Q144 / 150 · MHT CET 2025 (21 Apr Shift 1) · Physics

Three spheres, each of mass ' $m$ ' and radius ' $r$ ' are placed as shown in figure. Consider an axis $\mathrm{YY}^1$, 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 $\mathrm{YY}^1$ axis is Figure

  • A
    $\frac{7}{5} \mathrm{mr}^2$
  • B
    $\frac{2}{5} \mathrm{mr}^2$
  • Correct answer
    C
    $\frac{16}{5} \mathrm{mr}^2$
  • D
    $\frac{\mathrm{mr}^2}{2}$

Solution

Moment of inertia of Sphere 1: Since axis $\mathrm{YY}^1$ passes through its diameter, $I_1 = \frac{2}{5}mr^2$.

Moment of inertia of Sphere 2 and Sphere 3: The axis is tangential to both spheres at distance $r$ from their centers. Using the parallel axis theorem, $I_2 = I_3 = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2$.

Summing the contributions: $I_{\text{total}} = I_1 + I_2 + I_3 = \frac{2}{5}mr^2 + \frac{7}{5}mr^2 + \frac{7}{5}mr^2 = \frac{16}{5}mr^2$.

The result corresponds to option C.

Hence, the correct answer is option: 'C'.

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