AP Physics 2 Exam Question

A cylindrical container with a height of 2.0 meters is filled with a fluid of density 1000 kg/m³ up to a height of 1.5 meters. The container has a small hole at the bottom that allows the fluid to flow out. Assuming the fluid flows out in a steady state, calculate the speed of the fluid as it exits the hole. Also, determine the mass flow rate of the fluid at the hole.

(g = 9.8 m/s², ignore any losses due to friction or viscosity)

Answer:

To solve this problem, we will use the principles of fluid statics and dynamics.

1. Determine the pressure at the hole: The pressure at the hole can be calculated using the equation: where is the pressure difference, is the density of the fluid, is the acceleration due to gravity, and is the change in height. Here, = 2.0 m - 1.5 m = 0.5 m:

   

   Therefore, .

2. Calculate the speed of the fluid as it exits the hole: According to Torricelli's theorem, the speed of the fluid as it exits the hole can be found using the equation: .

   Substituting the given values, we get: .

   Therefore, the speed of the fluid as it exits the hole is: (rounded to two decimal places).

3. Determine the mass flow rate of the fluid at the hole: The mass flow rate can be calculated using the equation: , where is the cross-sectional area of the hole.

   We need to calculate the value of first. Since the container is cylindrical, the area of the circular hole is given by: , where is the radius of the hole.

   To calculate , we use the fact that the height of the fluid remaining in the container is 0.5 m: , therefore, .

   Substituting the values into the equation, we get: .

   Therefore, the mass flow rate of the fluid at the hole is: (rounded to the nearest whole number).