Conical Pendulum — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Conical Pendulum 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 — Conical Pendulum · easy · theory
The period of revolution of a conical pendulum of string length L at semi-vertical angle θ is:
A. $T = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}$ ✓ Correct
B. $T = 2\pi\sqrt{\dfrac{L}{g}}$
C. $T = 2\pi\sqrt{\dfrac{L}{g\cos\theta}}$
D. $T = 2\pi\sqrt{\dfrac{L\sin\theta}{g}}$
Solution: For a conical pendulum, T = 2π√(L cosθ / g) = 2π√(h/g), where h = L cosθ is the height of the apex above the circle.
Q2 — Conical Pendulum · medium · theory
In a conical pendulum, the vertical component of the string tension:
A. Is zero
B. Balances the weight of the bob (T cosθ = mg) ✓ Correct
C. Provides the centripetal force
D. Equals the horizontal component
Solution: The bob moves in a horizontal circle, so the vertical component T cosθ balances the weight mg.
Q3 — Conical Pendulum · medium · theory
In a conical pendulum, the centripetal force needed for the circular motion is provided by:
A. The normal reaction
B. The weight of the bob
C. The horizontal component of the tension (T sinθ) ✓ Correct
D. The vertical component of the tension
Solution: The horizontal component T sinθ = mω²r = mv²/r supplies the centripetal force.
Q4 — Conical Pendulum · medium · theory
For a conical pendulum, the semi-vertical angle θ is related to the speed v and radius r by:
A. $\sin\theta = \dfrac{v^2}{rg}$
B. $\cos\theta = \dfrac{v^2}{rg}$
C. $\tan\theta = \dfrac{rg}{v^2}$
D. $\tan\theta = \dfrac{v^2}{rg}$ ✓ Correct
Solution: Dividing T sinθ = mv²/r by T cosθ = mg gives tanθ = v²/rg.
Q5 — Conical Pendulum · easy · theory
In a conical pendulum, the bob:
A. Moves in a vertical circle
B. Oscillates in a straight line
C. Moves in a horizontal circle while the string sweeps out a cone ✓ Correct
D. Remains stationary
Solution: The bob describes a horizontal circle and the string traces the surface of a cone — hence the name.
Q6 — Conical Pendulum · medium · theory
The period of a conical pendulum (for a given angle θ) is:
A. Inversely proportional to the mass
B. Directly proportional to the mass
C. Proportional to the square of the mass
D. Independent of the mass of the bob ✓ Correct
Solution: T = 2π√(L cosθ/g) contains no mass term, so the period is independent of the mass of the bob.
Q7 — Conical Pendulum · medium · numerical
A conical pendulum has the apex of its cone at a height 0.4 m above the plane of the circle. Its period is (g = 10 m/s², π = 3.14):
A. ≈ 0.4 s
B. ≈ 0.63 s
C. ≈ 2.5 s
D. ≈ 1.26 s ✓ Correct
Solution: T = 2π√(h/g) = 2π√(0.4/10) = 2π√0.04 = 2π × 0.2 = 0.4π ≈ 1.26 s.
Q8 — Conical Pendulum · medium · numerical
A conical pendulum bob moves in a circle of radius 2 m at a semi-vertical angle of 45° (g = 10 m/s²). Its speed is:
A. ≈ 2 m/s
B. ≈ 4.47 m/s ✓ Correct
C. ≈ 20 m/s
D. ≈ 10 m/s
Solution: tanθ = v²/rg ⇒ v = √(rg tanθ) = √(2 × 10 × tan45°) = √20 ≈ 4.47 m/s.
Q9 — Conical Pendulum · easy · numerical
A conical pendulum bob of mass 1 kg makes a semi-vertical angle of 60° with the vertical (g = 10 m/s²). The tension in the string is:
A. 5 N
B. 10 N
C. 17.3 N
D. 20 N ✓ Correct
Solution: T cosθ = mg ⇒ T = mg/cosθ = (1 × 10)/cos60° = 10/0.5 = 20 N.
Q10 — Conical Pendulum · medium · numerical
A conical pendulum has L cosθ = 0.1 m (g = 10 m/s²). Its angular velocity is:
A. 10 rad/s ✓ Correct
B. 5 rad/s
C. 1 rad/s
D. 100 rad/s
Solution: ω = √(g / L cosθ) = √(10/0.1) = √100 = 10 rad/s.
Q11 — Conical Pendulum · medium · numerical
A conical pendulum of length 0.8 m makes an angle of 60° with the vertical (g = 10 m/s²). The angular velocity of the bob is:
A. 2.5 rad/s
B. 25 rad/s
C. 10 rad/s
D. 5 rad/s ✓ Correct
Solution: h = L cosθ = 0.8 × cos60° = 0.4 m. ω = √(g/h) = √(10/0.4) = √25 = 5 rad/s.
Q12 — Conical Pendulum · medium · numerical
A conical pendulum has L cosθ = 0.25 m (g = 10 m/s², π = 3.14). Its frequency of revolution is about:
A. ≈ 0.5 Hz
B. ≈ 1.0 Hz ✓ Correct
C. ≈ 2.0 Hz
D. ≈ 6.3 Hz
Solution: n = (1/2π)√(g/L cosθ) = (1/2π)√(10/0.25) = (1/2π)√40 = 6.32/6.28 ≈ 1.0 Hz.
Q13 — Conical Pendulum · medium · numerical
A conical pendulum bob moves in a circle of radius 0.4 m at a speed of 2 m/s (g = 10 m/s²). The semi-vertical angle of the string is:
A. 15°
B. 60°
C. 30°
D. 45° ✓ Correct
Solution: tanθ = v²/rg = 2²/(0.4 × 10) = 4/4 = 1 ⇒ θ = 45°.
Q14 — Conical Pendulum · medium · numerical
A conical pendulum bob of mass 0.5 kg makes a semi-vertical angle of 30° (cos30° ≈ 0.866, g = 10 m/s²). The tension in the string is:
A. ≈ 2.5 N
B. ≈ 4.33 N
C. ≈ 10 N
D. ≈ 5.77 N ✓ Correct
Solution: T = mg/cosθ = (0.5 × 10)/0.866 = 5/0.866 ≈ 5.77 N.
Q15 — Conical Pendulum · medium · numerical
If the length of a conical pendulum is made 4 times (keeping the semi-vertical angle the same), its period becomes:
A. 2 times ✓ Correct
B. Half
C. 4 times
D. Unchanged
Solution: T = 2π√(L cosθ/g) ∝ √L (for fixed θ). Making L four times increases T by √4 = 2 times.