1.

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 .........

Answer»

`(ML^(2))/(24)`
`(ML^(2))/(12)`
`(ML^(2))/(6)`
`(sqrt(2)ML^(2))/(24)`

SOLUTION :
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)`


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