Physics 9702/12 — October/November 2016

Background Concept

Density $\rho$ is defined as mass per unit volume:

$$\rho = \frac{m}{V}$$

So the mass of a material occupying a volume $V$ is

$$m = \rho V$$

This only works cleanly if you use consistent SI units: $\rho$ in $\text{kg m}^{-3}$ and $V$ in $\text{m}^3$, giving $m$ in $\text{kg}$.

Understanding the Question

You are given the density of concrete, $\rho = 2400\ \text{kg m}^{-3}$, and a rectangular space with dimensions $8.0\ \text{cm} \times 90\ \text{cm} \times 110\ \text{cm}$. You must find the mass of concrete that would fill that space, then choose the matching option.

Approach

1. Convert all lengths from cm to m (because the density is in $\text{kg m}^{-3}$).
2. Find the volume of the rectangular block using $V = lwh$.
3. Use $m = \rho V$ to calculate the mass.
4. Round appropriately and match to the closest option.

Step-by-Step Reasoning

Convert each dimension:

• $8.0\ \text{cm} = 8.0 \times 10^{-2}\ \text{m} = 0.080\ \text{m}$
• $90\ \text{cm} = 0.90\ \text{m}$
• $110\ \text{cm} = 1.10\ \text{m}$

Calculate volume:

$$V = 0.080 \times 0.90 \times 1.10 = 0.0792\ \text{m}^3$$

Calculate mass:

$$m = \rho V = 2400 \times 0.0792 = 190.08\ \text{kg} \approx 190\ \text{kg}$$

This corresponds to option B.

Key Takeaways

• Always match units: $\text{kg m}^{-3}$ requires volume in $\text{m}^3$.
• For a rectangular solid, $V = lwh$.
• Mass from density is found using $m = \rho V$.

Common Mistakes

• Forgetting to convert cm to m, which makes the volume too large by a factor of $10^6$.
• Converting only one or two of the dimensions to metres.
• Using $\rho = mV$ or $m = \rho / V$ (wrong rearrangement).

Things to Be Careful About

• Converting centimetres to metres: divide by $100$ (i.e. multiply by $10^{-2}$).
• Volume unit after multiplication must be $\text{m}^3$.
• Rounding: $190.08\ \text{kg}$ rounds to $190\ \text{kg}$, not $1900\ \text{kg}$.