Three identical thin rods each of length `l` and mass `M` are joined together to from a letter `H`. What is the moment of inertia of the system about one of the sides of `H` ?

Options

A. `(2ML^(2))/(3)`
B. `(ML^(2))/(2)`
C. `(ML^(2))/(6)`
D. `(4ML^(2))/(3)`

Answer

D

Explanation

The given situation can be shown as

Let us take the moment of inertia of the system about rod `R_(1)` then total noment of inertia is
`l_(T) = l_(1) +l_(2) + l_(3) " "………(i)`
For rod `R_(1), l_(1) =0`
For rod `R_(2)` using perpendicular axis theorms, `l_(2) =(ML^(2))/(3)`
For rod `R_(3)` using parallel axis theorem,
`l_(3) = l_(CM) + l_((at L)) = 0 + ML^(2) = ML^(2)`
Now , putting the values of `l_(1),l_(2) and l_(3)` in Eq(i) we get
`I_(T) = 0 +(ML^(2))/(3) + ML^(2) rArr l_(T) = (4 ML^(2))/(3)`