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 = (V)/(t)`

`Q = (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 = √(2gh)`

`2.4 = √(2*9.8*h)`

`h = 0.294 m`

d) we need figure to find this part