Question

On another planet that you are exploring, a large tank is open to the atmosphere and contains ethanol. A horizontal pipe of cross-sectional area 9.0×10−4m2 has one end inserted into the tank just above the bottom of the tank. The other end of the pipe is open to the atmosphere. The viscosity of the ethanol can be neglected. You measure the volume flow rate of the ethanol from the tank as a function of the depth h of the ethanol in the tank. If you graph the volume flow rate squared as a function of h, your data lie close to a straight line that has slope 2.54×10−5m5/s2. What is the value of g, the acceleration of a free-falling object at the surface of the planet?

Answer 1

Using the volumteric flow rate of the ethanol, we can get an expression for the velocity.

where , the cross-sectional area of the hole.

We can then equate the velocity of the liquid to the velocity of the free fall, and then substitute to express the square of the flow rate as a function of height.

Here, we can see the slope is . Now, we can substitute to find the value of the acceleration due to gravity.