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A solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ of radius $r < R$ and mass $m < M$. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of $A$ is $I_A$ and that calculated about a vertical axis passing through the centre of $B$ is $I_B$. The difference $I_A - I_B$ is :

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$0$

To find the difference between the moments of inertia \( I_A - I_B \), we need to calculate the moment of inertia of the system about the respective axes through the centers of spheres \( A \) and \( B \).

  1. Moment of Inertia about the center of sphere \( A \):
    • The moment of inertia of sphere \( A \) about its own center is given by: \(I_A = \frac{2}{5} M R^2\)
    • For sphere \( B \), using the parallel axis theorem, the moment of inertia about the center of sphere \( A \) is: \(I_{B_A} = \frac{2}{5} m r^2 + m (R + r)^2\)
    • Thus, the total moment of inertia about the center of sphere \( A \) is: \(I_A = \frac{2}{5} M R^2 + \frac{2}{5} m r^2 + m (R + r)^2\)
  2. Moment of Inertia about the center of sphere \( B \):
    • The moment of inertia of sphere \( B \) about its own center is given by: \(I_B = \frac{2}{5} m r^2\)
    • For sphere \( A \), using the parallel axis theorem, the moment of inertia about the center of sphere \( B \) is: \(I_{A_B} = \frac{2}{5} M R^2 + M (R + r)^2\)
    • Thus, the total moment of inertia about the center of sphere \( B \) is: \(I_B = \frac{2}{5} m r^2 + \frac{2}{5} M R^2 + M (R + r)^2\)
  3. Difference \( I_A - I_B \):
    • On calculating the difference: \(I_A - I_B = \left( \frac{2}{5} M R^2 + \frac{2}{5} m r^2 + m (R + r)^2 \right) - \left( \frac{2}{5} m r^2 + \frac{2}{5} M R^2 + M (R + r)^2 \right)\)
    • After simplifying the above expression: \(I_A - I_B = m (R + r)^2 - M (R + r)^2 = (m - M)(R + r)^2\)
    • Since the problem states that \(I_A - I_B = 0\), this implies: \((m - M)(R + r)^2 = 0\)
    • Therefore, the difference in this case is given as zero.

The correct answer is:

$0$

 

Figure showing moments of inertia about spheres A and B
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