question_answer

A particle of mass m is attached to q a thin uniform rod of length a and mass 4 m. The distance of the particle from the centre of mass of the rod is a/4. The moment of inertia of the combination about an axis passing through 0 normal  to the rod is               

A) \[\frac{64}{48}m{{a}^{2}}\]

B) \[\frac{91}{48}m{{a}^{2}}\]

C) \[\frac{27}{48}m{{a}^{2}}\]

D) \[\frac{51}{48}m{{a}^{2}}\]

Correct Answer: B

Solution :

[b] Moment of inertia \[=m{{\left( \frac{3a}{4} \right)}^{2}}+{{m}_{1}}\frac{{{a}^{2}}}{3}\] For the centre of rod  \[\left( \frac{{{m}_{1}}{{a}^{2}}}{12}+\frac{{{m}_{1}}{{a}^{2}}}{4} \right)=\frac{{{m}_{1}}{{a}^{2}}}{3}\] \[\therefore \,\,\,{{m}_{1}}=4m\] Total \[I=m{{\left( \frac{3a}{4} \right)}^{2}}+\frac{4m{{a}^{2}}}{3}\] \[=\frac{9m{{a}^{2}}}{16}+\frac{4m{{a}^{2}}}{3}\] \[=\frac{(27+64)}{48}m{{a}^{2}}=\frac{91}{48}m{{a}^{2}}\]