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A hemispherical smooth bowl of radius R is set rotating about its axis of symmetry which is kept vertical. A small block kept in the bowl rotates with the bowl without slipping on its surface. If ' $θ$ ' is the angular position of block with the vertical, the angular speed of rotation of bowl is

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$\sqrt{\frac{gcos θ}{R}}$

$\sqrt{\frac{gsin θ}{R}}$

$\sqrt{\frac{g}{Rsin θ}}$

$\sqrt{\frac{g}{Rcos θ}}$

answer is D.

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Detailed Solution

$Ncos θ = mg Nsin θ = {mrω}^{2} Where r = Rsin θ$

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