Variable Mass Systems — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Variable Mass Systems MCQs with step-by-step solutions (8 questions). Part of Centre of Mass, Momentum & Collisions. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Variable Mass Systems · easy · theory
The thrust on a rocket is given by:
A. F = v²(dm/dt)
B. F = v_rel(dm/dt), from the momentum of the ejected exhaust ✓ Correct
C. F = mg
D. F = ma always
Solution: Expelling mass at relative speed v_rel carries momentum away at rate v_rel·dm/dt — the reaction is the thrust.
Q2 — Variable Mass Systems · medium · theory
A rocket accelerates in DEEP SPACE (no gravity, no drag) because:
A. It pushes exhaust backward and the exhaust pushes it forward (momentum conservation) ✓ Correct
B. Gravity assists it
C. It pushes against the air
D. It pushes against the launch pad
Solution: No external medium is needed: the rocket–exhaust system conserves momentum internally. "Nothing to push on" is the classic misconception.
Q3 — Variable Mass Systems · easy · numerical
A rocket ejects gas at 2 kg/s with relative speed 500 m/s. The thrust is:
A. 250 N
B. 500 N
C. 2000 N
D. 1000 N ✓ Correct
Solution: F = v_rel(dm/dt) = 500×2 = 1000 N.
Q4 — Variable Mass Systems · medium · numerical
A 600 kg rocket must hover (g = 10). With exhaust speed 1200 m/s, the required burn rate is:
A. 5 kg/s ✓ Correct
B. 0.5 kg/s
C. 10 kg/s
D. 2 kg/s
Solution: Thrust = weight: v_rel(dm/dt) = 6000 ⇒ dm/dt = 6000/1200 = 5 kg/s.
Q5 — Variable Mass Systems · hard · numerical
A rocket of mass 1000 kg ejects gas at 50 kg/s with v_rel = 400 m/s, starting from rest on the pad (g = 10). Its initial upward acceleration is:
A. 20 m/s²
B. 30 m/s²
C. 10 m/s² ✓ Correct
D. 5 m/s²
Solution: a = (thrust − mg)/m = (20000 − 10000)/1000 = 10 m/s². Trap: subtract the weight at lift-off.
Q6 — Variable Mass Systems · medium · numerical
Sand drops at 5 kg/s onto a conveyor belt moving at 4 m/s. The extra force needed to keep the belt's speed constant is:
A. 5 N
B. 10 N
C. 40 N
D. 20 N ✓ Correct
Solution: F = v(dm/dt) = 4×5 = 20 N (power needed = Fv = 80 W, half of which becomes sand KE).
Q7 — Variable Mass Systems · hard · numerical
A chain of linear density 1 kg/m falls onto a table. At an instant, 2 m of chain already lies on the table and the links are arriving at 6 m/s. The force on the table at that instant is:
A. 92 N
B. 36 N
C. 56 N ✓ Correct
D. 20 N
Solution: Weight of the landed part = λℓg = 1×2×10 = 20 N; momentum-destruction term = λv² = 1×36 = 36 N. Total = 56 N. Trap: forgetting either term.
Q8 — Variable Mass Systems · medium · numerical
In deep space a rocket (v_rel = 2 km/s) burns until its mass halves. Its velocity gain is (ln 2 ≈ 0.69):
A. ≈ 1.39 km/s ✓ Correct
B. ≈ 2 km/s
C. ≈ 1 km/s
D. ≈ 0.69 km/s
Solution: Δv = v_rel ln(m₀/m) = 2×ln2 ≈ 1.39 km/s — Tsiolkovsky's equation.