Q 11-06-100JEE MainJEE Main 2024 (31 Jan, Shift 2)Medium
Two identical spheres each of mass $2\ \text{kg}$ and radius $50\ \text{cm}$ are fixed at the ends of a light rod so that the separation between the centres is $150\ \text{cm}$. Then, the moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is $\dfrac{x}{20}\ \text{kg m}^2$, where the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 53
Each centre is $0.75\ \text{m}$ from the axis. Using the parallel axis theorem for each solid sphere:
$$I = 2\left[\tfrac25(2)(0.5)^2 + 2(0.75)^2\right] = 2[0.2 + 1.125] = 2.65\ \text{kg m}^2 = \frac{53}{20}\ \text{kg m}^2$$
$x = 53$.
Solution by Sreeraj P, M.Sc Physics