JEE Main 17 March 2021 Shift 2 Solved Paper

The free body diagram for the wooden block is shown belowUsing the condition of the equilibrium,
In the x-direction, the summation of all the forces is to be zero.
f−T‌‌=0
µR−T‌‌=0
T‌‌=0.5R...(i)
In the y-direction, , the summation of all the forces is to be zero.
T+R−90=0
⇒‌‌0.5R+R−90=0
R=60N
Hence, the normal force on the wooden block is 60N
Using the Eq. (i),
T=0.5(60)=30N
Hence, the maximum value of the tension in the rope,
so that wooden block will not move is 30N.