Physics 9702/11 — May/June 2017

Background Concept

A “reasonable estimate” means an order-of-magnitude value that matches everyday experience and simple physics.

For liquid pressure at a depth $h$, the hydrostatic pressure (gauge pressure due to the liquid) is

where:

• $\rho$ is the density of the liquid (for water $\rho \approx 1000\ \text{kg m}^{-3}$),
• $g \approx 9.8\ \text{m s}^{-2}$,
• $h$ is the vertical depth below the surface (in m),
• $p$ is in pascals (Pa), where $1\ \text{Pa} = 1\ \text{N m}^{-2}$.

This allows quick plausibility checks: each extra metre of water adds roughly $10^{4}\ \text{Pa}$.

Understanding the Question

You are given four everyday quantities with suggested values. Three should be realistic; one should clearly be far too large or too small.

The options A–C are common “everyday” values (current, mass, speed). Option D involves pressure at the bottom of a garden pond, which can be checked using $p = \rho g h$ and a typical pond depth.

Approach

Scan the options for the one most likely to be wildly wrong.

• A, B, C can be judged by general experience.
• D can be checked quantitatively by estimating pond depth and calculating the pressure, or by converting the stated pressure into an equivalent water depth.

Step-by-Step Reasoning

• Option A: A fan heater is a high-power appliance. On $230\ \text{V}$ mains, a current of order $10\ \text{A}$ corresponds to power $P \sim VI \sim 230 \times 12 \approx 2.8\ \text{kW}$, which is plausible.

• Option B: Adult mass $70\ \text{kg}$ is a standard typical value.

• Option C: Elite sprinters run $100\ \text{m}$ in about $10\ \text{s}$, giving an average speed about $10\ \text{m s}^{-1}$ (peak speed slightly higher). So $10\ \text{m s}^{-1}$ is reasonable.

• Option D: Use hydrostatic pressure.

  Take a typical garden pond depth $h \sim 1\ \text{m}$ (even a deep one might be a few metres).

  $$p = \rho g h \approx 1000 \times 9.8 \times 1 \approx 1 \times 10^{4}\ \text{Pa}$$

  The suggested value is $10^{6}\ \text{Pa}$, which is $\approx 100$ times larger.

  Convert $10^{6}\ \text{Pa}$ to an equivalent depth:

  $$h = \frac{p}{\rho g} \approx \frac{10^{6}}{1000 \times 9.8} \approx 1.0 \times 10^{2}\ \text{m}$$

  That is about $100\ \text{m}$ of water, completely unrealistic for a garden pond. Therefore D is not a reasonable estimate.

Key Takeaways

• Reasonableness checks often rely on order-of-magnitude thinking.
• For water, pressure increases by about $10^{4}\ \text{Pa}$ per metre depth.
• Converting a quantity into a more intuitive equivalent (here: pressure to depth) is a powerful plausibility test.

Common Mistakes

• Forgetting that pond depths are typically a few metres at most, not tens or hundreds.
• Mixing up units (using $h$ in cm instead of m in $p = \rho g h$).
• Confusing hydrostatic (gauge) pressure with absolute pressure; even adding atmospheric pressure ($\sim 10^{5}\ \text{Pa}$) would not make the bottom of a pond anywhere near $10^{6}\ \text{Pa}$.

Things to Be Careful About

• Use a sensible estimate for $h$ (a “garden pond” strongly suggests shallow water).
• Remember $\rho \approx 1000\ \text{kg m}^{-3}$ and $g \approx 10\ \text{m s}^{-2}$ are fine for quick estimates.
• Keep powers of ten consistent: $1000 \times 10 \times 1 \approx 10^{4}$, not $10^{6}$.