A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle θ to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection θ is then given by: (a) sin ⁻¹[(2gT²)/(π² R)]1/2 (b) sin ⁻¹[(π² R)/(2gT²)]1/2 (c) cos ⁻¹[(π R)/(2gT²)]1/2 (d) cos ⁻¹[(2gT²)/(π² R)]1/2

Answer: (a). Speed in the circle: v = (2π R)/(T). Maximum height: H = (v²sin ²θ)/(2g) = 4R ⇒ sin ²θ = (8gR)/(v²). sin ²θ = (8gR T²)/(4π²R²) = (2gT²)/(π² R) So the answer is (a) θ = sin ⁻¹[(2gT²)/(π² R)]1/2.

A particle moving in a circle of radius R with uniform speed takes… | JEE Main 2024 (1 Feb, Shift 1)

Q17JEE Main 2024 · 1 FEB · SHIFT 1Medium

Centripetal Acceleration & Force

A particle moving in a circle of radius $R$ with uniform speed takes time $T$ to complete one revolution.

If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to $4R$ . The angle of projection $\theta$ is then given by:

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Answer: (a)

Speed in the circle: $v = \frac{2\pi R}{T}$ .
Maximum height: $H = \frac{v^2\sin^2\theta}{2g} = 4R \Rightarrow \sin^2\theta = \frac{8gR}{v^2}$ .
$\sin^2\theta = \frac{8gR\,T^2}{4\pi^2R^2} = \frac{2gT^2}{\pi^2 R}$
So the answer is (a) $\theta = \sin^{-1}\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2}$ .

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