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
$\sin\theta = 0.6$, so $\cos\theta = 0.8$ and the height of the cone is $h = l\cos\theta = 0.8$ m.
$T\cos\theta = mg$ and $T\sin\theta = m\omega^2 r = m\omega^2 l\sin\theta$ give $\omega^2 = \dfrac{g}{l\cos\theta}$:
$$T_{period} = 2\pi\sqrt{\frac{l\cos\theta}{g}} = 2\pi\sqrt{\frac{0.8}{10}} = 2\pi\sqrt{0.08}\ \text{s} \approx 1.78\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics