Q 11-04-201JEE MainJEE Main 2017 (9 Apr)Easy
A conical pendulum of length $l$ makes an angle $\theta = 45^\circ$ with respect to $Z$-axis and moves in a circle in the $XY$ plane. The radius of the circle is $0.4$ m and its center is vertically below $O$. The speed of the pendulum, in its circular path, will be (Take $g = 10\ \text{m s}^{-2}$):
Answer: (C) $2\ \text{m s}^{-1}$
Resolving the string tension: $T\cos\theta = mg$ and $T\sin\theta = \dfrac{mv^2}{r}$. Dividing,
$$\tan\theta = \frac{v^2}{rg} \;\Rightarrow\; v^2 = rg\tan45^\circ = 0.4\times10\times1 = 4$$
$$v = 2\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics