JEE Main 25 Feb 2021 Shift 2 Solved Paper

Given, radius of sphere is a, mass of sphere is m, horizontal speed of sphere is v0.
Moment of inertia of solid sphere, I=2∕5ma2
Let, h be the height of inclined plane and ∕ be the length of inclined plane.By using law of conservation of energy,
Total energy at 1= Total energy at 2
∴‌‌PE1+KE1=PE2+KE2
⇒‌‌0+1∕2mv02+1∕2∕ω2=mgh. . . (i)
As, ‌‌v0=aω⇒ω=‌

v0

a

∴‌‌‌

1

2

mv02+‌

1

2

(‌

2

5

ma2)‌

v02

a2

=mgh
⇒‌‌‌

1

2

mv02(1+2∕5)=mgh
⇒‌‌‌

7v02

10

=gh . . . (ii)
As, ‌‌sin‌θ=h∕l
⇒‌‌h=I‌sin‌θ
Put this value in Eq. (ii), we get
⇒ ‌

7

10

v02=gl‌sin‌θ⇒l=‌

7v02

10g‌sin‌θ