Question

A solid sphere \(A\) of radius \(R\) and mass \(M\) is attached to a smaller solid sphere \(B\) of radius \(r\) (\(r\lt R\)) and mass \(m\) (\(m\lt M\)). The centres lie on the same horizontal line. The moments of inertia about the vertical axes passing through the centres of \(A\) and \(B\) are \(I_A\) and \(I_B\), respectively. The value of \(I_A-I_B\) is:

• A. \((M-m)(R+r)^2\)
• B. \((M-m)(R-r)^2\)
• C. \((m-M)(R+r)^2\)
• D. \((m-M)(R-r)^2\)

Hint:

For composite bodies, first find the moment of inertia of each component about its own centre and then use the parallel axis theorem wherever necessary.

Answer 1

The Correct Option is C

Concept:

• Moment of inertia of a solid sphere about a diameter: \[ I=\frac{2}{5}MR^2 \]

• Parallel axis theorem: \[ I=I_{cm}+Md^2 \]

Step 1: Calculate \(I_A\)
Axis passes through the centre of sphere \(A\). For sphere \(A\), \[ I_A^{(A)} = \frac{2}{5}MR^2 \] For sphere \(B\), distance from the axis is \[ R+r \] Hence, \[ I_A^{(B)} = \frac{2}{5}mr^2 + m(R+r)^2 \] Therefore, \[ I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m(R+r)^2 \]

Step 2: Calculate \(I_B\)
Similarly, \[ I_B^{(B)} = \frac{2}{5}mr^2 \] and \[ I_B^{(A)} = \frac{2}{5}MR^2 + M(R+r)^2 \] Therefore, \[ I_B = \frac{2}{5}mr^2 + \frac{2}{5}MR^2 + M(R+r)^2 \]

Step 3: Find \(I_A-I_B\)
Subtracting, \[ I_A-I_B = m(R+r)^2 - M(R+r)^2 \] \[ I_A-I_B = (m-M)(R+r)^2 \]

Step 4: Write the final answer
\[ \boxed{ I_A-I_B = (m-M)(R+r)^2 } \] Hence, \[ \boxed{\text{Option (C)}} \]