Question

A water tank standing on the floor has two small holes vertically one above the other punched on one side. The holes are $h_1$ cm and $h_2$ cm above the floor. How high does water stand in the tank, when the jets from the holes hit the floor at the same point?

• A. $(h_2-h_1)$
• B. $(h_2+h_1)$
• C. $(h_2^2-h_1^2)$
• D. $\frac{h_2}{h_2-h_1}$

Answer 1

The correct option is B $(h_2+h_1)$

Time taken by water at (1) to reach the ground = $\sqrt{\frac{2h_2}{g}}$

velocity of water at (1) = $\sqrt{2gH}$

So,

$d=\frac{\sqrt{2gH\times 2P_n}}{g}$-------------(1)

Similarly for point ,

Time taken = $\sqrt{\frac{2h_1}{g}}$

velocity = $\sqrt{2g(H+h_2-h_1)}$

So, $d=\sqrt{2g(H+h_2-h_1)\times \frac{2h_1}{g}}$----------(2)

Equation (1) and (2),

$\Rightarrow (H+h_2-h_1)h_1=H{h}_2$

$\Rightarrow H{h}_1+h_2{h}_1-h_1^2=H{h}_2$

$\Rightarrow H(h_1-h_2)=h_1(h_1-h_2)$

So,$\Rightarrow H=h_1$

Height of water = $h_1+h_2$