Question 1
1
A metal wire has a cross-section of diameter approximately 0.8 mm .
structured4 marks
Question 1(a)
1(a)
State what instrument should be used to measure the diameter of the wire.
Easystructured1 marks
Answer
micrometer/screw gauge/digital callipers B1 [1]
Question 1(b)
1(b)
State how the instrument in (a) is
structured3 marks
Question 1(b)(i)
1(b)(i)
checked so as to avoid a systematic error in the measurements,
Easystructured1 marks
Answer
look/check for zero error \(\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \quad\) B1
Question 1(b)(ii)
1(b)(ii)
used so as to reduce random errors.
Easystructured2 marks
Answer
take several readings M1 around the circumference/along the wire A1
Question 1
1
A digital voltmeter with a three-digit display is used to measure the potential difference across a resistor. The manufacturers of the meter state that its accuracy is \(\pm 1 \%\) and \(\pm 1\) digit. The reading on the voltmeter is 2.05 V .
structured4 marks
Question 1(a)
1(a)
For this reading, calculate, to the nearest digit,
structured2 marks
Question 1(a)(i)
1(a)(i)
a change of 1 % in the voltmeter reading, change = V
Easystructured1 marks
Answer
1 % of \(\pm 2.05\) is \(\pm 0.02\)
Question 1(a)(ii)
1(a)(ii)
the maximum possible value of the potential difference across the resistor. maximum value = V
Easystructured1 marks
Answer
max. value is 2.08 V
Question 1(b)
1(b)
The reading on the voltmeter has high precision. State and explain why the reading may not be accurate.
Easystructured2 marks
Answer
there may be a zero error/calibration error/systematic error which makes all readings either higher or lower than true value
Question 1
1
A well has a depth of 36 m from ground level to the surface of the water in the well, as shown in Fig. 1.1. A student wishes to find the depth of the well. The student plans to drop a stone down the well and record the time taken from releasing the stone to hearing the splash made by the stone as it enters the water.
structured7 marks
Question 1(b)
1(b)
The time recorded by the student using a stop-watch is not equal to the time in (a). Suggest three possible reasons, other than the effect of air resistance, for this difference. 1 2 3
Mediumstructured3 marks
Answer
reaction time between hearing the splash and stopping the stop-watch - the sound (of the splash) takes time to reach the student or the stone hits the water at a different time to the sound being heard or the sound (of the splash) has to travel to the student - the student might not let go of the stone from ground level - the student might not let go of the stone and start the stop-watch at the same time - stop-watch may not be properly calibrated / has a zero error - (local value of) g is not (exactly) \(9.81\left(\mathrm{~m} \mathrm{~s}^{-2}\right)\) - stone given initial velocity / initial velocity not zero - stone does not fall (exactly) vertically / in a straight line Any three points, 1 mark each B3
Question 1(c)
1(c)
The student repeats the experiment three times and uses the results to calculate the depth of the well. The values are shown in Table 1.1. The true depth of the well is 36.0 m . Explain why these results may be described as precise but not accurate.
Mediumstructured2 marks
Answer
precise: results are close together / have little scatter B1 not accurate: the values are not close to / 50\% different / (very) different from the true value B1
Question 1
1
The rate of flow Q of a liquid along a narrow pipe of length L and radius r is given by where \(\alpha\) is a constant. An experiment is carried out to determine the value of \(\alpha\). The data from the experiment are shown in Table 1.1.
structured5 marks
Question 1(b)
1(b)
Show that the percentage uncertainty in \(\alpha\) is 15 %.
Mediumstructured1 marks
Answer
(percentage uncertainty = ) \(3+4+2 \times 4=15\) (\%) A1
Question 1(c)
1(c)
Calculate \(\alpha\) with its absolute uncertainty. Give your answer to an appropriate number of significant figures.
Mediumstructured3 marks
Answer
\[ \begin{aligned} \alpha =Q L / r^{4} =2.72 \times 10^{-8} \times 2.5 \times 10^{-2} /\left(7.1 \times 10^{-5}\right)^{4} =2.7 \times 10^{7} \end{aligned} \] C1 absolute uncertainty \(=0.15 \times\left[2.7 \times 10^{7}\right]\) \[ =0.4 \times 10^{7} \] C1 \(\alpha=(2.7 \pm 0.4) \times 10^{7} \mathrm{~s}^{-1}\) A1