In a cylindrical water tank there are two small holes `Q` and `P` on the wall at a depth of `h_(1)` from the upper level of water and at a height of `h_(2)` from the lower end of the tank, respectively, as shown in the figure. Water coming out from both the holes strike the ground at the same point. The ratio of `h_(1)` and `h_(2)` is

Options

A. `1`
B. `2`
C. `gt1`
D. `lt1`

Answer

A

Explanation

The two streams strike at the same point on the ground.
`R_(1)=R_(2)=R`
`u_(1)t_(1)=u_(2)t_(2)`………i
Where `u_(1)=` velocity of efflux at `Q=sqrt((2gh_(1)))` and `u_(2)=` velocity of efflux at `P=sqrt([2g(H-h_(2))])`
`t_(1)=`time of fall of water stream through `Q` is
`=sqrt((2(H-h_(1)))/g)`
`t_(2)=` time of fall of the water stream through `P=sqrt((2h_(2))/g)`
Putting these values is eqn i we get
`(H-h_(1))h_(1)=(H-h_(1))h_(2)`
or `[H-(h_(1)+h_(2))][h_(1)-h_(2)]=0`
`H=h_(1)+h_(2)` is irrelevant because the holes are at two different heights. Therefore `h_(1)=h_(2)` or `(h_(1))/(h_(2))=1`