CBSE Class 11 Physics Question Paper Set 15 with Solutions

Answer: (b) 4.9 × 10⁻¹
The block is at rest, so friction balances the component of weight along the incline: f = mg sin 30° = 0.1 × 9.8 × 0.5 = 0.49 N.

Teacher's Note:
1. For a body at rest on an incline, friction equals mg sin θ, not μmg cos θ.
2. Remember to convert grams to kilograms before substituting.

Answer: (b) 70 J
By the first law, ΔQ = ΔU + W, so W = 110 − 40 = 70 J.

Teacher's Note:
1. Write the first law as ΔQ = ΔU + W, where W is work done by the gas.
2. Check the sign convention: heat added is positive and work done by the system is positive.

Answer: (a) The escape velocity on the Moon (about 2.4 km/s) is small, so the gas molecules, whose rms speeds are larger, escape into space.

Teacher's Note:
1. Link the escape velocity of a planet with the molecular speeds of gases.
2. Smoothness of the surface and distance from the earth have no role.

Answer: (c) Viscous force
Work done by viscous force depends on the path and dissipates energy as heat; the other three are conservative.

Teacher's Note:
1. A conservative force does zero work over a closed path.
2. Friction and viscous forces are the standard non-conservative examples.

Answer: (b) 4W
Work = surface tension × increase in surface area. Volume ∝ r³, so a volume 8V has radius 2r. Area ∝ r² becomes 4 times, so the work is 4W.

Teacher's Note:
1. Work done in forming a bubble depends on its surface area, not its volume.
2. Find how the radius changes first, then square it.

Answer: (b) ΔQ = 0
In an adiabatic process no heat enters or leaves the system.

Teacher's Note:
1. Match each process with its condition: isothermal ΔT = 0, isochoric ΔV = 0, adiabatic ΔQ = 0.
2. Do not confuse adiabatic with isothermal.

Answer: (d) (0, 1)
The centre of mass of a uniform square lies at the midpoint of its diagonal: ((−2 + 2)/2, (0 + 2)/2) = (0, 1).

Teacher's Note:
1. For a symmetric uniform body, the centre of mass is at the geometric centre.
2. The midpoint formula is the quickest route.

Answer: (a) P = P₀ + ρgh
Pressure increases by ρgh over the surface pressure P₀ at depth h.

Teacher's Note:
1. This is the basic hydrostatic pressure relation.
2. P = ρgh alone is only the gauge pressure.

Answer: (d) 18.75 N m
W = ½k(x₂² − x₁²) = ½ × 5000 × (0.10² − 0.05²) = 2500 × 0.0075 = 18.75 J.

Teacher's Note:
1. Use the difference of the stored energies at the two extensions, not ½k(0.05)².
2. Convert centimetres to metres first.

Answer: (b) 1 m/s
½Iω² = ½ × 3 × 4 = 6 J. Then ½ × 12 × v² = 6 gives v² = 1, so v = 1 m/s.

Teacher's Note:
1. Equate rotational kinetic energy ½Iω² with translational kinetic energy ½mv².
2. Do not forget to square ω and v.

Answer: (a) F/4
F = Gm₁m₂/r². New force = G(m₁/2)(m₂/2)/r² = F/4.

Teacher's Note:
1. The force depends on the product of the masses, so each halving counts.
2. The distance is unchanged here.

Answer: (d) conservation of mass
By the equation of continuity, A₁v₁ = A₂v₂, so a smaller cross-section means a larger speed. This is mass conservation for an incompressible fluid.

Teacher's Note:
1. The speed change comes from the continuity equation; Bernoulli's principle then explains the pressure change.
2. Read what the question asks: speed, not pressure.

Answer: (a) Torque = r × F. Near the hinge r is small, so a large force is needed for the same torque. Hence R correctly explains A.

Teacher's Note:
1. Torque equals force times perpendicular distance from the axis.
2. Always test whether R actually explains A.

Answer: (d) Friction is non-conservative, so R is false. It always opposes motion and does negative work on a round trip, so A is also false.

Teacher's Note:
1. Only conservative forces do zero work over a closed path.
2. Friction acts opposite to motion on both the forward and return paths.

Answer: (d) In uniform circular motion the direction of velocity changes continuously, so velocity is not constant (R is false). Momentum changes direction, so its rate of change equals the centripetal force, which is not zero (A is false).

Teacher's Note:
1. Speed can be constant while velocity changes, because velocity is a vector.
2. A non-zero centripetal force means a non-zero rate of change of momentum.

Answer: 1. Acceleration a = (30 − 20)/2 = 5 m/s².
2. Distance s = ut + ½at² = 20 × 2 + ½ × 5 × 4 = 40 + 10.
Distance = 50 m. (Check: average speed 25 m/s × 2 s = 50 m.)

Teacher's Note:
1. Find the acceleration first, then use s = ut + ½at².
2. The uniform acceleration is an assumption needed to use the equations of motion.

Answer: 1. Dimensions of v = [L T⁻¹].
2. [P] = [M L⁻¹ T⁻²], [ρ] = [M L⁻³], so [P/ρ] = [L² T⁻²].
3. Hence [√(P/ρ)] = [L T⁻¹], the same as the dimensions of v.
The equation is dimensionally correct.

Teacher's Note:
1. Find the dimensions of each side separately and compare them.
2. Dimensional correctness does not prove the equation is exactly right; numerical factors are not checked.

Answer: 1. Frequency ν = 1000/60 rev/s, so ω = 2πν = 2π × 1000/60 = 104.7 rad/s.
2. Radius r = 0.20 m, so v = rω = 0.20 × 104.7 = 20.9 m/s.
Angular velocity ≈ 104.7 rad/s (≈ 105 rad/s); linear velocity ≈ 20.9 m/s.

Teacher's Note:
1. Use the radius, which is half of the diameter, in v = rω.
2. The mass given is not needed for this question.

Answer: (a) Blowing over the paper increases the air speed above it, so by Bernoulli's principle the pressure above falls below the atmospheric pressure below. The net upward force keeps the paper up. Blowing under it lowers the pressure below and the paper would be pushed down.
(b) For water, the adhesive force between water and glass molecules is greater than the cohesive force among water molecules, so water wets glass and the angle is acute. For mercury, cohesion is greater than adhesion, so it does not wet glass and the angle is obtuse (about 140°).

Teacher's Note:
1. Apply Bernoulli's principle: higher speed means lower pressure.
2. Compare adhesive and cohesive forces to decide the shape of the meniscus.

Answer: 1. Force on the large piston F₂ = mg = 1350 × 9.8 = 13230 N.
2. By Pascal's law, F₁/A₁ = F₂/A₂, so F₁ = F₂ (r₁/r₂)² = 13230 × (5/15)² = 13230/9 = 1470 N.
3. Pressure P = F₁/A₁ = 1470/(π × 0.05²) = 1470/(7.854 × 10⁻³) ≈ 1.9 × 10⁵ Pa (1.87 × 10⁵ Pa).

Teacher's Note:
1. Pressure is the same in the two pistons, so the force ratio equals the area ratio.
2. Area depends on the square of the radius, so convert cm to m before computing pressure.

Answer: 1. Take the initial direction as positive: u = +12 m/s, v = −20 m/s, m = 0.15 kg.
2. Change in momentum = m(v − u) = 0.15 × (−20 − 12) = −4.8 kg m/s.
3. Average force = change in momentum / time = −4.8/0.01 = −480 N.
Magnitude 480 N, directed opposite to the initial motion of the ball.

Teacher's Note:
1. The velocity reverses, so the change in velocity is 12 + 20, not 20 − 12.
2. Convert grams to kilograms before using F = Δp/Δt.

Answer: 1. Force F = 3 × 9.8 = 29.4 N; L = 1.5 m; A = 1 × 10⁻⁶ m².
2. Y = FL/(AΔL), so ΔL = FL/(AY) = (29.4 × 1.5)/(10⁻⁶ × 2 × 10¹¹) = 44.1/(2 × 10⁵).
ΔL = 2.2 × 10⁻⁴ m = 0.22 mm.

Teacher's Note:
1. Convert mm² to m² (1 mm² = 10⁻⁶ m²) and kgf to newton.
2. Use Y = stress/strain rearranged for ΔL.

Answer: Terminal velocity: the constant maximum velocity acquired by a body falling through a viscous fluid, when the net force on it becomes zero.
Derivation:
1. Forces on the sphere: weight downward W = (4/3)πr³ρg; buoyant force upward U = (4/3)πr³σg; viscous drag upward F = 6πηrv (Stokes' law).
2. At terminal velocity v_t the net force is zero: W = U + F.
3. (4/3)πr³(ρ − σ)g = 6πηr v_t.
4. Therefore v_t = 2r²(ρ − σ)g / (9η).

Teacher's Note:
1. Write all three forces (weight, buoyancy, viscous drag) before equating them.
2. The question as printed gave only one density; the sphere and the liquid need separate symbols.

Answer: 1. The lamina is uniform, so each square has mass 1 kg, placed at its centre: C₁ (0.5, 0.5), C₂ (1.5, 0.5), C₃ (0.5, 1.5).
2. X = (1 × 0.5 + 1 × 1.5 + 1 × 0.5)/3 = 2.5/3 = 5/6 m.
3. Y = (1 × 0.5 + 1 × 0.5 + 1 × 1.5)/3 = 2.5/3 = 5/6 m.
Centre of mass = (5/6 m, 5/6 m) ≈ (0.83 m, 0.83 m).

Teacher's Note:
1. Split the plate into simple parts, treat each as a point mass at its own centre, and then use the centre of mass formula.
2. The answer lies on the line y = x because the shape is symmetric about it.

Answer: Compare with y = a sin(ωt + kx + φ): a = 3.0 cm, ω = 36 rad/s, k = 0.018 cm⁻¹, φ = π/4.
(a) It is a travelling wave (t and x appear in the combination ωt + kx). The plus sign shows it travels in the negative x direction (right to left). Speed v = ω/k = 36/0.018 = 2000 cm/s = 20 m/s.
(b) Amplitude = 3.0 cm; frequency ν = ω/2π = 36/(2π) = 5.73 Hz.
(c) Initial phase at the origin = π/4.
(d) Least distance between successive crests = wavelength λ = 2π/k = 2π/0.018 ≈ 349 cm = 3.49 m.

Teacher's Note:
1. Match the equation with the standard form and read off a, ω, k and φ.
2. A plus sign between ωt and kx means the wave moves towards negative x.

Answer: 1. Amplitude A = 0.10 m; ω = √(k/m) = √50 ≈ 7.07 rad/s.
2. At t = 0: x = A and v = 0. Potential energy = ½kx² = ½ × 50 × 0.01 = 0.25 J; kinetic energy = 0.
3. Total energy E = ½kA² = 0.25 J (constant, as the surface is frictionless).
4. At time t, with x = A cos ωt: PE = 0.25 cos²(ωt) J and KE = 0.25 sin²(ωt) J, and their sum is always 0.25 J.

Teacher's Note:
1. Released from rest at the extreme position, all the energy is potential at t = 0.
2. The total energy ½kA² stays constant, and KE and PE exchange with time.

Answer: Either:
(a) Capillarity is the rise or fall of a liquid in a narrow tube (capillary) dipped in it. Application: rise of oil in a wick, or of water in soil and plants (any one).
(b) Let the liquid have surface tension S, density ρ, and angle of contact θ in a tube of radius r. The surface tension acts along the circle of contact, 2πr, at angle θ with the wall. The upward force = S cos θ × 2πr. The weight of the liquid column of height h = πr²hρg (taking the meniscus volume as negligible). In equilibrium, 2πrS cos θ = πr²hρg, so h = 2S cos θ/(rρg).
(c) For mercury the angle of contact is obtuse (cos θ is negative), so h is negative. The meniscus is convex and the pressure just below it is greater, which pushes the level down in the capillary.
OR:
(a) Consider a drop of radius r and surface tension S. Let the excess pressure inside be ΔP. Increase the radius by a small dr. Work done by the excess pressure = ΔP × 4πr² dr. The increase in surface energy = S × (4π(r + dr)² − 4πr²) = S × 8πr dr. Equating, ΔP × 4πr² dr = 8πrS dr, so ΔP = 2S/r.
(b) A soap bubble has two free surfaces. Increase in area = 2 × 4π(3² − 2²) = 2 × 4π × 5 = 40π cm². Work done W = S × ΔA = 30 × 40π = 1200π ≈ 3770 erg = 3.77 × 10⁻⁴ J.

Teacher's Note:
1. A neat diagram of the meniscus and the force balance earns most of the marks in the derivation.
2. A soap bubble has two surfaces, so its area is doubled; this is the common mistake in the numerical part.

Answer: Either:
(a) The force on a mass m at the surface is GMm/R². This equals mg, so g = GM/R².
(b) At height h: g_h = GM/(R + h)² = g(1 + h/R)⁻² ≈ g(1 − 2h/R) for h ≪ R, so g decreases with altitude. At depth d: g_d = g(1 − d/R), so g decreases linearly with depth and becomes zero at the centre of the earth. Maximum value is at the surface.
(c) At h = R/4: g′ = g/(1 + 1/4)² = g × (4/5)² = 16g/25 = 0.64 g. Weight falls to 64%, so the percentage decrease = 36%.
OR:
(a) Escape velocity is the minimum speed with which a body must be projected from the surface of a planet so that it escapes its gravitational field and never returns. Required: kinetic energy at the surface + potential energy = 0 at infinity: ½mv_e² − GMm/R = 0, so v_e = √(2GM/R) = √(2gR). For the earth, v_e ≈ 11.2 km/s.
(b) v_e′ = √(2G(3M)/(3R)) = √(2GM/R) = v_e. So the escape velocity is 11.2 km/s, the same as for the earth.

Teacher's Note:
1. Use the binomial approximation only when the height is much less than R; in part (c) use the exact formula.
2. Percentage decrease means (original − new)/original × 100.

Answer: Angular projectile: a body projected into the air at an angle θ (0 < θ < 90°) with the horizontal, which then moves under gravity alone along a parabola. Example: a football kicked at an angle, or a javelin thrown.
Let the initial speed be u at angle θ with the horizontal. Components: u_x = u cos θ, u_y = u sin θ. Acceleration is g downward; a_x = 0.
1. Time of flight T: the vertical displacement is zero at the end, so 0 = (u sin θ)T − ½gT², giving T = 2u sin θ/g.
2. Maximum height H: at the top the vertical velocity is zero, so 0 = u²sin²θ − 2gH, giving H = u²sin²θ/(2g).
3. Horizontal range R: R = (u cos θ) × T = u cos θ × 2u sin θ/g = u² × 2 sin θ cos θ/g, so R = u² sin 2θ/g.
The range is maximum for θ = 45°, where R_max = u²/g.

Teacher's Note:
1. Treat horizontal and vertical motions separately, since horizontal velocity stays constant.
2. Draw a labelled parabola with u, θ, H and R marked, as it helps in an exam answer.

Answer: (i) (a) 0.49. R = mg = 10 × 10 = 100 N, so μ = F/R = 49/100 = 0.49.
(ii) (d) 0.577. μ = tan 30° = 1/√3 = 0.577.
(iii) (a) 0.45 kgf. Limiting friction = μR = 0.3 × 1.5 = 0.45 kgf. The applied force of 1.2 kgf is greater than this, so the block slides and friction is 0.45 kgf.
(iv) (d) 4.9 m/s². Net force = 1.2 − 0.45 = 0.75 kgf = 0.75 × 9.8 N. a = 0.75 × 9.8/1.5 = 4.9 m/s².

Teacher's Note:
1. Check whether the applied force exceeds the limiting friction before deciding which value of friction to use.
2. Convert kgf to N by multiplying by 9.8 before using F = ma.

Answer: (a) Kinetic energy ½mω²A² sin²(ωt + φ) and potential energy ½mω²A² cos²(ωt + φ) vary periodically with time (with half the period of the oscillation), exchanging with each other. Their sum, the total energy ½mω²A² = ½kA², stays constant.
(b) v = dx/dt = −Aω sin(ωt + φ). Equivalently, v = ±ω√(A² − x²).
(c) Here ω = 2π rad/s, so a = −ω²x = −(2π)² × 5.0 cos(2π × 1.5 + π/4) = −197.4 × cos(3π + π/4). Now cos(3π + π/4) = −cos(π/4) = −0.7071. Hence a = −197.4 × (−0.7071) ≈ +139.6 m/s² (about 1.4 × 10² m/s², with x in metres).
OR
(d) y = sin ωt + cos ωt = √2 [sin ωt cos(π/4) + cos ωt sin(π/4)] = √2 sin(ωt + π/4). Time period T = 2π/ω, amplitude = √2.

Teacher's Note:
1. Kinetic and potential energies keep exchanging while their sum stays constant.
2. In part (c) take the angle in radians, and remember a = −ω²x so the sign depends on the position of the body.