Question

The large container shown in the cross section above is filled with a liquid of density 1.1 x 103 kg/m3 . a small hole of area 2.5 x 10-6 m2 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 2.0 minutes is 7.2 x 10-4 m3 .
a. calculate the volume rate of flow of liquid from the hole in m3/s .

b. calculate the speed of the liquid as it exits from the hole.

c. calculate the height h of liquid needed above the hole to cause the speed you determined in part (b).

d. suppose that there is now less liquid in the container so that the height h is reduced to h/2. in relation to the beaker, where will the liquid hit the tabletop? left of the beaker in the

Answer 1

Part a)

Volume flow rate is given as

$Q = \frac{V}{t}$

$Q = \frac{7.2 * 10^{-4}}{120}$

$Q = 6 * 10^{-6} m^3/s$

Part b)

Volume flow rate = area * speed

$Q = A*v$

$6 * 10^{-6} = 2.5 * 10^{-6} * v$

$v = 2.4 m/s$

Part c)

for the speed of efflux we know that

$v = \sqrt{2gh}$

$2.4 = \sqrt{2*9.8*h}$

$h = 0.294 m$

d) we need figure to find this part