JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, mass of block, m=1kg
Acceleration due to gravity, g=9.8ms−2
Inclination, θ=30∘
Electric field, E=200N∕C
Coefficient of friction, µ=0.2
Charge, q=5mC=5×10−3C
Let friction force, f=µN
where, N be the normal reaction.
Since, net force is zero along the perpendicular direction of incline. Therefore, force along Y-axis will be zero.
⇒‌‌N‌‌=mg‌cos‌30∘+qE‌sin‌30∘
⇒‌‌N‌‌=1×9.8×√3∕2+5×10−3×200×1∕2
‌‌=8.49+0.5
‌‌=8.99N∼eq9N
∴‌‌f=µN=‌

2

10

×9=‌

18

10

=1.8N
Now, total force along the plane of incline,
‌mg‌sin‌30∘−f−qE‌cos‌30∘=ma‌
⇒‌1×9.8×1∕2−1.8−5×10−3×200√3∕2‌=a
⇒‌5−1.8−1.732∕2‌=a
⇒‌a=2.34ms−2
Since, initial velocity of body, u=0ms−1 and distance along incline, s=h∕sin‌30∘=1∕sin‌30∘=2
By using second equation of motion,
‌s‌‌=ut+1∕2at2
⇒‌2‌‌=0+1∕2×2.34×t2
⇒‌t2‌‌=‌

4

2.34

⇒‌t=‌

2

√2.34