PYQ NEET- Motion In A Plane L-4

Question: A particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ 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 $4 R$. The angle of projection $\theta$ is then given by#

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

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

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

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

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

Sol:#

Given, the radius of the circular path $=R$

The time taken by the particle to complete one revolution $=T$

When the particle is projected with the same speed (by which it is moving in circular orbit) at angle $\theta$ to the horizontal, the maximum height attained it is given as

$$ \begin{aligned} & H_{\max }=\frac{u^2 \sin ^2 \theta}{2 g} \ & H_{\max }=4 R \end{aligned} $$ (given) Also, we know that, speed of the particle in circular path, $$ u=\frac{2 \pi R}{T} $$

Substituting the values in the Eq. (i), we get $$ \begin{aligned} & 4 R & =\frac{\left(\frac{2 \pi R}{T}\right)^2 \sin ^2 \theta}{2 g} \ \Rightarrow \quad & \sin \theta & =\left(\frac{2 g T^2}{\pi^2 R}\right)^{1 / 2} \ \Rightarrow \quad & \theta & =\sin ^{-1}\left(\frac{2 g T^2}{\pi^2 R}\right)^{1 / 2} \end{aligned} $$