Question

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

• A. $1$
• B. $2$
• C. $>1$
• D. $<1$

Answer 1

The correct option is A $1$

For orifice $A$

$V=\sqrt{2g{h}_1}$

time taken by the liquid drop to fall $T=\sqrt{\frac{2(h+h_2)}{g}}$

Now,

Range $=\sqrt{2g{h}_1}\left(\sqrt{\frac{2(h+h_2)}{g}}\right)$

Range $=\sqrt{4(h+h_2)h_1}$

For orifice $B$

$V=\sqrt{2g(h_1+h)}$

Time taken by the liquid to fall $T=\sqrt{\frac{2h_2}{g}}$

Now

Range $=\sqrt{\frac{2h_2}{g}}\left(\sqrt{2g(h_1+h)}\right)=\sqrt{4\left[h_2(h_1+h)\right]}$

As range for both are same So,

$\sqrt{4(h+h_2)h_1}=\sqrt{4(h_1+h)h_2}$

$h{h}_1={hh}_2$

$h_1=h_2$

So, $\frac{h_1}{h_2}=1$