A hemispherical 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 the surface of bowl is smooth, and the angle made by the radius through the block with the vertical is $θ$ , the angular speed at which the bowl is rotating is

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$\sqrt{\frac{g}{Rcos θ}}$

$\sqrt{\frac{gcos θ}{R}}$

$\sqrt{\frac{gsin θ}{R}}$

$\sqrt{\frac{g}{Rsin θ}}$

answer is A.

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Detailed Solution

$Nsin θ = Mr ω^{2}; Ncos θ = Mg$
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$\begin{array}{l} \tan θ = \frac{r ω^{2}}{g} = Rsin θ \frac{ω^{2}}{g}; \\ ω = \sqrt{\frac{g}{Rcos θ}} \end{array}$

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