Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 $4R$ . The angle of projection, $\theta$ is then given by:

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

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

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

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

Step-by-Step Solution

Analyze Circular Motion: The particle moves in a circle of radius $R$ with uniform speed $v$ and time period $T$ . The speed is distance divided by time: $v = \frac{2\pi R}{T}$
Analyze Projectile Motion: The particle is projected with this same speed $v$ at an angle $\theta$ . The maximum height $H$ attained in projectile motion is given by: $H = \frac{v^2 \sin^2 \theta}{2g}$
Apply Given Condition: We are given that the maximum height equals $4R$ . So, $H = 4R$ . Substituting the expression for $H$ : $4R = \frac{v^2 \sin^2 \theta}{2g}$
Substitute and Solve: Substitute the expression for $v$ derived in step 1 into the equation: $4R = \frac{\left(\frac{2\pi R}{T}\right)^2 \sin^2 \theta}{2g}$ $4R = \frac{\frac{4\pi^2 R^2}{T^2} \sin^2 \theta}{2g}$ $4R = \frac{2\pi^2 R^2 \sin^2 \theta}{g T^2}$ Rearranging to solve for $\sin^2 \theta$ : $\sin^2 \theta = \frac{4R \cdot g T^2}{2\pi^2 R^2}$ $\sin^2 \theta = \frac{2g T^2}{\pi^2 R}$ Taking the square root: $\sin \theta = \left(\frac{2g T^2}{\pi^2 R}\right)^{1/2}$ $\theta = \sin^{-1}\left[\left(\frac{2g T^2}{\pi^2 R}\right)^{1/2}\right]$

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