BITSAT 2019 Slot 2 Solved Paper

Block does not move upto a maximum applied force of 3N down the inclined plane.
Equation of motion is 3+mg‌sin‌30°=F
3+mg‌sin‌30°=µmg‌cos‌30° .......(i) Similarly, block also does not move upto a maximum applied force of 12 N up the plane.
Now, equation of motion is
mg‌sin‌30°+F=12
or, mg‌sin‌30°+µmg‌cos‌30°=12 .......(ii)
Solving Eqs. (i) and (ii), we have
mg‌sin‌30°=

9

2

and µmg‌cos‌30°=

15

2

Hence, dividing
∴

1

µ

‌tan‌30°=

9∕2

15∕2

=

3

2

⇒µ=

5

3√3