question_answer

Moment of inertia of a ring having mass ?m? and radius ?r? about an axis which is tangent in the plane of the ring will be:

A)  \[2m{{r}^{2}}\]  

B)                     \[\frac{m{{r}^{2}}}{2}\]

C)  \[\frac{3}{2}m{{r}^{2}}\]       

D)     \[m{{r}^{2}}\]

Correct Answer: C

Solution :

\[\therefore {{I}_{x}}\,+{{I}_{y}}\,={{I}_{z}}\]             \[{{I}_{x}}={{I}_{y}}\]             \[\Rightarrow \,\,2{{I}_{x}}\,={{I}_{z}}\]             \[\Rightarrow \,{{I}_{x}}\,=\frac{{{I}_{z}}}{2}\,=\frac{m{{r}^{2}}}{2}\]             Require \[{{I}_{PQ}}\,={{I}_{x}}\,+m{{r}^{2}}\,=\frac{3}{2}\,m{{r}^{2}}\]