Question

The large container shown in the cross section is filled with a liquid of density 1.1 x 10³ kg/m³.  hole of area 2.5 x 10^-6m². A small  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^–4m³.

(a) Calculate the volume rate of flow of liquid from the hole in m³s.

(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 collection beaker, where will the liquid hit the tabletop? 

____ Left of the beaker ____ In the beaker ____ Right of the beaker 

Answer 1

(a) Volume flow rate = Q = V/t = 7.2 x 10^–4/(2min * 60 sec/min) = 6 x 10^–6m³/s 

(b) Your first thought is probably Bernoulli, but there are too many unknowns so this does not work. We can use the volume flow rate above the find the velocity. 

Q = Av 

6 x 10^–6 = (2.5 x 10^–6)v 

v = 2.4 m/s 

(c) Use Bernoulli, … v₂ = √2gh (2.4) = √2(9.8)h h=0.29m 

(d) Left of beaker. Based on the formula derived above, the exit velocity is dependent on the height and with less horizontal exit velocity the range will be less (d_x = v_xt). This makes sense because less height would result in less pressure and decrease the speed the fluid is ejected at, thus lessening the range.