question_answer

A uniform 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 = \[\frac{M{{L}^{2}}}{12}+\frac{M{{R}^{2}}}{4}=\frac{M{{R}^{2}}}{2}\] R           

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

D)        \[L=\sqrt{\frac{3}{2}}R\]

Correct Answer: B

Solution :

(I) Moment of inertia of a cylinder about its centre and parallel to its length \[\beta \] (II) Moment of inertia about its centre and perpendicular to its length \[M=\left( \frac{{{L}^{2}}}{12}+\frac{{{L}^{2}}}{4} \right)\] According to question \[\frac{L}{R}\] or          \[\left( \frac{L}{R} \right)\]