question_answer

The moment of inertia of a disc of mass m and radius R about an axis, which is tangential to the circumference of the disc and parallel to its diameter is:                                                                            [BHU PMT-2004]

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

B)  \[\frac{2}{3}m{{R}^{2}}\]

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

D)  \[\frac{4}{5}m{{R}^{2}}\]

Correct Answer: C

Solution :

                                Key Idea: Use parallel axis theorem. From the theorem of parallel axis, the moment of inertia \[\left( I \right)\] of a body about given is equal to its moment of inertia \[I\] about its diameter, plus the product of the mass \[M\] of the body and square of perpendicular distance between the two axes. That is                                                 \[I={{I}_{d}}+M{{R}^{2}}\]                                                 \[=\frac{1}{4}M{{R}^{2}}+M{{R}^{2}}=\frac{5}{4}M{{R}^{2}}\]