A tank is filled by water (`rho = 10^3 kg//m^3`). A small hole is made in the side wall of the tank at depth 10 m below water surface. A water jet emerges horizontal from the hole and falls at horizontal distance R from it. The amount of extra pressure (in terms of atmospheric pressure) that must be applied on the water surface, so that range becomes 3R on the on the ground will be (cross section area of hole is negligible and 1 atm = 105 Pa, g = 10 m/`s^2`)

Explanation

`rho g h=1//2 rhov_(1)^(2)...(1)`
`DeltaP=rho gh=1//2 rhov_(2)^(2).....(2)`
`v_(2)=3v_(1) rArr v_(2)^(2) =9 v_(1)^(2)`
`rArr 1/2 rhov_(2)^(2)=9(1/2 rhov_(1)^(2))rArr DeltaP+rhogh=9 rho g h`
`DeltaP=8 rho g h =8xx10^(3) xx10xx10=8 xx10^(5)` pascal =8 atm