2022 A/L Physics Past Paper · Question 34 (English Medium)

Answer: (4) $v = \sqrt{2g(h_2 + h_3)}$

Given information

What we know:

The tube is full of liquid. It rises out of the container over a bend, then falls to the open end $C$.

The distances in the figure are $h_1$ from the liquid surface up to the top of the bend, $h_2$ from the liquid surface down to the next level, and $h_3$ from there down to $C$. So $C$ is $h_2 + h_3$ below the liquid surface.

The container is wide, the tube is narrow, and the flow is steady with no viscosity.

The question has no numbers. So we keep the letters and check with sample values $h_1 = 0.3\ \text{m}$, $h_2 = 0.2\ \text{m}$, $h_3 = 0.8\ \text{m}$ and $g = 10\ \text{m s}^{-2}$.

What the question asks

We want: the speed $v$ of the liquid as it leaves the tube at $C$.

What each letter means:

$v$ ($\text{m s}^{-1}$): the speed of the liquid at $C$.

$g$ ($\text{m s}^{-2}$): the acceleration due to gravity.

$h_2 + h_3$ ($\text{m}$): how far $C$ is below the liquid surface in the container.

$\rho$ ($\text{kg m}^{-3}$): the density of the liquid.

$p$ ($\text{Pa}$): the pressure in the liquid.

Relevant theory

What is happening:

The tube is full, so the liquid in it is one long unbroken column.

The open end $C$ is lower than the liquid surface in the container. The liquid in the long outside arm is pulled down by gravity and runs out at $C$.

As it runs out, the air pressure on the surface in the container pushes more liquid up into the tube. So the liquid keeps going up over the bend and down to $C$, all the time.

The surface in the container sinks only very slowly, because the container is wide.

The liquid speeds up as it goes down. The lower $C$ is below the surface, the faster it comes out.

On any point along the same streamline of a steady flow of an incompressible fluid of negligible viscosity, the sum of the pressure, the potential energy per unit volume and the kinetic energy per unit volume is a constant.

$$p + \frac{1}{2}\rho v^2 + \rho gh = K$$ (K is a constant)

We know from the equation of continuity that when the cross – section gets narrow the velocity of flow increases.

We follow one streamline from the liquid surface in the container, point $S$, to the open end $C$.

$$p_S + \dfrac{1}{2}\rho v_S^2 + \rho g h_S = p_C + \dfrac{1}{2}\rho v_C^2 + \rho g h_C$$

In words: the pressure, the kinetic energy per unit volume and the potential energy per unit volume add up to the same value at $S$ and at $C$.

Both $S$ and $C$ are open to the air, so the pressure is the same at both and cancels.

The container is wide and the tube is narrow, so the surface moves very slowly. We take $v_S = 0$.

The height of $S$ above $C$ is $h_2 + h_3$. Then:

$$\rho g (h_2 + h_3) = \dfrac{1}{2}\rho v^2$$

In words: the potential energy lost in the fall becomes the kinetic energy of the liquid at $C$.

The height $h_1$ does not appear. The question tells us the flow is steady, so we do not need it.

For example, make a small hole in the side of a bucket of water, near the bottom. The water squirts out fast, while the water at the top of the bucket hardly moves. The deeper the hole is below the water surface, the faster the jet.

The open end $C$ is like that hole. Its depth below the surface is $h_2 + h_3$.

Required equations

$p_S + \dfrac{1}{2}\rho v_S^2 + \rho g h_S = p_C + \dfrac{1}{2}\rho v_C^2 + \rho g h_C$

$\rho g (h_2 + h_3) = \dfrac{1}{2}\rho v^2$

$v = \sqrt{2g(h_2 + h_3)}$

Applying it to this question

$\sqrt{\text{m s}^{-2} \times \text{m}} = \text{m s}^{-1}$, which is a speed. The density $\rho$ cancels, so the answer does not depend on the liquid.

$1.0$

$h_2 + h_3 = 0.2 + 0.8$

$20$

$v^2 = 2g(h_2 + h_3) = 2 \times 10 \times 1.0$

$4.47$

$v = \sqrt{20}$

Reasoning and calculation

The speed at $C$ is $v = \sqrt{2g(h_2 + h_3)}$.

That is option (4).

Checks

$sqrt(2 * g * h)$

Does it make sense?

The liquid falls a height $h_2 + h_3$ from the surface to $C$. A body falling freely through that height would reach $\sqrt{2g(h_2 + h_3)}$. The same expression comes out.

The liquid has to go up $h_1$ on the way, but it gets that height back on the way down. So $h_1$ does not change the speed at $C$.

Conclusion

$v = \sqrt{2g(h_2 + h_3)}$

The speed at $C$ is $v = \sqrt{2g(h_2 + h_3)}$.

The answer is (4).

Measure the height from the liquid surface in the container down to the outlet. That total drop is what sets the speed.

The top of the bend is not part of the drop. In this question the flow is steady, so it does not change the speed.

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