Model Question Paper Set - III Solutions
Physics
Paper Info:
Time: 3 Hours | Total Marks: 70
Note: All symbols have their usual meaning unless otherwise stated.
Time: 3 Hours | Total Marks: 70
Note: All symbols have their usual meaning unless otherwise stated.
SECTION – I
Q.1. Multiple Choice Questions
[7 Marks]
i. Particle A of mass M is revolving along a circle of radius R. Particle B of mass m is revolving in another circle of radius r. If they take the same time to complete one revolution, then the ratio of their angular velocities is _______.
Answer: (C) 1:1
Explanation: Angular velocity is given by \(\omega = \frac{2\pi}{T}\). Since both particles take the same time \(T\) to complete one revolution, their angular velocities must be equal. Ratio \(\omega_A : \omega_B = 1:1\).
ii. Angular momentum of a system of particles changes when _______.
Answer: (B) torque acts on a body
Explanation: According to the law of rotation, the rate of change of angular momentum is equal to the external torque acting on the system (\(\tau = \frac{dL}{dt}\)).
iii. The unit of quantity g/G where symbols have usual meaning is _______.
Answer: (C) kg/m²
Explanation: We know \(g = \frac{GM}{R^2}\). Therefore, \(\frac{g}{G} = \frac{M}{R^2}\). The unit of Mass (M) is kg and \(R^2\) is \(m^2\). Hence, unit is \(kg/m^2\).
iv. Which is a true statement about black body radiation?
Answer: (A) Intensity is different for all wavelengths.
Explanation: The distribution of radiant energy is not uniform. The intensity of radiation varies with wavelength at a given temperature.
v. The value of surface tension of a liquid at critical temperature is _______.
Answer: (A) zero
Explanation: At critical temperature, the interface between liquid and vapour disappears, and surface tension becomes zero.
vi. The harmonics which are present in a pipe open at one end are _______.
Answer: (A) odd harmonics
Explanation: For a pipe closed at one end (open at the other), the frequencies are \(n, 3n, 5n, \dots\), which corresponds to odd harmonics.
vii. A simple pendulum performs simple harmonic motion about x = 0 with an amplitude A and time period T. The speed of the pendulum at x = A/2 will be _______.
Answer: (A) \(\frac{\pi A \sqrt{3}}{T}\)
Explanation: Velocity in SHM is \(v = \omega \sqrt{A^2 - x^2}\).
Here \(x = A/2\) and \(\omega = \frac{2\pi}{T}\).
\(v = \frac{2\pi}{T} \sqrt{A^2 - \frac{A^2}{4}} = \frac{2\pi}{T} \sqrt{\frac{3A^2}{4}} = \frac{2\pi}{T} \cdot \frac{A\sqrt{3}}{2} = \frac{\pi A \sqrt{3}}{T}\).
Here \(x = A/2\) and \(\omega = \frac{2\pi}{T}\).
\(v = \frac{2\pi}{T} \sqrt{A^2 - \frac{A^2}{4}} = \frac{2\pi}{T} \sqrt{\frac{3A^2}{4}} = \frac{2\pi}{T} \cdot \frac{A\sqrt{3}}{2} = \frac{\pi A \sqrt{3}}{T}\).
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Q.2. Attempt any SIX
[12 Marks]
i. Obtain an expression for time period of a satellite orbiting very close to earth’s surface in terms of mean density. Show that \(T = \sqrt{\frac{3\pi}{G\rho}}\).
Solution:
For a satellite orbiting very close to the Earth, the orbital radius \(r \approx R\) (Radius of Earth).
The critical velocity is given by \(v_c = \sqrt{\frac{GM}{R}}\).
The time period \(T\) is given by: $$T = \frac{\text{Circumference}}{\text{Velocity}} = \frac{2\pi R}{v_c}$$ Substituting \(v_c\): $$T = \frac{2\pi R}{\sqrt{\frac{GM}{R}}} = 2\pi \sqrt{\frac{R^3}{GM}}$$ We know density \(\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{M}{\frac{4}{3}\pi R^3} \Rightarrow M = \frac{4}{3}\pi R^3 \rho\).
Substitute \(M\) in the equation for \(T\): $$T = 2\pi \sqrt{\frac{R^3}{G (\frac{4}{3}\pi R^3 \rho)}} = 2\pi \sqrt{\frac{3}{4\pi G \rho}}$$ $$T = \sqrt{4\pi^2 \times \frac{3}{4\pi G \rho}} = \sqrt{\frac{3\pi}{G\rho}}$$ Hence proved.
For a satellite orbiting very close to the Earth, the orbital radius \(r \approx R\) (Radius of Earth).
The critical velocity is given by \(v_c = \sqrt{\frac{GM}{R}}\).
The time period \(T\) is given by: $$T = \frac{\text{Circumference}}{\text{Velocity}} = \frac{2\pi R}{v_c}$$ Substituting \(v_c\): $$T = \frac{2\pi R}{\sqrt{\frac{GM}{R}}} = 2\pi \sqrt{\frac{R^3}{GM}}$$ We know density \(\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{M}{\frac{4}{3}\pi R^3} \Rightarrow M = \frac{4}{3}\pi R^3 \rho\).
Substitute \(M\) in the equation for \(T\): $$T = 2\pi \sqrt{\frac{R^3}{G (\frac{4}{3}\pi R^3 \rho)}} = 2\pi \sqrt{\frac{3}{4\pi G \rho}}$$ $$T = \sqrt{4\pi^2 \times \frac{3}{4\pi G \rho}} = \sqrt{\frac{3\pi}{G\rho}}$$ Hence proved.
ii. A mass of 4 kg is tied at the end of a string 1.2 m long, revolving in a horizontal circle. If the breaking tension in the string is 200 N, find the maximum number of revolutions per minute the mass can make.
Solution:
Given: \(m = 4\) kg, \(l = r = 1.2\) m, \(T_{max} = 200\) N.
The tension provides the centripetal force: $$T = mr\omega^2$$ $$200 = 4 \times 1.2 \times \omega^2$$ $$\omega^2 = \frac{200}{4.8} = 41.67$$ $$\omega = \sqrt{41.67} \approx 6.455 \text{ rad/s}$$ Frequency in rpm: $$n = \frac{\omega \times 60}{2\pi} = \frac{6.455 \times 60}{6.283} \approx 61.6 \text{ rev/min}$$
Given: \(m = 4\) kg, \(l = r = 1.2\) m, \(T_{max} = 200\) N.
The tension provides the centripetal force: $$T = mr\omega^2$$ $$200 = 4 \times 1.2 \times \omega^2$$ $$\omega^2 = \frac{200}{4.8} = 41.67$$ $$\omega = \sqrt{41.67} \approx 6.455 \text{ rad/s}$$ Frequency in rpm: $$n = \frac{\omega \times 60}{2\pi} = \frac{6.455 \times 60}{6.283} \approx 61.6 \text{ rev/min}$$
iii. A simple harmonic progressive wave of frequency 5 Hz is travelling along the positive X direction with a velocity of 40 m/s. Calculate the phase difference between two points separated by a distance of 0.8 m.
Solution:
Given: \(n = 5\) Hz, \(v = 40\) m/s, path difference \(\Delta x = 0.8\) m.
Wavelength \(\lambda = \frac{v}{n} = \frac{40}{5} = 8\) m.
Phase difference \(\Delta \phi\) is given by: $$\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x$$ $$\Delta \phi = \frac{2\pi}{8} \times 0.8 = 0.2\pi \text{ rad}$$ Or \(\Delta \phi = 36^\circ\).
Given: \(n = 5\) Hz, \(v = 40\) m/s, path difference \(\Delta x = 0.8\) m.
Wavelength \(\lambda = \frac{v}{n} = \frac{40}{5} = 8\) m.
Phase difference \(\Delta \phi\) is given by: $$\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x$$ $$\Delta \phi = \frac{2\pi}{8} \times 0.8 = 0.2\pi \text{ rad}$$ Or \(\Delta \phi = 36^\circ\).
iv. Define frequency of S.H.M. Discuss its unit and dimension.
Solution:
Definition: The number of oscillations or vibrations performed by a particle performing Simple Harmonic Motion (S.H.M.) per unit time is called the frequency of S.H.M.
It is denoted by \(n\) or \(f\) and is the reciprocal of the time period (\(n = 1/T\)).
Unit:
Definition: The number of oscillations or vibrations performed by a particle performing Simple Harmonic Motion (S.H.M.) per unit time is called the frequency of S.H.M.
It is denoted by \(n\) or \(f\) and is the reciprocal of the time period (\(n = 1/T\)).
Unit:
- SI Unit: Hertz (Hz) or per second (s\(^{-1}\)).
- CGS Unit: vibrations per second.
v. Distinguish between deforming force and stress.
Solution:
| Deforming Force | Stress |
|---|---|
| An external applied force that changes the shape or size of a body is called a deforming force. | The internal restoring force developed per unit area of the body to oppose deformation is called stress. |
| It is the cause of deformation. | It is the consequence (reaction) to the deformation. |
| Unit: Newton (N). | Unit: N/m² or Pascal (Pa). |
vi. A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 9 kg. It rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.
Solution:
Given: Bullet \(m = 10 \text{ g} = 0.01\) kg, \(v = 500\) m/s.
Door \(M = 9\) kg, Width \(L = 1.0\) m.
Axis of rotation is at the hinge (edge). Bullet hits at center, so \(r = L/2 = 0.5\) m.
Initial Angular Momentum (of bullet about hinge): $$L_i = mvr = 0.01 \times 500 \times 0.5 = 2.5 \text{ kg m}^2/\text{s}$$ Moment of Inertia of system after impact:
\(I_{door}\) (about edge) = \(\frac{ML^2}{3} = \frac{9 \times 1^2}{3} = 3 \text{ kg m}^2\).
\(I_{bullet}\) (at center) = \(mr^2 = 0.01 \times (0.5)^2 = 0.0025 \text{ kg m}^2\).
Total \(I = 3 + 0.0025 = 3.0025 \text{ kg m}^2\).
By Conservation of Angular Momentum: $$L_i = L_f \Rightarrow 2.5 = I \omega$$ $$\omega = \frac{2.5}{3.0025} \approx 0.8326 \text{ rad/s}$$
Given: Bullet \(m = 10 \text{ g} = 0.01\) kg, \(v = 500\) m/s.
Door \(M = 9\) kg, Width \(L = 1.0\) m.
Axis of rotation is at the hinge (edge). Bullet hits at center, so \(r = L/2 = 0.5\) m.
Initial Angular Momentum (of bullet about hinge): $$L_i = mvr = 0.01 \times 500 \times 0.5 = 2.5 \text{ kg m}^2/\text{s}$$ Moment of Inertia of system after impact:
\(I_{door}\) (about edge) = \(\frac{ML^2}{3} = \frac{9 \times 1^2}{3} = 3 \text{ kg m}^2\).
\(I_{bullet}\) (at center) = \(mr^2 = 0.01 \times (0.5)^2 = 0.0025 \text{ kg m}^2\).
Total \(I = 3 + 0.0025 = 3.0025 \text{ kg m}^2\).
By Conservation of Angular Momentum: $$L_i = L_f \Rightarrow 2.5 = I \omega$$ $$\omega = \frac{2.5}{3.0025} \approx 0.8326 \text{ rad/s}$$
vii. State Wien’s displacement law. State its significance.
Solution:
Statement: The wavelength for which the emissive power of a black body is maximum (\(\lambda_{max}\)) is inversely proportional to the absolute temperature (T) of the black body.
Formula: \(\lambda_{max} \propto \frac{1}{T}\) or \(\lambda_{max} T = b\) (Wien's constant).
Significance:
Statement: The wavelength for which the emissive power of a black body is maximum (\(\lambda_{max}\)) is inversely proportional to the absolute temperature (T) of the black body.
Formula: \(\lambda_{max} \propto \frac{1}{T}\) or \(\lambda_{max} T = b\) (Wien's constant).
Significance:
- It allows us to determine the temperature of celestial bodies like the Sun and stars by analyzing the spectrum of light emitted by them.
- It explains why the color of a heated object changes from red to yellow to white as temperature increases.
viii. A uniform wire under tension is fixed at its ends. If the ratio of tensions in the wire to the square of its length is 320 dyne/cm² and fundamental frequency of vibration of wire is 400 Hz, find its linear density.
Solution:
Given: \(n = 400\) Hz.
Ratio \(\frac{T}{l^2} = 320\) dyne/cm². (Note: The unit dyne/cm² suggests Force/Area, but the question says "ratio of tensions to square of length". Tension is Force (dyne). So \(\frac{T}{l^2} = 320\) in CGS units of Force/Length²).
Formula for fundamental frequency: $$n = \frac{1}{2l} \sqrt{\frac{T}{m}}$$ Square both sides: $$n^2 = \frac{1}{4l^2} \frac{T}{m}$$ Rearrange to isolate linear density \(m\): $$m = \frac{1}{4n^2} \left( \frac{T}{l^2} \right)$$ Substitute values (using CGS): $$m = \frac{1}{4 \times (400)^2} \times 320$$ $$m = \frac{320}{4 \times 160000} = \frac{320}{640000} = \frac{32}{64000}$$ $$m = 0.0005 \text{ g/cm}$$
Given: \(n = 400\) Hz.
Ratio \(\frac{T}{l^2} = 320\) dyne/cm². (Note: The unit dyne/cm² suggests Force/Area, but the question says "ratio of tensions to square of length". Tension is Force (dyne). So \(\frac{T}{l^2} = 320\) in CGS units of Force/Length²).
Formula for fundamental frequency: $$n = \frac{1}{2l} \sqrt{\frac{T}{m}}$$ Square both sides: $$n^2 = \frac{1}{4l^2} \frac{T}{m}$$ Rearrange to isolate linear density \(m\): $$m = \frac{1}{4n^2} \left( \frac{T}{l^2} \right)$$ Substitute values (using CGS): $$m = \frac{1}{4 \times (400)^2} \times 320$$ $$m = \frac{320}{4 \times 160000} = \frac{320}{640000} = \frac{32}{64000}$$ $$m = 0.0005 \text{ g/cm}$$
Q.3. Attempt any THREE
[9 Marks]
i. Define linear velocity. Derive the relation between linear velocity and angular velocity.
Solution:
Definition: Linear velocity is the rate of change of displacement of a particle with respect to time. It is a vector quantity tangent to the path.
Derivation:
Consider a particle moving in a circle of radius \(r\) with constant angular velocity \(\omega\).
Let the particle move from point A to B in a small time interval \(\delta t\). Let the angular displacement be \(\delta \theta\).
The arc length \(\delta s\) is given by: \(\delta s = r \cdot \delta \theta\).
Dividing both sides by \(\delta t\): $$\frac{\delta s}{\delta t} = r \frac{\delta \theta}{\delta t}$$ Taking limits as \(\delta t \rightarrow 0\): $$\lim_{\delta t \to 0} \frac{\delta s}{\delta t} = v \text{ (Linear Velocity)}$$ $$\lim_{\delta t \to 0} \frac{\delta \theta}{\delta t} = \omega \text{ (Angular Velocity)}$$ Therefore, the relation is: $$v = r\omega$$ In vector form: \(\vec{v} = \vec{\omega} \times \vec{r}\).
Definition: Linear velocity is the rate of change of displacement of a particle with respect to time. It is a vector quantity tangent to the path.
Derivation:
Consider a particle moving in a circle of radius \(r\) with constant angular velocity \(\omega\).
Let the particle move from point A to B in a small time interval \(\delta t\). Let the angular displacement be \(\delta \theta\).
The arc length \(\delta s\) is given by: \(\delta s = r \cdot \delta \theta\).
Dividing both sides by \(\delta t\): $$\frac{\delta s}{\delta t} = r \frac{\delta \theta}{\delta t}$$ Taking limits as \(\delta t \rightarrow 0\): $$\lim_{\delta t \to 0} \frac{\delta s}{\delta t} = v \text{ (Linear Velocity)}$$ $$\lim_{\delta t \to 0} \frac{\delta \theta}{\delta t} = \omega \text{ (Angular Velocity)}$$ Therefore, the relation is: $$v = r\omega$$ In vector form: \(\vec{v} = \vec{\omega} \times \vec{r}\).
ii. A set of 24 tuning forks is arranged in the increasing order of frequencies. Each tuning fork produces ‘y’ beats per second with previous one. The last fork is an octave of the first. The 8th tuning fork has a frequency 120 Hz. Find y and hence find the frequency of last fork.
Solution:
Let \(n_1, n_2, \dots, n_{24}\) be the frequencies.
Common difference (beats) = \(y\).
This forms an Arithmetic Progression where \(n_k = n_1 + (k-1)y\).
Given:
1. Last fork is octave of first: \(n_{24} = 2n_1\).
Using AP formula: \(n_{24} = n_1 + 23y\).
So, \(2n_1 = n_1 + 23y \Rightarrow n_1 = 23y\).
2. 8th fork is 120 Hz: \(n_8 = 120\).
\(n_8 = n_1 + 7y = 120\).
Substitute \(n_1 = 23y\) into the second equation:
\(23y + 7y = 120\)
\(30y = 120 \Rightarrow y = 4\).
Frequency of first fork: \(n_1 = 23(4) = 92\) Hz.
Frequency of last fork: \(n_{24} = 2n_1 = 2 \times 92 = 184\) Hz.
Let \(n_1, n_2, \dots, n_{24}\) be the frequencies.
Common difference (beats) = \(y\).
This forms an Arithmetic Progression where \(n_k = n_1 + (k-1)y\).
Given:
1. Last fork is octave of first: \(n_{24} = 2n_1\).
Using AP formula: \(n_{24} = n_1 + 23y\).
So, \(2n_1 = n_1 + 23y \Rightarrow n_1 = 23y\).
2. 8th fork is 120 Hz: \(n_8 = 120\).
\(n_8 = n_1 + 7y = 120\).
Substitute \(n_1 = 23y\) into the second equation:
\(23y + 7y = 120\)
\(30y = 120 \Rightarrow y = 4\).
Frequency of first fork: \(n_1 = 23(4) = 92\) Hz.
Frequency of last fork: \(n_{24} = 2n_1 = 2 \times 92 = 184\) Hz.
iii. What is surface energy? Obtain the relation between surface tension and surface energy.
Solution:
Surface Energy: The extra potential energy possessed by the molecules in the surface layer of a liquid compared to the molecules in the interior is called surface energy.
Relation:
Consider a rectangular frame of wire with a movable side PQ of length \(L\). A soap film is formed on it.
Surface tension \(T\) exerts force inwards. Since the film has two surfaces, total force \(F = 2TL\).
To increase the surface area, the wire PQ is pulled outwards by a distance \(dx\) against this force.
Work done \(dW = F \cdot dx = 2TL \cdot dx\).
Increase in area \(dA = 2 \cdot L \cdot dx\) (factor of 2 for two surfaces).
Therefore, \(dW = T(2L dx) = T \cdot dA\).
This work is stored as Surface Energy (E).
$$E = T \cdot dA \quad \text{or} \quad T = \frac{\text{Surface Energy}}{\text{Area}}$$ Thus, Surface Tension is numerically equal to Surface Energy per unit area.
Surface Energy: The extra potential energy possessed by the molecules in the surface layer of a liquid compared to the molecules in the interior is called surface energy.
Relation:
Consider a rectangular frame of wire with a movable side PQ of length \(L\). A soap film is formed on it.
Surface tension \(T\) exerts force inwards. Since the film has two surfaces, total force \(F = 2TL\).
To increase the surface area, the wire PQ is pulled outwards by a distance \(dx\) against this force.
Work done \(dW = F \cdot dx = 2TL \cdot dx\).
Increase in area \(dA = 2 \cdot L \cdot dx\) (factor of 2 for two surfaces).
Therefore, \(dW = T(2L dx) = T \cdot dA\).
This work is stored as Surface Energy (E).
$$E = T \cdot dA \quad \text{or} \quad T = \frac{\text{Surface Energy}}{\text{Area}}$$ Thus, Surface Tension is numerically equal to Surface Energy per unit area.
iv. A steel wire of length 6 m and cross section 1 mm² is hung from a rigid support with a steel weight of volume 2000 cc hanging from the other end. Find the decrease in the length of wire, when steel weight is completely immersed in water. [Ysteel = 2 × 10¹¹ N/m², Density of water = 1 g/cc]
Solution:
Given: \(L = 6\) m, \(A = 1 \text{ mm}^2 = 10^{-6} \text{ m}^2\).
Volume of weight \(V = 2000 \text{ cc} = 2 \times 10^{-3} \text{ m}^3\).
\(Y = 2 \times 10^{11} \text{ N/m}^2\).
The decrease in length corresponds to the decrease in tension due to Upthrust (Buoyant force).
Change in Force \(\Delta F = \text{Upthrust} = V \rho_w g\).
\(\rho_w = 1000 \text{ kg/m}^3\).
\(\Delta F = 2 \times 10^{-3} \times 1000 \times 9.8 = 19.6\) N.
Formula for extension: \(\Delta l = \frac{FL}{AY}\).
Decrease in length \(\Delta l_{dec} = \frac{\Delta F \cdot L}{AY}\).
$$\Delta l_{dec} = \frac{19.6 \times 6}{10^{-6} \times 2 \times 10^{11}}$$ $$\Delta l_{dec} = \frac{117.6}{2 \times 10^5} = 58.8 \times 10^{-5} \text{ m}$$ $$\Delta l_{dec} = 0.588 \text{ mm}$$
Given: \(L = 6\) m, \(A = 1 \text{ mm}^2 = 10^{-6} \text{ m}^2\).
Volume of weight \(V = 2000 \text{ cc} = 2 \times 10^{-3} \text{ m}^3\).
\(Y = 2 \times 10^{11} \text{ N/m}^2\).
The decrease in length corresponds to the decrease in tension due to Upthrust (Buoyant force).
Change in Force \(\Delta F = \text{Upthrust} = V \rho_w g\).
\(\rho_w = 1000 \text{ kg/m}^3\).
\(\Delta F = 2 \times 10^{-3} \times 1000 \times 9.8 = 19.6\) N.
Formula for extension: \(\Delta l = \frac{FL}{AY}\).
Decrease in length \(\Delta l_{dec} = \frac{\Delta F \cdot L}{AY}\).
$$\Delta l_{dec} = \frac{19.6 \times 6}{10^{-6} \times 2 \times 10^{11}}$$ $$\Delta l_{dec} = \frac{117.6}{2 \times 10^5} = 58.8 \times 10^{-5} \text{ m}$$ $$\Delta l_{dec} = 0.588 \text{ mm}$$
Q.4. Attempt the following
[7 Marks]
A. Derive an expression for the kinetic energy of a body rotating with uniform angular velocity. [4]
Solution:
Consider a rigid body rotating about an axis with uniform angular velocity \(\omega\).
The body consists of \(n\) particles of masses \(m_1, m_2, \dots, m_n\) at perpendicular distances \(r_1, r_2, \dots, r_n\) from the axis of rotation.
The linear velocity of the first particle is \(v_1 = r_1 \omega\).
Kinetic Energy of first particle: \(E_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 (r_1 \omega)^2 = \frac{1}{2} m_1 r_1^2 \omega^2\).
Similarly, for other particles: \(E_i = \frac{1}{2} m_i r_i^2 \omega^2\).
Total Rotational K.E. of the body is the sum of K.E. of all particles:
$$E = E_1 + E_2 + \dots + E_n$$ $$E = \sum_{i=1}^n \frac{1}{2} m_i r_i^2 \omega^2$$ $$E = \frac{1}{2} \omega^2 \left( \sum_{i=1}^n m_i r_i^2 \right)$$ By definition, Moment of Inertia \(I = \sum m_i r_i^2\).
Therefore, $$E_{rot} = \frac{1}{2} I \omega^2$$
Consider a rigid body rotating about an axis with uniform angular velocity \(\omega\).
The body consists of \(n\) particles of masses \(m_1, m_2, \dots, m_n\) at perpendicular distances \(r_1, r_2, \dots, r_n\) from the axis of rotation.
The linear velocity of the first particle is \(v_1 = r_1 \omega\).
Kinetic Energy of first particle: \(E_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 (r_1 \omega)^2 = \frac{1}{2} m_1 r_1^2 \omega^2\).
Similarly, for other particles: \(E_i = \frac{1}{2} m_i r_i^2 \omega^2\).
Total Rotational K.E. of the body is the sum of K.E. of all particles:
$$E = E_1 + E_2 + \dots + E_n$$ $$E = \sum_{i=1}^n \frac{1}{2} m_i r_i^2 \omega^2$$ $$E = \frac{1}{2} \omega^2 \left( \sum_{i=1}^n m_i r_i^2 \right)$$ By definition, Moment of Inertia \(I = \sum m_i r_i^2\).
Therefore, $$E_{rot} = \frac{1}{2} I \omega^2$$
B. A body describes S.H.M in a path 0.16 m long. Its velocity at the centre of the line is 0.12 m/s. Find the period and magnitude of velocity at a distance 1.7 × 10⁻² m from the mean position. [3]
Solution:
Path length = \(2A = 0.16\) m \(\Rightarrow\) Amplitude \(A = 0.08\) m.
Velocity at center (mean position) is \(v_{max} = A\omega = 0.12\) m/s.
1. Find Period (T):
\(0.08 \times \omega = 0.12 \Rightarrow \omega = \frac{0.12}{0.08} = 1.5\) rad/s.
\(T = \frac{2\pi}{\omega} = \frac{2\pi}{1.5} = \frac{4\pi}{3} \approx 4.19\) s.
2. Velocity at x = 1.7 × 10⁻² m = 0.017 m:
\(v = \omega \sqrt{A^2 - x^2}\)
\(v = 1.5 \sqrt{(0.08)^2 - (0.017)^2}\)
\(v = 1.5 \sqrt{0.0064 - 0.000289} = 1.5 \sqrt{0.006111}\)
\(v = 1.5 \times 0.07817 \approx 0.117\) m/s.
Path length = \(2A = 0.16\) m \(\Rightarrow\) Amplitude \(A = 0.08\) m.
Velocity at center (mean position) is \(v_{max} = A\omega = 0.12\) m/s.
1. Find Period (T):
\(0.08 \times \omega = 0.12 \Rightarrow \omega = \frac{0.12}{0.08} = 1.5\) rad/s.
\(T = \frac{2\pi}{\omega} = \frac{2\pi}{1.5} = \frac{4\pi}{3} \approx 4.19\) s.
2. Velocity at x = 1.7 × 10⁻² m = 0.017 m:
\(v = \omega \sqrt{A^2 - x^2}\)
\(v = 1.5 \sqrt{(0.08)^2 - (0.017)^2}\)
\(v = 1.5 \sqrt{0.0064 - 0.000289} = 1.5 \sqrt{0.006111}\)
\(v = 1.5 \times 0.07817 \approx 0.117\) m/s.
OR
Q.4. A. State Kirchhoff’s law of radiation and give its theoretical proof. [4]
Solution:
Statement: At a given temperature, the coefficient of absorption of a body is equal to its coefficient of emission (emissivity). (i.e., \(a = e\)).
Theoretical Proof:
Consider an ordinary body \(O\) and a perfectly black body \(B\) placed in an enclosure at constant temperature \(T\).
Let \(Q\) be the radiant energy incident per unit area per unit time on both bodies.
For body \(O\): Energy absorbed = \(aQ\). Energy emitted = \(E\) (emissive power).
At equilibrium, Energy Absorbed = Energy Emitted.
\(\therefore E = aQ \Rightarrow Q = E/a\). ... (1)
For Black body \(B\): Absorption coefficient \(a=1\). Energy absorbed = \(1 \cdot Q\). Emissive power = \(E_b\).
At equilibrium: \(E_b = Q\). ... (2)
From (1) and (2): \(\frac{E}{a} = E_b \Rightarrow a = \frac{E}{E_b}\).
But by definition, Emissivity \(e = \frac{E}{E_b}\).
Therefore, \(a = e\).
Statement: At a given temperature, the coefficient of absorption of a body is equal to its coefficient of emission (emissivity). (i.e., \(a = e\)).
Theoretical Proof:
Consider an ordinary body \(O\) and a perfectly black body \(B\) placed in an enclosure at constant temperature \(T\).
Let \(Q\) be the radiant energy incident per unit area per unit time on both bodies.
For body \(O\): Energy absorbed = \(aQ\). Energy emitted = \(E\) (emissive power).
At equilibrium, Energy Absorbed = Energy Emitted.
\(\therefore E = aQ \Rightarrow Q = E/a\). ... (1)
For Black body \(B\): Absorption coefficient \(a=1\). Energy absorbed = \(1 \cdot Q\). Emissive power = \(E_b\).
At equilibrium: \(E_b = Q\). ... (2)
From (1) and (2): \(\frac{E}{a} = E_b \Rightarrow a = \frac{E}{E_b}\).
But by definition, Emissivity \(e = \frac{E}{E_b}\).
Therefore, \(a = e\).
Q.4. B. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 50 Hz. The mass of the wire is 4.4 × 10⁻² kg and its linear mass density is 4.0 × 10⁻² kg m⁻¹. What is (i) the speed of a transverse wave on the string? (ii) the tension in the string? [3]
Solution:
Given: \(n = 50\) Hz, Mass \(M = 4.4 \times 10^{-2}\) kg, \(\mu (m) = 4.0 \times 10^{-2}\) kg/m.
Length of wire \(L = \frac{\text{Mass}}{\text{Linear Density}} = \frac{4.4 \times 10^{-2}}{4.0 \times 10^{-2}} = 1.1\) m.
i. Speed of wave (v):
For fundamental mode, \(L = \lambda/2 \Rightarrow \lambda = 2L\).
\(v = n \lambda = n(2L) = 50 \times 2(1.1) = 110\) m/s.
ii. Tension (T):
Formula: \(v = \sqrt{\frac{T}{m}} \Rightarrow T = v^2 m\).
\(T = (110)^2 \times (4.0 \times 10^{-2})\)
\(T = 12100 \times 0.04 = 484\) N.
Given: \(n = 50\) Hz, Mass \(M = 4.4 \times 10^{-2}\) kg, \(\mu (m) = 4.0 \times 10^{-2}\) kg/m.
Length of wire \(L = \frac{\text{Mass}}{\text{Linear Density}} = \frac{4.4 \times 10^{-2}}{4.0 \times 10^{-2}} = 1.1\) m.
i. Speed of wave (v):
For fundamental mode, \(L = \lambda/2 \Rightarrow \lambda = 2L\).
\(v = n \lambda = n(2L) = 50 \times 2(1.1) = 110\) m/s.
ii. Tension (T):
Formula: \(v = \sqrt{\frac{T}{m}} \Rightarrow T = v^2 m\).
\(T = (110)^2 \times (4.0 \times 10^{-2})\)
\(T = 12100 \times 0.04 = 484\) N.
SECTION – II
Q.5. Multiple Choice Questions
[7 Marks]
i. Velocity of light in air is ‘c’. Its velocity in a medium of refractive index 1.5 will be _______.
Answer: (B) c/1.5
Explanation: Refractive index \(\mu = \frac{c}{v} \Rightarrow v = \frac{c}{\mu} = \frac{c}{1.5}\).
ii. If the lengths of two wires of same material are in the ratio 2 : 1, then the ratio of their specific resistances will be _______.
Answer: (D) 1 : 1
Explanation: Specific resistance (resistivity) depends only on the nature of the material and temperature, not on dimensions. Since the material is the same, the ratio is 1:1.
iii. The phenomenon of paramagnetism is a consequence of _______.
Answer: (B) orientation effect
Explanation: Paramagnetism arises due to the alignment (orientation) of permanent magnetic dipoles in the direction of the external field.
iv. If ‘R’ and ‘L’ stand for the resistance and inductance respectively, then among the following the one having the dimensions of frequency is _______.
Answer: (A) R/L
Explanation: The time constant of an LR circuit is \(\tau = L/R\). Frequency is inverse of time. Hence \(R/L\) has dimensions of frequency.
v. For what value of velocity of electrons, the stopping potential will be able to stop them?
Answer: (D) all speeds.
Explanation: The stopping potential is determined by the maximum kinetic energy of the photoelectrons. If the potential is sufficient to stop the fastest electron, it automatically stops all electrons with lower speeds (all speeds).
vi. In an amplitude modulated wave, the power content of the carrier is maximum for which value of ‘m’?
Answer: (A) zero
Explanation: The fraction of total power contained in the carrier is given by \(\frac{P_c}{P_t} = \frac{1}{1 + m^2/2}\). This fraction is maximum (100%) when \(m=0\) (no modulation).
vii. In Boolean algebra, \(\overline{A + B} =\) _______.
Answer: (D) \(\overline{A} \cdot \overline{B}\)
Explanation: This is De Morgan's First Law: The complement of a sum is equal to the product of the complements.
Q.6. Attempt any SIX
[12 Marks]
i. Distinguish between interference and diffraction.
Solution:
| Interference | Diffraction |
|---|---|
| It is due to the superposition of waves from two separate coherent sources. | It is due to the superposition of secondary wavelets from different parts of the same wavefront. |
| Interference fringes (bright and dark) are usually of equal width. | Diffraction fringes are never of equal width (central max is widest). |
| All bright fringes have equal intensity. | Intensity of bright fringes decreases as order increases. |
ii. Explain working of Van de Graaff generator.
Solution:
Principle: It works on the principle of corona discharge (action of points) and the property that charge given to a hollow conductor resides on its outer surface.
Working:
Principle: It works on the principle of corona discharge (action of points) and the property that charge given to a hollow conductor resides on its outer surface.
Working:
- An electric motor drives a belt made of insulating material over two pulleys.
- A spray comb near the bottom pulley is connected to a high positive potential. It sprays positive charge onto the belt via corona discharge.
- The belt carries this charge upwards. A collecting comb near the top collects the charge and transfers it to the hollow metal sphere (dome).
- The charge spreads over the outer surface of the dome. As this process repeats, the potential of the dome rises to millions of volts (approx 10⁷ V).
iii. Resistances in a Wheatstone’s bridge are 45 Ω, 90 Ω, 25 Ω and a series combination of 5 Ω and X Ω. If the bridge is balanced, calculate value of X.
Solution:
Let resistances be \(P = 45\Omega\), \(Q = 90\Omega\), \(R = 25\Omega\).
The fourth arm \(S\) is a series combination of 5 and X, so \(S = 5 + X\).
For a balanced bridge: $$\frac{P}{Q} = \frac{R}{S}$$ $$\frac{45}{90} = \frac{25}{5 + X}$$ $$\frac{1}{2} = \frac{25}{5 + X}$$ $$5 + X = 50$$ $$X = 45 \Omega$$
Let resistances be \(P = 45\Omega\), \(Q = 90\Omega\), \(R = 25\Omega\).
The fourth arm \(S\) is a series combination of 5 and X, so \(S = 5 + X\).
For a balanced bridge: $$\frac{P}{Q} = \frac{R}{S}$$ $$\frac{45}{90} = \frac{25}{5 + X}$$ $$\frac{1}{2} = \frac{25}{5 + X}$$ $$5 + X = 50$$ $$X = 45 \Omega$$
iv. What is the effect of heat on a ferromagnetic substance?
Solution:
When a ferromagnetic substance is heated, the thermal agitation of domains increases, disturbing their alignment.
As temperature increases, ferromagnetism decreases. At a specific temperature called the Curie Temperature, the domain structure collapses, and the ferromagnetic substance transforms into a paramagnetic substance.
When a ferromagnetic substance is heated, the thermal agitation of domains increases, disturbing their alignment.
As temperature increases, ferromagnetism decreases. At a specific temperature called the Curie Temperature, the domain structure collapses, and the ferromagnetic substance transforms into a paramagnetic substance.
v. A long solenoid of length 2 m has 1000 turns of wire closely wound on it. If the current in the winding is 2 A, calculate the magnetic field at the centre of solenoid.
Solution:
Given: \(L = 2\) m, \(N = 1000\), \(I = 2\) A.
Number of turns per unit length \(n = \frac{N}{L} = \frac{1000}{2} = 500\) turns/m.
Magnetic field at center \(B = \mu_0 n I\).
\((\mu_0 = 4\pi \times 10^{-7} \text{ T m/A})\)
$$B = 4\pi \times 10^{-7} \times 500 \times 2$$ $$B = 4 \times 3.142 \times 10^{-7} \times 1000$$ $$B = 12.568 \times 10^{-4} \text{ T}$$ or \(1.26 \times 10^{-3} \text{ T}\).
Given: \(L = 2\) m, \(N = 1000\), \(I = 2\) A.
Number of turns per unit length \(n = \frac{N}{L} = \frac{1000}{2} = 500\) turns/m.
Magnetic field at center \(B = \mu_0 n I\).
\((\mu_0 = 4\pi \times 10^{-7} \text{ T m/A})\)
$$B = 4\pi \times 10^{-7} \times 500 \times 2$$ $$B = 4 \times 3.142 \times 10^{-7} \times 1000$$ $$B = 12.568 \times 10^{-4} \text{ T}$$ or \(1.26 \times 10^{-3} \text{ T}\).
vi. The electric current in a circuit is increased uniformly from 2 A to 10 A during a time interval of 2 seconds. Another coil of resistance 40 Ω and mutual inductance 1 H is placed near it. Calculate the induced current in the second coil.
Solution:
Given: \(dI_1 = 10 - 2 = 8\) A, \(dt = 2\) s, \(R = 40 \Omega\), \(M = 1\) H.
Induced EMF in second coil: $$|e| = M \frac{dI_1}{dt}$$ $$|e| = 1 \times \frac{8}{2} = 4 \text{ V}$$ Induced Current \(I_2 = \frac{e}{R}\): $$I_2 = \frac{4}{40} = 0.1 \text{ A}$$
Given: \(dI_1 = 10 - 2 = 8\) A, \(dt = 2\) s, \(R = 40 \Omega\), \(M = 1\) H.
Induced EMF in second coil: $$|e| = M \frac{dI_1}{dt}$$ $$|e| = 1 \times \frac{8}{2} = 4 \text{ V}$$ Induced Current \(I_2 = \frac{e}{R}\): $$I_2 = \frac{4}{40} = 0.1 \text{ A}$$
vii. Calculate the de Broglie wavelength of waves associated with a beam of neutrons of energy 0.05 eV. [h = 6.63 × 10⁻³⁴ Js, mn = 1.66 × 10⁻²⁷ kg]
Solution:
Given: \(E = 0.05 \text{ eV} = 0.05 \times 1.6 \times 10^{-19} = 8 \times 10^{-21}\) J.
Formula: \(\lambda = \frac{h}{\sqrt{2mE}}\).
$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.66 \times 10^{-27} \times 8 \times 10^{-21}}}$$ $$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{26.56 \times 10^{-48}}}$$ $$\lambda = \frac{6.63 \times 10^{-34}}{5.15 \times 10^{-24}}$$ $$\lambda \approx 1.28 \times 10^{-10} \text{ m} = 1.28 \text{ \AA}$$
Given: \(E = 0.05 \text{ eV} = 0.05 \times 1.6 \times 10^{-19} = 8 \times 10^{-21}\) J.
Formula: \(\lambda = \frac{h}{\sqrt{2mE}}\).
$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.66 \times 10^{-27} \times 8 \times 10^{-21}}}$$ $$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{26.56 \times 10^{-48}}}$$ $$\lambda = \frac{6.63 \times 10^{-34}}{5.15 \times 10^{-24}}$$ $$\lambda \approx 1.28 \times 10^{-10} \text{ m} = 1.28 \text{ \AA}$$
viii. Define the following terms.
a. Demodulation
b. Attenuation
a. Demodulation
b. Attenuation
Solution:
a. Demodulation: The process of recovering the original information signal (audio/video) from the modulated carrier wave at the receiver end is called demodulation or detection.
b. Attenuation: The loss of strength (power/intensity) of a signal while propagating through a transmission medium is called attenuation.
a. Demodulation: The process of recovering the original information signal (audio/video) from the modulated carrier wave at the receiver end is called demodulation or detection.
b. Attenuation: The loss of strength (power/intensity) of a signal while propagating through a transmission medium is called attenuation.
Q.7. Attempt any THREE
[9 Marks]
i. Two parallel plate air capacitors have their plate areas 100 cm² and 400 cm² respectively. If they have the same charge and potential and the distance between the plates of the first capacitor is 0.4 mm, what is the distance between the plates of the second capacitor?
Solution:
Given: \(A_1 = 100\), \(A_2 = 400\), \(d_1 = 0.4\) mm.
Same Charge (\(Q\)) and Potential (\(V\)) implies Capacitance \(C = Q/V\) is the same.
\(C_1 = C_2\).
For parallel plate: \(C = \frac{\varepsilon_0 A}{d}\).
$$\frac{\varepsilon_0 A_1}{d_1} = \frac{\varepsilon_0 A_2}{d_2}$$ $$\frac{100}{0.4} = \frac{400}{d_2}$$ $$d_2 = \frac{400 \times 0.4}{100} = 4 \times 0.4 = 1.6 \text{ mm}$$
Given: \(A_1 = 100\), \(A_2 = 400\), \(d_1 = 0.4\) mm.
Same Charge (\(Q\)) and Potential (\(V\)) implies Capacitance \(C = Q/V\) is the same.
\(C_1 = C_2\).
For parallel plate: \(C = \frac{\varepsilon_0 A}{d}\).
$$\frac{\varepsilon_0 A_1}{d_1} = \frac{\varepsilon_0 A_2}{d_2}$$ $$\frac{100}{0.4} = \frac{400}{d_2}$$ $$d_2 = \frac{400 \times 0.4}{100} = 4 \times 0.4 = 1.6 \text{ mm}$$
ii. An alternating e.m.f is applied to a circuit containing resistance. Discuss the behaviour of current in the circuit.
Solution:
Consider a pure resistor of resistance \(R\) connected to an AC source \(e = e_0 \sin \omega t\).
According to Ohm's law, instantaneous current \(i = e/R\).
\(i = \frac{e_0 \sin \omega t}{R}\).
Let \(i_0 = e_0/R\) be the peak current.
Then \(i = i_0 \sin \omega t\).
Comparison:
Consider a pure resistor of resistance \(R\) connected to an AC source \(e = e_0 \sin \omega t\).
According to Ohm's law, instantaneous current \(i = e/R\).
\(i = \frac{e_0 \sin \omega t}{R}\).
Let \(i_0 = e_0/R\) be the peak current.
Then \(i = i_0 \sin \omega t\).
Comparison:
- Equation of EMF: \(e = e_0 \sin \omega t\)
- Equation of Current: \(i = i_0 \sin \omega t\)
iii. With the help of a neat circuit diagram, explain working of P-N junction diode as a half-wave rectifier.
Solution:
Circuit: AC input is connected to the primary of a transformer. Secondary is connected to a P-N diode and Load Resistance (\(R_L\)) in series.
Working:
Circuit: AC input is connected to the primary of a transformer. Secondary is connected to a P-N diode and Load Resistance (\(R_L\)) in series.
Working:
- Positive Half Cycle: The P-end of the diode becomes positive wrt N-end. The diode is forward biased and conducts current. Output voltage appears across \(R_L\).
- Negative Half Cycle: The P-end becomes negative wrt N-end. The diode is reverse biased and does not conduct current (ideal case). No output across \(R_L\).
iv. A resistance of 5Ω is connected in parallel to a galvanometer of resistance 495Ω. Find the fraction of the total current, passing through galvanometer.
Solution:
Given: Galvanometer \(G = 495 \Omega\), Shunt \(S = 5 \Omega\).
Total current \(I\). Current through galvanometer \(I_g\).
Since \(G\) and \(S\) are in parallel, Potential Difference is same:
\(I_g G = (I - I_g) S\).
Or using direct formula for fraction:
$$\frac{I_g}{I} = \frac{S}{S + G}$$ $$\frac{I_g}{I} = \frac{5}{5 + 495} = \frac{5}{500} = \frac{1}{100}$$ Fraction is 0.01 or 1%.
Given: Galvanometer \(G = 495 \Omega\), Shunt \(S = 5 \Omega\).
Total current \(I\). Current through galvanometer \(I_g\).
Since \(G\) and \(S\) are in parallel, Potential Difference is same:
\(I_g G = (I - I_g) S\).
Or using direct formula for fraction:
$$\frac{I_g}{I} = \frac{S}{S + G}$$ $$\frac{I_g}{I} = \frac{5}{5 + 495} = \frac{5}{500} = \frac{1}{100}$$ Fraction is 0.01 or 1%.
Q.8. Attempt the following
[7 Marks]
A. Give the theory of interference bands and hence obtain an expression for bandwidth. [4]
Solution:
Consider two coherent sources \(S_1\) and \(S_2\) separated by distance \(d\). A screen is placed at distance \(D\) (\(D \gg d\)).
Let a point \(P\) be on the screen at distance \(y\) from the center \(O\).
Path difference \(\Delta x = S_2P - S_1P\).
From geometry approximation: \(\Delta x = \frac{y d}{D}\).
For Bright Bands (Constructive Interference):
\(\Delta x = n \lambda\) (where \(n = 0, 1, 2 \dots\))
\(\frac{y_n d}{D} = n \lambda \Rightarrow y_n = \frac{n \lambda D}{d}\).
Bandwidth (X): Distance between two consecutive bright bands.
\(X = y_{n+1} - y_n = \frac{(n+1)\lambda D}{d} - \frac{n \lambda D}{d}\).
$$X = \frac{\lambda D}{d}$$ (Similarly for dark bands, the bandwidth is same).
Consider two coherent sources \(S_1\) and \(S_2\) separated by distance \(d\). A screen is placed at distance \(D\) (\(D \gg d\)).
Let a point \(P\) be on the screen at distance \(y\) from the center \(O\).
Path difference \(\Delta x = S_2P - S_1P\).
From geometry approximation: \(\Delta x = \frac{y d}{D}\).
For Bright Bands (Constructive Interference):
\(\Delta x = n \lambda\) (where \(n = 0, 1, 2 \dots\))
\(\frac{y_n d}{D} = n \lambda \Rightarrow y_n = \frac{n \lambda D}{d}\).
Bandwidth (X): Distance between two consecutive bright bands.
\(X = y_{n+1} - y_n = \frac{(n+1)\lambda D}{d} - \frac{n \lambda D}{d}\).
$$X = \frac{\lambda D}{d}$$ (Similarly for dark bands, the bandwidth is same).
B. The photoelectric work function for a surface is 2.5 eV. Light of wavelength 6000 Å shines on the surface. Find the frequency of the incident light and also the threshold frequency. Will there be photoelectric emission or not? [3]
Solution:
Given: Work function \(\phi_0 = 2.5\) eV, \(\lambda = 6000 \text{ \AA} = 6 \times 10^{-7}\) m.
\(h = 6.63 \times 10^{-34}\) Js, \(1 \text{ eV} = 1.6 \times 10^{-19}\) J.
1. Frequency of incident light (\(\nu\)):
\(\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{6 \times 10^{-7}} = 0.5 \times 10^{15} = 5 \times 10^{14}\) Hz.
2. Threshold frequency (\(\nu_0\)):
\(\phi_0 = h \nu_0 \Rightarrow \nu_0 = \frac{\phi_0}{h}\).
\(\phi_0 \text{ in Joules} = 2.5 \times 1.6 \times 10^{-19} = 4.0 \times 10^{-19}\) J.
\(\nu_0 = \frac{4.0 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 6.03 \times 10^{14}\) Hz.
3. Emission?
Since incident frequency (\(5 \times 10^{14}\)) is less than threshold frequency (\(6.03 \times 10^{14}\)), photoelectric emission will NOT take place.
Given: Work function \(\phi_0 = 2.5\) eV, \(\lambda = 6000 \text{ \AA} = 6 \times 10^{-7}\) m.
\(h = 6.63 \times 10^{-34}\) Js, \(1 \text{ eV} = 1.6 \times 10^{-19}\) J.
1. Frequency of incident light (\(\nu\)):
\(\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{6 \times 10^{-7}} = 0.5 \times 10^{15} = 5 \times 10^{14}\) Hz.
2. Threshold frequency (\(\nu_0\)):
\(\phi_0 = h \nu_0 \Rightarrow \nu_0 = \frac{\phi_0}{h}\).
\(\phi_0 \text{ in Joules} = 2.5 \times 1.6 \times 10^{-19} = 4.0 \times 10^{-19}\) J.
\(\nu_0 = \frac{4.0 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 6.03 \times 10^{14}\) Hz.
3. Emission?
Since incident frequency (\(5 \times 10^{14}\)) is less than threshold frequency (\(6.03 \times 10^{14}\)), photoelectric emission will NOT take place.
OR
Q.8. A. Explain origin of spectral line and obtain Bohr’s formula. [4]
Solution:
According to Bohr's third postulate, when an electron jumps from a higher energy orbit (\(E_n\)) to a lower energy orbit (\(E_p\)), the energy difference is emitted as a photon.
$$h\nu = E_n - E_p$$ The energy of electron in \(n^{th}\) orbit is \(E_n = -\frac{me^4}{8\varepsilon_0^2 h^2 n^2}\).
Frequency \(\nu = \frac{E_n - E_p}{h} = \frac{me^4}{8\varepsilon_0^2 h^3} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)\).
Wave number \(\bar{\nu} = \frac{1}{\lambda} = \frac{\nu}{c}\).
$$\frac{1}{\lambda} = \frac{me^4}{8\varepsilon_0^2 h^3 c} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$$ The constant \(\frac{me^4}{8\varepsilon_0^2 h^3 c}\) is Rydberg's constant (R).
So, \(\frac{1}{\lambda} = R \left( \frac{1}{p^2} - \frac{1}{n^2} \right)\). This is Bohr's formula.
According to Bohr's third postulate, when an electron jumps from a higher energy orbit (\(E_n\)) to a lower energy orbit (\(E_p\)), the energy difference is emitted as a photon.
$$h\nu = E_n - E_p$$ The energy of electron in \(n^{th}\) orbit is \(E_n = -\frac{me^4}{8\varepsilon_0^2 h^2 n^2}\).
Frequency \(\nu = \frac{E_n - E_p}{h} = \frac{me^4}{8\varepsilon_0^2 h^3} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)\).
Wave number \(\bar{\nu} = \frac{1}{\lambda} = \frac{\nu}{c}\).
$$\frac{1}{\lambda} = \frac{me^4}{8\varepsilon_0^2 h^3 c} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$$ The constant \(\frac{me^4}{8\varepsilon_0^2 h^3 c}\) is Rydberg's constant (R).
So, \(\frac{1}{\lambda} = R \left( \frac{1}{p^2} - \frac{1}{n^2} \right)\). This is Bohr's formula.
B. If the difference in velocities of light in glass and water is 0.24 × 10⁸ m/s, find the velocity of light in air [Given : µg = 3/2 , µw = 4/3] [3]
Solution:
Let \(v_g\) and \(v_w\) be velocities in glass and water.
\(v_g = c / \mu_g = c / (3/2) = 2c/3\).
\(v_w = c / \mu_w = c / (4/3) = 3c/4\).
Given difference: \(v_w - v_g = 0.24 \times 10^8\) (since \(3/4 > 2/3\)).
$$\frac{3c}{4} - \frac{2c}{3} = 0.24 \times 10^8$$ $$c \left( \frac{9 - 8}{12} \right) = 0.24 \times 10^8$$ $$c \left( \frac{1}{12} \right) = 0.24 \times 10^8$$ $$c = 12 \times 0.24 \times 10^8$$ $$c = 2.88 \times 10^8 \text{ m/s}$$ (Note: Standard c is \(3 \times 10^8\), but based on the given difference, this is the calculated value).
Let \(v_g\) and \(v_w\) be velocities in glass and water.
\(v_g = c / \mu_g = c / (3/2) = 2c/3\).
\(v_w = c / \mu_w = c / (4/3) = 3c/4\).
Given difference: \(v_w - v_g = 0.24 \times 10^8\) (since \(3/4 > 2/3\)).
$$\frac{3c}{4} - \frac{2c}{3} = 0.24 \times 10^8$$ $$c \left( \frac{9 - 8}{12} \right) = 0.24 \times 10^8$$ $$c \left( \frac{1}{12} \right) = 0.24 \times 10^8$$ $$c = 12 \times 0.24 \times 10^8$$ $$c = 2.88 \times 10^8 \text{ m/s}$$ (Note: Standard c is \(3 \times 10^8\), but based on the given difference, this is the calculated value).