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

Explanation

Let H be the total height of the water column in cylindrical tank. Let `upsilon_1, upsilon_2` = velocity of efflux at P and Q respectively. Then `upsilon_1 = sqrt(2 gh_1) and upsilon_2 =sqrt(2g(H - h_2))` Time taken by water stream to fall from P to S on the ground is ,`t_1 = (sqrt(2(H - h_1))/(g))` Time taken by water stream to fall from Q to S on the ground is, `t_2 = (sqrt(2h_2))/(g)` Since, the two water streams from P and Q reach the same point S on the ground, so `R_1 = R_2 or upsilon_1 t_2 or sqrt(2gh_1) xx sqrt((2(H -h_1))/(g)) = sqrt(2g (H - h_2)) xx sqrt((2h_2)/(g))`
ro `h_1 (H - h_1) = (H - h_2) h _2 or H(h_(1) - h_(2) - h_1^2 + h_2^2 = 0`
or `[H - (h_1 + h_2] [h_1 - h_2] = 0` Since `h_1+ h_(2) =H` is not possible as the holes are at different height so `h_1 = h_2.` Hence `h_1//h_2 =1`