Q.1 The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height h from rest without sliding is

(a) 107gh−−−−√$\sqrt{\frac{10}{7}gh}$

(b) gh−−√$\sqrt{gh}$

(c) 65gh−−−−√$\sqrt{\frac{6}{5}gh}$

(d) 43gh−−−−√

Answer:-

Here the sphere is rolling ,                               Hence we will use formula ,     

V² = (2gh)/(1+K²/R²)

Here the term K²/R² , is the multiplier of

moment of inertia,(MI).

The MI for the solid sphere is 2/5 mR²

Therefore value of the multiplier in this case 

will be 2/5.

You will get extra information from the page given below.

After calculation we will conclude option a.
as correct one.