Energy & Work in Circular Paths — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Energy & Work in Circular Paths MCQs with step-by-step solutions (8 questions). Part of Circular Motion. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Energy & Work in Circular Paths · easy · theory
In UNIFORM circular motion, the kinetic energy of the particle:
A. Increases steadily
B. Stays constant, since speed is constant ✓ Correct
C. Oscillates
D. Is zero
Solution: KE = ½mv² depends only on speed. The centripetal force does no work, so KE cannot change.
Q2 — Energy & Work in Circular Paths · medium · theory
In VERTICAL circular motion (string), the kinetic energy of the bob is maximum at:
A. The same everywhere
B. The topmost point
C. The lowest point ✓ Correct
D. The horizontal position
Solution: Energy conservation: KE + mgh = const; KE peaks where h is least — the bottom.
Q3 — Energy & Work in Circular Paths · easy · numerical
A 2 kg stone moves in a horizontal circle of radius 2 m at 4 m/s. Its kinetic energy is:
A. 16 J ✓ Correct
B. 4 J
C. 32 J
D. 8 J
Solution: KE = ½×2×16 = 16 J (constant in UCM).
Q4 — Energy & Work in Circular Paths · medium · numerical
A 1 kg bob just completes a vertical circle of radius 1 m. The difference between its KE at the bottom and at the top is:
A. 10 J
B. 40 J
C. 5 J
D. 20 J ✓ Correct
Solution: ΔKE = mg(2r) = 1×10×2 = 20 J — equal to the PE change, whatever the speeds.
Q5 — Energy & Work in Circular Paths · hard · numerical
A particle of mass 1 kg in a vertical circle (r = 2 m) has speed 8 m/s at the bottom. Its KE at the horizontal position (height r) is:
A. 12 J ✓ Correct
B. 32 J
C. 20 J
D. 52 J
Solution: KE = ½×64 − mgr = 32 − 20 = 12 J.
Q6 — Energy & Work in Circular Paths · medium · numerical
The ratio of kinetic energies at the bottom and top for a bob JUST completing a vertical loop is:
A. 2 : 1
B. 4 : 1
C. 6 : 1
D. 5 : 1 ✓ Correct
Solution: v_b² : v_t² = 5gr : gr = 5 : 1 ⇒ same KE ratio.
Q7 — Energy & Work in Circular Paths · hard · numerical
A 1 kg particle on a string (r = 1 m) is given v = 6 m/s at the bottom. The height (above the bottom) at which its speed becomes zero is:
A. 3.6 m
B. 1.0 m
C. 2.0 m
D. 1.8 m — it never completes the loop and the string slackens before this ✓ Correct
Solution: Naive energy: h = v²/2g = 1.8 m (< 2r = 2 m ⇒ fails to loop). Since v_b = 6 < √(5gr) ≈ 7.07, the string goes slack above the horizontal and the bob becomes a projectile — the trap this question tests.
Q8 — Energy & Work in Circular Paths · medium · numerical
A car of mass 500 kg moves on a circular track with speed increasing as v = 2t. The power delivered by the tangential force at t = 5 s is:
A. 5 kW
B. 10 kW ✓ Correct
C. 2 kW
D. 20 kW
Solution: a_t = 2 m/s², F_t = 1000 N, v(5) = 10 m/s ⇒ P = F_t·v = 10⁴ W. (The centripetal force contributes no power.)