Question-44 Three spheres, each of mass ‘m’ and radius ‘r’ are placed as shown in figure. Consider an axis ‘YY’ which is touching two spheres and passing through the diameter of third sphere. The moment of inertia of the system consisting of these three spheres about YY’ axis is
(A) \frac{7}{5}mr^2
(B) \frac{2}{5}mr^2
(C) \frac{16}{5}mr^2
(D) \frac{mr^2}{2}
Answer : (C)
Explanation:
Moment of inertia of a solid sphere = \frac{2}{5}mr^2
Moment of inertia of the upper sphere = \frac{2}{5}mr^2
For each lower sphere M.I. = \frac{2}{5}mr^2 + mr^2
= \frac{7}{5}mr^2
Total moment of inertia I = \frac{2}{5}mr^2 + 2 \times \frac{7}{5}mr^2
= \frac{16}{5}mr^2