Question bank/JEE Main/Physics

A small block of mass m is released from rest from point A inside
a smooth hemisphere bowl of radius R, which is fixed on ground such that
OA is horizontal. The ratio (x) of magnitude of centripetal force and
normal reaction on the block at any point B varies with $\theta$ as

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$\frac{mv^{2}}{R} = N - mg\ sin\theta$
By energy conservation,
$mgR\ sin\theta = \frac{1}{2}{mv}^{2}$
$\frac{{mv}^{2}}{R} = 2mg\ sin\theta$
$N = 3mg\ sin\theta$
$Ratio = \frac{{mv}^{2}}{RN} = \frac{2}{3}$ (Constant)
$x = \frac{2}{3}$

Source: Grass-G / P-16 Physics paper 25 Nov First.docx