# Fluids Again (AP Physics)

The large container shown in the section above if filled with a liquid of density $1.10 * 10^3 kg/m^3$. A small hole of area $2.50*10^{–6} m^2$ is opened in the side of the container a distance $h$ below the liquid surface, which allows a stream of liquid to flow through the hole and into a beaker placed to the right of the container. At the same time, liquid is also added to the container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.00 minutes is $7.20*10^{–4} m^3$. A. Calculate the speed of the liquid as it exits from the hole. B. Calculate the height h of liquid needed above the hole to cause the speed you determined in part (b) C. Suppose that there is now less liquid in the beaker so that the height h is reduced to $h/2$. In relation to the beaker, where will the liquid hit the tabletop? a. Left of the beaker ___ b. In the beaker___ c. Right of the beaker___ Justify your answer.

A) the position pressure of a column of liquid of height $H$ must be balanced by the dynamic pressure of the liquid at the exit of the hole.

$\rho*g*H = \rho*V^2/2$

where V is the speed of the liquid exiting the hole.

Now the volume rate that exit the hole is

Volume rate $= 7.2*10^{-4}/(2*60) = 6*10^{-6} m^3/sec$

This volume rate is equal to the speed of the liquid multiplied bu the area of the hole (S)

Volume rate $= S*V$

V = volume rate/S $=6*10^-6/2.5*10^{-6} = 2.4 m/sec$

B) from the equation

$\rho*g*H = \rho*v^2/2$

$H = v^2/2/g = 2.4^2/2/9.81 = 0.122 m$

C) If there is less liquid then the exiting speed of the liquid from the hole will be smaller than the initial speed (as the above equation shows).

$\rho*g*H/2 = \rho*v1^2/2$

$\rho*v1^2 = \rho*v^2/2$

$v1 = v/\sqrt{2} = 1.697 m/s$

$v1$ is the new speed of the liquid and $v$ is the initial speed of the liquid.

The liquid will hit the tabletop to the left of the baker.