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NEET PHYSICSMedium

A particle moving in a circle of radius RR with a uniform speed takes a time TT to complete one revolution. If this particle were projected with the same speed at an angle θ\theta to the horizontal, the maximum height attained by it equals 4R4R. The angle of projection, θ\theta, is then given by :

A

θ=cos⁡−1(gT2π2R)1/2\theta = \cos^{-1} \left( \frac{gT^2}{\pi^2 R} \right)^{1/2}

B

θ=cos⁡−1(π2RgT2)1/2\theta = \cos^{-1} \left( \frac{\pi^2 R}{gT^2} \right)^{1/2}

C

θ=sin⁡−1(π2RgT2)1/2\theta = \sin^{-1} \left( \frac{\pi^2 R}{gT^2} \right)^{1/2}

D

θ=sin⁡−1(2gT2π2R)1/2\theta = \sin^{-1} \left( \frac{2gT^2}{\pi^2 R} \right)^{1/2}

Step-by-Step Solution

To complete a circular path of radius RR, time period is TT. So speed of particle U=2πRTU = \frac{2\pi R}{T}. Now the particle is projected with same speed at angle θ\theta to horizontal. Maximum Height H=U2sin⁡2θ2gH = \frac{U^2 \sin^2 \theta}{2g}. Given H=4RH = 4R, so U2sin⁡2θ2g=4R\frac{U^2 \sin^2 \theta}{2g} = 4R. Substituting U=2πRTU = \frac{2\pi R}{T}, we get 12g(4π2R2T2)sin⁡2θ=4R\frac{1}{2g} \left( \frac{4\pi^2 R^2}{T^2} \right) \sin^2 \theta = 4R. Simplifying, sin⁡2θ=8gRT24π2R2=2gT2π2R\sin^2 \theta = \frac{8gR T^2}{4\pi^2 R^2} = \frac{2gT^2}{\pi^2 R}. Thus, θ=sin⁡−1(2gT2π2R)1/2\theta = \sin^{-1} \left( \frac{2gT^2}{\pi^2 R} \right)^{1/2}.

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