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

In the arrangement shown in figure, there is a friction force between

1.	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
Answer» `(1)/(2)` `(1)/(sqrt(2))` `(1)/(4)` `(1)/(3)` Solution :( 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)`