Question

The diameter of a fly wheel is $R$. Its coefficient of linear expansion is $\alpha$ . If its temperature is increased by $\Delta T$, the percentage increase in its moment of inertia is:

• A. $200\alpha \Delta T$
• B. $100\alpha \Delta T$
• C. $50\alpha \Delta T$
• D. $150\alpha \Delta T$

Answer 1

The correct option is A $200\alpha \Delta T$

$I=M{R}^2$

So $dI=2MR\Delta R$...(1)

also we $\Delta R=R\alpha \Delta T$ putting this in equation 1

We get $dI=2M{R}^2\alpha \Delta T=2I\alpha \Delta T$

so $\frac{dI}{I}=2\alpha \Delta T$

So percentage increase is $2\alpha \Delta T\times 100=200\alpha \Delta T$