Question

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 $\alpha$ be the coefficient of linear expansion of the rod, then the value of $\frac{\Delta I}{I}$ is

• A. $2 \alpha \Delta t$
• B. $\alpha \Delta t$
• C. $\frac{\alpha \Delta t}{2}$
• D. $\frac{\Delta t}{\alpha}$

Answer 1

Answer: (A)

$2 \alpha \Delta t$

Answer 2

Moment of inertia of a rod
$\, \, \, \, \, \, \, \, \, \, I=\frac{M\lambda^2}{12}$
$\, \, \, \, \, \, \, \, \, \, dI =\frac{M}{12} 2\lambda d\lambda \, \, \, \, \, \, \, \, \,$ (on differenciate)
$\, \, \, \, \, \, \, \, \, \, \frac{dI}{I}=2 \frac{d \lambda}{\lambda}=2 \alpha \Delta t$