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The moment of inertia $I$ of uniform rod about a perpendicular bisector increases to $I+\Delta I$, if the temperature is increased slightly by $\Delta T$. If the coefficient of linear expansion is $\alpha$, then $\frac{\Delta I}{I}$ is
$$
\left(\text { Assume } \frac{\Delta T}{T} \ll 1\right)
$$
Options:
$$
\left(\text { Assume } \frac{\Delta T}{T} \ll 1\right)
$$
Solution:
2108 Upvotes
Verified Answer
The correct answer is:
$2 \alpha \Delta T$
Let initial moment of inertia of rod about perpendicular bisector,
where, $M=$ mass of rod and $\quad L=$ initial length of rod. After increasing temperature by $\Delta T$.
Now, moment of inertia,
$$
\begin{gathered}
I+\Delta I=\frac{1}{12} M(L+\Delta L)^2 \\
I\left(1+\frac{\Delta I}{I}\right)=\frac{1}{12} M L^2\left(1+\frac{\Delta L}{L}\right)^2
\end{gathered}
$$
[By thermal linear expansion, $\Delta L=L \alpha \Delta T$
$$
\left.\frac{\Delta L}{L}=\alpha \Delta t\right]
$$
$$
\begin{array}{rlrl}
\therefore I\left(1+\frac{\Delta I}{I}\right) =I\left(1+2 \frac{\Delta L}{L}\right) \\
1+\frac{\Delta I}{I} =1+2 \alpha \Delta T \\
\therefore \frac{\Delta I}{I} =2 \alpha \Delta T
\end{array}
$$
where, $M=$ mass of rod and $\quad L=$ initial length of rod. After increasing temperature by $\Delta T$.
Now, moment of inertia,
$$
\begin{gathered}
I+\Delta I=\frac{1}{12} M(L+\Delta L)^2 \\
I\left(1+\frac{\Delta I}{I}\right)=\frac{1}{12} M L^2\left(1+\frac{\Delta L}{L}\right)^2
\end{gathered}
$$
[By thermal linear expansion, $\Delta L=L \alpha \Delta T$
$$
\left.\frac{\Delta L}{L}=\alpha \Delta t\right]
$$
$$
\begin{array}{rlrl}
\therefore I\left(1+\frac{\Delta I}{I}\right) =I\left(1+2 \frac{\Delta L}{L}\right) \\
1+\frac{\Delta I}{I} =1+2 \alpha \Delta T \\
\therefore \frac{\Delta I}{I} =2 \alpha \Delta T
\end{array}
$$
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