The moment of inertia of a rod about an axis passing through its centre and perpendicular to it is `(1)/(12) ML^(2)`(where, M is the mass and L the length of the rod.) The rod is bent in the middle so that the two halves make an angle of `30^(@)`. The moment of inertia of the bent rod about the same axis would be

Options

A. `(1)/(48) ML^(2)`
B. `(1)/(12) ML^2`
C. `(1)/(24) ML^2`
D. `(ML^2)/(8sqrt3)`

Answer

B

Explanation

Since, rod is bent at the middle , so each part of it will have same length `((L)/(2))` and mass `((M)/(2))` as shown.

Moment of inertia of each part through its one end = `(1)/(3) ((M)/(2)) ((L)/(2))^(2)`
Hence net moment of inertia through its middle point O is
`I = (1)/(3) ((M)/(2)) ((L)/(2))^(2) + (1)/(3) ((M)/(2)) ((L)/(2))^(2) = (1)/(3) [ (ML^(2))/(8) + (ML^(2))/(8)] = (ML^(2))/(12)`