A barrel of water with radius of 2.0 m sits on top of an 8-meter-high wall. A hole of radius 1 cm forms about 5 m down from the level of the water in the barrel. What is the velocity of the water as it leaves the hole? Use a water flow approach. Remember that Bernoulli’s equation is:

p1 + pgy1 + (1/2)pv1^{2} = p2 +pgy2 + (1/2)pv2^{2}

Loading ...

Correct Answer: **B.** *10 m/s*

First apply the flow equation: A1v1 = A2v2, where A2 and v2 are the cross section and velocity of the top surface of the water:

A1v1 = A2v2

v1*pi* (1 * 10^{-2} m)^{2} = v2*pi*(2m^{2})

v1 = 4m^{2}/ (1 * 10^{-4} m^{2}) * v2

Solving for v1 gives us v1 = (4 * 10^{4} m^{2}/s) v2

Though this does not give us an exact answer yet, it DOES give us important info! That is the velocity of the top surface of the water (v2) is insignificant compared to the velocity of the water coming out of the hole (v1). This allows us to set v2 = 0 in our second equation, Bernoulli’s equation:

p1 + pgy1 + (1/2)pv1^{2} = p2 +pgy2 + (1/2)pv2^{2}

Not only can we set v2 = 0, but we can also recognize that in a 5 meter-high column of water, the pressure at the bottom (p2) and bottom (p1) will be essentially the same, due to the fact that the additional pressure added by 5 meters of water is insignificant compared with the atmospheric pressure at the top (p1). Thus, the pressure terms can also be neglected, and the Bernoulli equation simplifies to give us:

pgy1 + (1/2)pv1^{2} = pgy2

v1^{2} = 2g(y2 – y1)

Solving for v1 gives:

v1 = Sqrt(2(10m/s^{2})*5m)) = 10 m/s, or answer B.

Get it right? Tweet at us:

**Subscribe below** to receive the MCAT Question of the Day delivered straight to your inbox every morning.