A tunnel is dug along a chord of non rotating earth at a distance $d = \frac{R}{2}$ [R = radius of the earth] from its centre. A small block is released in the tunnel from the surface of the earth. The block comes to rest at the centre (C) of the tunnel. Assume that the friction coefficient between the block and the tunnel wall remains constant at $μ$ .

Calculate $μ$ .

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$\frac{3}{4}$

$\frac{2}{3}$

$\frac{\sqrt{3}}{4}$

$\frac{\sqrt{3}}{2}$

answer is B.

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Detailed Solution

$N o r m a l r e a c t i o n o n t h e b l o c k a t a n y p o s i t i o n i n t h e t u n n e l, N = m g d / R = \frac{m g}{2} T h e r e f o r e f o r c e o f f r i c t i o n f = μ m g / 2 D i s p l a c e m e n t o f t h e b l o c k u p t o p o i n t C = \sqrt{R^{2} - {(\frac{R}{2})}^{2}} = \frac{\sqrt{3} R}{2} U_{c} = - \frac{G M m}{2 R} (3 - \frac{1}{4}) = - \frac{11 G M m}{8 R} a n d U_{s} = - \frac{G M m}{R} T h e r e f o r e, - (\frac{μ m g}{2}) \times \frac{\sqrt{3} R}{2} = - \frac{11 G M m}{8 R} + \frac{G M m}{R} \Rightarrow μ = \frac{\sqrt{3}}{2}$

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