A diatomic molecule is formed by two atoms which may be treated as mass points $m_{1}$ and $m_{2}$ joined by a massless rod of length $r$ . Then the moment of inertia of molecule about an axis passing through center of mass and perpendicular to the rod is:

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(m_{1} + m_{2}) r^{2}

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(\frac{m_{1} m_{2}}{m_{1} + m_{2}}) r^{2}

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(\frac{m_{1} + m_{2}}{m_{1} m_{2}}) r^{2}

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Solution

The correct option is C $(\frac{m_{1} m_{2}}{m_{1} + m_{2}}) r^{2}$
Let the center of mass be at $x$ distance from $m_{1}$
so $m_{1} x = m_{2} (r - x)$
so $x = \frac{m_{2} r}{m_{1} + m_{2}}$
$I = m_{1} x^{2} + m_{2} (r - x)^{2} = (\frac{m_{1} m_{2}}{m_{1} + m_{2}}) r^{2}$

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