2019 A/L Physics Past Paper · Question 39 (English Medium)

Answer: (5) $2A\sqrt{\frac{2gh}{3}}$

What the question asks

The columns in the two tubes differ in height by $h$.

We want the flow rate $Q$ in terms of $A$, $g$ and $h$.

Relevant theory

Volume of fluid entering the tube through area $A_1$.

Since the flow is continuous, the volume of fluid flowing past the cross section $A_2$ per unit is the same.

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.

Required equations

$Q = Av_1 = 2Av_2$

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$

$p_2 - p_1 = \rho g h$

Reasoning and calculation

The letters:

$v_1$, $p_1$: speed and pressure in the narrow part, area $A$.

$v_2$, $p_2$: speed and pressure in the wide part, area $2A$.

$\rho$: the density of the liquid.

$h$: the height difference of the columns.

$Q$: the volume of liquid that flows each second.

The same liquid passes both parts each second.

The wide part is twice as big, so $v_2 = \dfrac{v_1}{2}$.

The narrow part is faster. So its pressure is lower, and its column is lower.

The pressure difference is the weight of the height $h$ of liquid:

$$p_2 - p_1 = \rho g h$$

$\rho g h = \frac{1}{2}\rho\left(v_1^2 - \frac{v_1^2}{4}\right) = \frac{3}{8}\rho v_1^2$

$v_1 = \sqrt{\frac{8gh}{3}}$

$Q = A\sqrt{\frac{8gh}{3}} = 2A\sqrt{\frac{2gh}{3}}$

Reading the options

The answer has the form $2A\sqrt{\frac{2gh}{3}}$.

That is option (5).

Checks

Check the units.

$A\sqrt{gh}$ has the dimensions $L^2 \cdot LT^{-1} = L^3T^{-1}$.

That is a volume each second, a flow rate. ✓

Check the limit.

If $h = 0$, then $Q = 0$ ✓.

No flow, no pressure difference.

If the two parts were equal ($2A \to A$), the speeds would be equal and $h$ would be 0.

The height difference comes from the change in area.

Conclusion

$Q = 2A\sqrt{\frac{2gh}{3}}$

$$\frac{3}{8}v_1^2 = gh$$

$$Q = Av_1 = 2A\sqrt{\frac{2gh}{3}}$$

So the answer is option (5).