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A wire of length l and mass m is bent in...

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 wife frame about the side `BC` is

A

`11/(252) m I^(2)`

B

`8/(203)m I^(2)`

C

`5/136 m I^(2)`

D

`7/(162) mI^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`(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)`
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