Question

The area of cross-section of a large tank is $0.5{\text{m}}^2$. It has an opening near bottom having area of cross-section $1{\text{cm}}^2$. A load of $20\text{kg}$ is applied on the water at the top. Find the velocity of the water coming out of the opening at the time when the height of water level is $50\text{cm}$ above the bottom. (Take $g=10{\text{m/s}}^2$)

• A. $3.3\text{m/s}$
• B. $5.0\text{m/s}$
• C. $6.6\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is A $3.3\text{m/s}$

As the area of cross-section of the tank is large compared to that of the opening, the speed of water in the tank will be very small (i.e. $v_A=0$) as compared to the speed at the opening. The pressure at the surface of water in the tank is that due to the atmosphere plus due to the load
$P_A=P_o+\frac{mg}{A}$
$P_A=P_o+\frac{\left(20\right)\times \left(10\right)}{0.5}$
$=P_o+\frac{200}{0.5}$
$=P_o+400{\text{N/m}}^2$
At the opening, the pressure is due to the atmosphere.
Using Bernoulli's equation,
$P_A+\rho gh+\frac{1}{2}\rho v_A^2=P_B+\frac{1}{2}\rho v_B^2$
$(P_0+400)+(1000\times 10\times 0.5)+0=P_0+\frac{1}{2}\left(1000\right)v_B^2$
$5400=\left(500\right)v_B^2$
$v_B=3.3\text{m/s}$