A body of mass M kg is on the top point of a smooth hemisphere of radius 5m. It is released to slide down the surface of hemisphere. It leaves the surface when its velocity is 5ms^–1. At this instant the angle made by the radius vector of the body with the vertical is (g = 10 ms^–2)

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30⁰

45⁰

60⁰

90⁰

answer is C.

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Detailed Solution

At the point of leaving contact, normal force is equal to the centripetal force.

$m g \cos θ = \frac{m v^{2}}{r}$
$∴ \cos θ = \frac{v^{2}}{r g} = \frac{1}{2}$
$∴ θ = 60^{\circ}$

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