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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

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

c

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

d

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

answer is D.

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Detailed Solution

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

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