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A rod of length L is composed of a uniform length 1/2 L of wood whose mass is m_w and a uniform length 1/2 L of brass whose mass is m_b. The moment of inertia I of the rod about an axis perpendicular to the rod and through its centre is equal to [AMU 2012]

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$(m_{w} + m_{b}) L^{2} / 2$

$(m_{w} + m_{b}) L^{2} / 12$

$(m_{w} + m_{b}) L^{2} / 6$

$(m_{w} + m_{b}) L^{2} / 3$

answer is A.

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Detailed Solution

The given situation as shown in the figure.

Moment of inertia for the wooden part,

$M_{1} = m_{w} \times \frac{(L / 2)^{2}}{3} = m_{w} \frac{L^{2}}{12}$

Moment of inertia for the brass part,

$M_{2} = m_{b} \times \frac{(L / 2)^{2}}{3} = m_{b} \times \frac{L^{2}}{12}$

Total moment of inertia,

$M = M_{1} + M_{2} = m_{w} \frac{L^{2}}{12} + m_{b} \times \frac{L^{2}}{12} = (m_{w} + m_{l}) \frac{L^{2}}{12}$

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