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(B) The moment of inertia $I$ of a body is given by $I = Mk^2$,where $M$ is the mass and $k$ is the radius of gyration.Since mass $M$ remains constant

Vedclass · Accepted Answer

$2 \alpha \Delta T$

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(B) The moment of inertia $I$ of a body is given by $I = Mk^2$,where $M$ is the mass and $k$ is the radius of gyration.Since mass $M$ remains constant,we have $I \propto k^2$.Taking the natural logarithm on both sides: $\ln I = \ln M + 2 \ln k$.Differentiating both sides,we get $\frac{dI}{I} = 2 \frac{dk}{k}$.For thermal expansion,the radius of gyration $k$ changes with temperature as $k' = k(1 + \alpha \Delta T)$,which implies $\frac{\Delta k}{k} = \alpha \Delta T$.Substituting this into the expression for the change in moment of inertia:$\frac{\Delta I}{I} = 2 \left( \frac{\Delta k}{k} \right) = 2 \alpha \Delta T$.

When the temperature of a body 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 material of the body,then $\frac{\Delta I}{I}$ is (neglect higher orders of $\alpha$ ).

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