There is a thin rod of uniform cross-section of mass M and length L. If this rodis is bent at `90^(@)` from the mid-point, then the moment of inertia about the axis passing through mid-point and perpendicular to the plane which includes both parts of rod is .........

Options

A. `(ML^(2))/(24)`
B. `(ML^(2))/(12)`
C. `(ML^(2))/(6)`
D. `(sqrt(2)ML^(2))/(24)`

Answer

B

Explanation

After the rod bent at right angle
mass of its energy part = `(M)/(2)`
Length of every part = `(L)/(2)`
Now total moment of inertia,
I = sum of moment of inertia of both the part
`therefore I=(1)/(3)((M)/(2))((L)/(2))^(2)+(1)/(3)((M)/(2))((L)/(2))^(2)`
`=2((1)/(3)((M)/(2))((L)/(2))^(2))`
`therefore I=(ML^(2))/(12)`