Note: This answer key provides solutions to selected questions from the Physics paper. Students should attempt any 5 questions from Section 1, any 5 from Section 2, any 6 from Section 3, and any 3 from Section 5 as per instructions.
Section 1: Questions 1–7 (Choose any 5, 1 mark each)
Question 1: Significant Figures
Find the number of significant figures in:
Answer:
(a) 0.0006032 → 4 significant figures (zeros before the first non-zero digit are not significant)
(b) 6.3200 → 5 significant figures (trailing zeros after decimal are significant)
Question 2: Velocity-Time Graph
The velocity-time graph for a particle moving on a straight line is shown in the figure. Choose the correct statement:
Answer:
Option (a): The particle has a constant acceleration
The graph shows a straight line with constant slope, which indicates constant acceleration.
Question 3: Angular Velocity in Circular Motion
If a particle executes uniform circular motion in the x-y plane in clockwise direction, then the angular velocity is in:
Answer:
Option (c): -z direction
Clockwise circular motion in the x-y plane has angular velocity pointing into the plane (right-hand rule), which is the -z direction.
Question 4: Work Done by Conservative Force
Work done by the conservative force for a closed path is:
Answer:
Option (b): Zero
For a closed path, the work done by a conservative force is zero because it depends only on initial and final positions.
Question 5: Centre of Mass
The centre of mass of a system of particles does not depend upon:
Answer:
Option (d): Forces acting on particle
The centre of mass depends on mass distribution and positions of particles, not on forces acting on them.
Question 6: Kinetic Energy in Elliptical Orbit
The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are \( K_A \), \( K_B \) and \( K_C \) respectively. Choose the correct relation:
Answer:
Option (d): \( K_B > K_A > K_C \)
In an elliptical orbit, kinetic energy is maximum at perihelion (closest to Sun) and minimum at aphelion (farthest from Sun).
Question 7: Wetting of a Surface
The ability of a liquid to wet a surface depend primarily on:
Answer:
Option (b): Angle of contact between surface and liquid
Wetting depends on adhesive forces between liquid and surface, characterized by the contact angle.
Section 2: Questions 8–14 (Choose any 5, 2 marks each)
Question 8: Definitions
Define the following: a) Average acceleration b) Instantaneous acceleration.
Answer:
(a) Average acceleration: Change in velocity divided by the time interval over which the change occurs.
\[ \text{Average acceleration} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i} \]
(b) Instantaneous acceleration: Acceleration at a particular instant of time, given by the derivative of velocity with respect to time.
\[ \text{Instantaneous acceleration} = \frac{dv}{dt} \]
Question 9: Body Released from String
If the string shown in the figure is released, the body will fly along the ______ (radius/tangent). Why?
Answer:
The body will fly along the tangent.
Reason: When released, the centripetal force (tension in string) ceases to act. Due to inertia, the body continues to move in the direction of its instantaneous velocity, which is tangent to the circular path.
Question 10: Inelastic Collision
a) What is inelastic collision? b) In which way it is different from elastic collision?
Answer:
(a) An inelastic collision is one where total kinetic energy is not conserved, but linear momentum is conserved. Some kinetic energy is converted to other forms like heat, sound, or deformation energy.
(b) Unlike elastic collisions where both momentum and kinetic energy are conserved, inelastic collisions conserve only momentum, not kinetic energy.
Question 11: Torque and Angular Momentum
Derive the relation between torque and angular momentum.
Answer:
Starting from the definition of angular momentum: \( \vec{L} = \vec{r} \times \vec{p} \)
Differentiating with respect to time:
\[ \frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p}) = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt} \]
Since \( \frac{d\vec{r}}{dt} = \vec{v} \) and \( \vec{v} \times \vec{p} = \vec{v} \times m\vec{v} = 0 \), we get:
\[ \frac{d\vec{L}}{dt} = \vec{r} \times \frac{d\vec{p}}{dt} = \vec{r} \times \vec{F} = \vec{\tau} \]
Thus, the relation is: \( \vec{\tau} = \frac{d\vec{L}}{dt} \)
Question 12: Acceleration Due to Gravity
a) Write down the expression for acceleration due to gravity of Earth. b) If both the mass and radius of the Earth are doubled, then the acceleration due to gravity g' is:
Answer:
(a) Expression for acceleration due to gravity:
\[ g = \frac{GM}{R^2} \]
where G = gravitational constant, M = mass of Earth, R = radius of Earth.
(b) If mass and radius are doubled: \( M' = 2M \), \( R' = 2R \)
\[ g' = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{1}{2} \cdot \frac{GM}{R^2} = \frac{g}{2} \]
So the correct option is (ii) g/2.
Question 13: Equation of Continuity
a) State equation of continuity. b) Water is flowing through a tube as shown in figure. Which among the following is true?
Answer:
(a) Equation of continuity: For steady flow of an incompressible fluid, the product of cross-sectional area and velocity remains constant.
\[ A_1 v_1 = A_2 v_2 \]
(b) From the figure, if \( A_A > A_B \), then according to the equation of continuity, \( v_A < v_B \).
So the correct option is (ii) \( V_A < V_B \)**.
Question 14: Stefan-Boltzmann Law
a) State Stefan-Boltzmann law. b) For a perfect radiator, emissivity = ______
Answer:
(a) Stefan-Boltzmann law states that the total energy radiated per unit surface area of a black body per unit time is proportional to the fourth power of its absolute temperature.
\[ E = \sigma T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, \text{W m}^{-2} \text{K}^{-4} \)).
(b) For a perfect radiator (black body), emissivity = 1.
Section 3: Questions 15–21 (Choose any 6, 3 marks each)
Question 15: Dimensional Analysis for Vibrating String
Assuming that the frequency \( v \) of a vibrating string may depend on i) Applied force (F) ii) Length (l) iii) Mass per unit length (\(\mu\)). Prove that \( v \propto \frac{1}{l} \sqrt{\frac{F}{\mu}} \) applying Principle of Homogeneity of dimensions.
Answer:
Let \( v \propto F^a l^b \mu^c \)
Dimensions: [v] = T⁻¹, [F] = MLT⁻², [l] = L, [μ] = ML⁻¹
Equating dimensions: T⁻¹ = (MLT⁻²)^a (L)^b (ML⁻¹)^c
This gives: M: a + c = 0; L: a + b - c = 0; T: -2a = -1
Solving: a = ½, c = -½, b = -1
\[ v \propto F^{1/2} l^{-1} \mu^{-1/2} \propto \frac{1}{l} \sqrt{\frac{F}{\mu}} \]
Hence proved.
Question 16: Projectile Motion
a) Derive an expression for horizontal range for a particle in projectile motion. b) Two objects are projected at angles \( 30^\circ \) and \( 60^\circ \) respectively with respect to the horizontal direction. The range of two objects are denoted as \( R_{30} \) and \( R_{60} \), choose the correct relation.
Answer:
(a) For a projectile launched with velocity u at angle θ:
Time of flight: \( T = \frac{2u \sin \theta}{g} \)
Horizontal range: \( R = u \cos \theta \times T = u \cos \theta \times \frac{2u \sin \theta}{g} \)
\[ R = \frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{u^2 \sin 2\theta}{g} \]
(b) For complementary angles (θ and 90°-θ), range is the same: \( R_\theta = R_{90°-\theta} \)
Since 30° and 60° are complementary: \( R_{30} = R_{60} \)
Correct option: (i) \( R_{30} = R_{60} \)**
Question 17: Conservation of Linear Momentum
a) State law of conservation of linear momentum. b) A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s. What is the recoil speed of the gun?
Answer:
(a) Law of conservation of linear momentum: In the absence of external forces, the total linear momentum of a system remains constant.
(b) Using conservation of momentum:
\[ m_{\text{shell}} \times v_{\text{shell}} = m_{\text{gun}} \times v_{\text{gun}} \]
\[ 0.020 \times 80 = 100 \times v_{\text{gun}} \]
\[ v_{\text{gun}} = \frac{1.6}{100} = 0.016 \, \text{m/s} \]
Recoil speed of gun = 0.016 m/s
Question 18: Couple and Moment
a) A couple produces i) Pure rotation ii) Pure translation iii) Rotation and translation iv) No motion. b) Derive principle of moments for the lever shown in figure.
Answer:
(a) A couple produces (i) Pure rotation.
(b) Principle of moments (for a lever in equilibrium):
Sum of clockwise moments about pivot = Sum of anticlockwise moments about pivot
\[ F_1 \times d_1 = F_2 \times d_2 \]
where \( d_1 \) and \( d_2 \) are perpendicular distances from pivot to lines of action of forces \( F_1 \) and \( F_2 \).
Section 4: Questions 23–25
Question 23: Work Done
a) Define work done by a force. b) Work done by frictional force is negative (True/False). c) Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m.
Answer:
(a) Work done by a force is defined as the product of the force and the displacement of the point of application in the direction of the force.
\[ W = \vec{F} \cdot \vec{s} = F s \cos \theta \]
where θ is the angle between force and displacement.
(b) True - Work done by frictional force is usually negative as it opposes motion (θ = 180°, cosθ = -1).
(c) Work done = Force × displacement = 30 N × 10 m = 300 J
Question 24: Young's Modulus
a) Arrive at an expression for Young's modulus of a material. b) One end of a wire of 2 m long and 0.2 cm² in cross section is fixed in a ceiling and a load of 4.8 kg is attached to the free end. Find the extension of the wire. Young's modulus of steel is \(2 \times 10^{11} \, \text{N/m}^2\). Take \(g = 10 \, \text{m/s}^2\).
Answer:
(a) Young's modulus (Y) = Stress / Strain
\[ Y = \frac{F/A}{\Delta L / L} = \frac{FL}{A \Delta L} \]
where F = applied force, A = cross-sectional area, L = original length, ΔL = extension.
(b) Given: L = 2 m, A = 0.2 cm² = 0.2 × 10⁻⁴ m² = 2 × 10⁻⁵ m², m = 4.8 kg, g = 10 m/s², Y = 2 × 10¹¹ N/m²
Force F = mg = 4.8 × 10 = 48 N
\[ \Delta L = \frac{FL}{AY} = \frac{48 \times 2}{(2 \times 10^{-5}) \times (2 \times 10^{11})} \]
\[ \Delta L = \frac{96}{4 \times 10^{6}} = 2.4 \times 10^{-5} \, \text{m} \]
Extension = 2.4 × 10⁻⁵ m or 0.024 mm
Question 25: Isothermal Process
a) If heat is supplied to an ideal gas in an isothermal process, which is true? b) Derive an expression for work done by the gas in an isothermal process.
Answer:
(a) (ii) The gas will do positive work
In an isothermal process, temperature is constant, so internal energy (which depends only on temperature for ideal gas) remains constant. Heat supplied equals work done by gas.
(b) For isothermal expansion of ideal gas:
From first law: dQ = dU + dW, but dU = 0 (constant T)
So dQ = dW
\[ W = \int_{V_i}^{V_f} P \, dV = \int_{V_i}^{V_f} \frac{nRT}{V} \, dV \]
\[ W = nRT \ln \frac{V_f}{V_i} \]
Section 5: Questions 26–29 (Choose any 3, 5 marks each)
Question 26: Vector Resultant
a) Arrive at an expression for the resultant of two vectors \(\vec{A}\) and \(\vec{B}\) making an angle \(\theta\) with each other. b) Two vectors \(\vec{A}\) and \(\vec{B}\) of magnitude 5 units and 7 units respectively make an angle \(60^\circ\) with each other. Find the magnitude of the resultant vector.
Answer:
(a) Using law of cosines:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \]
Direction: \( \tan \phi = \frac{B \sin \theta}{A + B \cos \theta} \), where φ is angle of resultant with A.
(b) A = 5 units, B = 7 units, θ = 60°
\[ R = \sqrt{5^2 + 7^2 + 2 \times 5 \times 7 \times \cos 60^\circ} \]
\[ R = \sqrt{25 + 49 + 35} = \sqrt{109} \approx 10.44 \, \text{units} \]
Question 27: Banked Road
a) Redraw the figure and mark the normal reaction and frictional force acting on the car taking a curve on a banked road. b) Arrive at an expression for the maximum safe speed the car can take while executing a curve on a banked road.
Answer:
(a) In the figure, mark:
- Normal reaction (N) perpendicular to road surface
- Frictional force (f) parallel to road surface, directed downward along the incline
- Weight (mg) vertically downward
(b) For maximum safe speed without skidding:
Balancing forces: \( N \sin \theta + f \cos \theta = \frac{mv^2}{r} \) (centripetal force)
and \( N \cos \theta - f \sin \theta = mg \) (vertical equilibrium)
Maximum frictional force: \( f_{\max} = \mu N \)
Solving these equations gives:
\[ v_{\max} = \sqrt{\frac{rg(\mu + \tan \theta)}{1 - \mu \tan \theta}} \]
For frictionless case (μ = 0): \( v = \sqrt{rg \tan \theta} \)
Question 28: Escape Velocity
a) Define escape velocity. b) Arrive at an expression for escape velocity of a body of mass m from the surface of earth. c) Calculate the escape velocity from moon. The mass of the moon = \( 7.4 \times 10^{22} \) kg and radius of the moon = \( 1740 \) km, G = \( 6.67 \times 10^{-11} \) Nm²/kg².
Answer:
(a) Escape velocity is the minimum velocity required for an object to escape the gravitational field of a planet/moon without any further propulsion.
(b) Using energy conservation: Initial KE + Initial PE = 0 (at infinity)
\[ \frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 \]
\[ v_e = \sqrt{\frac{2GM}{R}} \]
(c) For moon: M = 7.4 × 10²² kg, R = 1740 km = 1.74 × 10⁶ m
\[ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.4 \times 10^{22}}{1.74 \times 10^6}} \]
\[ v_e = \sqrt{\frac{9.8716 \times 10^{12}}{1.74 \times 10^6}} = \sqrt{5.673 \times 10^6} \approx 2382 \, \text{m/s} \]
Escape velocity from moon ≈ 2.38 km/s
Question 29: Terminal Velocity
a) Arrive at an expression for terminal velocity of a sphere of radius \( a \) and density \( \rho \) falling through a fluid of density \( \sigma \). b) The terminal velocity of copper ball of radius 2 mm falling through a tank of oil at \( 20^\circ C \) is 6.5 cm/s. Compute the viscosity of the oil at \( 20^\circ C \). Density of oil is \( 1.5 \times 10^3 \) kg/m³, density of copper is \( 8.9 \times 10^3 \) kg/m³.
Answer:
(a) At terminal velocity, net force = 0:
Weight = Buoyant force + Viscous drag
\[ \frac{4}{3}\pi a^3 \rho g = \frac{4}{3}\pi a^3 \sigma g + 6\pi \eta a v_t \]
Solving for \( v_t \):
\[ v_t = \frac{2a^2 (\rho - \sigma) g}{9\eta} \]
(b) Given: a = 2 mm = 2 × 10⁻³ m, ρ = 8.9 × 10³ kg/m³, σ = 1.5 × 10³ kg/m³, v_t = 6.5 cm/s = 6.5 × 10⁻² m/s, g = 9.8 m/s²
\[ \eta = \frac{2a^2 (\rho - \sigma) g}{9v_t} \]
\[ \eta = \frac{2 \times (2 \times 10^{-3})^2 \times (8.9 - 1.5) \times 10^3 \times 9.8}{9 \times 6.5 \times 10^{-2}} \]
\[ \eta = \frac{2 \times 4 \times 10^{-6} \times 7.4 \times 10^3 \times 9.8}{0.585} \approx 0.99 \, \text{Pa·s} \]
Viscosity of oil ≈ 0.99 Pa·s
Important Note: Some figures and diagrams referenced in the original question paper are not reproduced here. For complete understanding, refer to the diagrams in the question paper. This answer key assumes standard physics principles and may have rounding differences in numerical answers.
