A buket full of water is revolved in a vertical circle of radius 1m . What is the minimum frequency of revolution , required to prevent the water from failing down ? `[ g= 10 m//s^2 ]`

Answer

A

Explanation

At the highest point of the vertical circle , the water is not falling down
for this the C.F force `(mv^2)/(r )` = the weight (mg)
` therefore (mv^2)/(r ) = mg therefore v^(2) = rg therefore v= sqrt( rg)`
But `v= r omega = 2 pi n r `
` therefore 2 pi n r = sqrt( r g) `
` " or " n= ( sqrt( rg))/( 2pi r) = ( 1xx10)/( 2pi xx1) = ( sqrt(10))/( 2pi ) Hz `