Block A of mass m and block B of mass `2m` are placed on a fixed traingular wedge by means of a massless inextensible string and a frictionless pulley as shown in figure The wedge is inclined at `45^(@)`to the horizontal on both sides .The coefficient of friction between blocks A and the wedge is `2//3` and that between block B and is released from rest find the following

The tension in the string is

Options

A. `(3)/(sqrt(5))mg`
B. `(5)/(3sqrt(2))mg`
C. `(2sqrt(2)mg)/(3)`
D. `(mg)/(5)`

Answer

c

Explanation

`f_(l_2) lt2 mg sin 45^(@)` hence friction only will not able to prevent slipping of `2m` mass .So on `2m` mass, friction will be maximum i.e. `f_(l_2)`
`T =(2sqrt(2))/(3) = mg sin 45^(@) rArr T = (2sqrt(2))/(3)mg`