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NEET 2021
Exam
NEET
Year
2021
Subject
Physics
Chapter
Motion in Two Dimensions

NEET 2021 · Physics · Motion in Two Dimensions

A particle moving in a circle of radius R with a uniform speed...

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___

Options

  1. A\theta = \cos^{- 1}(gT2π2R)12\left(\dfrac{g{T}^{2}}{{\pi}^{2}R}\right)\dfrac{1}{2}
  2. B\theta = \cos^{- 1}{\left(\dfrac{{\pi}^{2}R}{g{T}^{2}}\right)}^{\dfrac{1}{2}}
  3. C\theta = \sin^{- 1}{\left(\dfrac{{\pi}^{2}R}{g{T}^{2}}\right)}^{\dfrac{1}{2}}
  4. D\theta = \sin^{- 1}(2gT2π2R)12\left(\dfrac{2g{T}^{2}}{{\pi}^{2}R}\right)\dfrac{1}{2}

Correct answer

D. \theta = \sin^{- 1}(2gT2π2R)12\left(\dfrac{2g{T}^{2}}{{\pi}^{2}R}\right)\dfrac{1}{2}

Step-by-step solution

T=2πω=2\piRv⇒v=2\piRTT = \dfrac{2\pi}{\omega} = \dfrac{2\piR}{v} ⇒ v = \dfrac{2\piR}{T}

H=v2sin⁡2θ2g⇒4R=4π2R2sin⁡2θ2gT2H = \dfrac{{v}^{2}\sin^{2}\theta}{2g} ⇒ 4R = \dfrac{4{\pi}^{2}{R}^{2}\sin^{2}\theta}{2g{T}^{2}}

⇒sin⁡2θ=2gT2π2R⇒ \sin^{2}\theta = \dfrac{2g{T}^{2}}{{\pi}^{2}R}

⇒sinθ=(2gT2π2R)1/2⇒ sin\theta = {\left(\dfrac{2g{T}^{2}}{{\pi}^{2}R}\right)}^{1 / 2}

⇒θ=sin⁡−1(2gT2π2R)12⇒ \theta = \sin^{- 1}\left(\dfrac{2g{T}^{2}}{{\pi}^{2}R}\right)\dfrac{1}{2}

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