Q 11-04-040JEE MainMedium
A small ball is tied to a string $1$ m long and whirled as a conical pendulum so that it moves in a horizontal circle of radius $0.6$ m. The time period of revolution is ($g = 10\ \text{m/s}^2$)
Answer: (A) $2\pi\sqrt{0.08}$ s
Concept. Conical pendulum: $T_{\text{period}} = 2\pi\sqrt{\dfrac{l\cos\theta}{g}}$.
Step 1. Angle. $\sin\theta = 0.6$, $\cos\theta = 0.8$.
Step 2. Period. $2\pi\sqrt{\dfrac{0.8}{10}} = 2\pi\sqrt{0.08}$ s.
Answer. $2\pi\sqrt{0.08}$ s (Option A).
Solution by Sreeraj P, M.Sc Physics