STD 11 - 6. system of particles and rotational motion — NEET STD 11 Science — Question

 $(i)$ For solid sphere, the moment of inertia about the diameter is $I _{ s }=\frac{2}{5} MR ^2$

Now $I = MK ^2$ for any body, where $K$ is radius of gyration of that body.

so $MK ^2=\frac{2}{5} MR ^2 \Rightarrow K = R \sqrt{2 / 5}$

$(ii)$ The moment of inertia of disc about an axis passing through its centre and perpendicular to plane is

$I _{ d }=\frac{ MR ^2}{2}= MK ^2 \Rightarrow K = R \sqrt{1 / 2}$

Now acceleration of any body which is rolling on an inclined plane is

$a=\frac{g \sin \theta}{1+K^2 / R^2}$

For same R, the acceleration of the body depends only on radius of gyration $K$, [see eq$(iii)$] so solid sphere will reach earlier to bottom of an inclined plane than disc.