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A weightless rod of length l carries two...

A weightless rod of length l carries two equal masses m one fixed at the end and other in the middle ofhte rod. The rod can revolve in a vertical plane about A. Then, horizontal velocity which must be imparted to end C of rod to deflect it to horizontal position is

A

`sqrt(12)/5`gl

B

`sqrt(3gl)`

C

`sqrt(16/5)`gl

D

`sqrt(2gl)`

Text Solution

Verified by Experts

a) Loss in kinetic energy = Gain in potential energy
`therefore 1/2 Iomega^(2) = mgI/2 + mgI`………….(i)
Here, `I = (ml^(2))/4 + mlk^(2) = 5/4 ml^(2)`
`v_(c) = Iomega`
From Eq. (i),
`1/2(5/4ml^(2))omega^(2) = (mgl)/2 + mgl = 3/2 mgl`
`ml^(2)omega^(2) = 12/5mgl` `therefore` v_(C) = sqrt((12gl)/5))`
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