JEE Main 2023 — Rotational Motion Question with Solution

The problem requires calculating the moment of inertia of the system consisting of two identical solid spheres fixed at the ends of a light rod. We need to find the moment of inertia about an axis perpendicular to the rod and passing through its midpoint.

First, let’s identify the moment of inertia of each solid sphere about its own center, which is given by the formula:

$$I_{\text{sphere}}=\frac{2}{5}m{r}^2$$

Where:

• $$m$$ is the mass of the sphere = $$2\mathrm{kg}$$
• $$r$$ is the radius of the sphere = $$0.1\mathrm{m}$$

Substituting the values:

$$I_{\text{sphere}}=\frac{2}{5}\times 2\mathrm{kg}\times (0.1\mathrm{m}{)}^2=\frac{4}{5}\times 0.01\mathrm{kg}{\mathrm{m}}^2=0.008\mathrm{kg}{\mathrm{m}}^2$$

Now, we need the moment of inertia of the two spheres about the axis passing through the midpoint of the rod. This requires using the parallel axis theorem, which states:

$$I_{\text{total}}=I_{\text{sphere}}+m{d}^2$$

Where:

• $$d$$ is the distance from the center of the sphere to the axis through the rod’s midpoint = $$0.2\mathrm{m}$$

Calculating the additional inertia due to the parallel axis theorem for one sphere:

$$I_{\text{parallel}}=m{d}^2=2\mathrm{kg}\times (0.2\mathrm{m}{)}^2=2\mathrm{kg}\times 0.04{\mathrm{m}}^2=0.08\mathrm{kg}{\mathrm{m}}^2$$

The total moment of inertia for one sphere about the midpoint of the rod is:

$$I_{\text{one sphere, total}}=I_{\text{sphere}}+I_{\text{parallel}}=0.008\mathrm{kg}{\mathrm{m}}^2+0.08\mathrm{kg}{\mathrm{m}}^2=0.088\mathrm{kg}{\mathrm{m}}^2$$

Since there are two identical spheres, the total moment of inertia of the system is:

$$I_{\text{system}}=2\times 0.088\mathrm{kg}{\mathrm{m}}^2=0.176\mathrm{kg}{\mathrm{m}}^2$$

Converting the result to the given form:

$$0.176\mathrm{kg}{\mathrm{m}}^2=176\times 10^{-3}\mathrm{kg}{\mathrm{m}}^2$$

So, the moment of inertia of the system about the given axis is:

$$176\times 10^{-3}\mathrm{kg}{\mathrm{m}}^2$$