A thin wire of mass M and length L is turned into a ring. The moment of inertia about its axis will be:

Options

A. `(1)/(4pi^(2))ML^(2)`
B. `(1)/(12)ML^(2)`
C. `(1)/(3pi^(2))ML^(2)`
D. `(1)/(pi^(2))ML^(2)`

Answer

A

Explanation

`2pi R=L`
` R=(L)/(2pi)`
`i=Mr^(2) =M((L)/(2pi))^(2) =(ML^(2))/(4pi ^(2))`