Verified Solution • 2025–26 Session moderate

QNEET-2021-Ph-652: NEET 2021 Physics - Class 11 NEET-2021-Ph-652 Solution: A particle moving in a circle of radius R with a uniform speed…

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 'θ\theta' to the horizontal the maximum height attained by it equals 4R. The angle of projection θ\theta is then given by: θ=sin⁡−1(2gT2π2R)1/2\theta = \sin^{-1}(\frac{2gT^2}{\pi^2R})^{1/2} θ=cos⁡−1(π2RgT2)1/2\theta = \cos^{-1}(\frac{\pi^2R}{gT^2})^{1/2} θ=cos⁡−1(gT2π2R)1/2\theta = \cos^{-1}(\frac{gT^2}{\pi^2R})^{1/2} θ=sin⁡−1(π2RgT2)1/2\theta = \sin^{-1}(\frac{\pi^2R}{gT^2})^{1/2}

Multiple Choice Options

  • A θ=sin⁡−1(2gT2π2R)1/2\theta = {\sin ^{ - 1}}{\left( {{{2g{T^2}} \over {{\pi ^2}R}}} \right)^{1/2}} _ = sin _ 1 ( � 2 R 2 g T 2 _ ) 1/2
  • B θ=cos⁡−1(gT2π2R)1/2\theta = {\cos ^{ - 1}}{\left( {{{g{T^2}} \over {{\pi ^2}R}}} \right)^{1/2}} _ = cos _ 1 ( � 2 R g T 2 _ ) 1/2
  • C θ=cos⁡−1(π2RgT2)1/2\theta = {\cos ^{ - 1}}{\left( {{{{\pi ^2}R} \over {g{T^2}}}} \right)^{1/2}} _ = cos _ 1 ( g T 2 � 2 R _ ) 1/2
  • D θ=sin⁡−1(π2RgT2)1/2\theta = {\sin ^{ - 1}}{\left( {{{{\pi ^2}R} \over {g{T^2}}}} \right)^{1/2}} _ = sin _ 1 ( g T 2 � 2 R _ ) 1/2

Step-by-Step Verified Solution

T = 2 � R V V = 2 � R T H = u 2 sin 2 _ 2 g 4 R = 4 � 2 R 2 sin 2 _ T 2 2 g sin 2 _ = 8 R T 2 g 4 � 2 R 2 sin 2 _ = 2 T 2 g � 2 R _ = sin _ 1 ( 2 T 2 g � 2 R ) 1 / 2

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