A wire of length l and mass m is bent in the form of a rectangle ABCD with `(AB)/(BC)=2.` The moment of inertia of this wire frarne about the side BC is

Options

A. `11/(252) m I^(2)`
B. `8/(203)m I^(2)`
C. `5/136 m I^(2)`
D. `7/(162) mI^(2)`

Answer

D

Explanation

`(AB)/(BC)=2 therefore AB = DC =1/3` and `BC = AD = 1/6`
Similarly, `m_(AB) = m_(DC)=m/3` and `m_(BC) = m_(AD) = m/6`
Now, `I=2I_(AB) + I_(AD) + I_(BC)`
`=2{m/3 (1/3)^(2) .1/3}+ (m/6)(1/3)^(2) +0=7/162 mI^(2)`