In the arrangement shown in Fig there is no friction between of mass `2m` and ground but there is friction between the block of masses `m` is stationary with respect to block of mass `2m` The value of friction between `m` and `2m`is

Options

A. `(1)/(2)`
B. `(1)/(sqrt(2))`
C. `(1)/(4)`
D. `(1)/(3)`

Answer

C

Explanation

( c) Maximum acceleration due to friction of mass m over mass 2m can be `mug`. Now, for the whole system
`a=("Net pulling force")/("Total mass")`
`therefore" " mug=(mg)/(4m)" or "mu=(1)/(4)`