The moment of inertia of two equal masses each of mass m at separtion L connected by a rod of mass M about an axis passing through centre and perpendicuar to length of a rod is

Options

A. `((M+3m)L^(2))/(12)`
B. `((M+6m)L^(2))/(12)`
C. `(ML^(2))/(4)`
D. `(ML^(2))/(12)`

Answer

B

Explanation

Moment of inertia of rod about XY, `l_(1)=(ML^(2))/(12)`

`AO=OB=L//2`
Moment of inertia of two masses,
`l_(2)=m((L)/(2))^(2)+m((L)/(2))^(2)=(mL^(2))/(2)`
`rArr" "l=l_(1)+l_(2)=(ML^(2))/(12)+(mL^(2))/(2)`
`=((M+6m)L^(2))/(12)`