question_answer

A thin rod of length I, coefficient of linear expansion \[\alpha \] is heated, so that its temperature changes by AT. Fractional increase in moment of inertia of rod about an axis perpendicular to its length and passing through one of its end is

A) \[\alpha \,\Delta \,T\]                  

B)  \[\frac{3}{2}\alpha \,\Delta T\]

C) \[\frac{2}{3}\alpha \,\Delta T\]               

D)                                             \[\frac{4}{3}\alpha \,\Delta T\]

Correct Answer: C

Solution :

Initially , \[{{I}_{1}}=\frac{M{{L}^{2}}}{3}\]

When temperature is increased, length of rod becomes

\[l'=l\left( 1+\alpha \,\Delta T \right)\]

So, moment of inertia will be

\[{{I}_{2}}\frac{M}{3}{{\left( l\left( 1+\alpha \,\Delta T \right) \right)}^{2}}\]

\[=\frac{M}{3}({{l}^{2}}+{{l}^{2}}{{(\alpha \,\Delta T)}^{2}}+2{{l}^{2}}\alpha \Delta T)\]

Neglecting \[{{l}^{2}}{{a}^{2}}\Delta {{T}^{2}}\]  as it is very small, we have

\[{{I}_{2}}=\frac{M{{l}^{2}}}{3}+\frac{2}{3}a\Delta T\]

So,  \[{{I}_{2}}-{{I}_{1}}=\frac{2}{3}M{{l}^{2}}\alpha \Delta T\]

and \[\frac{{{I}_{2}}-{{I}_{1}}}{{{I}_{1}}}=\frac{2}{3}\alpha \Delta T\]