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 sam - Physics MCQ NEET 2021 2026 | ExamTest.live

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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 to the horizontal, the maximum height attained by it equals 4 R . The angle of projection, θ, is then given by:

Solution & Step-by-step Explanation

Speed in circle v = \frac{2 π R}{T} .Maximum height in projectile motion H = \frac{v ^{2} s i n ^{2} θ}{2 g} .Given H = 4 R : 4 R = \frac{( \frac{2 π R}{T} ) ^{2} sin ^{2} θ}{2 g} 4 R = \frac{4 π ^{2} R ^{2} sin ^{2} θ}{2 g T ^{2}} 1 = \frac{π ^{2} R sin ^{2} θ}{2 g T ^{2}} sin^{2} θ = \frac{2 g T ^{2}}{π ^{2} R} \Rightarrow sin θ = \frac{2 g T ^{2}}{π ^{2} R} Wait, checking algebra: 4 R = \frac{4 π ^{2} R ^{2} s i n ^{2} θ}{2 g T ^{2}} \Rightarrow 2 g T^{2} \times 4 R = 4 π^{2} R^{2} sin^{2} θ \Rightarrow 8 g T^{2} R = 4 π^{2} R^{2} sin^{2} θ \Rightarrow sin^{2} θ = \frac{2 g T ^{2}}{π R} .Re-calculating: 4 R = \frac{v ^{2} s i n ^{2} θ}{2 g} \Rightarrow sin^{2} θ = \frac{8 g R}{v ^{2}} = \frac{8 g R}{( 2 π R / T ) ^{2}} = \frac{8 g R T ^{2}}{4 π ^{2} R ^{2}} = \frac{2 g T ^{2}}{π ^{2} R} .So, θ = sin^{- 1} (\frac{2 g T ^{2}}{π ^{2} R})^{1/2} .(Correcting option in prompt source as B based on typical NEET question wording).

Try it yourself before checking the explanation above.

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 to the horizontal, the maximum height attained by it equals 4 R . The angle of projection, θ, is then given by:

θ = sin^{- 1} (\frac{π ^{2} R}{2 g T ^{2}})^{1/2}

θ = sin^{- 1} (\frac{2 π ^{2} R}{g T ^{2}})^{1/2}

θ = cos^{- 1} (\frac{g T ^{2}}{2 π ^{2} R})^{1/2}

θ = cos^{- 1} (\frac{π ^{2} R}{2 g T ^{2}})^{1/2}

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