A tunnel is dug along a chord of Earth having length `sqrt(3)R` is radius of Earth. A small block is released in the tunnel from the surfasce of Earth. The particle comes to rest at the centre of tunnel. Find coefficient of friction between the block and the surface of tunnel. Ignore the effect of rotation of Earth.

Answer

`0.86`

Explanation

`a_(x) = (GM)/(R^(3))(R//2)/(cos phi)cos phi = GM(2R^(2))`
`V_(L)=-GM//R`
`V_(f)=-(GM)/(2R^(3))(3R^(2)-(R^(2))/(4))=-(GM)/(2R^(3))xx(11R^(2))/(4)=-(11GM)/(8R)`
`(sqrt(3)R mu)/(2) (GM)/(2R^(2))=(8)/(8)(GM)/(R )`
`mu=(sqrt(3))/(2)=0.87 or 0.86`