question_answer

A uniform solid cylinder has a radius R and length L. If the moment of inertia of this cylinder about an axis passing through its centre and normal to its circular face is equal to the moment of inertia of the same cylinder about an axis passing through its centre and perpendicular to its length, then

A) L=R      

B) \[L=\sqrt{3}R\]

C) \[L=\frac{R}{\sqrt{3}}\]

D)   \[L=\sqrt{\frac{3}{2}}R\]

Correct Answer: B

Solution :

[b]:      (I) Moment of inertia of a cylinder about an axis passing through its centre and normal to its circular race \[=\frac{M{{R}^{2}}}{2}\] (II) Moment of inertia of the same cylinder about an axis passing through its centre and perpendicular to its length \[=M\left( \frac{{{L}^{2}}}{12}+\frac{{{R}^{2}}}{4} \right)\] According to the question, \[\frac{M{{L}^{2}}}{12}+\frac{M{{R}^{2}}}{4}=\frac{M{{R}^{2}}}{2}\]or\[L=\sqrt{3}R\]