A stone is thrown horizontally from height h with initial velocity u. Its horizontal range is x. If height is doubled and velocity doubled, new range is:
Answer: B
Step 1: Establish the original range formula
For a projectile launched horizontally from height h with initial velocity u, the time to fall is found from:
h=21gt2⟹t=g2h
The horizontal range is therefore:
x=ug2h
Step 2: Apply the new conditions
The new height is 2h and the new velocity is 2u. The new range x′ is:
x′=2ug2(2h)=2ug4h=2u⋅2gh=4ugh
**Step 3: Express the new range in terms of x
Rewrite 4ugh to match the original expression ug2h:
x′=4ugh=4u⋅21g2h=24⋅ug2h=22⋅x
Result
x′=22x
Q.2Hard
A block of mass m slides down from height h on a smooth incline. Final velocity is v. If height is doubled and friction coefficient μ is introduced, velocity becomes:
Answer: C
Without friction: v² = 2gh. With friction and 2h: friction opposes motion, so final velocity < √(4gh) = √2v
Q.3Hard
In an elastic collision between equal masses where one is at rest, the velocities after collision are:
Answer: B
For equal masses in elastic collision with one at rest: v₁' = 0, v₂' = u₁ (velocities exchange)
Q.4Hard
A rotating body has moment of inertia I and angular acceleration α. The rotational kinetic energy increases at rate:
Answer: B
KE = ½Iω². dKE/dt = Iω(dω/dt) = Iωα. Since τ = Iα and dKE/dt = τω = Iαω
Q.5Hard
Three objects with masses 1 kg, 2 kg, and 3 kg are connected in a straight line by rigid massless rods. The center of mass from the 1 kg mass is at distance:
Answer: C
Assuming equal spacing of 1 m: x_cm = (1×0 + 2×1 + 3×2)/(1+2+3) = 68 = 34 m. If spacing is different, need clarification. Standard answer assumes x_cm = 2 m
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Q.6Hard
A solid sphere rolls without slipping down an incline of angle θ. The acceleration of center of mass is:
Answer: B
For rolling sphere: a = g sin θ/(1 + I/(mR²)) = g sin θ/(1 + 52) = (75)g sin θ
Q.7Hard
The coefficient of restitution between two colliding balls is 0.8. If they approach with velocities 5 m/s and 3 m/s, the relative velocity of separation is:
Answer: B
e = relative velocity of separation/relative velocity of approach. 0.8 = v_sep/(5-3), so v_sep = 0.8 × 2 = 1.6 m/s
Q.8Hard
A hollow sphere and solid sphere of equal mass roll down the same incline without slipping. Which reaches the bottom first?
Answer: B
Solid sphere has smaller moment of inertia (I = 52 mR²) vs hollow sphere (I = 32 mR²). Smaller I means faster acceleration, so solid sphere reaches first.
Q.9Hard
Two particles have equal kinetic energies but different momenta. Particle A has momentum p_A and particle B has momentum p_B. If m_A > m_B, then:
Answer: B
KE = p²/(2m). If KE_A = KE_B, then p_A²/m_A = p_B²/m_B. Since m_A > m_B, we need p_A < p_B
Q.10Hard
Two satellites orbit at radii r₁ and r₂ from Earth's center where r₂ = 4r₁. The ratio of their orbital periods T₁:T₂ is:
Answer: C
By Kepler's third law: T² ∝ r³. (T₁/T₂)² = (r₁/r₂)³ = (41)³ = 641. T₁/T₂ = 81. So T₁:T₂ = 1:8
Q.11Hard
A wedge of mass M = 10 kg with angle 30° is on a frictionless surface. A block of mass m = 5 kg slides down the wedge. What is the acceleration of the wedge in the horizontal direction? (g = 10 m/s²)
Answer: D
Using center of mass concept or constraint analysis: a_wedge = (mg sinθ cosθ)/(M + m sin²θ) = (5×10×sin30°×cos30°)/(10 + 5×sin²30°) = (50×0.5×0.866)/(10 + 1.25) ≈ 1.25 m/s²
Q.12Hard
Three identical conducting rods are arranged in series between two heat reservoirs at 100°C and 0°C. At steady state, what is the temperature at the junction between the second and third rod?
Answer: A
In series arrangement with identical rods, temperature difference is equally distributed. ΔT_total = 100°C, so ΔT per rod = 3100 = 33.3°C. Second junction = 100 - 2(33.3) = 33.3°C
Q.13Hard
A gas undergoes a cyclic process ABCA where AB is isothermal, BC is adiabatic, and CA is isochoric. If work is done on the gas in the cycle, what can be concluded?
Answer: A
If W_net < 0 (work done on gas), then from first law: ΔU_cycle = 0 = Q - W, so Q = W < 0, meaning net heat flows out
Q.14Hard
Two bodies at temperatures T₁ = 400 K and T₂ = 300 K are brought into thermal contact. If entropy change of universe is 0.575 J/K and heat capacity of both bodies is 1000 J/K, what is the final equilibrium temperature? (Assume no heat loss to surroundings)
Answer: B
Heat lost by body 1: Q = C(T₁ - T_f) = 1000(400 - T_f). Heat gained by body 2: Q = 1000(T_f - 300). ΔS_univ = C ln(T_f/T₁) + C ln(T_f/T₂) = 1000[ln(T_f/400) + ln(T_f/300)] = 0.575. Solving: T_f = 350 K
Q.15Hard
If a reversible process occurs in an isolated system, the entropy of the system:
Answer: C
For a reversible process in an isolated system, ΔS_total = ΔS_sys + ΔS_surr = 0 (no heat exchange with surroundings). Therefore entropy remains constant at its initial value.
Q.16Hard
An ideal gas undergoes a process where PV^n = constant. If n = 1, this process is _____ and if n = γ, this process is _____
Answer: A
When n = 1: PV = constant (isothermal). When n = γ = Cp/Cv: PV^γ = constant (adiabatic process).
Q.17Hard
Consider a system where entropy decreases by 50 J/K. Which statement must be true?
Answer: B
By second law, ΔS_total ≥ 0. If ΔS_sys = -50 J/K, then ΔS_surr ≥ +50 J/K to maintain ΔS_total ≥ 0.
Q.18Hard
A polytropic process with polytropic index n = 1.3 is performed on 1 mole of air (diatomic). The work done when volume changes from 1 m³ to 0.5 m³ at initial pressure 100 kPa is:
Answer: A
For polytropic process: W = [P₁V₁ - P₂V₂]/(n-1). Using PVⁿ = const: P₂ = 100 × (01.5)^1.3 ≈ 245.7 kPa. W = [100×1 - 245.7×0.5]/0.3 ≈ 81.2 kJ
Q.19Hard
For a van der Waals gas with equation (P + a/V²m)(Vm - b) = RT, at the critical point, which relationship is valid?
Answer: A
At the critical point, both first and second derivatives of pressure with respect to volume (at constant T) are zero: (∂P/∂V)T = 0 and (∂²P/∂V²)T = 0. This defines the critical point.
Q.20Hard
Two moles of an ideal diatomic gas undergo a process where temperature changes from 400 K to 800 K. If the process is such that PV^(1.4) = constant, then the work done by the gas is approximately (R = 8.314 J/mol·K):
Answer: C
For polytropic process: W = nR(T_i − T_f)/(γ − n) = 2 × 8.314 × (400 − 800)/(1.4 − 1.4). Since n = γ, this is adiabatic: W = nCvΔT = 2 × (25) × 8.314 × (400 − 800) = 5 × 8.314 × (−400) ≈ −16,628 J. But if heating occurs, work done is positive. Actual calculation needs clarification on process direction.
