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(C) Let the body leave the surface of the hemisphere at point $P$. At this point,the radius vector makes an angle $	heta$ with the vertical.The force

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$60$

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(C) Let the body leave the surface of the hemisphere at point $P$. At this point,the radius vector makes an angle $	heta$ with the vertical.The forces acting on the body at point $P$ are the gravitational force $mg$ (acting downwards) and the normal reaction $R$ (acting radially outwards).The component of gravitational force towards the center is $mg \cos 	heta$.The net centripetal force required for circular motion is provided by the difference between the radial component of gravity and the normal reaction:$mg \cos 	heta - R = \frac{mv^2}{r}$When the body leaves the surface,the normal reaction $R$ becomes zero.Therefore,$mg \cos 	heta = \frac{mv^2}{r}$.Substituting the given values $v = 5 \ m/s$,$r = 5 \ m$,and $g = 10 \ m/s^2$:$\cos 	heta = \frac{v^2}{rg} = \frac{5^2}{5 	imes 10} = \frac{25}{50} = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,we have $	heta = 60^{\circ}$.

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