JEE Main 17 March 2021 Shift 1 Solved Paper

For rolling without slipping on an inclined plane, we can write
Rmg‌sin‌θ=(mK2+mR2)α
⇒‌‌Rmg‌sin‌θ=m(K2+R2)α
⇒‌‌α=‌

Rg‌sin‌θ

K2+R2

⇒‌

α

R

=‌

g‌sin‌θ

1+‌ K2 R2

⇒‌‌a=‌

g‌sin‌θ

1+‌ K2 R2

  [∵a=‌

α

R

]... (i)
Time period, t=√‌

2s

a

...(ii)
From Eqs. (i) and (ii), we get
t=√‌

2s

g‌sin‌θ

(1+‌

k2

R2

)
For least time acceleration a should be maximum and K should be minimum and we know that K is least for solid sphere. So, time will be least for sphere.
It means the body which will reach first at the bottom of the inclined plane is 4 , i.e. solid sphere.