AP Physics 2 Exam Question

A cylindrical tank of height 1.5 m is filled with a liquid of density 800 kg/m^3. The area of the bottom of the tank is 0.5 m^2. A small hole of diameter 0.01 m is made 1.2 m above the bottom of the tank. What is the speed at which the liquid exits the hole?

Answer:

To solve this problem, we can use the principles of fluid dynamics, particularly the Bernoulli's equation.

According to Bernoulli's equation, the sum of the pressure energy, kinetic energy, and potential energy per unit volume of fluid remains constant along a streamline.

Let's consider two points: Point A at the top surface of the tank where the liquid is still, and Point B at the hole where the liquid is exiting.

At Point A (top surface of the liquid), the velocity of the liquid is zero, so the kinetic energy per unit volume is zero.

Point B (hole) can be considered at the same height level as Point A, so the potential energy per unit volume of both points is the same.

Using Bernoulli's equation, we can equate the pressure energy and potential energy at Point A and Point B:

Where:

• $P_A$ and $P_B$ are the pressures at Point A and Point B, respectively.
• $\rho$ is the density of the liquid.
• $g$ is the acceleration due to gravity.
• $h_A$ and $h_B$ are the heights at Point A and Point B, respectively.

Since the liquid is open to the atmosphere, we can assume the pressure at Point B is atmospheric pressure, which is $P_B = P_{\text{atm}} = 1.013 \times 10^5 \, \text{N/m}^2$. The height at Point A is $h_A = 0$ since it is at the top surface (reference point).

At Point B, the pressure is slightly higher due to the liquid column above the hole, so we will assume $P_A = P_{\text{atm}}$.

Substituting these values, we have:

Next, let's find the height at Point B, which is the difference in height between the hole and the bottom of the tank:

$h_B = \text{height of hole from bottom} = 1.2 \, \text{m}$

Substituting this value into the equation:

$0 = \frac{P_B}{\rho g} + 1.2$

Now, let's solve for the pressure at Point B:

$P_B = -1.2 \rho g$

The negative sign indicates that the pressure at Point B is lower than atmospheric pressure.

Finally, we can use the equation for the speed of fluid exiting a hole in a tank:

$v = \sqrt{\frac{2 \cdot g \cdot h_B}{2 \cdot h_A - h_B}}$

where:

• $v$ is the speed at which the liquid exits the hole.
• $g$ is the acceleration due to gravity.
• $h_B$ is the height at the hole from the bottom of the tank.
• $h_A$ is the height at the top surface of the liquid.

Substituting the given values, we have:

$v = \sqrt{\frac{2 \cdot 9.8 \cdot 1.2}{2 \cdot 0 - 1.2}} \approx \sqrt{23.76}$

$v \approx 4.875 \, \text{m/s}$

Therefore, the speed at which the liquid exits the hole is approximately 4.875 m/s.