Chain of mass `M` length `L` kept on rough sphere. `mu` is coefficient of friction between sphere and chain `F` is minimum force required to slide chain

Options

A. `( 2 mu Mg)/(pi)`
B. `mu` Mg
C. `gt ( 2mu Mg)/(pi)`
D. `2 mu Mg`

Answer

C

Explanation

If we consider

dN = `Delta "mg cos" theta`
`F = mu dN = mu(M)/(pi)g underset((-pi)/(2))overset(pi//2)(int) "cos" theta d theta`
`F = (2 mu "Mg")/(pi)` but dN = `Delta "mg cos" theta + T d theta`
So , the frictional force F `gt ( 2 mu mg)/(pi)` hence `F gt ( 2mu mg)/(pi)`