2023 A/L Physics Past Paper · Question 49 (English Medium)

Answer: (1) $a^5$

What the question asks

We want: how the power of the viscous force at the terminal velocity grows with the radius $a$.

What each letter means:

$a$ ($\text{m}$): the radius of the sphere.

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

$\eta$ ($\text{N s m}^{-2}$): the coefficient of viscosity of the liquid.

$W$, $U$, $F$ ($\text{N}$): the weight, the upthrust and the viscous force on the sphere.

$v$ ($\text{m s}^{-1}$): the terminal velocity.

$P$ ($\text{W}$): the power of the viscous force, the rate of doing work.

Relevant theory

The resultant force on the sphere becomes zero.

The sphere now begins to move with a uniform velocity.

The relationship between the viscous frictional

force F acting on a small sphere moving freely in

a viscous liquid and the terminal velocity it attains

is revealed by Stokes' law.

power = force × velocity

The first quote says that at the terminal velocity the resultant force on the sphere is zero, and it moves with a uniform velocity.

The second quote names Stokes' law, the link between the viscous force $F$ and the terminal velocity $v$: $F = 6\pi\eta a v$.

The third quote gives the power, the rate of doing work: $$power = force$$ $\times$ velocity.

For example, drop a small pearl and a big marble into a jar of honey. Both soon sink at a steady speed. The marble is heavier for its size, so it sinks faster, and the honey has to push back harder on it.

Required equations

$F = W - U = \dfrac{4}{3}\pi a^3 (\rho - \sigma) g$

$F = 6\pi\eta a v$

$v = \dfrac{2 a^2 (\rho - \sigma) g}{9\eta}$

$P = F v$

Reasoning and calculation

Separate what changes from what stays fixed.

Fixed: the liquid ($\sigma$, $\eta$), the material of the sphere ($\rho$) and $g$.

Changing: the radius $a$. With it, the force $F$ and the terminal velocity $v$ both change.

Write each part as a power of $a$.

The force equals the weight less the upthrust, and both go with the volume. So $F \propto a^3$.

The terminal velocity has $a^2$ on top. So $v \propto a^2$.

The power is the product: $P = Fv \propto a^3 \times a^2 = a^5$.

$8$

$\dfrac{F_2}{F_1} = 2^3$

$4$

$\dfrac{v_2}{v_1} = 2^2$

$32$

$\dfrac{P_2}{P_1} = 8 \times 4$

Simplify: $32 = 2^5$. Doubling the radius makes the power $2^5$ times larger, so $P \propto a^5$.

$(8 * 4)$

Reading the options

Now compare $a^5$ with the five options.

Option (1) is $a^5$. It matches.

Option (3), $a^3$, is how the force grows. Option (4), $a^2$, is how the velocity grows.Options (2) and (5) do not come from $F \times v$.

Checks

Check the units with a second route. Put $F = 6\pi\eta a v$ into $P = Fv$: $P = 6\pi\eta a v^2$.

Units: $\text{N s m}^{-2} \times \text{m} \times \text{m}^2\,\text{s}^{-2} = \text{N m s}^{-1} = \text{W}$. That is a power.

And with $v \propto a^2$, $P \propto a \times a^4 = a^5$ again.

Conclusion

$P \propto a^5$

At the terminal velocity the viscous force goes as $a^3$ and the terminal velocity as $a^2$, so the rate of work $P = Fv$ is proportional to $a^5$.

The answer is (1).

$$The\,quotes\,give\,the\,terminal\,velocity,\; Stokes'\,law\,and\,power = force$$ $\times$ velocity. The balance $F = W - U$ and the expression for $v$ are our working from them.

Teacher's note

For a sphere at terminal velocity: the viscous force goes as $a^3$, the speed as $a^2$, so the power goes as $a^5$. Check it by doubling $a$: the power grows $32$ times.

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