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Find out the increase in moment of inertia $I$ of a uniform rod (coefficient of linear expansion a) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$.
Solution
Suppose mass of rod is $\mathrm{M}$ and moment of inertia is $\mathrm{I}$. Moment of inertia of rod about perpendicular bisector,
$I=\frac{M l^{2}}{l^{2}}$
Increase in length due to increase in temperature $\Delta \mathrm{T}$, $\Delta l=l \alpha \Delta \mathrm{T}$
$\therefore$ Now, new moment of inertia of rod,
$\mathrm{I}^{\prime} =\frac{\mathrm{M}}{12}(l+\Delta l)^{2}$
$=\frac{\mathrm{M}}{12}\left(l^{2}+2 l \Delta l+\Delta l^{2}\right)$
But, $\Delta l$ is very small, $\Delta l^{2}$ is neglected.
$\mathrm{I}^{\prime}=\frac{\mathrm{M}}{12}\left(l^{2}+2 l \Delta l\right)$
$=\frac{\mathrm{M} l^{2}}{12}+\frac{\mathrm{M} l \Delta l}{6}$
$=\mathrm{I}+\frac{\mathrm{M} l \Delta l}{6}$
Similar Questions
A student records the initial length $l$, change in temperature $\Delta T$ and change in length $\Delta l$ of a rod as follows :
| S.No. | $l(m)$ | $\Delta T{(^o}C)$ | $\Delta l(m)$ |
| $(1)$ | $2$ | $10$ | $4\times 10^{-4}$ |
| $(2)$ | $1$ | $10$ | $4\times 10^{-4}$ |
| $(3)$ | $2$ | $20$ | $2\times 10^{-4}$ |
| $(4)$ | $3$ | $10$ | $6\times 10^{-4}$ |
If the first observation is correct, what can you say about observations $2,\,3$ and $4$.
