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(A) The moment of inertia of a rod about an axis passing through its center and perpendicular to its length is given by:$I = \frac{1}{12} ML^2$  --- $

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$2\alpha \Delta t$

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(A) The moment of inertia of a rod about an axis passing through its center and perpendicular to its length is given by:$I = \frac{1}{12} ML^2$  --- $(i)$where $M$ is the mass and $L$ is the length of the rod.Since the mass $M$ remains constant during thermal expansion,we differentiate the expression with respect to $L$:$\Delta I = \frac{1}{12} M (2L \Delta L) = 2 (\frac{1}{12} ML^2) \frac{\Delta L}{L} = 2I \frac{\Delta L}{L}$  --- $(ii)$Dividing equation $(ii)$ by equation $(i)$,we get:$\frac{\Delta I}{I} = 2 \frac{\Delta L}{L}$  --- $(iii)$We know that the change in length due to thermal expansion is given by $\Delta L = L \alpha \Delta t$,which implies:$\frac{\Delta L}{L} = \alpha \Delta t$Substituting this into equation $(iii)$,we get:$\frac{\Delta I}{I} = 2 \alpha \Delta t$

When the temperature of a rod increases from $t$ to $(t + \Delta t)$,its moment of inertia increases from $I$ to $(I + \Delta I)$. If $\alpha$ is the coefficient of linear expansion of the rod,then the value of $(\frac{\Delta I}{I})$ is

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