A thin uniform rod mass `M` and length `L` is hinged at its upper end. And released from rest from a horizontal position. The tension at a point located at a distance `L//3` from the hinge point, when the rod become vertical will

Answer

`2 mg`

Explanation

From energy conservation

Loss of `P.E`. = Gain of `K.E`
`Mg `(L)/(2) = (1)/(2) I omega^(2) rArr Mg `(L)/(2) =(1)/(2) ((1)/(3) ML^(2)) omega^(2)`
`omega^(2) =(3g)/(L) rArr omega = sqrt((3g)/(L))` ....(1)
for centripetal force is
`T-(2Mg)/(3) =((2M)/(3)) omega^(2) .((2L)/(3))`...(2)
`T=(2Mg)/(3)+(4Momega^(2)L)/(9)`...(2)
from (1) & (2)
`T = 2 Mg`.