Question

The range of water flowing out of a small hole made at a depth $10m$ below water surface in a large tank is $R$. Find the extra pressure (in $atm$) applied on the water surface so that range becomes $2R$. Take $1atm={10}^5$ Pa.

Answer 1

initial velocity of efflux: $v_1=\sqrt{2gh}$
For range to become double, velocity of efflux should be doubled:

$v_2=2_{v_1}=2\sqrt{2gh}$

Let extra pressure applied be $P$, then
$(p_0+p_0+pgh)-p_0=\frac{1}{2}p{v}_2^2$

$\Rightarrow p+pgh=\frac{1}{2}p8gh$

$\Rightarrow p+3\rho gh=3\times {10}^3\times 10\times 10$

$=3\times {10}^{5}Pa=3atm$