Previous Year NEET Question- Optics L-2

Question: A particle moving in a circle of radius $\mathrm{R}$ with a uniform speed takes a time $\mathrm{T}$ to complete one revolution.#

If this particle were projected with the same speed at an angle ’ $\theta$ ’ to the horizontal, the maximum height attained by it equals $4 \mathrm{R}$. The angle of projection, $\theta$, is then given by :

A) $\theta=\sin ^{-1}\left(\frac{2 g T^2}{\pi^2 R}\right)^{1 / 2}$

B) $\theta=\cos ^{-1}\left(\frac{g T^2}{\pi^2 R}\right)^{1 / 2}$

C) $\theta=\cos ^{-1}\left(\frac{\pi^2 R}{g T^2}\right)^{1 / 2}$

D) $\theta=\sin ^{-1}\left(\frac{\pi^2 R}{g T^2}\right)^{1 / 2}$

Answer: $\theta=\sin ^{-1}\left(\frac{2 g T^2}{\pi^2 R}\right)^{1 / 2}$#

Sol:#

$\begin{aligned} & T=\frac{2 \pi R}{V} \ & V=\frac{2 \pi R}{T} \ & H=\frac{u^2 \sin ^2 \theta}{2 g} \ & 4 R=\frac{4 \pi^2 R^2 \sin ^2 \theta}{T^2 2 g} \ & \sin ^2 \theta=\frac{8 R T^2 g}{4 \pi^2 R^2} \ & \sin ^2 \theta=\sqrt{\frac{2 T^2 g}{\pi^2 R}} \ & \theta=\sin ^{-1}\left(\frac{2 T^2 g}{\pi^2 R}\right)^{1 / 2} \end{aligned}$