Question

The temperature of a thin uniform rod increase by $\Delta t$ If moment of intertia 1 about an axis perpendicular to its length then its moment of increase by

• A. 0
• B. $\alpha I\Delta t$
• C. $2\alpha I\Delta t$
• D. ${\alpha }^{2}I\Delta t$

Answer 1

The correct option is C $2\alpha I\Delta t$

$\begin{array}{c}\text{Given -}\ast \text{Temperatwe of Rod increased by}\Delta t\\ \text{Let the Length of the Rod be (L) and its}\\ \text{coefficient of linear expansion be (}\alpha \text{)}\\ \text{Therefore its change in Length will be}\\ \Delta L=L_0\cdot \alpha \Delta t\end{array}$

$\begin{array}{c}\text{we have moment of Inertia of Rod about its}\\ \text{centre perpendicular to its Length}=I\\ I=\frac{M{L}^2}{12}\end{array}$

$\begin{array}{c}\Rightarrow \text{By Error analysis we get-}\\ \frac{\Delta I}{I}=2\frac{\Delta L}{L}\Delta I=\text{change in moment of}\\ \text{Inertia}\end{array}$

$\begin{array}{cc}\Rightarrow \Delta I& =\frac{2\Delta L}{L_0}\cdot I\\ \Delta I& =2(\frac{L_0\alpha \Delta t)}{L_0}\cdot I=2\alpha I\Delta t\\ \Delta I& =2\alpha I\alpha t\\ \text{Hence, option}\left(C\right)\text{is correct.}& \end{array}$