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)

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