question_answer

A diatomic molecule is formed by two atoms which may be treated as mass points \[{{m}_{1}}\] and \[{{m}_{2}},\] joined by a massless rod of length r. Then the moment of inertia of the molecule about an axis passing through the centre of mass and perpendicular to rod is

A)  zero         

B)  \[({{m}_{1}}+{{m}_{2}}){{r}^{2}}\]

C)  \[\left( \frac{{{m}_{1}}+{{m}_{2}}}{{{m}_{1}}{{m}_{2}}} \right){{r}^{2}}\]

D)  \[\left( \frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{r}^{2}}\]

Correct Answer: D

Solution :

Total moment of inertia will be equal to the sum of moment of inertia due to individual masses. \[I=\underset{l=1}{\mathop{\overset{n}{\mathop{\Sigma }}\,}}\,{{I}_{i}}\] where  \[{{I}_{1}}={{m}_{1}}{{r}_{1}}^{2},\,\,\,\,\,\,\,\,{{I}_{2}}{{m}_{2}}{{r}_{2}}^{2}.\] Given,          \[r={{r}_{1}}+{{r}_{2}}\] \[{{m}_{1}}\,\,{{r}_{1}}={{m}_{2}}{{r}_{2}}\] \[\therefore \] \[{{m}_{1}}\,{{r}_{1}}={{m}_{2}}(r-{{r}_{1}})\] \[\Rightarrow \] \[{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{1}}={{m}_{2}}r\] \[\Rightarrow \] \[{{r}_{1}}({{m}_{1}}+{{m}_{2}})={{m}_{2}}r\] \[\Rightarrow \] \[{{r}_{1}}=\frac{{{m}_{2}}r}{({{m}_{1}}+{{m}_{2}})}\] Also,        \[{{r}_{2}}=r-{{r}_{1}}\] \[{{r}_{2}}=r-\frac{{{m}_{2}}r}{({{m}_{1}}+{{m}_{2}})}=\frac{{{m}_{1}}r}{{{m}_{1}}+{{m}_{2}}}\] \[\therefore \] \[I={{I}_{1}}+{{I}_{2}}={{m}_{1}}{{r}_{1}}^{2}+{{m}_{2}}{{r}_{2}}^{2}\] \[I={{m}_{1}}\frac{m_{2}^{2}\,{{r}^{2}}}{{{({{m}_{1}}+{{m}_{2}})}^{2}}}+{{m}_{2}}\frac{{{m}_{1}}^{2}{{r}^{2}}}{({{m}_{1}}+{{m}_{2}})}.\] \[I=\frac{{{m}_{1}}{{m}_{2}}{{r}^{2}}}{({{m}_{1}}+{{m}_{2}})}=\frac{{{m}_{1}}{{m}_{2}}}{({{m}_{1}}+{{m}_{2}})}.{{r}^{2}}\]