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IB DP Physics Mock Exam HL Paper 1B Set 3 - 2025 Syllabus

IB DP Physics Mock Exam HL Paper 1B Set 1

Prepare for the IB DP Physics Mock Exam HL Paper 1B Set 3 with our comprehensive mock exam set 3. Test your knowledge and understanding of key concepts with challenging questions covering all essential topics. Identify areas for improvement and boost your confidence for the real exam

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Question 

A student measures the resistivity \(\rho\) of a metal wire with a cylindrical shape. To do this, the student records the following experimental data:
Length \(L\) of the wire \(= (462 \pm 2)\,mm\)
Measurements of the wire diameter \(d\) are taken as follows:
Reading123456
\(d / mm\)\(0.18\)\(0.20\)\(0.21\)\(0.22\)\(0.26\)\(0.18\)
Resistance \(R\) of the wire \(= 13.7\,\Omega \pm 1.5\%\)
(a) Explain why the diameter of the wire was measured at several different positions along its length.
(b) Suggest appropriate instruments to measure:
(i) \(L\)
(ii) \(d\)
(c) (i) Show that the average diameter of the wire is approximately \(0.2\,mm\).
(ii) Determine the fractional uncertainty in the diameter measurement.
(iii) Determine the fractional uncertainty in the measured length of the wire.
(d) The resistivity of the metal is given by \(\rho = \frac{RA}{L}\), where \(A\) is the cross-sectional area of the wire. Calculate the value of \(\rho\) and its absolute uncertainty. Give your final answers using a suitable number of significant figures.
▶️ Answer/Explanation
Detailed solution

(a)
Measurements are taken at different places to check for uniformity along the wire and to reduce the impact of random measurement errors by averaging the values.

(b)
(i) A metre ruler or measuring tape (with sufficient length).
(ii) A micrometer screw gauge or a Vernier caliper.

(c)
(i) The value \(0.26\,mm\) is anomalous and should be excluded. The mean of the remaining readings is:
\[\frac{0.18 + 0.20 + 0.21 + 0.22 + 0.18}{5} = 0.198\,mm \approx 0.2\,mm\]
(ii) The range is \(0.22 – 0.18 = 0.04\,mm\), so the absolute uncertainty is \(0.02\,mm\). The fractional uncertainty is:
\[\frac{0.02}{0.20} = 0.10\]
(iii) The fractional uncertainty in the length is:
\[\frac{2}{462} \approx 0.004\]

(d)
Using \(d = 0.20\,mm = 2.0 \times 10^{-4}\,m\):
\[A = \pi \left(\frac{d}{2}\right)^2 \approx 3.14 \times 10^{-8}\,m^2\]
\[\rho = \frac{RA}{L} \approx 9.3 \times 10^{-7}\,\Omega m\]
Combining uncertainties:
\[\frac{\Delta \rho}{\rho} = 0.015 + 0.004 + 2(0.10) = 0.219\]
\[\rho = (9 \pm 2) \times 10^{-7}\,\Omega m\]

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