The density of a metal sphere is determined using a digital caliper and a mass balance.
The digital caliper is used to measure the diameter \(D\) of the sphere by placing the sphere in the jaws of the digital caliper. This reading is shown.
The sphere is then removed and another reading is taken immediately afterwards with the jaws closed.
(a)
(i) Calculate \(D\).
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(ii) The manufacturer of the digital caliper states that the uncertainty in the device reading is \(\pm 0.1\,\mathrm{mm}\).
Calculate the percentage uncertainty in \(D\).
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(b) State one way in which the procedure for the measurement of \(D\) can be improved using the same digital caliper.
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(c) The mass \(M\) of the sphere is \((54.0 \pm 0.2)\,\mathrm{g}\).
The density of the sphere \(\rho\) is calculated to be \(11.3\times10^{3}\,\mathrm{kg\,m^{-3}}\), using
\(\rho=\dfrac{6M}{\pi D^{3}}\)
(i) Calculate the percentage uncertainty in \(\rho\).
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(ii) State the value of \(\rho\), including the absolute uncertainty of \(\rho\).
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(a)(i) Correct Answer: \( \boxed{20.9\,\mathrm{mm}} \)
The first caliper reading is \(20.6\,\mathrm{mm}\), while the reading with the jaws closed is \(-0.3\,\mathrm{mm}\).
Therefore, the diameter is
\(D=20.6-(-0.3)\)
\(D=20.9\,\mathrm{mm}\)
Hence, \( \boxed{D=20.9\,\mathrm{mm}} \).
(a)(ii) Correct Answer: \( \boxed{1\%} \)
The uncertainty in each caliper reading is \(0.1\,\mathrm{mm}\). Since \(D\) is found by subtracting two readings, the absolute uncertainties are added:
\(\Delta D=0.1+0.1=0.2\,\mathrm{mm}\)
Percentage uncertainty is
\(\dfrac{0.2}{20.9}\times100=0.96\%\)
To an appropriate number of significant figures, \( \boxed{1\%} \).
(b) Correct Answer:
Make multiple measurements of \(D\) and calculate the average value.
Alternatively, measure the sphere across different diameters and calculate an average.
Another valid improvement is to zero the digital caliper before taking the measurement.
(c)(i) Correct Answer: \( \boxed{3.4\%} \)
The percentage uncertainty in the mass is
\(\dfrac{0.2}{54.0}\times100=0.37\%\)
Since
\(\rho=\dfrac{6M}{\pi D^{3}}\)
the percentage uncertainty in \(D\) must be multiplied by \(3\).
Therefore, the overall percentage uncertainty is
\(0.37+3(1.0)=3.37\%\)
Hence, \( \boxed{3.4\%} \).
(c)(ii) Correct Answer: \( \boxed{(11.3\pm0.4)\times10^{3}\,\mathrm{kg\,m^{-3}}} \)
The absolute uncertainty is calculated from the percentage uncertainty:
\(\Delta\rho=\dfrac{3.4}{100}\times11.3\times10^{3}\)
\(\Delta\rho\approx0.4\times10^{3}\,\mathrm{kg\,m^{-3}}\)
Therefore,
\(\boxed{\rho=(11.3\pm0.4)\times10^{3}\,\mathrm{kg\,m^{-3}}}\)
