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A small object is released from rest from the topmost point of a smooth hemispherical surface having radius R. Height from the centre at which the object leaves the surface will be

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$\frac{2 R}{3}$

$\frac{R}{2}$

$\frac{R}{3}$

$\frac{R}{4}$

answer is C.

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Detailed Solution

$v^{2} = u^{2} + 2 gh = 2 gR (1 - cosθ) . . . . . . . (1)$

$mgcosθ - N = \frac{{mv}^{2}}{R} . . . . . . . . . . . . (2)$

After Solving,

$cos θ = \frac{2}{3} \Rightarrow h = R (1 - cosθ) (from top) From center, h = Rcosθ = \frac{2 R}{3}$

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