A cone of radius r and height r is under a liquid of density d.its base is parallel to the free surface of the liquid at a depth H from it as shown.What is the net force due to liquid on its curved surface? (neglect atmospheric pressure)

Options

A. `pir^(2)dg(H-(r)/(3))`
B. `2pir^(2)dg(H-(2r)/(3))`
C. `pir^(2)dg(2H-(r)/(3))`
D. `pir^(2)dg(2H-(2r)/(3))`

Answer

A

Explanation

Force due to liquid on curved surface `=F`
`F=` weight of liquid above the cone
`F=(pir^(2)H-(1)/(3)pir^(3))dg=pir^(2)dg[H-(r)/(3)]`