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Q.

A small glass marble of mass ‘m’ oscillates between the two edges, inside a hemispherical glass bowl of radius ‘r’. If ‘V’ is the speed of marble at the lowest position, the normal reaction at that position is

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a

\frac{{mv}^{2}}{r}

b

\frac{{2mv}^{2}}{r}

c

\frac{{3mv}^{2}}{r}

d

\frac{{4mv}^{2}}{r}

answer is C.

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Detailed Solution

Question Image

$v^{2} - u^{2} = 2 gh$

$v^{2} = 2 gh$

$v^{2} = 2 gR (1 - \cos θ)$

$θ = 90 °$

$v^{2} = 2 gR$

$g = \frac{v^{2}}{2R}$

$N - mg = \frac{{mv}^{2}}{r}$

$N = mg + \frac{{mv}^{2}}{r}$

$= \frac{{mv}^{2}}{2R} + \frac{{mv}^{2}}{R}$

$N = \frac{{3mv}^{2}}{r}$

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