IB DP Physics Mock Exam HL Paper 1B Set 1 - 2025 Syllabus

(a)
To ensure accuracy, the student should read the volume at the bottom of the meniscus. Alternatively, ensure the cylinder is placed on a horizontal surface so it is vertical, or ensure eyes are level with the reading to avoid parallax error.

(b)
(i) First, calculate the density \(\rho\):
\[\rho = \frac{m}{V} = \frac{10.82}{10.6} \approx 1.02\,g\,cm^{-3}\]
Next, calculate the uncertainty. The fractional uncertainty in mass \(\frac{0.01}{10.82}\) is negligible compared to the volume. We use the fractional uncertainty of the volume:
\[\frac{\Delta \rho}{\rho} \approx \frac{\Delta V}{V} = \frac{0.2}{10.6}\]
\[\Delta \rho = 1.02 \times \frac{0.2}{10.6} \approx 0.019 \approx 0.02\,g\,cm^{-3}\]
So, \(\rho = (1.02 \pm 0.02)\,g\,cm^{-3}\).

(ii) Converting to \(kg\,m^{-3}\) (multiplying by \(1000\)):
\[\rho = 1020 \pm 20\,kg\,m^{-3}\]
The answer is stated to the correct precision (matching the uncertainty).

(c)
Using the provided graph, the density of pure water at \(35^{\circ}C\) is approximately \(994 \pm 1\,kg\,m^{-3}\). The calculated density of the sample is \(1020 \pm 20\,kg\,m^{-3}\) (range \(1000\) to \(1040\)). Since the value for pure water lies outside the uncertainty range of the sample, the sample is not pure.