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Increase in moment of inertia of a uniform rod of length $l$ ( $α$ =coefficient of linear expansion) about its perpendicular bisector when its temperature is slightly increased by $Δ T$ is

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$\frac{1}{2} l α Δ T$

$2 l α Δ T$

$\frac{2}{3} l α Δ T$

$l α Δ T$

answer is C.

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Detailed Solution

Moment of inertia $I = \frac{1}{12} m l^{2} \Rightarrow I \propto l^{2}$

$\frac{Δ I}{I} = 2 \frac{Δ l}{l}$ But $\frac{Δ l}{l} = α Δ t$

$∴ \frac{Δ T}{I} = 2 \times Δ t (o r) Δ I = 2 α I ∆ T$

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