The moment of inertia of a rod about an axis through its centre and perpendicular to it is 1/12ML² (where M is the mass and L is te length of the rod). The rod is bent in the middle so that the two halves make an angle of 60°. 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))/(8sqrt(3))`

Answer

b

Explanation

Since, rod is bent at the middle , so each per of it will have same length `((L)/(2))` and mass` ((M)/(2))` as shown moment of inertia of each pert through its one end
`= (1)/(2)((M)/(2))((L)/(2))^(2)`
Hence , net moment of inertia through middle poind O is
`1 = (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)`