1.

A thin wire of length l and mass m is turned into the form of a semicircle. Its moment of inertia about an axis joining its free ends will be

Answer»

`mpil^(2)`
`ML^(2)//pi^(2)`
`(ml^(2))/(2pi)`
`(ml^(2))/(2pi^(2))`

Solution :Circumference `2piR` = 2l `thereforeR=l/pi`
MOMENT of mertta of nng about a DIAMETER =`(mR^(2))/2`
M.I. of SEMICIRCLE = `1/2[m.(1/pi)^(2)]=(ml^(2))/(2pi^(2))`


Discussion

No Comment Found

Related InterviewSolutions