Energy & Work in Circular Paths — NEET Physics MCQs with Solutions
Free NEET 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. Stays constant, since speed is constant ✓ Correct
B. Oscillates
C. Is zero
D. Increases steadily
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 topmost point
B. The same everywhere
C. The horizontal position
D. The lowest point ✓ Correct
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. 4 J
B. 8 J
C. 16 J ✓ Correct
D. 32 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. 5 J
C. 40 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. 20 J
B. 12 J ✓ Correct
C. 52 J
D. 32 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. 4 : 1
B. 2 : 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 · easy · numerical
A cyclist (total mass 100 kg) rides a circular track at constant 10 m/s. The net work done on him per lap is:
A. Zero ✓ Correct
B. 10⁴ J
C. 2πr × 100 J
D. 5000 J
Solution: Constant speed ⇒ ΔKE = 0 ⇒ W_net = 0 per lap (work–energy theorem).
Q8 — Energy & Work in Circular Paths · medium · numerical
A 0.5 kg bob swings in a vertical circle of radius 2 m. Between the top and bottom, gravity does work equal to:
A. 20 J ✓ Correct
B. 5 J
C. 40 J
D. 10 J
Solution: W = mg(2r) = 0.5×10×4 = 20 J.