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A particle is kept at rest at the top of a sphere of diameter 42 m. When disturbed slightly, it slides down. At what height 'h' from the bottom, the particle will leave the sphere

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14 m

28 m

35 m

7 m

answer is C.

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Detailed Solution

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Let ‘h’ be the height from centre.

mg(R – h) = $\frac{1}{2} m v^{2}$

$V^{2} = 2 g (R - h) \to (1)$

$ C o s θ = \frac{R - h}{R}$

To leave the hemisphere,

$\frac{m V^{2}}{R} = m g C o s θ$

$\frac{2 g (R - h)}{R} = g (\frac{R - h}{R})$

$h = \frac{2 R}{3}$

For full sphere

R + $\frac{2 R}{3} = \frac{5 R}{3}$

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