question_answer

The moment of inertia of a rod about an axis through its centre and perpendicular to it is \[\frac{1}{12}\,M{{L}^{2}}\] where M is the mass and L the length of the rod). The rod is bent in the middle so that the two halves make an angle of \[60{}^\circ \]. The moment of inertia of the bent rod about the same axis would be

A) \[\frac{1}{48}\,\,M{{L}^{2}}\]

B) \[\frac{1}{12}\,\,M{{L}^{2}}\]

C) \[\frac{1}{24}\,\,M{{L}^{2}}\]            

D) \[\frac{M{{L}^{2}}}{8\sqrt{3}}\,\,\]

Correct Answer: B

Solution :

Since rod is bent at the middle, so each part of it will have the same length \[\left( \frac{L}{2} \right)\] and mass \[\left( \frac{M}{2} \right)\] as shown. Moment of inertia of each part through its one end \[=\,\,\frac{1}{3}\,\left( \frac{M}{2} \right)\,\,{{\left( \frac{L}{2} \right)}^{2}}\] Hence, net moment of inertia through its middle point O is \[l=\,\,\frac{1}{3}\,\left( \frac{M}{2} \right)\,\,{{\left( \frac{L}{2} \right)}^{2}}+\frac{1}{3}\,\left( \frac{M}{2} \right)\,\,{{\left( \frac{L}{2} \right)}^{2}}\] \[=\frac{1}{3}\left[ \frac{M{{L}^{2}}}{8}+\frac{M{{L}^{2}}}{8} \right]=\frac{M{{L}^{2}}}{12}\]