On the opposite sides of a wide vertical vessel filled with water, two identical holes are opened, each having cross sectiona area a. The height dference between them is equal to `h`. Find the resultant force of reaction of water flowing out of vessel.

Explanation

Let `A` and `B` be two opening each off area a the velocity of water coming out of opening are `V_(A)` and `V_(B)` respectively. Volume of water emerging per second through `A` is `aV_(A)`. Volume of water emerging per second through `B` is `aV_(B)`
force of reaction due to water coming out at `A` is given by
`F_(A)=` Rate of change of momentum per second at `A`
`(aVArho)V_(A)=arhogV^(A)`
Similarly force of reaction at `B` is given by `F_(B)=arhoV_(B)^(2)`

Therefore net force on vessel is
`F=F_(B)-F_(A)=arho(V_(B)^(2)-V_(A)^(2))`..........i
Now, applying bernoulli's theorem at `A` and `B`, we get
`p_(a)+1/2rhoV_(A)^(2)+(h_(2)+h)rhog=p_(0)+1/2rhoV_(B)^(2)+rhogh`........ii
where `p_(a)` is atmospheric pressure
`:. V_(B)^(2)-V_(A)^(2)=2gh`
Substituting this value in eqn i we get `F=arho2gh`
or `F=2rhog ah`