The moment of inertia of a ring about one of its diameter is `2 kg m^(2)` . What will be its moment of inertia about a tangent parallel to the diameter?

Options

A. `2 kg m^(2)`
B. `6 kg m^(2)`
C. `4 kh m^(2)`
D. `5 kg m^(2)`

Answer

B

Explanation

`I_(d)=(MR^(2))/(2)=2 " " therefore I_(t)=(3)/(2) MR^(2)`