of the following: (24)

Question 15.

Derive laws of reflection of light using Huygen’s principle.

Answer:

Reflection of a plane wave front of light at a plane surface.

Where MN : Plane mirror,

RA and QC : Incident rays,

AP : Normal to MN

AB : Incident wavefront

CE : Angle of incident

CE : Reflected wavefront

r : Angle of reflection.

When wavefront AB is incident on the mirror; at first, point A becomes a secondary source and emits secondary waves in the same medium. If T is the time taken by the incident wavefront to travel from B to C, then BC = vT. During this time, the secondary wave originating at A covers the same distance, so that the secondary wave originating at A covers the same distance, So that the secondary spherical wavelet has a radius vT at time T.

To construct the reflected wavefront, a hemisphere of radius vT is down from point A.

Draw a tangent FC to the secondary wavelet.

The arrow AE shows the direction of propagation of the reflected wave.

AP is the normal to MN at A.

∠RAP = i = angle of incidence and ∠PAE = r = angle of reflection

In ∆ABC and ∆AEC,

AE = BC and ∠ABC = ∠AEC = 90°

∠RAP = ∠BAC = i …….(i)

Also, as AE is perpendicular to CE and AP is perpendicular to AO.

∠ACE = ∠PAE = r ………(ii)

∴ From equations, (i) and (ii),

i = r

Thus, the angle of incidence is equal to the angle of reflection. Thus, it is the first law of reflection. Also, it can be seen from the figure that the incident ray and reflected ray lie on the opposite sides of the normal to the reflecting surface at the point of surface at the point of incidence and all of them lie., this is the second law of reflection. Thus, the laws of reflection of light can be disused by construction of a plane wavefront

Question 16.

State postulates of Bohr’s atomic model

Answer:

The postulates of Bohr’s atomic model (for the hydrogen atom).

1. The electrons revolve around the nucleus in circular orbits. This is the same assumptions in Rutherford’s model and the centripetal force necessary for the circular motion is provided by the electrostatic force of attraction between the electron and the nucleus.

2. The radius of the orbit of an electron can only take certain fixed values such that the angular momentum of the electron in these orbits is an integral multiple of \(\frac{h}{2 \pi}\), h being planck’s constant such orbits are called stable orbits or stable states of the electrons and electrons in these orbits do not emit radiation as is demanded by classical physics.

3. An electron can make a transition from one of its orbits to other orbit having lower energy. In doing so, it emits a photon of energy equal to the difference in its energies.

Question 17.

Define and state unit and dimensions of:

(a) Magnetization

(b) Explain magnetic susceptibility.

Answer:

(a)

- The ratio of magnetic moment to the volume of the material is called magnetization.
- Unit: Am
^{-1}in SI system - Dimensions: [M°L
^{-1}T°I^{1}]

(b)

(i) The mathematical definition of magnetic susceptibility is the ratio of magnetisation to applied magnetizing field intensity. This is a dimensionless quantity.

χ = M/H where, χ = magnetic susceptibility

M = Magnetization

H = Field intensity

Question 18.

With neat labelled circuit diagram, describe an experiment to study the characteristics of photoelectric effect.

Answer:

- A laboratory experimental setup for the photoelectric effect consists of an evacuated glass tube with a quartz window.
- The glass tube contains photosensitive metal plates. One is the emitter E and another plate is the collector C.

Schematic of experimental set-up for the photoelectric effects. - 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.
- A potential difference of V, as measured by the voltmeter is maintained between the emitter E and collector C, Generalty, C (the anode) is at a positive potential with respect to the emitter E (the cathode).
- When the anode potential (V) is positive it accelerates the elements. 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.
- A source S of monochromatic light of sufficiently high frequency (short wavelength ≤ 10
^{-7}m) is used.

Question 19.

Explain the use of potentiometer to determine internal resistance of a cell.

Answer:

The cell of EMF. E (internal resistance ‘r’) is connected across a resistance box (R) through keg K_{2}

When K_{2} is open, balance; length is obtained at length

AN_{1} = l_{1}

E = φl_{1}

When K_{2} is closed

AN_{2} = l_{2}

V = φl_{1}

∴ \(\frac{E}{V}\) = \(\frac{\mathrm{I}_1}{\mathrm{I}_2}\) ……. (1)

E = I(r + R) ∵ [V = IR]

\(\frac{E}{V}\) = \(\frac{r+R}{R}\) ……. (2)

From (1) and (2),

\(\frac{R+r}{R}\) \(=\frac{I_1}{I_2}\)

\(\frac{E}{V}\) = \(\frac{I_1}{I_2}\) [∵ r = R(\(\frac{E}{V}\) – 1)]

Question 20.

Explain the working of n-p-n transistor in common base configuration.

Answer:

The n-p-n transistor consists of two n-type semiconductor that sandwich a p-type semiconductor. Here electrons are the majority charge carrier, while tides are the minority charge carrier. The n-p-n transistor is represented, as shown below.

Construction of n-p-n transistor:

The n-p-n transistor is mode of semiconductor material like silicon or germanium. When a p-type semiconductor material is fused between two n-type semiconductor material an n-p-n transistor is formed.

Working of n-p-n transistor

When the emitter voltage is applied, as it is forward Biased, the electrons from the negative terminal repel the emitter electrons and current flows through the emitter and base to the collector to contribute collector current.

Question 21.

State the differential equation of linear S.H.M. Hence, obtain expression for:

(a) acceleration

(b) velocity.

Answer:

In any S.H.M., there always exists a restoring force, which tries of bring the object back to mean position.

This force causes acceleration in the object

Hence, F = -kx

by Newton’s second Law of motion

F = ma

a = \(\frac{\mathrm{F}}{\mathrm{~m}}\)

a = \(\frac{-k x}{m}\)

Let \(\frac{k}{m}\) be ω^{2}

We have,

a = -ω^{2}x

This is the expression for acceleration.

a = \(\frac{d^2 x}{d t^2}\)

\(\frac{d^2 x}{d t^2}\) + ω^{2}x = 0

We can also write, a = \(\frac{d v}{d t}\)

Question 22.

Two tuning forks of frequencies 320 Hz and 340 Hz are sounded together to produce sound wave. The velocity of sound in air is 326.4 m/s. Calculate the difference in wavelengths of these waves.

Answer:

Given, f_{1} = 320 Hz

V_{1} = 326.5 m/s

f_{1} = 340Hz

v_{2} = 326.4 m/s

v_{1} = v_{2} = v

λ_{1} = \(\frac{v}{f_1}\) = \(\frac{326.4}{320}\) = 0.98 m

λ_{2} = \(\frac{v}{f_2}\) = \(\frac{326.4}{340}\) = 0.96 m

Difference between wavelength

= λ_{1} – λ_{2}

= 0.98 – 0.96 = 0.02 m

Question 23.

In a biprism experiment, the fringes are observed in the focal plane of the eye-piece at a distance of 1.2 m from the slit. The distance between the central bright band and the 20th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eye-piece, 90 cm from the eye-piece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used.

Answer:

Given: D = 1.2 m

= 0.4 × 10^{-2}m

ω = \(\frac{Y_{20}}{20}\) = \(\frac{0.4}{20}\) × 10^{-2} = 2 × 10^{-4}m

d_{1} = 0.9 cm = 0.9 × 10^{-2} m, v_{1} = 90 cm = 0.9 m

v_{1} = D – v_{1} = 1.2 m – 0.9 m = 0.3 m

Question 24.

Calculate the current flowing through two long parallel wires carrying equal currents and separated by a distance of 1.35 cm experiencing a force per unit length of 4.76 × 10^{-2}N/m.

Answer:

Question 25.

An alternating voltage given by e= 140 sin (314.2 t) is connected across a pure resistor of 50Ω.

Calculate:

(i) the frequency of the source

(ii) the r.m.s. current through the resistor

Answer:

Given,

V = 140 sin 314.2 t

R = 50Ω

V = V_{0}sinωt

V_{0} = 140V

(i) ω = 314.2

2πv = 314.2

v = \(\frac{314.2}{2 \times 3.14}\)Hz = 50Hz

(ii) V_{rms} = \(\frac{v_0}{\sqrt{2}}\) = \(\frac{140}{\sqrt{2}}\) = 98.99V

I_{rms} = \(\frac{V_{\mathrm{rms}}}{R}\) = \(\frac{98.99}{50}\) = 1.97A

Question 26.

An electric dipole consists of two opposite charges each of magnitude lpC, separated by 2 cm. The dipole is placed in an external electric field of 10s N/C

Calculate the:

(i) maximum torque experienced by the dipole and

(ii) work done by the external field to turn the dipole through 180°

Answer:

Given, two opposite charge of magnitude = 1μC

= 1 × 10^{-6}C

Distance = 2 cm

External field = 10^{5} N/C

Max. torque on the dipole =?

q = 1 × 10^{-6}C, 2a = 2 cm

∴ P = q × 2a

= (1 × 10^{-6}) × 0.02

= 2 × 10^{-8}m

(i) Z_{max} = PE = (2 × 10^{-8}) (10^{5})

= 2 × 10^{-3} Nm

(ii) W = \(\int_0^{180^{\circ}} \mathrm{PE} \sin \theta d \theta\)

= PE\([-\cos \theta]_0^{180^{\circ}}\)

= 4 × 10^{-3} J

**Section – D**

Attempt any THREE questions of the following: (12)

Question 27.

On the basis of kinetic theory of gases obtain an expression for pressure exerted by gas molecules enclosed in a container on its walls.

Answer:

(1) Consider a cubicle box of side L let v_{x}, v_{y} and v_{z} be the velocity of particle away X, Y and Z axis respectively.

(2) Consider that the particle is moving along X-axis then change in momentum of particle will be ∆px.

(3) Final momentum – initial momentum = – mv – mv = – 2mv

(4) Time for the molecule to travel shaded wall is

∆t = \(\frac{2 L}{v_x}\)

(5) Average force exerted on the shaded wall by molecule 1.

Average rate of change of momentum.

= \(=\frac{2 m v_{x_1}}{\frac{2 L}{v_{x_1}}}\) where, v_{x1} 1 = x component of velocity of molecule 1

= \(\frac{m v \times 1^2}{L}\)

(6) If we consider all molecules, the total average force exerted on shaded wall will be, total average force

(7)

(8) The average of the square of the component of the velocities is given by

(9) \(\overline{v^2}\) = \(\bar{v}_a^2+\bar{v}_y^2+\bar{v}_{x^2}\)

(10) By symmetry, \(\bar{v}_a^2\) = \(\bar{v}_y^2+\bar{v}_z^2\) = \(\frac{1}{3} \bar{v} v^3\) as the molecules have no preferred direction to move.

(11) Therefore, average pressure. P = \(\frac{1}{3} \frac{N}{v} \overline{m v^2}\)

Question 28.

Derive an expression for energy stored in the magnetic field in terms of induced current.

A wire 5 m long is supported horizontally at a height of 15 m along east-west direction. When it is about to hit the ground, calculate the average e.m.f induced in it. (g = 10 m/s^{2}).

Answer:

Expression for energy stored in the magnetic field.

We know that, e = -L\(\frac{d i}{d t}\)

The work done in moving a charge dq against this emf is

∴ This is the energy stored (U_{B}) in magnetic field.

Given:

h = 15 m, L = 5 m, g = 10 m/s^{2}, e = ?

Formula: e = Blv

The magnetic flux density (B) due to the earth’s magnetic field is about 3.6 × 10^{-5}T.

Assuming that the wire is falling vertically under the influence of gravity, we can use free fall equation to find the velocity (v).

∴ v = \(\sqrt{2 g h}\) = \(\sqrt{2 \times 10 \times 15}\) = 17.32 m/s

Now, e = Blv = 3.6 × 10^{-5} × 5 × 17.32

∴ e = 3.1176 × 10^{-3} volts.

Question 29.

Derive an expression for the work done during an isothermal process.

104 J of work is done on certain volume of a gas. If the gas releases 125 kJ of heat, calculate the change in internal energy of the gas.

Answer:

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

dW = pdV

The total work done in bringing out 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.

Given: Q = 125 kJ, W = 104 kJ, ∆U = ?

Formula: ∆U = Q – W

∆U = 125 – 104 = 21 kJ

Question 30.

Obtain the relation between surface energy and surface tension.

Calculate the work done in blowing a soap bubble to a radius of 1 cm. The surface tension of soap solution is 2.5 × 10^{-2} N/m.

Answer:

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

The magnitude of force due to surface tension is,

F: 2Tl [∵ T = F/l]

∴ Work done by force,

dW = F’dx = 2Tldx

2ldx = dA

[Increase in the area of two surface of film]

∴ Work done in stretching

Surface energy = TdA

\(\frac{E}{d A}\) = T

Given: T = 2.5 × 10^{-2} N/m

r_{1} = 0 m

r_{2} = 10^{-2}m

w = 2Tda

= 2 × 2.5 × 10^{-2} × 4n[\(r_1{ }^2\) – \(r_1{ }^2\)]

= 6.283 × 10^{-5J}

Question 31.

Derive expressions for linear velocity at lowest position, mid-way position and the top-most position fora particle revolving in a vertical circle, if it has to just complete circular motion without string slackening at top.

Answer:

Consider a bob (treated as a point mass) tied to a (practically) massless inextensible and flexible string. It is whirled along vertical inlc. so that the bob performs a vertical circular motion and the string rotates in a vertical plane. At any position of the bob, there are only two forces outing on the bob.

(i) Its weight mg, vertically downwards, which is constant.

(ii) The force due to the tension along the string directed along the string and towards the centre. Its magnitude changes periodically with time and location.

Top position (A),

Mid way [at C and D]

Total energy at C = K.E + P.E.

= \(\frac{1}{2} m v_e^2\) + mgr

Total energy at (B) = \(\frac{1}{2} m v_B^2\) + 0

= \(\frac{1}{2} m v_B^2\) = \(\frac{5}{2} m g r\)

From Law of conservation of energy

T.E at C = T.E. at B

\(\frac{1}{2} m v_c^2\) + mgr = \(\frac{5}{2} m g r\)

\(v_c^2\) + 2rg = 5rg

V_{c} = \(\sqrt{3 r g}\)