Three point masses, each of mass m, are placed at the corner of an equilateral triangle of side 1. Then the moment of inertia of this system about an axis along one side of the triangle is

Options

A. `3ml^2`
B. `ml^2`
C. `(3)/(4) ml^2`
D. `(3)/(2) ml^2`

Answer

C

Explanation

Moment of inertia about BC is `I_(BC) = m(0)^(2) + m(0)^(2) + m(AD)^(2)`
Since
`AD = sqrt(AB^2 - ((BC)/(2))^(2)) = sqrt(l^2 - (l^2)/(4) ) = (sqrt3)/(2) l`
`therefore I_(BC) = m ((sqrt3)/(2) l)^(2) = (3)/(4) ml^(2)`
Similarly , we find the moment of inertia about AB is `I_(AB) = m (0)^(2) + m(0)^(2) + m (CE)^(2)`
In right angle `Delta BEC`
`sin 60^(@) = (CE)/(BC) = (CE)/(l) or CE = l sin 60^(@) = (sqrt3)/(2) l`
`therefore I_(AB) = m ((sqrt3)/(2) l)^(2) = (3)/(4) ml^(2)`
Moment of inertia along AC is
`I_(AC) = m(0)^(2) + m(0)^(2) + m(BF)^(2)`
In right angle `Delta BFC`
`sin 60^(@) = (BF)/(BC) = (BF)/(l)` or `BF = l sin 60^(@) = l ""(sqrt3)/(2)`
`therefore I_(AC) = m ((sqrt3)/(2)""l )^(2) = (3)/(4) ml^(2)`