2020 A/L Physics Past Paper · Question 41 (English Medium)

Answer: (3) $\frac{3}{4}h$

What the question asks

We must find: once the tap is opened and the flow is steady, the height of the water level in the vertical tube at $B$ — that shows the pressure at $B$: $p_B = p_0 + h_B\rho g$.

Relevant theory

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.

$$p + \frac{1}{2}\rho v^2 + \rho gh = K$$

$$p_1 - p_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$

$$\therefore A_1v_1 = A_2v_2$$

$$p = p_0 + h\rho g$$

Required equations

$p_0 + 0 + \rho g h = p_0 + \tfrac{1}{2}\rho v_C^2 + 0$

$A_B v_B = A_C v_C$

$p_B - p_0 = \tfrac{1}{2}\rho(v_C^2 - v_B^2)$

$p_B = p_0 + h_B\rho g$

Reasoning and calculation

Model: the tank is large — the speed at its surface is $\approx 0$, the pressure there is $p_0$.

At $C$ the water leaves into the open atmosphere, so the pressure there is also $p_0$.

The pipe is horizontal; take its axis as the height level ($h$ is measured from it).

Symbols: $v_B$, $v_C$ — the speeds at $B$, $C$; $A_B$, $A_C$ — the areas; $h_B$ — the water level in the vertical tube.

$\rho g h = \tfrac{1}{2}\rho v_C^2 \Rightarrow v_C^2 = 2gh$

$2A_C\, v_B = A_C\, v_C \Rightarrow v_B^2 = \tfrac{1}{4}v_C^2 = \tfrac{1}{2}gh$

$p_B - p_0 = \tfrac{1}{2}\rho\left(2gh - \tfrac{1}{2}gh\right) = \tfrac{3}{4}\rho g h$

$h_B\rho g = \tfrac{3}{4}\rho g h \Rightarrow h_B = \tfrac{3}{4}h$

Reading the options

The result $h_B = \tfrac{3}{4}h$ matches choice (3).

$h$ (the rest level) and $\tfrac{4}{3}h$ both forget that the pressure falls during flow.

$\tfrac{1}{4}h$ is only the kinetic-energy part.

Checks

Unit check:

the unit of $\tfrac{1}{2}\rho v^2$ is $\text{kg m}^{-3}(\text{m s}^{-1})^2 = \text{N m}^{-2}$ — pressure ✓.

$\rho g h_B$ $is\,also\,N\,m^{-2}$ ✓.

$h_B$ is a length ✓.

Limiting cases:

if $A_B = A_C$ then $v_B = v_C$, $p_B = p_0$, level $0$ ✓ (like an outlet nozzle of constant width).

if $A_B \gg A_C$ then $v_B \to 0$, level $\to h$ ✓ (the rest state).

At $A_B = 2A_C$, in between: $\tfrac{3}{4}h$.

$(2 * g * h - 2 * g * h / 4) / (2 * g)$

Second way (surface → $B$ directly): $p_0 + \rho g h = p_B + \tfrac{1}{2}\rho v_B^2 \Rightarrow p_B - p_0 = \rho g h - \tfrac{1}{4}\rho g h = \tfrac{3}{4}\rho g h$ — the same answer.

Conclusion

$h_B = \tfrac{3}{4}h$

Water at $B$ moves slower than at $C$, so the pressure there is $p_0 + \tfrac{3}{4}\rho g h$; the level in the vertical tube is $\tfrac{3}{4}h$ — the answer is (3).

Thinking the level stays at $h$ even after the tap is opened.

The pressure in flowing water is lower than in water at rest: $p_B = p_0 + \rho g h - \tfrac{1}{2}\rho v_B^2$.