Question

A very small hole is made at the bottom of a tank filled with water (density $=1000{\text{kg/m}}^3$). If the total pressure at the bottom of the tank is $3\text{atm}$ $(1\text{atm}={10}^5{\text{N/m}}^2)$, then the velocity of efflux is:
Take $g=10{\text{m/s}}^2$.

• A. $20\text{m/s}$
• B. $\sqrt{300}\text{m/s}$
• C. $\sqrt{600}\text{m/s}$
• D. $\sqrt{500}\text{m/s}$

Answer 1

The correct option is A $20\text{m/s}$

Pressure at hole, $P_2=3\text{atm}$
Pressure at open end of tank is $P_1=1\text{atm}$
Since the hole is also open to atmosphere, excess pressure $P_2-P_1=2\text{atm}$ at the hole is due to hydrostatic pressure of water column above the hole.
$\Rightarrow \rho gh=P_2-P_1=2\text{atm}$
Or, $\rho gh=2\times {10}^5{\text{N/m}}^2$
$\therefore h=\frac{2\times {10}^5}{{10}^3\times 10}=20\text{m}$
Cross-sectional area of tank is very large compared to area of the hole (very small), i.e $A_1>>A_2$
Hence, we can apply formula for velocity of efflux at the bottom of the tank:
$\Rightarrow v=\sqrt{2gh}$
Or, $v=\sqrt{2\times 10\times 20}$
$\therefore v=20\text{m/s}$