JEE Physics 2023 Question with Solution

Given: A solid sphere has mass $5 \, \text{kg}$ and a disc has mass $4 \, \text{kg}$. Both have the same radius $R$.

Find: The value of $x$ if the ratio of the moment of inertia of the disc about a tangent in its plane to that of the sphere about its tangent is $\frac{x}{7}$.

For the solid sphere about a tangent, by the parallel axis theorem,

Substituting $m_1 = 5 \, \text{kg}$,

For the disc about a tangent in its plane,

Substituting $m_2 = 4 \, \text{kg}$,

Now,

Comparing with $\frac{x}{7}$, we get

Therefore, the required value of $x$ is $5$.

Given: A solid sphere and a disc have equal radius $R$, with masses $5 \, \text{kg}$ and $4 \, \text{kg}$ respectively.

Find: The number $x$ in the ratio $\frac{x}{7}$.

Use the known results:

• Solid sphere about its centre: $I_{\text{CM}} = \frac{2}{5}mR^2$
• Disc about a diameter in its plane: $I = \frac{1}{4}mR^2$

For each body, shift the axis from the centre to the tangent using the parallel axis theorem.

Sphere:

With $m_1 = 5$,

Disc:

With $m_2 = 4$,

Hence,

So the numerator is $x = 5$. The correct answer is $5$.