JEE Main 17 March 2021 Shift 2 Solved Paper

As we know the moment of inertia of the solid sphere, I=‌2/5mR2Acceleration of sphere on inclined plane, a=‌

g‌sin‌θ

1+‌ I mR2

Substituting the values in the above equation, we get
a=‌

g‌sin‌30∘

1+‌ ‌ 2 5 mR2 mR2

=‌

10+1/2

(1+2/5)

=‌

5×5

7

a=3.5m/s2
At the top of incline,
v=u+at⇒0=1−3.5t
t=1/3.5s
∴ Total time = time of ascent + time of decent
As, time of ascent = time of decent
⇒ Total time =2 (time of ascent)
Total time =2(1/3.5)s
Total time =0.57s
Hence, the total time taken to sphere to return the starting point A is 0.57s.