Question

If the temperature of a uniform steel rod of mass $12\text{kg}$ and length $1\text{m}$ is increased by $20{}^{\circ }\text{C}$, its moment of inertia about its perpendicular bisector increases by -
$[\alpha =11\times {10}^{-6}/{}^{\circ }\text{C}]$

• A. $2.2\times {10}^{-4}{\text{kg-m}}^2$
• B. $4.4\times {10}^{-4}{\text{kg-m}}^2$
• C. $1.1\times {10}^{-4}{\text{kg-m}}^2$
• D. $6.6\times {10}^{-4}{\text{kg-m}}^2$

Answer 1

The correct option is B $4.4\times {10}^{-4}{\text{kg-m}}^2$

Given:
Change in temperature, $\Delta T={20}^{\circ }\text{C}$
Co-efficient of linear expansion of steel rod, $\alpha =11\times {10}^{-6}/{}^{\circ }\text{C}$
Length of rod, $L=1\text{m}$
Mass of rod, $m=12\text{kg}$
We know that change in moment of inertia with change in temperature is given by
$\Delta I=2\alpha I\Delta T$
$\Rightarrow \Delta I=2\times 11\times {10}^{-6}\times \frac{12\times 1^2}{12}\times 20$
$[I_{rod}=\frac{m{L}^2}{12}$ about perpendicular bisector]
$\Rightarrow \Delta I=4.4\times {10}^{-4}{\text{kg-m}}^2$