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: [AIPMT 1999]

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

\[I={{I}_{d}}+M{{R}^{2}}\]

\[=\frac{1}{4}M{{R}^{2}}+M{{R}^{2}}\]

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