Consider the diagram shown in which two masses of m and 2m are placed on a fixed triangular wedge. The coefficient of friction between block A and the wedge is 2/3, while that for block B and the wedge is 1/3. If the whole system is released from rest, then acceleration of block A is?

Correct option is A. Zero
Given that μa=2/3 and μb=1/3
Let the acceleration of the whole system be a
Then
As per the diagram
For the block B
2mgcos45° + 2ma = T + μbNb ————————-(1)
Nb = 2mgsin45°
And for block A
T = mgcos45° + μaNa
Na = mgsin45°
So, T = mgcos45° + μa mgsin45° ————————-(2)
From eq (1) and (2)
2mgcos45° + 2ma = mgcos45° + μa mgsin45° + 2μb mgsin45°
After solving we get
a = 0
So, after acceleration of block A will be zero