2021 A/L Physics Past Paper · Question 26 (English Medium)

Answer: (5) All (A), (B) and (C) are true.

What the question asks

Find which of the statements (A), (B) and (C) are true.

They say how the terminal velocity $v$ depends on $r$, $\rho$ and $\eta$.

The plan: find $v$ in terms of $r$, $\rho$ and $\eta$, then test each statement.

Relevant theory

The sphere now begins to move with a uniform velocity.

This expression is referred to as Stokes’ law.

Reading the options

The options are combinations of (A), (B) and (C).

At the terminal velocity the weight equals the viscous force:

$$\tfrac{4}{3}\pi r^3 \rho g = 6\pi\eta r v$$

So:

$$v = \dfrac{2 r^2 \rho g}{9\eta}$$

We test each statement on this result.

Reasoning and calculation

It is directly proportional to $r^2$.

$r$ appears as $r^2$ in the top of the fraction. The other symbols stay the same. So $v \propto r^2$.

A

It is directly proportional to $\rho$.

$\rho$ is in the top of the fraction, to the power $1$. So $v \propto \rho$.

B

It is inversely proportional to $\eta$.

$\eta$ is in the bottom of the fraction. So $v \propto \dfrac{1}{\eta}$.

C

All three statements follow from $v = \dfrac{2 r^2 \rho g}{9\eta}$.

So (A), (B) and (C) are all true.

Why the other options fail

(B) and (C) are true too, so (A) alone is not enough.

(A) and (C) are true too, so (B) alone is not enough.

(C) is true too, because $\eta$ is in the bottom of the fraction.

(A) is true too, because $v \propto r^2$.

Conclusion

$v = \dfrac{2 r^2 \rho g}{9\eta}$.

(A), (B) and (C) are all true. The answer is (5).

At terminal velocity the forces balance. Write weight $=$ Stokes force, solve for $v$, and read off each dependence.

Look at which symbols sit on top and which sit at the bottom.

Exam tip