Vertical Circular Motion — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Vertical Circular Motion MCQs with step-by-step solutions (15 questions). Part of Circular Motion. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Vertical Circular Motion · medium · theory
For a mass tied to a string moving in a vertical circle of radius r, the minimum speed at the highest point is:
A. $\sqrt{2gr}$
B. $\sqrt{3gr}$
C. $\sqrt{5gr}$
D. $\sqrt{gr}$ ✓ Correct
Solution: At the top, for minimum speed the tension is zero and gravity alone provides the centripetal force: mg = mv²/r ⇒ v_top = √(gr).
Q2 — Vertical Circular Motion · medium · theory
The minimum speed at the lowest point for a mass on a string to just complete a vertical circle of radius r is:
A. $\sqrt{3gr}$
B. $\sqrt{2gr}$
C. $\sqrt{gr}$
D. $\sqrt{5gr}$ ✓ Correct
Solution: By energy conservation between top and bottom (height 2r): v_bottom² = v_top² + 4gr = gr + 4gr = 5gr ⇒ v_bottom = √(5gr).
Q3 — Vertical Circular Motion · medium · theory
For a body completing a vertical circle on a string, the difference between the tension at the lowest and highest points is:
A. mg
B. 2mg
C. 6mg ✓ Correct
D. 3mg
Solution: T_bottom − T_top = m(v_b² − v_t²)/r + 2mg = m(4gr)/r + 2mg = 4mg + 2mg = 6mg (independent of speed, as long as the circle is completed).
Q4 — Vertical Circular Motion · medium · theory
For a mass attached to a light rigid rod moving in a vertical circle, the minimum speed at the highest point is:
A. $\sqrt{gr}$
B. $\sqrt{5gr}$
C. $\sqrt{2gr}$
D. Zero ✓ Correct
Solution: A rigid rod can push as well as pull, so it can support the weight at the top; the minimum speed there can be zero.
Q5 — Vertical Circular Motion · medium · theory
For a mass on a light rigid rod, the minimum speed at the lowest point to just complete a vertical circle of radius r is:
A. $\sqrt{gr}$
B. Zero
C. $\sqrt{4gr}$ ✓ Correct
D. $\sqrt{5gr}$
Solution: With v_top = 0, energy conservation gives v_bottom² = 0 + 4gr ⇒ v_bottom = √(4gr) = 2√(gr).
Q6 — Vertical Circular Motion · easy · theory
As a body moves up along a vertical circle, its speed and the tension in the string:
A. Remain constant
B. Speed increases while tension decreases
C. Both increase towards the top
D. Both decrease (both are maximum at the lowest point) ✓ Correct
Solution: By energy conservation the speed decreases with height, and the tension is greatest at the lowest point and least at the top.
Q7 — Vertical Circular Motion · easy · numerical
A stone tied to a string moves in a vertical circle of radius 0.9 m. The minimum speed at the top (g = 10 m/s²) is:
A. 6 m/s
B. 1 m/s
C. 9 m/s
D. 3 m/s ✓ Correct
Solution: v_top = √(gr) = √(10 × 0.9) = √9 = 3 m/s.
Q8 — Vertical Circular Motion · medium · numerical
For a body just completing a vertical circle of radius 0.8 m on a string, the minimum speed at the bottom (g = 10 m/s²) is:
A. ≈ 6.32 m/s ✓ Correct
B. ≈ 4.9 m/s
C. ≈ 2.83 m/s
D. ≈ 8 m/s
Solution: v_bottom = √(5gr) = √(5 × 10 × 0.8) = √40 ≈ 6.32 m/s.
Q9 — Vertical Circular Motion · medium · numerical
A ball just loops a vertical circle of radius 1.2 m. Its speed at the mid-level (height r) is (g = 10 m/s²):
A. 12 m/s
B. 3 m/s
C. 4 m/s
D. 6 m/s ✓ Correct
Solution: v_mid = √(3gr) = √(3 × 10 × 1.2) = √36 = 6 m/s.
Q10 — Vertical Circular Motion · easy · numerical
A 2 kg mass moves in a vertical circle on a string. The difference between the tensions at the lowest and highest points (g = 10 m/s²) is:
A. 120 N ✓ Correct
B. 20 N
C. 60 N
D. 12 N
Solution: T_bottom − T_top = 6mg = 6 × 2 × 10 = 120 N.
Q11 — Vertical Circular Motion · medium · numerical
A mass on a light rigid rod just completes a vertical circle of radius 2.5 m. The minimum speed at the lowest point (g = 10 m/s²) is:
A. 5 m/s
B. 20 m/s
C. 10 m/s ✓ Correct
D. 15 m/s
Solution: v_bottom = √(4gr) = √(4 × 10 × 2.5) = √100 = 10 m/s.
Q12 — Vertical Circular Motion · medium · numerical
A 0.5 kg stone just completes a vertical circle of radius r on a string. The tension at the lowest point (with T_top = 0, g = 10 m/s²) is:
A. 5 N
B. 15 N
C. 30 N ✓ Correct
D. 60 N
Solution: For the just-completing case T_top = 0, so T_bottom = 6mg = 6 × 0.5 × 10 = 30 N.
Q13 — Vertical Circular Motion · easy · numerical
The minimum speed required at the top of a vertical circle of radius 10 m (string, g = 10 m/s²) is:
A. 10 m/s ✓ Correct
B. 5 m/s
C. 1 m/s
D. 100 m/s
Solution: v_top = √(gr) = √(10 × 10) = √100 = 10 m/s.
Q14 — Vertical Circular Motion · medium · numerical
A mass just loops a vertical circle of radius 1.6 m on a string. The minimum speed at the lowest point (g = 10 m/s²) is:
A. ≈ 8.94 m/s ✓ Correct
B. ≈ 12.6 m/s
C. ≈ 6.9 m/s
D. ≈ 4 m/s
Solution: v_bottom = √(5gr) = √(5 × 10 × 1.6) = √80 ≈ 8.94 m/s.
Q15 — Vertical Circular Motion · medium · numerical
A 1 kg ball moves at 10 m/s at the lowest point of a vertical circle of radius 2 m. The tension in the string there (g = 10 m/s²) is:
A. 110 N
B. 10 N
C. 60 N ✓ Correct
D. 50 N
Solution: At the bottom, T = mv²/r + mg = 1 × 10²/2 + 1 × 10 = 50 + 10 = 60 N.