Question

If $I$ is the moment of inertia of a solid body having $\alpha$-coefficient of linear expansion then the change in $I$ corresponding to a small change in temperature $\Delta T$ is

• A. $\alpha I\Delta T$
• B. $\frac{1}{2}\alpha I\Delta T$
• C. $2\alpha I\Delta T$
• D. $3\alpha I\Delta T$

Answer 1

Answer: (C)

$2\alpha I\Delta T$

$I=M{R}^2$

Now $R_1=R(1+\alpha \Delta T)$

$R_1^2=R^2(1+\alpha \Delta T{)}^2=R^2(1+2\alpha \Delta T)$ , we can neglect ${\alpha }^2$ as it is very small.

so new moment of Inertia $I_1=M{R}^2(1+2\alpha \Delta T)$

so $I_1-I=2M{R}^2\alpha \Delta T=2\alpha I\Delta T$