In the figure shown

find the range R.

Explanation

Appling Bernouli's equation at point 1 and 2.
`p_(1)+(1)/(2)(2 rho)v_(1)^(2)+(2 rho)gh_(1)=p_(2)+(1)/(2)(2 rho)v_(2)^(2)+(2 rho)gh_(2)`
Here, `v_(1)~~0, v_(2)=v, h_(1)=h_(2), p_(2)=p_(0)`
and `p_(1)=p_(0)+(rho)gH+2rhogH`
`=p_(0)+3rhogH`
Subsituting in the above equation we have ,
`p_(0)+3rho gH=p_(0)+rho v^(2)`
`:. v-sqrt(3 gH)`
Now this velocity is horizontal. So, taken by the liquid to fall to the ground is free fall time or,
`t=sqrt((2h)/(g))` (where,` h=H)`
or, `t=sqrt((2H)/(g))`
`:. R=vt=(sqrt(3gh))(sqrt((2H)/(g)))`
`=sqrt(6)H`.