Question

Water is filled upto height H units in a tank placed on the ground and whose side walls are vertical. A hole is made in one of the vertical wall such that the emerging stream of water strikes the ground at the maximum range. If the level in the tank is changing at the rate of R units per second, at that instant the rate at which range will be changing will be:

• A. R/2 units per second
• B. R units per second
• C. 2R units per second
• D. Zero

Answer 1

The correct option is D Zero

According to to Torricelli's theorem, velocity of efflux at point A ( shown in the figure ) is
$v_A=\sqrt{\left(2gh\right)}$
After emerging from the orifice the water adopts parabolic path. if it takes time t secs in falling a vertical height (H=h) and covers a horizontal range $R_1$, then
$H-h=\frac{1}{2}g{t}^2$
$\Rightarrow t=\sqrt{\frac{2(H-h)}{g}}$
$R_1=v_{A}t=\sqrt{\left(2gh\right)}\ast \sqrt{\frac{2(H-h)}{g}}=2\sqrt{h(H-h)}$ ....(1) where $R_1$ is the range.
The above results show that Range is same whether the hole is made at distance h from the top or at (H-h) from the top.
$\therefore h=H-h$
$\Rightarrow h=\frac{1}{2}H$ ....(2)
If h is changing at the rate $\frac{dh}{dt}=R$
differentiating eqn (1) wrt t, we get
$\frac{d{R}_1}{dt}=\left(h\right(H-h\left){)}^{-\frac{1}{2}}\right(\frac{dh}{dt}(H-h)+h(-\frac{dh}{dt}))=(h(H-h){)}^{-\frac{1}{2}}\left(R\right(H-h)-Rh)$ ....(3)
Substituting $h=\frac{H}{2}$ from eqn(2) in eqn(3) we get
$\frac{d{R}_1}{dt}=\left(\frac{H}{2}{)}^{-1}R\right(H-2h)=(\frac{H}{2}{)}^{-1}R(H-H)=0$