JEE Physics 2024 Question with Solution

Given: A ring and a solid sphere roll down the same inclined plane without slipping. Both start from rest and have identical radii.

Find: The required numerical value in the ratio of their kinetic energies.

In pure rolling motion, static friction does no work, so the loss of gravitational potential energy is converted completely into total kinetic energy.

Since both bodies start from rest from the same height on the same inclined plane, each loses the same potential energy. Therefore, their total kinetic energies at any given vertical drop are equal.

So the ratio of kinetic energies is

the solution expresses this ratio as

Hence,

Therefore, the required answer is $7$.

Given: The same ring and solid sphere roll without slipping from rest.

Find: Check the consistency of the alternate working shown in the source.

The first approach computes

and

then forms the ratio

This comparison uses the same symbol $v$ for both bodies, which is not valid because the ring and the solid sphere do not reach the same speed after rolling down the incline. Their moments of inertia are different, so their accelerations and final speeds differ.

Hence that method is inconsistent for comparing the actual total kinetic energies of the two bodies after descending the incline.

Using energy conservation for each body from the same initial height gives equal total kinetic energy instead. That agrees with the final answer provided on the solution's:

Therefore, the correct answer is $7$.