Answer:
option E
Explanation:
given,
I is moment of inertia about an axis tangent to its surface.
moment of inertia about the center of mass
[tex]I_{CM} = \dfrac{2}{5}mR^2[/tex].....(1)
now, moment of inertia about tangent
[tex]I= \dfrac{2}{5}mR^2 + mR^2[/tex]
[tex]I= \dfrac{7}{5}mR^2[/tex]...........(2)
dividing equation (1)/(2)
[tex]\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}[/tex]
[tex]\dfrac{I_{CM}}{I}=\dfrac{2}{7}[/tex]
[tex]I_{CM}=\dfrac{2}{7}I[/tex]
the correct answer is option E