On the wall of a cylindrical water tank, two holes are made as shown in figure. Water coming out from these holes hits the ground at the same point. What will be the ration `h_(1)/h_(2)?`

Explanation

h=total height of water column in tank
`v_(1)`= Velocity of efflux at A
`v_(2)`= Velocity of efflux at B
`v_(2)=sqrt(2g (h-h_(1))`
`v_(1)=sqrt(2gh_(2))`
`t_(1)`= time taken by water to fall from A to R.
`t_(1)=sqrt((2h_(1))/(g))`
Similar `t_(2)`= time taken by water stream to fall from B to R. `t_(2)=sqrt(2(h-h_(2))/(g))`
As the water streams are striking at the same point on the ground.
Therefore, horizontal distance covered by streams coming from A and B are same.
Let `R_(1) and R_(2)` are horizontal ranges.
`R_(1)=R_(2)`
`v_(1) t_(1)=v_(2)t_(2)`,
Putting values we get
`sqrt(2g(h-h_(1)) sqrt((2h_(1))/(g))=sqrt((2gh_(2)) (2(h-h_(2))/(g))`
`(h-h_(1)) h_(1)=h_(2) (h-h_(2))`
`h h_(1)-h_(1)^(2)=h h_(2)-h_(2)^(2)`
`hh_(1)-h_(1)^(2)-hh_(2)+h_(2)^(2)=0`
`h(h_(1)-h_(2))+h_(2)^(2)-h_(1)^(2)=0`
`h(h_(1)-h_(2)) +(h_(2)-h_(1)) (h_(2)+h_(1))=0`
`h(h_(1)-h_(2)) -(h_(1)-h_(2)) (h_(1)+h_(2))=0`
`(h_(1)-h_(2)) (h-(h_(1)+h_(2)) =0`
Case (i) `h_(1)-h_(2)=0 or h_(1)=h_(2)`
Case (ii) `(h-(h_(1)+h_(2))=0 or h_(1)+h_(2)=h`
But `h_(1)+h_(2) ne h`
Case (ii) is not possible
We can say, from case (i) that `h_(1)=h_(2)`
Therefore, the ratio is `(h_(1))/(h_(2))=1`