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

Step 1. Set up energy conservation.

The sphere starts from rest at height $h$ and rolls without slipping to the bottom. No energy is lost to friction (static friction does no work when there's no slipping at the contact point), so all the lost gravitational PE converts into kinetic energy — both translational (of the center of mass) and rotational (spin about the center):

Step 2. Substitute the solid sphere's moment of inertia and the rolling condition.

For a solid sphere, $I = \dfrac25 mR^2$, and pure rolling means $\omega = v/R$. Substituting:

Step 3. Combine the kinetic-energy terms.

Step 4. Solve for $v$.

The mass $m$ cancels from both sides:

Step 5. Match against the printed options.

Among the four choices — $\sqrt{4gh/3}$, $\sqrt{10gh/7}$, $\sqrt{2gh}$, $\sqrt{gh/2}$ — the derivation above lands exactly on $\sqrt{10gh/7}$, i.e. option (b).

Step 6. Rule out the other options physically. …

Energy conservation for rolling without slipping: mgh = ½mv² + ½Iω², with …

• Forgetting the rotational kinetic energy term entirely and using plain free-fall $v=\sqrt{2gh}$ — that's only correct for frictionless sliding, not rolling. …