Physics is where JEE ranks are usually won or lost, since a single conceptual slip changes the whole answer. Questions here span mechanics, rotational motion, thermodynamics, waves and oscillations, electrostatics, current electricity, magnetism, optics, and modern physics. Numerical problems carry full derivations so you can see which step you skipped, not just which option was right.
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
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.
