JEE Advanced 2021 Full Test 2 Paper 1

For the impending motion, block A must slip up and block C down the inclined plane. Since the normal force between A and B is less than that between block B and C, the maximum frictional force (limiting friction) will be reached first between A and B while B and C will stay together.
Writing equilibrium equations:

∑Fy=0
NA−WA‌cos‌30°=0
NA=WA‌cos‌30°
NA=20√3N
Also, for impending motion if FAB is frictional force between blocks A and B, then
FAB=µs.NA=20√3µsN ........(1)
∑Fx=0
T−WA‌sin‌30°−FAB=0
T−40

1

2

−20√3µs=0
T=20(1+√3µs)
From FBD of block B and C combinedWriting equilibrium equation
∑Fy=0
Nc−NA−(WB+Wc)‌cos‌30°=0
Nc−20√3−110

√3

2

=0
Nc=75√3N
Also, for impending motion:
Fc=µs.Nc=75√3µs .......(3)
For ∑Fx=0, we have
TA+(FBA+Fc)−(WB+WC)‌sin‌30°=0
T+[20√3+75√3µs]−

110

2

=0
T=(55−95√3µs)
Since tension is same, so from (2) and (4), we get
20(1+√3µs)=(55−95√3µs)
Solving for µs we get, 115√3µs=35
or µs=

35

115√3

=0.1757
∴ Minimum µs=0.1757