Maharashtra Board HSC Physics Question Paper Solution 2024
Subject: Physics (54) | Date: 2024 | Max Marks: 70
SECTION – A
Q. 1. Select and write the correct answer for the following multiple choice type of questions : [10]
(i) The moment of inertia (MI) of a disc of radius R and mass M about its central axis is _____.
Explanation: The standard formula for the moment of inertia of a uniform circular disc about an axis passing through its centre and perpendicular to its plane (central axis) is \( I = \frac{1}{2}MR^2 \).
(ii) The dimensional formula of surface tension is _____.
Explanation: Surface Tension \( T = \frac{\text{Force}}{\text{Length}} \).
Dimensions of Force = \( [M^1L^1T^{-2}] \)
Dimensions of Length = \( [L^1] \)
Dimensions of T = \( \frac{[M^1L^1T^{-2}]}{[L^1]} = [L^0M^1T^{-2}] \).
(iii) Phase difference between a node and an adjacent antinode in a stationary wave is _____.
Explanation: The distance between a node and an adjacent antinode is \( \frac{\lambda}{4} \).
Phase difference \( \phi = \frac{2\pi}{\lambda} \times \text{path difference} = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \) rad.
(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 _____.
Explanation: This is the standard definition of electric potential at a point.
(v) To convert a moving coil galvanometer into an ammeter we need to connect a _____.
Explanation: An ammeter requires very low resistance and must pass most current through a shunt. A galvanometer is converted into an ammeter by connecting a low resistance (shunt) in parallel.
(vi) If the frequency of incident light falling on a photosensitive material is doubled, then kinetic energy of the emitted photoelectron will be _____.
Explanation: Einstein's photoelectric equation: \( K_{max} = h\nu - \phi_0 \).
Let initial KE be \( K_1 = h\nu - \phi_0 \).
If frequency is doubled (\( 2\nu \)), new KE is \( K_2 = h(2\nu) - \phi_0 = 2h\nu - \phi_0 \).
Rewrite \( K_2 = 2(h\nu - \phi_0) + \phi_0 = 2K_1 + \phi_0 \).
Since work function \( \phi_0 > 0 \), \( K_2 > 2K_1 \).
(vii) In a cyclic process, if \( \Delta U = \) internal energy, W = work done, Q = Heat supplied then
Explanation: In a cyclic process, the system returns to its initial state, so the change in internal energy \( \Delta U = 0 \). From the first law of thermodynamics \( Q = \Delta U + W \), we get \( Q = 0 + W \Rightarrow Q = W \).
(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 _____.
Explanation: Formula: \( |e| = L \frac{di}{dt} \).
\( L = 20 \) H, \( di = 50 - 10 = 40 \) A, \( dt = 0.1 \) s.
\( e = 20 \times \frac{40}{0.1} = 20 \times 400 = 8000 \) V.
(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 _____.
Explanation: For a second's pendulum, \( T = 2 \) s. \( \omega = \frac{2\pi}{T} = \pi \) rad/s.
Velocity \( v = \omega \sqrt{A^2 - x^2} \).
\( v = \pi \sqrt{10^2 - 6^2} = \pi \sqrt{100 - 36} = \pi \sqrt{64} = 8\pi \) cm/s.
(x) In biprism experiment, the distance of 20th bright band from the central bright band is 1.2 cm. Without changing the experimental set-up, the distance of 30th bright band from the central bright band will be _____.
Explanation: Distance of \( n \)-th bright band \( x_n = n \frac{\lambda D}{d} \).
\( x_{20} = 20 \beta = 1.2 \) cm \(\Rightarrow \beta = \frac{1.2}{20} = 0.06 \) cm.
\( x_{30} = 30 \beta = 30 \times 0.06 = 1.8 \) cm.
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Q. 2. Answer the following questions : [8]
(i) Define centripetal force.
(ii) Why a detergent powder is mixed with water to wash clothes?
(iii) What is the resistance of an ideal voltmeter?
(iv) Write the formula for torque acting on rotating current carrying coil in terms of magnetic dipole moment, in vector form.
(where \( \vec{m} \) is the magnetic dipole moment and \( \vec{B} \) is the magnetic field).
(v) What is binding energy of a hydrogen atom?
(vi) What is surroundings in thermodynamics?
(vii) In a photoelectric experiment, the stopping potential is 1.5V. What is the maximum kinetic energy of a photoelectron?
Given \( V_s = 1.5 \) V.
\( K_{max} = 1.5 \) eV (electron-volt) or \( 1.5 \times 1.6 \times 10^{-19} = 2.4 \times 10^{-19} \) J.
(viii) Two capacitors of capacities 5µF and 10µF respectively are connected in series. Calculate the resultant capacity of the combination.
SECTION – B
Attempt any EIGHT questions of the following : [16]
Q. 3. Explain the change in internal energy of a thermodynamic system (the gas) by heating it.
When heat is supplied to a gas (thermodynamic system):
- The kinetic energy of the gas molecules increases. Since internal energy is a function of temperature (for an ideal gas, it depends solely on kinetic energy), the internal energy of the system increases (\( \Delta U > 0 \)).
- According to the First Law of Thermodynamics \( Q = \Delta U + W \), part of the heat supplied increases the internal energy, and the rest may be used to do work against external pressure (expansion).
- If the volume is kept constant (Isochoric process), no work is done (\( W=0 \)), and all the heat supplied goes into increasing the internal energy (\( Q = \Delta U \)).
Q. 4. Explain the construction of a spherical wavefront by using Huygens’ principle.
Construction of Spherical Wavefront:
- Consider a point source of light 'S' in a homogeneous isotropic medium. At time \( t=0 \), the disturbance is at the source.
- The disturbance travels in all directions with speed 'c'. After time 't', the locus of points reaching the same phase is a sphere of radius \( ct \). This is the primary spherical wavefront.
- According to Huygens' principle, every point on this primary wavefront acts as a secondary source of light, emitting secondary wavelets in all directions.
- To find the new position of the wavefront after a further time \( \Delta t \), draw spheres of radius \( c\Delta t \) with centers at various points on the primary wavefront.
- The forward envelope (tangential surface) of these secondary wavelets gives the new position of the spherical wavefront at time \( t + \Delta t \).
Q. 5. Define magnetization. State its SI unit and dimensions.
Definition: Magnetization (or intensity of magnetization) is defined as the net magnetic dipole moment per unit volume of the material.
$$ M_z = \frac{m_{net}}{V} $$
SI Unit: Ampere per meter (A/m).
Dimensions: \( [L^{-1}M^0T^0I^1] \) or \( [L^{-1}A] \).
Q. 6. Obtain the differential equation of linear simple harmonic motion.
In linear Simple Harmonic Motion (SHM), the restoring force \( F \) is directly proportional to the displacement \( x \) from the mean position and is directed opposite to it.
$$ F = -kx $$ (where \( k \) is the force constant).
According to Newton's second law, \( F = ma = m \frac{d^2x}{dt^2} \).
Equating the forces:
$$ m \frac{d^2x}{dt^2} = -kx $$
$$ m \frac{d^2x}{dt^2} + kx = 0 $$
Dividing by \( m \):
$$ \frac{d^2x}{dt^2} + \frac{k}{m}x = 0 $$
Substituting \( \frac{k}{m} = \omega^2 \) (where \( \omega \) is angular frequency):
$$ \frac{d^2x}{dt^2} + \omega^2 x = 0 $$
This is the differential equation of linear SHM.
Q. 7. A galvanometer has a resistance of 30\( \Omega \) and its full scale deflection current is 20 microampere (\( \mu \)A). What resistance should be added to it to have a range 0-10 volt?
Given:
Galvanometer resistance \( G = 30 \, \Omega \)
Full scale current \( I_g = 20 \, \mu\text{A} = 20 \times 10^{-6} \, \text{A} \)
Voltage range \( V = 10 \, \text{V} \)
To convert a galvanometer into a voltmeter, a high resistance \( X \) (or \( R_s \)) is connected in series.
Formula: \( V = I_g (G + X) \)
$$ 10 = 20 \times 10^{-6} (30 + X) $$
$$ 30 + X = \frac{10}{20 \times 10^{-6}} $$
$$ 30 + X = \frac{10^6}{2} = 500,000 $$
$$ X = 500,000 - 30 = 499,970 \, \Omega $$
Answer: A resistance of \( 499,970 \, \Omega \) should be connected in series.
Q. 8. Explain Biot-Savart law.
The Biot-Savart law describes the magnetic field produced by a current-carrying element.
Consider a finite conductor carrying current \( I \). Let \( dl \) be a small element of the conductor. The magnetic field \( dB \) produced at a point P at a distance \( r \) from the element is:
- Directly proportional to the current \( I \).
- Directly proportional to the length of the element \( dl \).
- Directly proportional to the sine of the angle \( \theta \) between the element \( dl \) and the position vector \( r \).
- Inversely proportional to the square of the distance \( r \) (\( dB \propto 1/r^2 \)).
Mathematical form:
$$ dB = \frac{\mu_0}{4\pi} \frac{I \, dl \, \sin\theta}{r^2} $$
In vector form:
$$ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3} $$
Q. 9. What is a Light Emitting Diode? Draw its circuit symbol.
Definition: A Light Emitting Diode (LED) is a heavily doped p-n junction diode which emits spontaneous radiation (light) when it is forward biased. The energy is released in the form of photons due to the recombination of electrons and holes at the junction.
Circuit Symbol:
(Symbol shows a diode with two arrows pointing away, indicating light emission)
Q. 10. An aircraft of wing span of 60 m flies horizontally in earth’s magnetic field of \( 6\times10^{-5} \)T at a speed of 500 m/s. Calculate the e.m.f. induced between the tips of wings of aircraft.
This is a case of motional EMF.
Given:
Wing span (length) \( l = 60 \) m
Magnetic field \( B = 6 \times 10^{-5} \) T
Speed \( v = 500 \) m/s
Formula: \( e = Blv \)
$$ e = (6 \times 10^{-5}) \times 60 \times 500 $$
$$ e = 6 \times 10^{-5} \times 30000 $$
$$ e = 6 \times 10^{-5} \times 3 \times 10^4 $$
$$ e = 18 \times 10^{-1} = 1.8 \, \text{V} $$
Answer: The induced EMF is 1.8 V.
Q. 11. Derive an expression for maximum speed of a vehicle moving along a horizontal circular track.
Consider a vehicle of mass \( m \) moving with speed \( v \) on a horizontal circular track of radius \( r \).
The forces acting on the vehicle are:
1. Weight \( mg \) acting vertically downwards.
2. Normal reaction \( N \) acting vertically upwards (\( N = mg \)).
3. Force of static friction \( f_s \) between the tyres and the road, acting towards the centre.
The necessary centripetal force is provided by the static friction.
$$ \frac{mv^2}{r} = f_s $$
For maximum speed \( v_{max} \), the static friction reaches its limiting value \( f_{s,max} = \mu N = \mu mg \) (where \( \mu \) is the coefficient of static friction).
$$ \frac{mv_{max}^2}{r} = \mu mg $$
$$ v_{max}^2 = \mu rg $$
$$ v_{max} = \sqrt{\mu rg} $$
Q. 12. A horizontal force of 0.5N is required to move a metal plate of area \( 10^{-2} \text{m}^2 \) with a velocity of \( 3\times10^{-2} \text{m/s} \), when it rests on \( 0.5\times10^{-3} \text{m} \) thick layer of glycerin. Find the coefficient of viscosity of glycerin.
Given:
Force \( F = 0.5 \) N
Area \( A = 10^{-2} \, \text{m}^2 \)
Velocity \( dv = 3 \times 10^{-2} \) m/s
Thickness (separation) \( dx = 0.5 \times 10^{-3} \) m
Formula (Newton's law of viscosity): \( F = \eta A \frac{dv}{dx} \)
$$ \eta = \frac{F \cdot dx}{A \cdot dv} $$
$$ \eta = \frac{0.5 \times 0.5 \times 10^{-3}}{10^{-2} \times 3 \times 10^{-2}} $$
$$ \eta = \frac{0.25 \times 10^{-3}}{3 \times 10^{-4}} = \frac{2.5 \times 10^{-4}}{3 \times 10^{-4}} $$
$$ \eta = \frac{2.5}{3} \approx 0.833 \, \text{Ns/m}^2 \, (\text{or Pa}\cdot\text{s}) $$
Q. 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.
Given:
\( n_1 = 320 \) Hz, \( n_2 = 340 \) Hz
\( v = 340 \) m/s
Wavelength \( \lambda = \frac{v}{n} \)
\( \lambda_1 = \frac{340}{320} = \frac{34}{32} = 1.0625 \) m
\( \lambda_2 = \frac{340}{340} = 1.0 \) m
Difference in wavelength \( \Delta \lambda = \lambda_1 - \lambda_2 = 1.0625 - 1.0 = 0.0625 \) m.
Q. 14. Calculate the change in angular momentum of electron when it jumps from third orbit to first orbit in hydrogen atom.
According to Bohr's second postulate, angular momentum \( L = \frac{nh}{2\pi} \).
For \( n_1 = 3 \): \( L_3 = \frac{3h}{2\pi} \)
For \( n_2 = 1 \): \( L_1 = \frac{1h}{2\pi} \)
Change in angular momentum \( \Delta L = L_3 - L_1 = \frac{3h}{2\pi} - \frac{h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi} \).
Using \( h = 6.63 \times 10^{-34} \) Js and \( \pi = 3.142 \):
$$ \Delta L = \frac{6.63 \times 10^{-34}}{3.142} \approx 2.11 \times 10^{-34} \, \text{kg m}^2/\text{s} $$
SECTION – C
Attempt any EIGHT questions of the following : [24]
Q. 15. A circular coil of wire is made up of 200 turns, each of radius 10 cm. If a current of 0.5A passes through it, what will be the magnetic field at the centre of the coil?
Given: \( N = 200 \), \( R = 10 \text{ cm} = 0.1 \text{ m} \), \( I = 0.5 \text{ A} \).
Formula for magnetic field at the centre of a coil:
$$ B = \frac{\mu_0 N I}{2R} $$
$$ B = \frac{4\pi \times 10^{-7} \times 200 \times 0.5}{2 \times 0.1} $$
$$ B = \frac{4\pi \times 10^{-7} \times 100}{0.2} = \frac{400\pi \times 10^{-7}}{0.2} $$
$$ B = 2000\pi \times 10^{-7} = 2\pi \times 10^{-4} \text{ T} $$
$$ B \approx 2 \times 3.142 \times 10^{-4} = 6.284 \times 10^{-4} \text{ T} $$
Q. 16. Define photoelectric effect and explain the experimental set-up of photoelectric effect.
Definition: The phenomenon of emission of electrons from a metal surface when electromagnetic radiation (light) of appropriate frequency is incident on it is called the photoelectric effect.
Experimental Set-up:
- It consists of an evacuated glass tube containing two electrodes: a photosensitive plate (Emitter E) and a collector plate (Collector C).
- A quartz window allows ultraviolet or visible light to fall on the emitter.
- The plates are connected to a variable voltage source (battery) via a commutator to vary the potential and polarity.
- A microammeter (\( \mu A \)) is connected in series to measure the photoelectric current, and a voltmeter (V) measures the potential difference between the plates.
- When light falls on E, photoelectrons are emitted and attracted to C, causing a current flow measured by the microammeter.
Q. 17. Define the current gain \( \alpha_{DC} \) and \( \beta_{DC} \) for a transistor. Obtain the relation between them.
Definitions:
1. \( \alpha_{DC} \) (Common Base current gain): Ratio of collector current to emitter current. \( \alpha_{DC} = \frac{I_C}{I_E} \).
2. \( \beta_{DC} \) (Common Emitter current gain): Ratio of collector current to base current. \( \beta_{DC} = \frac{I_C}{I_B} \).
Relation:
We know \( I_E = I_C + I_B \).
Dividing by \( I_C \):
$$ \frac{I_E}{I_C} = 1 + \frac{I_B}{I_C} $$
Substituting the definitions (\( \frac{I_E}{I_C} = \frac{1}{\alpha} \) and \( \frac{I_B}{I_C} = \frac{1}{\beta} \)):
$$ \frac{1}{\alpha} = 1 + \frac{1}{\beta} = \frac{\beta + 1}{\beta} $$
$$ \alpha = \frac{\beta}{1 + \beta} $$
Or conversely: \( \beta = \frac{\alpha}{1 - \alpha} \).
Q. 18. Define surface energy of the liquid. Obtain the relation between the surface energy and surface tension.
Surface Energy: The extra potential energy possessed by the molecules in the surface layer of a liquid compared to the molecules in the bulk is called surface energy.
Relation:
Consider a rectangular frame with a movable side of length \( L \). A soap film is formed. Surface tension \( T \) exerts a force \( F = 2TL \) (factor of 2 for two surfaces).
If the wire is moved by a distance \( dx \) against this force, work done is:
$$ dW = F \cdot dx = 2TL \cdot dx $$
The increase in surface area \( dA = 2(L \cdot dx) \).
So, \( dW = T (2L \cdot dx) = T \cdot dA \).
This work is stored as surface energy \( E \).
$$ \text{Surface Energy } (E) = T \times \text{Change in Area } (dA) $$
$$ T = \frac{\text{Surface Energy}}{\text{Area}} $$
Q. 19. What is an isothermal process? Obtain an expression for work done by a gas in an isothermal process.
Isothermal Process: A thermodynamic process in which the temperature of the system remains constant throughout the change.
Work Done:
Work done \( W = \int_{V_i}^{V_f} P \, dV \).
For an ideal gas, \( PV = nRT \implies P = \frac{nRT}{V} \).
Since T is constant:
$$ W = \int_{V_i}^{V_f} \frac{nRT}{V} \, dV $$
$$ W = nRT \int_{V_i}^{V_f} \frac{1}{V} \, dV $$
$$ W = nRT [\ln V]_{V_i}^{V_f} $$
$$ W = nRT \ln\left(\frac{V_f}{V_i}\right) $$ (or \( 2.303 nRT \log_{10}\left(\frac{V_f}{V_i}\right) \)).
Q. 20. Derive an expression for equation of stationary wave on a stretched string. Show that the distance between two successive nodes or antinodes is \( \lambda/2 \).
Consider two identical progressive waves travelling in opposite directions:
\( y_1 = A \sin(\omega t - kx) \)
\( y_2 = A \sin(\omega t + kx) \)
By superposition principle: \( y = y_1 + y_2 \)
\( y = A [\sin(\omega t - kx) + \sin(\omega t + kx)] \)
Using \( \sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2}) \):
\( y = 2A \sin(\omega t) \cos(-kx) = 2A \cos(kx) \sin(\omega t) \).
This is the equation of a stationary wave.
Distance between nodes:
Nodes occur where amplitude is zero: \( \cos(kx) = 0 \).
\( kx = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2} \dots \)
\( \frac{2\pi}{\lambda} x = (2n-1)\frac{\pi}{2} \implies x = (2n-1)\frac{\lambda}{4} \).
Successive nodes at \( x_n = \frac{\lambda}{4} \) and \( x_{n+1} = \frac{3\lambda}{4} \).
Distance = \( \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2} \).
Q. 21. Derive an expression for the impedance of an LCR circuit connected to an AC power supply. Draw phasor diagram.
Consider a series LCR circuit with resistance \( R \), inductance \( L \), and capacitance \( C \).
Let current \( I = I_0 \sin(\omega t) \).
Voltages:
\( V_R \) is in phase with \( I \).
\( V_L \) leads \( I \) by \( 90^\circ \).
\( V_C \) lags \( I \) by \( 90^\circ \).
From Phasor diagram, total voltage \( V \) is the vector sum:
\( V = \sqrt{V_R^2 + (V_L - V_C)^2} \)
\( V = \sqrt{(IR)^2 + (IX_L - IX_C)^2} \)
\( V = I \sqrt{R^2 + (X_L - X_C)^2} \)
Impedance \( Z = \frac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2} \).
Q. 22. Calculate the wavelength of the first two lines in Balmer series of hydrogen atom.
Formula: \( \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \), where \( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \).
1. First line (\( H_\alpha \)): \( n = 3 \).
\( \frac{1}{\lambda_1} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) = 1.097 \times 10^7 \left( \frac{5}{36} \right) \)
\( \lambda_1 = \frac{36}{5 \times 1.097 \times 10^7} \approx 6.563 \times 10^{-7} \text{ m} = 6563 \, \mathring{A} \).
2. Second line (\( H_\beta \)): \( n = 4 \).
\( \frac{1}{\lambda_2} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) = 1.097 \times 10^7 \left( \frac{3}{16} \right) \)
\( \lambda_2 = \frac{16}{3 \times 1.097 \times 10^7} \approx 4.861 \times 10^{-7} \text{ m} = 4861 \, \mathring{A} \).
Q. 23. A current carrying toroid winding is internally filled with lithium having susceptibility \( \chi = 2.1 \times 10^{-5} \). What is the percentage increase in the magnetic field in the presence of lithium over that without it?
Magnetic field without core \( B_0 = \mu_0 n I \).
Magnetic field with lithium core \( B = \mu n I = \mu_0 (1 + \chi) n I \).
\( B = B_0 (1 + \chi) \).
Increase in field \( \Delta B = B - B_0 = B_0 (1 + \chi) - B_0 = B_0 \chi \).
Percentage increase = \( \frac{B - B_0}{B_0} \times 100\% = \chi \times 100\% \).
\( \% \text{ increase} = 2.1 \times 10^{-5} \times 100 = 2.1 \times 10^{-3} \% = 0.0021\% \).
Q. 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.
Ideally, for safe driving without relying on friction (optimum speed), the banking angle \( \theta \) is given by:
\( \tan\theta = \frac{v^2}{rg} \).
Given: \( r = 200 \) m, \( v = 25 \) m/s, \( g = 9.8 \) m/s\(^2\).
\( \tan\theta = \frac{25^2}{200 \times 9.8} = \frac{625}{1960} \).
\( \tan\theta \approx 0.3189 \).
\( \theta = \tan^{-1}(0.3189) \approx 17.69^\circ \) (or \( 17^\circ 41' \)).
Q. 25. Prove the Mayer’s relation : \( C_p - C_v = \frac{R}{J} \)
Consider one mole of an ideal gas.
Process 1 (Constant Volume): Heat supplied \( dQ_v = C_v dT \). Since \( dV=0 \), work \( dW=0 \). From 1st law, \( dU = dQ_v = C_v dT \).
Process 2 (Constant Pressure): Heat supplied \( dQ_p = C_p dT \). Work done \( dW = P dV \).
From 1st law, \( dQ_p = dU + dW \).
Substitute \( dU \) (it depends only on \( dT \)):
\( C_p dT = C_v dT + P dV \).
From ideal gas equation \( PV = RT \), differentiating at constant P: \( P dV = R dT \).
\( C_p dT = C_v dT + R dT \).
\( C_p = C_v + R \implies C_p - C_v = R \) (in work units).
If heat is in calories and work in ergs/joules, dividing by mechanical equivalent of heat \( J \):
$$ C_p - C_v = \frac{R}{J} $$
Q. 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.
Comparing with \( e = e_0 \sin(\omega t) \):
(i) Peak value (\( e_0 \)):
\( e_0 = 8 \) V.
(ii) Frequency (\( f \)):
\( \omega = 628.4 \) rad/s.
\( 2\pi f = 628.4 \Rightarrow f = \frac{628.4}{2 \times 3.142} = \frac{628.4}{6.284} = 100 \) Hz.
(iii) Instantaneous value at \( t = 10 \text{ ms} = 0.01 \text{ s} \):
\( e = 8 \sin(628.4 \times 0.01) = 8 \sin(6.284) \).
Since \( 6.284 \text{ rad} \approx 2\pi \text{ rad} \).
\( e = 8 \sin(2\pi) = 8 \times 0 = 0 \) V.
SECTION – D
Attempt any THREE questions of the following : [12]
Q. 27. What is a transformer? Explain construction and working of a transformer. Derive the equation for a transformer.
Definition: A transformer is an electrical device used to convert low alternating voltage at high current into high alternating voltage at low current and vice versa, based on mutual induction.
Construction: It consists of a soft iron core (laminated to reduce eddy currents) and two separate coils of insulated copper wire wound on it. The coil connected to the source is the Primary coil (\( N_p \) turns), and the coil connected to the load is the Secondary coil (\( N_s \) turns).
Working: When AC flows through the primary, it creates a changing magnetic flux in the core. This changing flux links with the secondary coil, inducing an EMF in it due to mutual induction.
Equation Derivation:
EMF in primary: \( e_p = -N_p \frac{d\phi}{dt} \).
EMF in secondary: \( e_s = -N_s \frac{d\phi}{dt} \).
Dividing the two:
$$ \frac{e_s}{e_p} = \frac{N_s}{N_p} $$
This is the transformer equation (Transformation ratio \( k \)).
Q. 28. Using the geometry of the double slit experiment, derive the expression for fringe width of interference bands.
Let \( S_1 \) and \( S_2 \) be two coherent sources separated by distance \( d \). Let the screen be at distance \( D \).
Path difference between waves reaching a point P at distance \( y \) from the center of the screen:
\( \Delta x = S_2P - S_1P \).
From geometry (assuming \( D \gg d \)):
\( S_2P^2 - S_1P^2 = (D^2 + (y+d/2)^2) - (D^2 + (y-d/2)^2) = 2yd \).
\( (S_2P - S_1P)(S_2P + S_1P) = 2yd \).
\( \Delta x \cdot 2D \approx 2yd \implies \Delta x = \frac{yd}{D} \).
For Constructive Interference (Bright band):
\( \Delta x = n\lambda \implies \frac{y_n d}{D} = n\lambda \implies y_n = \frac{n\lambda D}{d} \).
Fringe Width (\( X \) or \( \beta \)): Distance between successive bright bands.
\( X = y_{n+1} - y_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} \).
$$ X = \frac{\lambda D}{d} $$
Q. 29. Distinguish between an ammeter and a voltmeter. (Two points each).
The displacement of a particle performing simple harmonic motion is \( \frac{1}{3}\text{rd} \) of its amplitude. What fraction of total energy will be its kinetic energy?
Distinguish between Ammeter and Voltmeter:
| Ammeter | Voltmeter |
|---|---|
| Measures current in a circuit. | Measures potential difference. |
| Connected in series. | Connected in parallel. |
| Has very low resistance. | Has very high resistance. |
Problem Solution:
Given displacement \( x = \frac{A}{3} \).
Total Energy \( E = \frac{1}{2}kA^2 \).
Potential Energy \( U = \frac{1}{2}kx^2 = \frac{1}{2}k(\frac{A}{3})^2 = \frac{1}{9} (\frac{1}{2}kA^2) = \frac{E}{9} \).
Kinetic Energy \( K = E - U = E - \frac{E}{9} = \frac{8E}{9} \).
Fraction of Total Energy that is Kinetic = \( \frac{K}{E} = \frac{8}{9} \).
Q. 30. Draw a neat labelled diagram of Ferry’s perfectly black body. Compare the rms speed of hydrogen molecules at 227ºC with rms speed of oxygen molecule at 127ºC. Given that molecular masses of hydrogen and oxygen are 2 and 32 respectively.
Problem Solution:
RMS speed \( v_{rms} = \sqrt{\frac{3RT}{M}} \).
For Hydrogen (\( H_2 \)): \( T_H = 227+273 = 500 \) K, \( M_H = 2 \).
For Oxygen (\( O_2 \)): \( T_O = 127+273 = 400 \) K, \( M_O = 32 \).
Ratio:
$$ \frac{v_H}{v_O} = \sqrt{\frac{T_H}{M_H} \times \frac{M_O}{T_O}} $$
$$ \frac{v_H}{v_O} = \sqrt{\frac{500}{2} \times \frac{32}{400}} $$
$$ \frac{v_H}{v_O} = \sqrt{250 \times 0.08} = \sqrt{20} $$
$$ \frac{v_H}{v_O} = 2\sqrt{5} \approx 4.47 $$
Q. 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.
Derivation:
Consider a capacitor being charged. Let current charge be \( q \) and potential \( V = q/C \).
Work done to move additional charge \( dq \):
\( dW = V dq = \frac{q}{C} dq \).
Total work to charge from 0 to Q:
\( W = \int_0^Q \frac{q}{C} dq = \frac{1}{C} [\frac{q^2}{2}]_0^Q = \frac{Q^2}{2C} \).
This work is stored as energy \( U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 \).
Problem Solution:
Sphere radius \( R = 15 \) cm. Charge \( Q = 2 \mu C = 2 \times 10^{-6} \) C.
Distance \( r = 20 \) cm \( = 0.2 \) m.
Since \( r > R \), the point is outside the sphere. It behaves like a point charge at the center.
Electric Field \( E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \).
Using \( \frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \) Nm\(^2\)/C\(^2\).
$$ E = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{(0.2)^2} $$
$$ E = \frac{18 \times 10^3}{0.04} = \frac{18000}{0.04} $$
$$ E = 450,000 \, \text{N/C} = 4.5 \times 10^5 \, \text{N/C} $$