In the arrangement shown in figure, there is a friction force between the blocks of masses m and 2m. The mass of the suspended block is m. The block of mass m is stationary with respect to block of mass 2m. The minimum value of coefficient 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)`