question_answer

                The moment of inertia of a disc of mass M and radius R about a tangent to its rim in its plane is:

A)                                                                                                                                                                                                            \[\frac{2}{3}M{{R}^{2}}\]

B)                 \[\frac{3}{2}M{{R}^{2}}\]             

C)                 \[\frac{4}{5}M{{R}^{2}}\]             

D)                 \[\frac{5}{4}M{{R}^{2}}\]

Correct Answer: D

Solution :

                Moment of inertia of a disc about its diameter is                 \[{{I}_{d}}=\frac{1}{4}M{{R}^{2}}\]                 Now, according to perpendicular axis theorem moment of inertia of disc about a tangent passing through rim and in the plane of discs                                 \[I={{I}_{d}}+M{{R}^{2}}\]                 \[=\frac{1}{4}M{{R}^{2}}+M{{R}^{2}}\]                 \[=\frac{5}{4}M{{R}^{2}}\]