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