VCM — Mass on a Light String — IISER Physics MCQs with Solutions
Free IISER Physics VCM — Mass on a Light String 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 — VCM — Mass on a Light String · easy · theory
For a mass on a light STRING to complete a vertical circle of radius r, its minimum speed at the topmost point is:
A. √(gr) ✓ Correct
B. Zero
C. √(5gr)
D. √(2gr)
Solution: At the top the string can only pull; the critical case has T = 0, gravity alone centripetal: mg = mv²/r ⇒ v = √(gr).
Q2 — VCM — Mass on a Light String · medium · theory
For a body just completing a vertical circle on a string, the tension difference T(bottom) − T(top) equals:
A. mg
B. 2mg
C. 3mg
D. 6mg ✓ Correct
Solution: Using energy conservation (v_b² = v_t² + 4gr): T_b − T_t = m(v_b² − v_t²)/r + 2mg = 6mg — independent of the actual speeds.
Q3 — VCM — Mass on a Light String · easy · numerical
A stone on a string just loops a vertical circle of radius 1.6 m. Its speed at the top is:
A. 4 m/s ✓ Correct
B. 8 m/s
C. 2 m/s
D. 16 m/s
Solution: v_top = √(gr) = √16 = 4 m/s.
Q4 — VCM — Mass on a Light String · medium · numerical
For the same critical loop (r = 1.6 m), the speed at the LOWEST point is:
A. 4 m/s
B. √80 ≈ 8.9 m/s ✓ Correct
C. √48 ≈ 6.9 m/s
D. 16 m/s
Solution: v_bottom = √(5gr) = √80 ≈ 8.9 m/s.
Q5 — VCM — Mass on a Light String · hard · numerical
A 1 kg ball moves at 8 m/s at the bottom of a vertical circle of radius 2 m. The tension in the string at that point is:
A. 22 N
B. 32 N
C. 10 N
D. 42 N ✓ Correct
Solution: T − mg = mv²/r ⇒ T = 10 + 64/2 = 42 N. Trap: at the bottom, tension EXCEEDS the centripetal value by mg... careful: T = mg + mv²/r.
Q6 — VCM — Mass on a Light String · medium · numerical
A 0.5 kg stone just completes a vertical circle. The tension in the string at the LOWEST point is:
A. Zero
B. 30 N ✓ Correct
C. 5 N
D. 15 N
Solution: For the critical loop T_b = 6mg − T_t with T_t = 0 ⇒ T_b = 6mg = 30 N.
Q7 — VCM — Mass on a Light String · hard · theory
In vertical circular motion on a string, the mechanical energy is conserved but the SPEED varies because:
A. The mass changes
B. The tension does work
C. Energy is not conserved
D. Gravity does tangential work along the path while tension does none ✓ Correct
Solution: Tension ⊥ path (no work); the tangential component of gravity alternately decelerates (up) and accelerates (down) the bob, exchanging KE ↔ PE.
Q8 — VCM — Mass on a Light String · hard · numerical
A bob on a string swings in a vertical circle, JUST completing it. At the instant the string makes 60° with the upward vertical (cos60° = 0.5), its speed satisfies v² =
A. 5gr
B. 2gr ✓ Correct
C. gr
D. 3gr
Solution: Measuring θ from the top: v² = v_top² + 2gr(1 − cosθ) = gr + 2gr(1 − 0.5) = 2gr.