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Maharashtra Board HSC Physics Question Paper July 2016 Solutions

Maharashtra Board Class 12 Physics Question Paper July 2016 Solutions
Question Paper Page No. 1 Question Paper Page No. 2 Question Paper Page No. 3 Question Paper Page No. 4 Maharashtra Board HSC Physics Paper

Maharashtra Board HSC Physics Question Paper: July 2016 - Solutions

Note: All questions are solved below according to the original question paper structure. Mathematical derivations and calculations are presented clearly.

SECTION – I

Q.1. Select and write the most appropriate answer from the given alternatives for each sub-question:

i. The difference in tensions in the string at lowest and highest points in the path of the particle of mass ‘m’ performing vertical circular motion is:
Explanation: In a vertical circular motion, the tension at the lowest point ($T_L$) and the highest point ($T_H$) are given by: $$T_L = \frac{mv_L^2}{r} + mg$$ $$T_H = \frac{mv_H^2}{r} - mg$$ Using the principle of conservation of energy, $v_L^2 = v_H^2 + 4gr$. Substituting this into the tension difference equation: $$T_L - T_H = \frac{m}{r}(v_L^2 - v_H^2) + 2mg = \frac{m}{r}(4gr) + 2mg = 4mg + 2mg = 6mg$$ Correct Answer: (C) 6 mg
ii. The body is rotating with uniform angular velocity ($\omega$) having rotational kinetic energy (E). Its angular momentum (L) is:
Explanation: Rotational Kinetic Energy $E = \frac{1}{2}I\omega^2$ Angular Momentum $L = I\omega \Rightarrow I = \frac{L}{\omega}$ Substitute I in E: $$E = \frac{1}{2}\left(\frac{L}{\omega}\right)\omega^2 = \frac{1}{2}L\omega$$ $$\therefore L = \frac{2E}{\omega}$$ Correct Answer: (A) \(\frac{2E}{\omega}\)
iii. The S.I. unit of compressibility is ______.
Explanation: Compressibility is the reciprocal of Bulk Modulus (K). The SI unit of Bulk Modulus is $N/m^2$ (Pascal). Therefore, the unit of compressibility is the inverse: $m^2/N$ or $Pa^{-1}$. Correct Answer: (C) \(\frac{m^2}{N}\)
iv. The working of RADAR is based on ______.
Explanation: RADAR (Radio Detection And Ranging) uses the principle of reflection of radio waves and the Doppler effect to determine the velocity of moving objects. Correct Answer: (C) Doppler effect
v. If two capillary tubes of different diameters are partially dipped in the same liquid vertically, then the rise of liquid ______.
Explanation: The height of capillary rise is given by $h = \frac{2T\cos\theta}{r\rho g}$. Thus, $h \propto \frac{1}{r}$ (inversely proportional to radius/diameter). A smaller diameter means a larger rise. Correct Answer: (D) is more in the tube of smaller diameter.
vi. A sonometer wire vibrates with three nodes and two antinodes, the corresponding mode of vibration is ______.
Explanation: Ends are always nodes in a sonometer. Nodes = 3, Antinodes = 2 implies the pattern: N - A - N - A - N. This represents 2 loops. 1 loop = Fundamental mode. 2 loops = First overtone (or Second harmonic). Correct Answer: (A) First overtone
vii. Two gases exert pressure in the ratio 3:2 and their densities are in the ratio 2:3, then the ratio of their R.M.S. velocities is ______.
Explanation: RMS velocity $v_{rms} = \sqrt{\frac{3P}{\rho}}$. Ratio $\frac{(v_{rms})_1}{(v_{rms})_2} = \sqrt{\frac{P_1}{P_2} \times \frac{\rho_2}{\rho_1}}$ Given: $\frac{P_1}{P_2} = \frac{3}{2}$ and $\frac{\rho_1}{\rho_2} = \frac{2}{3} \Rightarrow \frac{\rho_2}{\rho_1} = \frac{3}{2}$. Ratio = $\sqrt{\frac{3}{2} \times \frac{3}{2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$. Correct Answer: (B) 3 : 2

HSC Physics Board Papers with Solution

Q.2. Attempt any SIX :

i. Draw a neat labelled diagram showing the various forces and their components acting on a vehicle moving along curved banked road.
[Diagram of a vehicle on a banked road inclined at angle $\theta$]
Key components to draw:
1. Weight ($mg$) acting vertically downwards.
2. Normal Reaction ($N$) perpendicular to the road surface.
3. Components of $N$: $N\cos\theta$ (vertical, balances components of mg and friction) and $N\sin\theta$ (horizontal towards center).
4. Force of friction ($f_s$) along the road surface downwards.
5. Components of friction: $f_s\cos\theta$ (horizontal) and $f_s\sin\theta$ (vertical).
ii. Obtain an expression for critical velocity of a satellite orbiting around the earth.
Consider a satellite of mass $m$ revolving around the Earth (mass $M$) in a circular orbit of radius $r$ (where $r = R + h$). The necessary centripetal force is provided by the gravitational force of attraction between the Earth and the satellite. $$F_{CP} = F_G$$ $$\frac{mv_c^2}{r} = \frac{GMm}{r^2}$$ $$v_c^2 = \frac{GM}{r}$$ $$v_c = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{R+h}}$$ This is the expression for critical velocity.
iii. Draw a neat labelled diagram of rise of liquid in capillary tube showing different components of tension (force).
[Diagram of a capillary tube dipped in liquid showing concave meniscus]
Key components to draw:
1. Tangent to the meniscus at the point of contact.
2. Surface Tension force ($T$) acting along the tangent.
3. Angle of contact ($\theta$).
4. Horizontal component: $T\sin\theta$ (acting outwards, cancelled by opposite side).
5. Vertical component: $T\cos\theta$ (acting upwards, responsible for rise).
6. Reaction force acting on the glass wall.
iv. State any four assumptions of kinetic theory of gases.
1. A gas consists of a large number of extremely small particles called molecules.
2. The molecules are in a state of constant, random motion, colliding with each other and the walls of the container.
3. The collisions between molecules and with the walls are perfectly elastic (kinetic energy is conserved).
4. The volume of the molecules themselves is negligible compared to the total volume of the gas.
5. There are no intermolecular forces of attraction or repulsion between the molecules (except during collision).
v. A tube open at both ends has length 47 cm. Calculate the fundamental frequency of air column. (Neglect end correction. Speed of sound in air is 3.3 × 10² m/s).
Given:
Length of tube ($L$) = 47 cm = 0.47 m
Velocity of sound ($v$) = \(3.3 \times 10^2\) m/s = 330 m/s
Formula: For a tube open at both ends, fundamental frequency $n = \frac{v}{2L}$.
Calculation:
$$n = \frac{330}{2 \times 0.47}$$ $$n = \frac{330}{0.94}$$ $$n \approx 351.06 \text{ Hz}$$ Answer: The fundamental frequency is approximately 351.06 Hz.
vi. A uniform solid sphere has a radius 0.1 m and density 6 × 10³ kg/m³. Find its moment of inertia about a tangent to its surface.
Given:
Radius $R = 0.1$ m
Density $\rho = 6 \times 10^3 \text{ kg/m}^3$
Step 1: Calculate Mass (M)
$$M = \text{Volume} \times \text{Density} = \frac{4}{3}\pi R^3 \times \rho$$ $$M = \frac{4}{3} \times 3.142 \times (0.1)^3 \times 6000$$ $$M = \frac{4}{3} \times 3.142 \times 0.001 \times 6000$$ $$M = 4 \times 3.142 \times 2 = 25.136 \text{ kg}$$ Step 2: Calculate Moment of Inertia (I) about tangent
By parallel axis theorem: $I_{\text{tangent}} = I_{\text{diameter}} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$
$$I = 1.4 \times 25.136 \times (0.1)^2$$ $$I = 1.4 \times 25.136 \times 0.01$$ $$I \approx 0.3519 \text{ kg m}^2$$ Answer: Moment of inertia is 0.3519 kg m².
vii. A particle executes S.H.M. with a period of 10 seconds. Find the time in which its potential energy will be half of its total energy.
Given: $T = 10$ s. Condition: $PE = \frac{1}{2}TE$.
$PE = \frac{1}{2}kx^2$ and $TE = \frac{1}{2}kA^2$.
$$\frac{1}{2}kx^2 = \frac{1}{2} \left( \frac{1}{2}kA^2 \right) \Rightarrow x^2 = \frac{A^2}{2} \Rightarrow x = \frac{A}{\sqrt{2}}$$ Displacement equation starting from mean position: $x = A \sin(\omega t)$ where $\omega = \frac{2\pi}{T}$.
$$\frac{A}{\sqrt{2}} = A \sin\left(\frac{2\pi t}{10}\right)$$ $$\sin\left(\frac{\pi t}{5}\right) = \frac{1}{\sqrt{2}} = \sin\left(\frac{\pi}{4}\right)$$ $$\frac{\pi t}{5} = \frac{\pi}{4} \Rightarrow t = \frac{5}{4} = 1.25 \text{ s}$$ Answer: The time required is 1.25 seconds.
viii. A stone of mass 2 kg is whirled in a horizontal circle attached at the end of 1.5 m long string. If the string makes an angle of 30° with vertical, compute its period. (g = 9.8 m/s²)
This is a Conical Pendulum.
Given: $l = 1.5$ m, $\theta = 30^\circ$, $g = 9.8 \text{ m/s}^2$. (Mass is not required for period).
Formula: $T = 2\pi \sqrt{\frac{l \cos\theta}{g}}$
Calculation:
$$T = 2\pi \sqrt{\frac{1.5 \times \cos 30^\circ}{9.8}}$$ $$T = 2 \times 3.142 \times \sqrt{\frac{1.5 \times 0.866}{9.8}}$$ $$T = 6.284 \times \sqrt{\frac{1.299}{9.8}} = 6.284 \times \sqrt{0.13255}$$ $$T = 6.284 \times 0.364 = 2.287 \text{ s}$$ Answer: The period is approximately 2.29 seconds.

Q.3. Attempt any THREE:

i. State Kepler’s laws of planetary motion.
1. Law of Orbit: Every planet revolves around the Sun in an elliptical orbit with the Sun situated at one of the foci.
2. Law of Area: The radius vector drawn from the Sun to the planet sweeps out equal areas in equal intervals of time (i.e., areal velocity is constant).
3. Law of Period: The square of the period of revolution of a planet around the Sun is directly proportional to the cube of the semimajor axis of the elliptical orbit ($T^2 \propto r^3$).
ii. Obtain an expression for torque acting on a body rotating with uniform angular acceleration.
Consider a rigid body rotating about an axis with uniform angular acceleration $\alpha$.
Let the body consist of $n$ particles of masses $m_1, m_2, ..., m_n$ at distances $r_1, r_2, ..., r_n$ from the axis.
The linear acceleration of particle 1 is $a_1 = r_1 \alpha$.
Force on particle 1: $f_1 = m_1 a_1 = m_1 r_1 \alpha$.
Torque on particle 1: $\tau_1 = f_1 r_1 = (m_1 r_1 \alpha) r_1 = m_1 r_1^2 \alpha$.
Similarly, $\tau_2 = m_2 r_2^2 \alpha$, etc.
Total torque $\tau = \tau_1 + \tau_2 + ... + \tau_n$
$\tau = (m_1 r_1^2 + m_2 r_2^2 + ... + m_n r_n^2) \alpha$
$\tau = (\sum m_i r_i^2) \alpha$
Since $\sum m_i r_i^2 = I$ (Moment of Inertia),
$$\tau = I \alpha$$
iii. A steel wire having cross-sectional area 2 mm² is stretched by 10 N. Find the lateral strain produced in the wire. (Given : Y for steel = \(2 \times 10^{11} \text{ N/m}^2\), Poisson’s ratio \(\sigma = 0.29\))
Given:
Area $A = 2 \text{ mm}^2 = 2 \times 10^{-6} \text{ m}^2$
Force $F = 10 \text{ N}$
Young's Modulus $Y = 2 \times 10^{11} \text{ N/m}^2$
Poisson's ratio $\sigma = 0.29$
Step 1: Calculate Longitudinal Stress
Stress = $F/A = 10 / (2 \times 10^{-6}) = 5 \times 10^6 \text{ N/m}^2$
Step 2: Calculate Longitudinal Strain
Strain$_{long}$ = Stress / Y = $(5 \times 10^6) / (2 \times 10^{11}) = 2.5 \times 10^{-5}$
Step 3: Calculate Lateral Strain
$\sigma = \text{Lateral Strain} / \text{Longitudinal Strain}$
Lateral Strain = $\sigma \times \text{Strain}_{long}$
Lateral Strain = $0.29 \times 2.5 \times 10^{-5}$
Lateral Strain = $0.725 \times 10^{-5}$ or $7.25 \times 10^{-6}$.
iv. A body cools from 62°C to 54°C in 10 minutes and to 48°C in the next 10 minutes. Find the temperature of the surroundings.
Using Newton's Law of Cooling: $\frac{d\theta}{dt} = K(\theta_{avg} - \theta_0)$
Let $\theta_0$ be the surrounding temperature.
Case 1 (62 to 54 in 10 min):
Rate = $(62 - 54) / 10 = 0.8$
Avg Temp = $(62 + 54) / 2 = 58$
Eq 1: $0.8 = K(58 - \theta_0)$

Case 2 (54 to 48 in 10 min):
Rate = $(54 - 48) / 10 = 0.6$
Avg Temp = $(54 + 48) / 2 = 51$
Eq 2: $0.6 = K(51 - \theta_0)$

Divide Eq 1 by Eq 2:
$$\frac{0.8}{0.6} = \frac{58 - \theta_0}{51 - \theta_0}$$ $$\frac{4}{3} = \frac{58 - \theta_0}{51 - \theta_0}$$ $$4(51 - \theta_0) = 3(58 - \theta_0)$$ $$204 - 4\theta_0 = 174 - 3\theta_0$$ $$\theta_0 = 204 - 174 = 30$$ Answer: The temperature of the surroundings is 30°C.

Q.4. A. Explain the formation of stationary wave by analytical method. Show that nodes and antinodes are equally spaced in a stationary wave.

Consider two identical simple harmonic progressive waves travelling in opposite directions along the same straight line.
$y_1 = a \sin \omega(t - x/v)$ and $y_2 = a \sin \omega(t + x/v)$
According to superposition principle, resultant displacement $y = y_1 + y_2$.
Using trigonometry ($\sin C + \sin D = 2 \sin((C+D)/2) \cos((C-D)/2)$):
$$y = 2a \sin(\omega t) \cos(\omega x / v)$$ $$y = 2a \sin(\omega t) \cos(2\pi x / \lambda)$$ Let Resultant Amplitude $A = 2a \cos(2\pi x / \lambda)$.
Then $y = A \sin(\omega t)$. This represents a stationary wave.

Nodes: Points of zero amplitude ($A=0$).
$\cos(2\pi x / \lambda) = 0 \Rightarrow 2\pi x / \lambda = \pi/2, 3\pi/2, ...$
$x = \lambda/4, 3\lambda/4, 5\lambda/4, ...$
Distance between consecutive nodes = $3\lambda/4 - \lambda/4 = \lambda/2$.

Antinodes: Points of maximum amplitude ($A = \pm 2a$).
$\cos(2\pi x / \lambda) = \pm 1 \Rightarrow 2\pi x / \lambda = 0, \pi, 2\pi, ...$
$x = 0, \lambda/2, \lambda, ...$
Distance between consecutive antinodes = $\lambda/2 - 0 = \lambda/2$.

Conclusion: Nodes and antinodes are equally spaced with distance $\lambda/2$.

Q.4. B. The speed limit for a vehicle on road is 120 km/hr. A policeman detects a drop of 10% in the pitch of horn of a car as it passes him. Is the policeman justified in punishing the car driver for crossing the speed limit? (Velocity of sound = 340 m/s)

Given: $v = 340 \text{ m/s}$. Drop in pitch is 10%.
Let actual frequency be $n_0$.
Apparent frequency approaching ($n_a$) = $n_0 (\frac{v}{v - v_s})$
Apparent frequency receding ($n_r$) = $n_0 (\frac{v}{v + v_s})$
The "drop in pitch as it passes" implies the receding frequency is 10% lower than the approaching frequency. Thus $n_r = 0.9 n_a$. (Drop of 10% means 90% remains).
$$n_0 \left(\frac{v}{v + v_s}\right) = 0.9 \times n_0 \left(\frac{v}{v - v_s}\right)$$ $$\frac{1}{v + v_s} = \frac{0.9}{v - v_s}$$ $$v - v_s = 0.9v + 0.9v_s$$ $$0.1v = 1.9v_s$$ $$v_s = \frac{0.1 \times 340}{1.9} = \frac{34}{1.9} \approx 17.89 \text{ m/s}$$ Convert to km/hr: $17.89 \times 3.6 \approx 64.4 \text{ km/hr}$.
Conclusion: The speed of the car is 64.4 km/hr, which is much less than the limit of 120 km/hr. The policeman is not justified.
OR

Q.4. A. Define practical simple pendulum. Show that motion of bob of pendulum with small amplitude is linear S.H.M. Hence obtain an expression for its period. What are the factors on which its period depends?

Definition: A heavy point mass suspended from a rigid support by a weightless, inextensible, and perfectly flexible string is called a simple pendulum.
Proof of SHM:
Forces on bob at angle $\theta$: Weight $mg$ and Tension $T$.
Resolve $mg$: $mg\cos\theta$ balances $T$. $mg\sin\theta$ provides restoring force.
$F = -mg\sin\theta$.
For small amplitudes, $\sin\theta \approx \theta = x/L$.
$F = -mg(x/L)$. Since $m, g, L$ are constant, $F \propto -x$. Thus, motion is linear SHM.
Period Expression:
Force constant $k = mg/L$.
$\omega^2 = k/m = g/L$.
$T = 2\pi/\omega = 2\pi\sqrt{L/g}$.
Factors: Period depends on Length ($L$) and Acceleration due to gravity ($g$). It is independent of mass and amplitude (for small $\theta$).

Q.4. B. The total free surface energy of a liquid drop is \(\pi\sqrt{2}\) times the surface tension of the liquid. Calculate the diameter of the drop in S.I. unit.

Given: Surface Energy $E = \pi\sqrt{2} \times T$.
Formula for Surface Energy of a drop: $E = T \times \Delta A$.
Area of drop $A = 4\pi r^2$.
$$T(4\pi r^2) = \pi\sqrt{2} T$$ $$4r^2 = \sqrt{2}$$ $$r^2 = \frac{\sqrt{2}}{4}$$ $$r = \frac{2^{1/4}}{2}$$ $$r = \frac{1.189}{2} = 0.5946 \text{ m}$$ Diameter $D = 2r = 2 \times \frac{2^{1/4}}{2} = 2^{1/4} = \sqrt[4]{2}$ m.
$\sqrt[4]{2} \approx 1.189$ m.
Answer: Diameter is 1.189 m.

SECTION – II

Q.5. Select and write the most appropriate answer from the given alternatives for each sub-question:

i. A parallel beam of light travelling in water is incident obliquely on a glass surface. After refraction its width ______.
Explanation: Light travels from water (rarer) to glass (denser). The ray bends towards the normal ($i > r$). Relationship between beam widths: $w' = w \frac{\cos r}{\cos i}$. Since $r < i$ (in range 0 to 90), $\cos r > \cos i$. Therefore, the ratio $> 1$. The width increases. Correct Answer: (B) increases
ii. If ‘a’ is the aperture of telescope and ‘\(\lambda\)’ is the wavelength of light then resolving power of telescope is ______.
Explanation: Resolving power is the reciprocal of the limit of resolution ($d\theta = 1.22\lambda/a$). $RP = \frac{a}{1.22\lambda}$. Correct Answer: (D) \(\frac{a}{1.22\lambda}\) (Note: The option in the image is (D) \( \frac{a}{1.22\lambda} \). The OCR text might have typos, but formula is standard).
iii. From earth’s surface, ionospheric layer of atmosphere lies between ______.
Explanation: The ionosphere extends roughly from 80 km to 400 km above the Earth's surface. Correct Answer: (C) 80 km to 400 km
iv. The kinetic energy of emitted photoelectrons is independent of ______.
Explanation: According to Einstein's photoelectric equation, max KE depends on frequency and work function. It is independent of the intensity of incident light (intensity affects the number of electrons, not their energy). Correct Answer: (B) intensity of incident radiation.
v. In hydrogen atom Balmer series in obtained when the electron jumps from ______.
Explanation: Balmer series lines correspond to transitions from higher energy levels ($n > 2$) to the second orbit ($n = 2$). Correct Answer: (C) higher orbit to the second orbit
vi. The fraction of the total current passing through the galvanometer is ______.
Explanation: For a galvanometer with resistance G and shunt S connected in parallel: $I_g G = (I - I_g) S$. $I_g (G+S) = I S$. $\frac{I_g}{I} = \frac{S}{S+G}$. Correct Answer: (A) \(\frac{S}{S+G}\)
vii. A meter gauge train is heading north with speed 54 km/hr in earth’s magnetic field \(3 \times 10^{-4}\) T. The e.m.f. induced across the axle joining the wheels is ______.
Explanation: $l = 1$ m. $v = 54$ km/hr = 15 m/s. $B_v = 3 \times 10^{-4}$ T (Vertical component is cut). $e = Bvl = 3 \times 10^{-4} \times 15 \times 1 = 45 \times 10^{-4}$ V = 4.5 mV. Correct Answer: (B) 4.5 mV

Q.6. Attempt any SIX:

i. Distinguish between intrinsic and extrinsic semiconductor. (Give any two points).
1. Purity: Intrinsic semiconductors are pure semiconductor crystals (e.g., pure Si or Ge). Extrinsic semiconductors are doped with impurities (e.g., Si doped with As or B).
2. Conductivity: Intrinsic has low electrical conductivity at room temperature. Extrinsic has significantly higher conductivity due to added charge carriers.
ii. Draw the block diagram of a receiver in communication system.
[Block Diagram Sequence]:
Receiving Antenna $\rightarrow$ Tuner/Filter $\rightarrow$ Amplifier (RF) $\rightarrow$ Demodulator/Detector $\rightarrow$ Amplifier (AF) $\rightarrow$ Output (Loudspeaker).
iii. A point is situated at 6.5 cm and 6.65 cm from two coherent sources. Find the nature of illumination at the point, if wavelength of light is 5000 Å.
Given: $x_1 = 6.5$ cm, $x_2 = 6.65$ cm, $\lambda = 5000 \text{ Å} = 5 \times 10^{-5}$ cm.
Path Difference ($\Delta x$): $6.65 - 6.5 = 0.15$ cm.
Check if $\Delta x$ is an integral multiple of $\lambda$:
$n = \frac{\Delta x}{\lambda} = \frac{0.15}{5 \times 10^{-5}} = \frac{15 \times 10^{-2}}{5 \times 10^{-5}} = 3 \times 10^3 = 3000$.
Since $n$ is an integer, constructive interference occurs.
Answer: The point will be bright (maximum illumination).
iv. Draw the diagrams showing the dipole moments in paramagnetic substance when external magnetic field is (a) absent (b) strong.
(a) Field Absent: Atomic dipoles are randomly oriented in all directions. Net magnetic moment is zero.
(b) Strong Field: Most atomic dipoles align parallel to the direction of the external magnetic field.
v. A voltmeter has a resistance of 100 \(\Omega\). What will be its reading when it is connected across a cell of e.m.f. 2 V and internal resistance 20 \(\Omega\)?
Given: $R_v = 100 \Omega$, $E = 2$ V, $r = 20 \Omega$.
Total Resistance in circuit = $R_v + r = 100 + 20 = 120 \Omega$.
Current $I = \frac{E}{R_{total}} = \frac{2}{120} = \frac{1}{60}$ A.
Voltmeter reading (Terminal Voltage) $V = I \times R_v$.
$V = \frac{1}{60} \times 100 = \frac{10}{6} = 1.67$ V.
Answer: Reading is 1.67 V.
vi. The susceptibility of magnesium at 300 K is \(1.2 \times 10^{-5}\). At what temperature will the susceptibility increase to \(1.8 \times 10^{-5}\)?
Using Curie's Law for paramagnetic substances: $\chi \propto \frac{1}{T}$.
$\chi_1 T_1 = \chi_2 T_2$.
$(1.2 \times 10^{-5}) \times 300 = (1.8 \times 10^{-5}) \times T_2$.
$T_2 = \frac{1.2 \times 300}{1.8} = \frac{2}{3} \times 300 = 200$ K.
Answer: Temperature is 200 K.
vii. What is de Broglie wavelength of an electron accelerated through 25000 volt?
Formula: $\lambda = \frac{12.27}{\sqrt{V}} \text{ Å}$.
$\lambda = \frac{12.27}{\sqrt{25000}} = \frac{12.27}{158.11}$.
$\lambda \approx 0.0776 \text{ Å}$.
Answer: The wavelength is 0.0776 Å.
viii. Draw the schematic symbols for AND, OR, NOT and NAND gate.
[Draw standard logic symbols]
AND: D-shape.
OR: Curved, arrow-head shape.
NOT: Triangle with a bubble at tip.
NAND: AND gate with a bubble at output.

Q.7. Attempt any THREE:

i. Using analytical method for interference bands, obtain an expression for path difference between two light waves.
Consider two coherent sources $S_1$ and $S_2$ separated by distance $d$. Screen is at distance $D$. Point P is at distance $x$ from the central bright fringe. Path difference $\Delta l = S_2P - S_1P$. From geometry: $(S_2P)^2 - (S_1P)^2 = (D^2 + (x+d/2)^2) - (D^2 + (x-d/2)^2) = 2xd$. $(S_2P - S_1P)(S_2P + S_1P) = 2xd$. Approximation: $S_2P + S_1P \approx 2D$. $\Delta l (2D) = 2xd$. Path difference $\Delta l = \frac{xd}{D}$.
ii. State law of radioactive decay. Hence derive the relation \(N = N_0e^{-\lambda t}\). Represent it graphically.
Law: The rate of decay of radioactive atoms at any instant is directly proportional to the number of atoms present at that instant. $$\frac{dN}{dt} \propto -N \Rightarrow \frac{dN}{dt} = -\lambda N$$ $\frac{dN}{N} = -\lambda dt$ Integrating both sides: $\ln N = -\lambda t + C$. At $t=0, N=N_0 \Rightarrow C = \ln N_0$. $\ln N - \ln N_0 = -\lambda t$. $\ln(N/N_0) = -\lambda t$. $N = N_0 e^{-\lambda t}$. Graph: An exponential decay curve starting at $N_0$ on y-axis and approaching 0 asymptotically on x-axis.
iii. Determine the change in wavelength of light during its passage from air to glass, if the refractive index of glass with respect to air is 1.5 and the frequency of light is \(5 \times 10^{14}\) Hz. Find the wave number of light in glass (velocity of light in air c = \(3 \times 10^8\) m/s).
Given: $\mu = 1.5$, $\nu = 5 \times 10^{14}$ Hz, $c = 3 \times 10^8$ m/s.
Wavelength in air $\lambda_a = \frac{c}{\nu} = \frac{3 \times 10^8}{5 \times 10^{14}} = 0.6 \times 10^{-6} \text{ m} = 6000 \text{ Å}$.
Wavelength in glass $\lambda_g = \frac{\lambda_a}{\mu} = \frac{6000}{1.5} = 4000 \text{ Å}$.
Change in wavelength $\Delta \lambda = \lambda_a - \lambda_g = 6000 - 4000 = 2000 \text{ Å}$.
Wave number in glass $\bar{\nu} = \frac{1}{\lambda_g} = \frac{1}{4000 \times 10^{-10}} = \frac{10^7}{4} = 2.5 \times 10^6 \text{ m}^{-1}$.
iv. Light of wavelength 3000 Å falls on a metal surface having work function 2.3 eV. Calculate the maximum velocity of ejected electrons. (Planck’s constant h = \(6.63 \times 10^{-34}\) Js, Velocity of light c = \(3 \times 10^8\) m/s, mass of an electron = \(9.1 \times 10^{-31}\) kg)
Energy of photon (E): $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3000 \times 10^{-10}}$ $E = \frac{19.89 \times 10^{-26}}{3 \times 10^{-7}} = 6.63 \times 10^{-19}$ J.
Work Function ($\phi$): $\phi = 2.3 \text{ eV} = 2.3 \times 1.6 \times 10^{-19} = 3.68 \times 10^{-19}$ J.
Max Kinetic Energy ($KE_{max}$): $KE_{max} = E - \phi = (6.63 - 3.68) \times 10^{-19} = 2.95 \times 10^{-19}$ J.
Velocity ($v$): $\frac{1}{2}mv^2 = 2.95 \times 10^{-19}$ $v = \sqrt{\frac{2 \times 2.95 \times 10^{-19}}{9.1 \times 10^{-31}}}$ $v = \sqrt{0.648 \times 10^{12}} \approx 0.805 \times 10^6 \text{ m/s}$.

Q.8. A. What is electromagnetic induction? Prove theoretically \(e = \frac{d\phi}{dt}\).

Definition: The phenomenon of production of induced EMF (and current) in a coil due to the change in magnetic flux linked with it is called electromagnetic induction.
Proof: Consider a rectangular loop of width $l$ moving out of a uniform magnetic field $B$ with velocity $v$. Flux $\phi = B \cdot A = B \cdot (lx)$. Rate of change of flux $\frac{d\phi}{dt} = \frac{d}{dt}(Blx) = Bl \frac{dx}{dt} = Blv$. We know induced EMF in a conductor moving in magnetic field is $e = Blv$. Comparing, magnitude $|e| = \frac{d\phi}{dt}$. By Lenz's law, direction opposes change, so $e = -\frac{d\phi}{dt}$.

Q.8. B. A potentiometer wire has length of 2 m and resistance 10 \(\Omega\). It is connected in series with resistance 990 \(\Omega\) and a cell of e.m.f. 2 V. Calculate the potential gradient along the wire.

Given: $L = 2$ m, $R_{wire} = 10 \Omega$, $R_{series} = 990 \Omega$, $E = 2$ V.
Total Resistance = $10 + 990 = 1000 \Omega$.
Current $I = \frac{E}{R_{total}} = \frac{2}{1000} = 0.002$ A.
Potential Difference across wire $V_{wire} = I \times R_{wire} = 0.002 \times 10 = 0.02$ V.
Potential Gradient $K = \frac{V_{wire}}{L} = \frac{0.02}{2} = 0.01$ V/m.
Answer: Potential gradient is 0.01 V/m.
OR

Q.8. A. With the help of a neat diagram, describe the construction and working of van de Graff generator.

[Diagram of Van de Graaff generator showing large dome, belt, pulleys, spray comb, collector comb]
Construction: Consists of a large hollow metal sphere mounted on insulating pillars. An insulating belt moves over two pulleys. Lower comb is connected to high voltage source (spray), upper comb connected to the sphere (collector).
Working: 1. Electric discharge at pointed conductor (Corona discharge) sprays positive charge onto the belt at the bottom. 2. Belt carries charge to the top. 3. Collector comb collects charge and transfers it to the outer surface of the sphere. 4. Potential rises to millions of volts ($V = Q/C$).

Q.8. B. A moving coil galvanometer has a resistance of 25 \(\Omega\) and gives a full scale deflection for a current of 10 mA. How will you convert it into a voltmeter having range 0 – 100 V?

To convert galvanometer to voltmeter, a high resistance ($R_s$) is connected in series.
Given: $G = 25 \Omega$, $I_g = 10 \text{ mA} = 0.01$ A, $V = 100$ V.
Formula: $R_s = \frac{V}{I_g} - G$.
$R_s = \frac{100}{0.01} - 25$
$R_s = 10000 - 25 = 9975 \Omega$.
Answer: Connect a resistance of 9975 \(\Omega\) in series with the galvanometer.
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