Question

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

• A. \(\frac{7}{5}\mathrm{mr}^{2}\)
• B. \(\frac{2}{5}\mathrm{mr}^{2}\)
• C. \(\frac{16}{5}\mathrm{mr}^{2}\)
• D. \(\frac{\mathrm{mr}^{2}}{2}\)

Hint:

Use the moment of inertia of a sphere about its diameter for the sphere on the axis and the parallel axis theorem for the two touching spheres.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
From the figure, the axis passes through the diameter of sphere 1 and is tangent to spheres 2 and 3. A sphere about its diameter has \(I=\dfrac25mr^2\).

Step 2: Sphere on the axis
\(I_1=\dfrac25mr^2\).

Step 3: Spheres touching the axis
The axis is at distance \(r\) from the centre of each, so by the parallel axis theorem \(I_2=I_3=\dfrac25mr^2+mr^2=\dfrac75mr^2\).

Step 4: Total
\(I=\dfrac25mr^2+\dfrac75mr^2+\dfrac75mr^2=\dfrac{16}{5}mr^2\).

Final Answer:
The moment of inertia of the system is \(\dfrac{16}{5}mr^2\). \[ \boxed{\text{(C) }\dfrac{16}{5}\mathrm{mr}^{2}} \]