Board Question Paper: July 2022 HSC Physics - Maharashtra Complete Solutions and Answer Key

BOARD QUESTION PAPER: JULY 2022 PHYSICS

Time: 3 Hrs. | Max. Marks: 70

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 contain Twelve short answer type of questions carrying Two marks each. (Attempt any Eight).
3. Section C: Q. No. 15 to Q. No. 26 contain Twelve short answer type of questions carrying Three marks each. (Attempt any Eight).
4. Section D: Q. No. 27 to Q. No. 31 contain Five long answer type of questions carrying Four marks each. (Attempt any Three).
5. Use of the log table is allowed. Use of calculator is not allowed.
6. Figures to the right indicate full marks.
7. For each multiple choice type of question, it is mandatory to write the correct answer along with its alphabet. e.g., (a)……./(b)……./(c)……./(d)……. No marks(s) shall be given, if ONLY the correct answer or the alphabet of the correct answer is written. Only the first attempt will be considered for evaluation.
8. Physical Constants:
   • (i) \( \mu_0 = 4\pi \times 10^{-7} \) Wb/Am
   • (ii) \( \sigma = 5.7 \times 10^{-8} \) J/m\(^2\)s K\(^4\)
   • (iii) \( g = 9.8 \) m/s\(^2\)
   • (iv) \( \pi = 3.142 \)

SECTION − A

Q.1. Select and write the correct answers for the following multiple choice type of questions: [10]

(i) Raindrops are spherical in shape because of _______.

• (a) surface tension
• (b) capillarity
• (c) downward motion
• (d) acceleration due to gravity

Answer:

(a) surface tension

Reasoning: Surface tension is the property of a liquid at rest by virtue of which its free surface tends to have minimum surface area. For a given volume, a sphere has the minimum surface area.

(ii) The average K.E. of a gas is _______.

• (a) directly proportional to absolute temperature of gas
• (b) directly proportional to square of absolute temperature of gas
• (c) directly proportional to square root of absolute temperature of gas
• (d) inversely proportional to absolute temperature of gas

Answer:

(a) directly proportional to absolute temperature of gas

Reasoning: The average kinetic energy per molecule of an ideal gas is given by \( E = \frac{3}{2}k_B T \), where \( T \) is the absolute temperature.

(iii) A graph of pressure versus volume for an ideal gas for different processes is as shown. In the graph curve OA represents _______.

(Note: The graph shows curve OA where Pressure increases while Volume remains constant on the V-axis at 0 offset, or represents a vertical line. Based on standard P-V diagrams, a vertical line represents constant volume.)

• (a) isochoric process
• (b) isothermal process
• (c) isobaric process
• (d) adiabatic process

Answer:

(a) isochoric process

Reasoning: In the graph, curve OA is a vertical line parallel to the pressure axis, indicating that volume remains constant while pressure changes. A process with constant volume is an isochoric process.

(iv) Standing waves are produced on a string fixed at both ends. In this case _______.

• (a) all particles vibrate in phase
• (b) all antinodes vibrate in phase
• (c) all alternate antinodes vibrate in phase
• (d) all particles between two consecutive antinodes vibrate in phase

Answer:

(c) all alternate antinodes vibrate in phase

Reasoning: In a stationary wave, particles in adjacent loops vibrate in opposite phases (phase difference of \(\pi\)). Consequently, consecutive antinodes vibrate in opposite directions. However, alternate antinodes (e.g., the 1st and 3rd) vibrate in the same direction at the same time, meaning they are in phase.
Note: Option (d) would be correct only if the text read "between two consecutive nodes". As written ("antinodes"), the region includes a node where phase changes, making the statement false.

(v) What changes are observed in a diffraction pattern if the whole apparatus is immersed in water?

• (a) the wavelength of light increases
• (b) width of central maximum increases
• (c) width of central maximum decreases
• (d) frequency of light decreases

Answer:

(c) width of central maximum decreases

Reasoning: When immersed in water (refractive index \(\mu > 1\)), the wavelength decreases (\( \lambda' = \lambda / \mu \)). The width of the central maximum is \( W = \frac{2\lambda D}{a} \). Since \(\lambda\) decreases, the width \(W\) decreases.

(vi) The magnitude of the magnetic field at the centre of a circular current carrying coil varies _______.

• (a) inversely with the square of the radius of the coil
• (b) directly with the radius of the coil
• (c) inversely with the radius of the coil
• (d) directly with the square of the radius of the coil

Answer:

(c) inversely with the radius of the coil

Reasoning: The magnetic field at the center of a circular coil is \( B = \frac{\mu_0 I}{2R} \). Thus, \( B \propto \frac{1}{R} \).

(vii) Balmer series is obtained when all transitions of electron terminate on _______.

• (a) 2nd orbit
• (b) 1st orbit
• (c) 3rd orbit
• (d) 4th orbit

Answer:

(a) 2nd orbit

Reasoning: The Balmer series corresponds to electron transitions from higher energy levels (\(n > 2\)) to the second orbit (\(n = 2\)).

(viii) A simple harmonic oscillator has amplitude 16 cm and period 4 seconds. The interval of time required by it to travel from x = 16 cm to x = 8 cm is _______.

• (a) \( \frac{1}{2} \) second
• (b) \( \frac{2}{3} \) second
• (c) \( \frac{5}{6} \) second
• (d) \( \frac{4}{3} \) second

Answer:

(b) \( \frac{2}{3} \) second

Reasoning: Motion starts from extreme position \(x = 16\) cm. The equation is \( x = A \cos(\omega t) \).
Here \( A = 16 \) cm, \( T = 4 \) s, so \( \omega = \frac{2\pi}{T} = \frac{\pi}{2} \) rad/s.
We need time to reach \( x = 8 \) cm.
\( 8 = 16 \cos(\frac{\pi}{2} t) \)
\( \frac{1}{2} = \cos(\frac{\pi}{2} t) \)
\( \cos(\frac{\pi}{3}) = \cos(\frac{\pi}{2} t) \)
\( \frac{\pi}{3} = \frac{\pi}{2} t \Rightarrow t = \frac{2}{3} \) s.

(ix) In potentiometer experiment, the cell balances at a length of 240 cm. When the cell is shunted by a resistance of 2 Ω, the balancing length becomes 120 cm. The internal resistance of the cell is _______.

• (a) 4 Ω
• (b) 2 Ω
• (c) 1 Ω
• (d) 0.5 Ω

Answer:

(b) 2 Ω

Reasoning: Formula for internal resistance: \( r = R \left( \frac{l_1 - l_2}{l_2} \right) \).
Given \( l_1 = 240 \) cm, \( l_2 = 120 \) cm, \( R = 2 \, \Omega \).
\( r = 2 \left( \frac{240 - 120}{120} \right) = 2 \left( \frac{120}{120} \right) = 2 \times 1 = 2 \, \Omega \).

(x) A small piece of metal wire is dragged across the gap between the pole pieces of magnet in 0.5 second. The magnetic flux between the pole pieces is \( 8 \times 10^{-4} \) weber. The emf induced in the wire is _______.

• (a) 1.6 millivolt
• (b) 16 millivolt
• (c) 1.6 volt
• (d) 16 volt

Answer:

(a) 1.6 millivolt

Reasoning: Induced emf \( e = \left| \frac{d\phi}{dt} \right| \).
\( d\phi = 8 \times 10^{-4} \) Wb, \( dt = 0.5 \) s.
\( e = \frac{8 \times 10^{-4}}{0.5} = 16 \times 10^{-4} \) V \( = 1.6 \times 10^{-3} \) V = 1.6 mV.

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Q.2. Answer the following questions: [8]

(i) In which thermodynamic process the total internal energy of system remains constant?

Isothermal process. (Since internal energy of an ideal gas depends only on temperature, and temperature is constant in an isothermal process).

(ii) State the law of conservation of angular momentum.

It states that if the resultant external torque acting on a rotating body is zero, then its angular momentum remains constant.

(iii) What is shunt?

A shunt is a small resistance connected in parallel with a galvanometer (or any device) to effectively lower the total resistance and allow a major portion of current to pass through it, thereby protecting the device or converting it into an ammeter.

(iv) What happens to the fringe width in diffraction pattern if the diameter of wire is increased?

The fringe width decreases. (Fringe width \( W \propto \frac{1}{a} \), where \( a \) is the diameter of the wire or slit width).

(v) What is perfectly black body?

A perfectly black body is a body that absorbs all the radiant energy incident upon it, for all wavelengths and at all angles of incidence.

(vi) State the formula for electric field intensity at a point outside an infinitely long charged cylindrical conductor.

\( E = \frac{\lambda}{2\pi \epsilon_0 r} \)
Where:
\( \lambda \) = linear charge density,
\( r \) = distance of the point from the axis of the cylinder,
\( \epsilon_0 \) = permittivity of free space.

(vii) The half-life of a nuclear species is 1.386 years. Calculate its decay constant per year.

Formula: \( \lambda = \frac{0.693}{T_{1/2}} \)
Given \( T_{1/2} = 1.386 \) years.
\( \lambda = \frac{0.693}{1.386} = 0.5 \text{ year}^{-1} \).

(viii) An automobile engine develops 62.84 kW while rotating at a speed of 1200 rpm. What torque does it deliver?

Given:
Power \( P = 62.84 \text{ kW} = 62840 \text{ W} \)
Speed \( N = 1200 \text{ rpm} \)
Angular speed \( \omega = \frac{2\pi N}{60} = \frac{2 \times 3.142 \times 1200}{60} = 40 \times 3.142 = 125.68 \text{ rad/s} \).
Formula: \( P = \tau \omega \Rightarrow \tau = \frac{P}{\omega} \)
Calculation: \( \tau = \frac{62840}{125.68} \approx 500 \text{ Nm} \).

SECTION − B

Attempt any EIGHT questions of the following: [16]

Q.3. What is capillarity? State any two uses of capillarity.

Definition: The phenomenon of rise or fall of a liquid inside a capillary tube when it is dipped in the liquid is called capillarity or capillary action.

Uses/Applications (Any two):

1. Oil rises up the wick of a lamp due to capillarity.
2. Sap and water rise up to the top of tall trees through fine capillaries in the stem.
3. Blotting paper absorbs ink via capillary pores.
4. Water rises in the crevices of rocks.

Q.4. Define: (i) Emissive power (ii) Co-efficient of emission

(i) Emissive power (Radiant power): The quantity of radiant energy emitted by a body per unit time per unit surface area at a given temperature is called its emissive power.

(ii) Co-efficient of emission (Emissivity): The ratio of the emissive power of a body at a given temperature to the emissive power of a perfectly black body at the same temperature is called the coefficient of emission.

Q.5. State any two characteristics of progressive waves.

Characteristics (Any two):

1. Energy is transferred from one point of the medium to another.
2. All particles of the medium vibrate with the same amplitude and period but differing phases.
3. The wave propagates through the medium with a certain velocity.
4. No particle of the medium remains permanently at rest.

Q.6. Draw neat, labelled diagram of a parallel plate capacitor with a dielectric slab between the plates.

(Diagram description: Two parallel plates separated by distance 'd'. A dielectric slab of thickness 't' (where t < d) is inserted between them. The plates are connected to a potential source. Charges +Q and -Q are on the plates. Induced charges -q and +q appear on the dielectric faces.)

[Diagram of Parallel Plate Capacitor with Dielectric Slab]
Shows plates with area A, separation d, dielectric slab of thickness t.

Q.7. State the formula for magnetic potential energy of a dipole and hence obtain the minimum and maximum magnetic potential energy.

Formula: The magnetic potential energy \( U \) of a magnetic dipole of moment \( \mathbf{M} \) in a uniform magnetic field \( \mathbf{B} \) is given by: \[ U = -\mathbf{M} \cdot \mathbf{B} = -MB \cos \theta \]

Minimum Potential Energy: Occurs when \( \theta = 0^\circ \) (dipole aligned with field). \[ U_{min} = -MB \cos(0^\circ) = -MB \]

Maximum Potential Energy: Occurs when \( \theta = 180^\circ \) (dipole anti-parallel to field). \[ U_{max} = -MB \cos(180^\circ) = -MB(-1) = +MB \]

Q.8. What is gyromagnetic ratio? Write the necessary expression.

Definition: The ratio of the magnetic dipole moment of an orbiting electron to its angular momentum is called the gyromagnetic ratio. It is a constant for an electron.

Expression: \[ \text{Gyromagnetic ratio} = \frac{e}{2m_e} \] Where \( e \) is the charge of the electron and \( m_e \) is its mass. Its value is approximately \( 8.8 \times 10^{10} \) C/kg.

Q.9. How does the wave theory of light fail to explain the observations from experiment on photoelectric effect. [Give any two points]

Failures (Any two):

1. Instantaneous emission: Wave theory predicts a time lag for energy accumulation before electron emission, but experimentally emission is instantaneous.
2. Threshold frequency: Wave theory cannot explain why no emission occurs below a certain frequency (threshold frequency), regardless of the intensity of light.
3. Independence of Max KE on Intensity: Wave theory predicts that increasing intensity should increase the kinetic energy of emitted electrons, but experimentally KE depends only on frequency, not intensity.

Q.10. A system releases 125 kJ of heat while 104 kJ of work is done on the system. Calculate the change in internal energy of the gas.

Given:
Heat released, \( Q = -125 \) kJ (negative because heat is lost).
Work done on system, \( W = -104 \) kJ (negative because work is done on the system, compression).
(Note: Sign convention \( \Delta U = Q - W \) where W is work done by system. If W is work done on system, \( \Delta U = Q + W_{\text{on}} \). Let's use standard: \( Q = \Delta U + W_{by} \). Here work is done on system, so \( W_{by} = -104 \) kJ.)

Calculation:
From First Law of Thermodynamics: \( Q = \Delta U + W_{by} \)
\( -125 = \Delta U + (-104) \)
\( \Delta U = -125 + 104 \)
\( \Delta U = -21 \) kJ.

Answer: The internal energy decreases by 21 kJ.

Q.11. A plane wavefront of light of wavelength 4000 Å is incident on two slits on a screen perpendicular to the direction of light ray. If the total separation of 10 bright fringes on a screen 2 m away is 2 cm, find the distance between the slits.

Given:
Wavelength \( \lambda = 4000 \text{ \AA} = 4 \times 10^{-7} \) m.
Distance to screen \( D = 2 \) m.
Separation of 10 bright fringes (assumed width of 10 fringes) \( 10 \beta = 2 \text{ cm} = 2 \times 10^{-2} \) m.
Alternatively, distance of 10th bright fringe from center \( x_{10} = 2 \) cm.

Formula:
Fringe width \( \beta = \frac{\lambda D}{d} \).
Width of 10 fringes \( = 10 \beta = \frac{10 \lambda D}{d} \).

Calculation:
\( 2 \times 10^{-2} = \frac{10 \times (4 \times 10^{-7}) \times 2}{d} \)
\( d = \frac{80 \times 10^{-7}}{2 \times 10^{-2}} \)
\( d = 40 \times 10^{-5} \text{ m} \)
\( d = 4 \times 10^{-4} \text{ m} = 0.4 \text{ mm} \).

Q.12. An emf of 96.0 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.20 A/s. What is the mutual inductance (M) of the coils?

Given:
Induced emf \( |e| = 96.0 \text{ mV} = 96 \times 10^{-3} \text{ V} \).
Rate of change of current \( \frac{di}{dt} = 1.20 \text{ A/s} \).

Formula: \( |e| = M \frac{di}{dt} \)

Calculation:
\( 96 \times 10^{-3} = M \times 1.20 \)
\( M = \frac{96 \times 10^{-3}}{1.2} \)
\( M = \frac{96}{1.2} \times 10^{-3} = 80 \times 10^{-3} \text{ H} \)
\( M = 80 \text{ mH} \).

Q.13. An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of \( 6 \times 10^{-5} \) T. Calculate the velocity required to generate an e.m.f. of 1.2 V between the tips of the wings of the aircraft.

Given:
Wing span (Length) \( l = 50 \) m.
Magnetic Field \( B = 6 \times 10^{-5} \) T (Assuming vertical component cuts the wings).
EMF \( e = 1.2 \) V.

Formula: \( e = Blv \)

Calculation:
\( 1.2 = (6 \times 10^{-5}) \times 50 \times v \)
\( 1.2 = 300 \times 10^{-5} \times v \)
\( 1.2 = 3 \times 10^{-3} \times v \)
\( v = \frac{1.2}{3 \times 10^{-3}} = \frac{1200}{3} = 400 \text{ m/s} \).

Q.14. The surface density of a uniform disc of radius 10 cm is 2 kg/m\(^2\). Find its MI about an axis passing through its centre and perpendicular to its plane.

Given:
Surface density \( \sigma = 2 \text{ kg/m}^2 \).
Radius \( R = 10 \text{ cm} = 0.1 \text{ m} \).

Step 1: Calculate Mass (M)
\( M = \text{Area} \times \sigma = \pi R^2 \times \sigma \)
\( M = 3.142 \times (0.1)^2 \times 2 \)
\( M = 3.142 \times 0.01 \times 2 = 0.06284 \text{ kg} \).

Step 2: Calculate Moment of Inertia (MI)
Formula: \( I = \frac{1}{2} M R^2 \)
\( I = \frac{1}{2} \times 0.06284 \times (0.1)^2 \)
\( I = 0.03142 \times 0.01 \)
\( I = 3.142 \times 10^{-4} \text{ kg m}^2 \).

SECTION − C

Attempt any EIGHT questions of the following: [24]

Q.15. State zeroth law of thermodynamics. What are the limitations of first law of thermodynamics?

Zeroth Law of Thermodynamics: If two systems are each in thermal equilibrium with a third system, then they are also in thermal equilibrium with each other.

Limitations of First Law of Thermodynamics:

• It does not indicate the direction of heat transfer (e.g., it doesn't say why heat flows from hot to cold, not vice versa).
• It does not tell about the extent to which heat can be converted into mechanical work (efficiency cannot be 100%).
• It does not explain why certain processes are irreversible.

Q.16. What is de-Broglie hypothesis? Obtain the relation for de-Broglie wavelength.

De-Broglie Hypothesis: Every moving material particle is associated with a wave, called a matter wave or de-Broglie wave.

Relation for Wavelength:
According to Planck's quantum theory, energy of a photon is \( E = h\nu \).
According to Einstein's mass-energy relation, \( E = mc^2 \).
Equating both: \( h\nu = mc^2 \Rightarrow h\frac{c}{\lambda} = mc^2 \).
\( \lambda = \frac{h}{mc} \).
For a material particle of mass \( m \) moving with velocity \( v \), momentum \( p = mv \).
Thus, the de-Broglie wavelength is: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \]

Q.17. Derive an expression for kinetic energy of a rotating body.

Consider a rigid body rotating with constant angular velocity \( \omega \) about an axis.
The body consists of \( n \) particles of masses \( m_1, m_2, \dots, m_n \) at distances \( r_1, r_2, \dots, r_n \) from the axis of rotation.
Linear velocity of particle \( i \) is \( v_i = r_i \omega \).
Kinetic energy of particle \( i \) is \( E_i = \frac{1}{2} m_i v_i^2 = \frac{1}{2} m_i (r_i \omega)^2 = \frac{1}{2} m_i r_i^2 \omega^2 \).
Total Rotational K.E. of the body is the sum of K.E. of all particles:
\( E = \sum \frac{1}{2} m_i r_i^2 \omega^2 = \frac{1}{2} \omega^2 \sum m_i r_i^2 \).
Since moment of inertia \( I = \sum m_i r_i^2 \),
\[ E_{\text{rot}} = \frac{1}{2} I \omega^2 \]

Q.18. Derive the laws of reflection of light using Huygens’ theory.

Derivation Outline:

1. Consider a plane wavefront AB incident on a reflecting surface MN at angle \( i \).
2. At time \( t=0 \), end A touches the surface. Point B is at a distance \( BC = v t \) from the surface.
3. According to Huygens' principle, A acts as a secondary source emitting spherical wavelets. In time \( t \), the wavelet from A travels a distance \( AE = v t \).
4. Draw a tangent EC from point C to the hemisphere of radius \( vt \) centered at A. EC represents the reflected wavefront.
5. In triangles \( \Delta ABC \) and \( \Delta AEC \):
   • \( \angle B = \angle E = 90^\circ \) (Rays perpendicular to wavefront).
   • Hypotenuse AC is common.
   • Side \( BC = AE = vt \).
6. Therefore, \( \Delta ABC \cong \Delta AEC \) (RHS congruence).
7. Hence, \( \angle BAC = \angle ECA \).
8. \( \angle BAC \) is the angle of incidence \( i \), and \( \angle ECA \) is the angle of reflection \( r \).
9. Thus, \( i = r \). Also, incident ray, reflected ray, and normal lie in the same plane.

Q.19. Derive an expression for orbital magnetic moment of an electron revolving around the nucleus in an atom. State the formula for the Bohr magneton.

Derivation:
Consider an electron of charge \( e \) revolving in a circular orbit of radius \( r \) with velocity \( v \).
Period of revolution \( T = \frac{2\pi r}{v} \).
Current \( I = \frac{e}{T} = \frac{e v}{2\pi r} \).
Magnetic moment \( M_{orb} = I \times A = I (\pi r^2) \).
\( M_{orb} = \left(\frac{e v}{2\pi r}\right) (\pi r^2) = \frac{e v r}{2} \).
In terms of angular momentum \( L = mvr \), \( v r = L/m \).
\( M_{orb} = \frac{e}{2m} L \).

Bohr Magneton Formula:
For the first orbit (\( n=1 \)), angular momentum \( L = \frac{h}{2\pi} \).
\( \mu_B = \frac{e h}{4\pi m_e} \).

Q.20. Explain the terms: (a) Capacitive reactance (b) Inductive reactance (c) Impedance

(a) Capacitive reactance (\( X_C \)): The opposition offered by a capacitor to the flow of alternating current. Given by \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \).

(b) Inductive reactance (\( X_L \)): The opposition offered by an inductor to the flow of alternating current. Given by \( X_L = \omega L = 2\pi f L \).

(c) Impedance (\( Z \)): The total opposition offered by an AC circuit containing combinations of resistor, inductor, and capacitor to the flow of alternating current. \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).

Q.21. Define \( \alpha_{dc} \) and \( \beta_{dc} \). Obtain the relation between them.

Definitions:
\( \alpha_{dc} \): The current gain in Common Base configuration. Ratio of collector current to emitter current (\( I_C / I_E \)).
\( \beta_{dc} \): The current gain in Common Emitter configuration. Ratio of collector current to base current (\( I_C / I_B \)).

Relation:
We know \( I_E = I_C + I_B \).
Divide by \( I_C \): \( \frac{I_E}{I_C} = 1 + \frac{I_B}{I_C} \).
Since \( \frac{I_C}{I_E} = \alpha \) and \( \frac{I_C}{I_B} = \beta \),
\( \frac{1}{\alpha} = 1 + \frac{1}{\beta} = \frac{\beta + 1}{\beta} \).
\( \alpha = \frac{\beta}{1 + \beta} \).
Or conversely, \( \beta = \frac{\alpha}{1 - \alpha} \).

Q.22. A pipe at both the ends has a fundamental frequency of 600 Hz. The first overtone of a pipe closed at one end has the same frequency as the first overtone of the open pipe. How long are the two pipes? [Velocity of sound in air = 330 m/s]

Given:
Open pipe fundamental freq \( n_o = 600 \) Hz.
Velocity \( v = 330 \) m/s.
Condition: 1st overtone of closed pipe = 1st overtone of open pipe.

Calculations:
1. Length of Open Pipe (\( L_o \)):
\( n_o = \frac{v}{2L_o} \Rightarrow L_o = \frac{v}{2n_o} \)
\( L_o = \frac{330}{2 \times 600} = \frac{330}{1200} = 0.275 \) m.

2. Length of Closed Pipe (\( L_c \)):
Frequency of 1st overtone of open pipe \( n'_o = 2n_o = 1200 \) Hz.
Frequency of 1st overtone of closed pipe \( n'_c = 3 \times (\text{fundamental closed}) = 3 \frac{v}{4L_c} \).
Given \( n'_c = n'_o \), so:
\( \frac{3v}{4L_c} = 2 \left( \frac{v}{2L_o} \right) \)
\( \frac{3}{4L_c} = \frac{1}{L_o} \)
\( L_c = \frac{3}{4} L_o = \frac{3}{4} \times 0.275 \)
\( L_c = 3 \times 0.06875 = 0.20625 \) m.

Answer: Open pipe length = 0.275 m, Closed pipe length = 0.206 m.

Q.23. A small particle carrying a negative charge of \( 1.6 \times 10^{-19} \) C is suspended in equilibrium between two horizontal metal plates 10 cm apart, having a potential difference of 4000 volts across them. Find the mass of the particle.

Given:
Charge \( q = 1.6 \times 10^{-19} \) C.
Distance \( d = 10 \text{ cm} = 0.1 \) m.
Voltage \( V = 4000 \) V.
Gravity \( g = 9.8 \) m/s\(^2\).

Formula:
For equilibrium, Electric force = Gravitational force.
\( qE = mg \)
Since \( E = V/d \), we have \( q \frac{V}{d} = mg \).

Calculation:
\( m = \frac{qV}{gd} \)
\( m = \frac{1.6 \times 10^{-19} \times 4000}{9.8 \times 0.1} \)
\( m = \frac{6.4 \times 10^{-16}}{0.98} \)
\( m \approx 6.53 \times 10^{-16} \) kg.

Q.24. A 1000 mH inductor, 36 µF capacitor and 12 Ω resistor are connected in series to 120 V, 50 Hz AC source. Calculate:
(i) impedance of the circuit at resonance
(ii) current at resonance
(iii) resonant frequency

Given:
\( L = 1000 \text{ mH} = 1 \) H.
\( C = 36 \mu\text{F} = 36 \times 10^{-6} \) F.
\( R = 12 \, \Omega \).
\( V_{rms} = 120 \) V.

(i) Impedance at resonance:
At resonance, \( X_L = X_C \), so \( Z = R \).
\( Z = 12 \, \Omega \).

(ii) Current at resonance:
\( I = \frac{V}{Z} = \frac{120}{12} = 10 \) A.

(iii) Resonant frequency:
\( f_r = \frac{1}{2\pi \sqrt{LC}} \)
\( f_r = \frac{1}{2 \times 3.142 \times \sqrt{1 \times 36 \times 10^{-6}}} \)
\( f_r = \frac{1}{6.284 \times 6 \times 10^{-3}} \)
\( f_r = \frac{1000}{37.704} \approx 26.52 \) Hz.

Q.25. A current of equal magnitude flows through two long parallel wires separated by 2 cm. If force per unit length of \( 4 \times 10^{-2} \) N/m acts on both the wires respectively, calculate the current through each wire.

Given:
Force per unit length \( F/L = 4 \times 10^{-2} \) N/m.
Separation \( d = 2 \text{ cm} = 2 \times 10^{-2} \) m.
Currents \( I_1 = I_2 = I \).

Formula:
\( \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{4\pi \times 10^{-7} I^2}{2\pi d} = \frac{2 \times 10^{-7} I^2}{d} \)

Calculation:
\( 4 \times 10^{-2} = \frac{2 \times 10^{-7} \times I^2}{2 \times 10^{-2}} \)
\( 4 \times 10^{-2} = 10^{-5} I^2 \)
\( I^2 = \frac{4 \times 10^{-2}}{10^{-5}} = 4000 \)
\( I = \sqrt{4000} \approx 63.25 \) A.

Q.26. Calculate the energy radiated in half a minute by a black body of surface area 200 cm\(^2\) at 127° C.

Given:
Time \( t = 30 \) s (half minute).
Area \( A = 200 \text{ cm}^2 = 200 \times 10^{-4} = 2 \times 10^{-2} \text{ m}^2 \).
Temp \( T = 127^\circ\text{C} = 127 + 273 = 400 \) K.
Stefan's constant \( \sigma = 5.7 \times 10^{-8} \).

Formula: \( Q = \sigma A T^4 t \)

Calculation:
\( Q = (5.7 \times 10^{-8}) \times (2 \times 10^{-2}) \times (400)^4 \times 30 \)
\( 400^4 = 256 \times 10^8 \).
\( Q = 5.7 \times 2 \times 30 \times 256 \times 10^{-8} \times 10^{-2} \times 10^8 \)
\( Q = 5.7 \times 60 \times 256 \times 10^{-2} \)
\( Q = 342 \times 2.56 \)
\( Q = 875.52 \) J.

SECTION − D

Attempt any THREE questions of the following: [12]

Q.27. Discuss analytically the composition of two linear SHMs having same period and along the same path. Obtain the expression for resultant amplitude. Find the resultant amplitude when the phase difference is (i) zero radians (ii) \( \frac{\pi}{2} \) radians.

Analytical Composition:
Let two SHMs be \( x_1 = A_1 \sin(\omega t + \phi_1) \) and \( x_2 = A_2 \sin(\omega t + \phi_2) \).
Resultant displacement \( x = x_1 + x_2 \).
Expanding and combining, we get \( x = R \sin(\omega t + \delta) \).

Resultant Amplitude Expression:
\[ R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\phi_1 - \phi_2)} \]

Cases:
(i) Phase difference \( (\phi_1 - \phi_2) = 0 \):
\( \cos(0) = 1 \).
\( R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2} = \sqrt{(A_1+A_2)^2} = A_1 + A_2 \).
(ii) Phase difference \( \frac{\pi}{2} \):
\( \cos(\frac{\pi}{2}) = 0 \).
\( R = \sqrt{A_1^2 + A_2^2} \).

Q.28. What is a transformer? With the help of a suitable diagram describe working of transformer.

Definition: A transformer is a device used to change the voltage of an alternating current. It works on the principle of mutual induction.

Diagram:

[Diagram of Transformer: Core with Primary and Secondary Windings]

Working:
1. An alternating voltage \( E_p \) is applied to the primary coil.
2. This creates a changing magnetic flux \( \phi \) through the soft iron core, which links with the secondary coil.
3. According to Faraday's law, an emf is induced in the secondary coil: \( E_s = -N_s \frac{d\phi}{dt} \).
4. Similarly for primary: \( E_p = -N_p \frac{d\phi}{dt} \).
5. Ratio: \( \frac{E_s}{E_p} = \frac{N_s}{N_p} \).
Depending on turns ratio, voltage is stepped up or down.

Q.29. Define angle of contact. State any two properties of angle of contact. Find the difference of pressure between inside and outside of a spherical water drop of radius 2 mm, if surface tension of water is \( 73 \times 10^{-3} \) N/m.

Definition: The angle between the tangent drawn to the free surface of the liquid and the surface of the solid at the point of contact, measured within the liquid.

Properties:
1. It is constant for a given solid-liquid pair.
2. It depends on the nature of the solid and liquid and impurities.

Problem:
Radius \( R = 2 \text{ mm} = 2 \times 10^{-3} \) m.
Surface Tension \( T = 73 \times 10^{-3} \) N/m.
For a drop (one surface): Excess Pressure \( \Delta P = \frac{2T}{R} \).
\( \Delta P = \frac{2 \times 73 \times 10^{-3}}{2 \times 10^{-3}} = 73 \) N/m\(^2\) (Pa).

Q.30. State Kirchhoff’s laws of electrical network. When two cells of emfs \( E_1 \) and \( E_2 \) are connected in series so as to assist each other, their balancing length on potentiometer wire is found to be 3.2 m. When two cells are connected in series so as to oppose each other, the balancing length is found to be 0.7 m. Compare the emfs of two cells.

Kirchhoff’s Laws:
1. Current Law (KCL): The algebraic sum of currents meeting at a junction is zero (\( \sum I = 0 \)).
2. Voltage Law (KVL): The algebraic sum of potential differences (products of current and resistance) and emfs in a closed loop is zero (\( \sum IR + \sum E = 0 \)).

Problem:
Assist: \( E_1 + E_2 \propto L_1 \) where \( L_1 = 3.2 \) m.
Oppose: \( E_1 - E_2 \propto L_2 \) where \( L_2 = 0.7 \) m.
Comparison Formula: \( \frac{E_1}{E_2} = \frac{L_1 + L_2}{L_1 - L_2} \).
\( \frac{E_1}{E_2} = \frac{3.2 + 0.7}{3.2 - 0.7} = \frac{3.9}{2.5} \).
\( \frac{E_1}{E_2} = \frac{39}{25} = 1.56 \).

Q.31. State the first and second postulate of Bohr’s atomic model. Compute the ratio of longest wavelengths of Lyman and Balmer series in hydrogen atom.

Postulates:
1. The electron revolves around the nucleus in circular orbits. The centripetal force is provided by the electrostatic force of attraction.
2. The electron revolves only in those orbits where its angular momentum is an integral multiple of \( h/2\pi \) (\( mvr = nh/2\pi \)).

Ratio Calculation:
Wavelength formula: \( \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \).
Longest wavelength corresponds to smallest energy gap (transition between adjacent levels).
Lyman Series (Longest): \( n_f=1, n_i=2 \).
\( \frac{1}{\lambda_L} = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \Rightarrow \lambda_L = \frac{4}{3R} \).
Balmer Series (Longest): \( n_f=2, n_i=3 \).
\( \frac{1}{\lambda_B} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36} \Rightarrow \lambda_B = \frac{36}{5R} \).
Ratio:
\( \frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{4}{3} \times \frac{5}{36} = \frac{1}{3} \times \frac{5}{9} = \frac{5}{27} \).