(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 $M$ is constant,we have $

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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 $M$ is constant,we have $I \propto k^2$.Taking the natural logarithm on both sides: $\ln I = \ln M + 2 \ln k$.Differentiating both sides: $\frac{dI}{I} = 2 \frac{dk}{k}$.For thermal expansion,the change in length (or any linear dimension like $k$) is given by $\Delta k = k \alpha \Delta T$,which implies $\frac{\Delta k}{k} = \alpha \Delta T$.Substituting this into the expression for $\frac{\Delta I}{I}$:$\frac{\Delta I}{I} = 2 \alpha \Delta T$.

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Thermal Expansion for Solid

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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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A
$\alpha \Delta T$
B
$2 \alpha \Delta T$
C
$\frac{\Delta T}{\alpha}$
D
$\frac{2 \alpha}{\Delta T}$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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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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