Conical Pendulum & Rotating Systems — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Conical Pendulum & Rotating Systems 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 — Conical Pendulum & Rotating Systems · easy · theory
In a conical pendulum, the centripetal force on the bob is supplied by:
A. The weight of the bob
B. Friction
C. The horizontal component of the string tension (T sinθ) ✓ Correct
D. The vertical component of the tension
Solution: T cosθ = mg balances gravity; T sinθ = mω²r drives the circular motion.
Q2 — Conical Pendulum & Rotating Systems · medium · theory
The period of a conical pendulum of length L at semi-vertical angle θ is:
A. 2π√(g/L cosθ)
B. 2π√(L sinθ/g)
C. 2π√(L/g)
D. 2π√(L cosθ/g) ✓ Correct
Solution: T = 2π√(h/g) with h = L cosθ (height of the support above the circle's plane). Note it is SHORTER than the simple-pendulum period.
Q3 — Conical Pendulum & Rotating Systems · easy · numerical
A conical pendulum's bob of mass 2 kg circles with the string at 60° to the vertical (cos60° = 0.5). The string tension is:
A. 20 N
B. 40 N ✓ Correct
C. 10 N
D. 34.6 N
Solution: T cosθ = mg ⇒ T = 20/0.5 = 40 N.
Q4 — Conical Pendulum & Rotating Systems · medium · numerical
A conical pendulum has L cosθ = 0.4 m. Its angular speed is:
A. 5 rad/s ✓ Correct
B. 25 rad/s
C. √5 rad/s
D. 2.5 rad/s
Solution: ω = √(g/L cosθ) = √(10/0.4) = √25 = 5 rad/s.
Q5 — Conical Pendulum & Rotating Systems · hard · numerical
A 1 kg bob on a 2 m string moves as a conical pendulum with a period of 2 s (π² ≈ 10). The semi-vertical angle satisfies cosθ =
A. 0.87 (θ = 30°)
B. 0.25
C. 0.71 (θ = 45°)
D. 0.5 (θ = 60°) ✓ Correct
Solution: T = 2π√(Lcosθ/g) ⇒ cosθ = gT²/(4π²L) = 10×4/(40×2) = 0.5 ⇒ θ = 60°.
Q6 — Conical Pendulum & Rotating Systems · medium · numerical
A bob circles as a conical pendulum with radius 0.3 m and speed 3 m/s. The tangent of the string's angle with the vertical is:
A. 1
B. 0.9
C. 3 ✓ Correct
D. 0.3
Solution: tanθ = v²/rg = 9/(0.3×10) = 3 ⇒ θ ≈ 72°.
Q7 — Conical Pendulum & Rotating Systems · hard · numerical
Two beads circle on frictionless horizontal circles inside a smooth hemispherical bowl of radius R, at heights giving semi-angle θ from the vertical axis. Their angular speed is:
A. ω = √(gR)
B. ω = √(g/(R cosθ)) ✓ Correct
C. ω = √(g tanθ/R)
D. ω = √(g/R)
Solution: The bowl's normal reaction acts like the string of a conical pendulum of length R: ω² = g/(R cosθ) — independent of the bead's mass.
Q8 — Conical Pendulum & Rotating Systems · medium · numerical
A 0.5 kg bob on a 1.25 m string circles as a conical pendulum with ω = 4 rad/s. cosθ equals:
A. 0.64
B. 0.8
C. 0.25
D. 0.5 ✓ Correct
Solution: cosθ = g/(ω²L) = 10/(16×1.25) = 0.5 ⇒ θ = 60°.