JEE Main 26 Feb 2021 Shift 1 Solved Paper

Given, mass of block, m=√3kg
Coefficient of friction, µ=1∕3√3
According to diagram,Let F be the force applied on the body,
w be the weight (=mg),
N be the normal reaction.
Friction force f=µN
For no movement of body along X-axis, net force along X-axis should be zero.
If, Fy be the net force along y-axis then it will also be zero because body is not accelerating at all.
∴‌N=F‌sin‌60∘+mg
⇒‌N=‌

√3

2

F+10√3 . . . (i)
∴‌N=F‌sin‌60∘
⇒‌N=‌

√3

2

F+1
‌ Similarly, ‌Fx=F‌cos‌60∘‌−µN=0
From Eq. (i), we get
⇒‌‌

F

2

−‌

1

3√3

(‌

√3

2

F+10√3)=0
⇒‌‌

F

2

=‌

F

6

+‌

10

3

⇒‌

F

2

−‌

F

6

=‌

10

3

⇒‌F=10N
‌ Given, ‌‌F=3x
⇒‌x=‌

10

3

=3.33