mistake when measuring and record a value that's too low, but becausc was measured, the mistake went on unnoticed.
1.4.3 Measure the diameters of a few ball bearings of different sizes and estimaghe their volumes. Mention uncertainty in each result. analyze and evaluate the abrom experiment and suggest improvements.
To measure the diameters of different-sized ball bearings and estimate their volumes, assume we have a caliper with a measurement uncertainty of \( \pm 0.01 \mathrm{~mm} \). Here's an example of measuring three ball bearings and estimating their volumes:
Ball Bearing 1:
Diameter measurement: \( 5.12 \mathrm{~mm} \pm 0.01 \mathrm{~mm} \) (using the caliper)
\[
\begin{array}{c}
\text { Radius }=\frac{\text { diameter } 5}{2}=2.56 \mathrm{~mm} \pm 0.005 \mathrm{~mm}=r+\Delta r \\
\text { Volume }=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi(r+\Delta r)^{3}
\end{array}
\]
\( \therefore \Delta r \) is very small as compared to \( r \) so therefore square and higher power will be neglected
\[
\begin{array}{c}
\text { Volume }=\frac{4}{3} \pi r^{3} \pm 4 \pi r^{2} \Delta r \\
\left.V \pm \Delta V=\frac{4}{3}(3.14)(2.56)^{3} \pm 4(3.14)(2.56)^{2}(0.005)\right) \\
=(70.24 \pm 0.41) \mathrm{mm}^{3}
\end{array}
\]
Volume of ball bearing 1 is found to be \( 70.24 \mathrm{~mm}^{3} \) with uncertainty of \( 0.41 \mathrm{~mm}^{3} \)
Ball Bearing 2:
\[
\begin{array}{c}
\text { Diameter measurement: } 3.78 \mathrm{~mm} \pm 0.01 \mathrm{~mm} \\
\text { Radius }=\text { diameter } / 2=1.89 \mathrm{~mm} \pm 0.005 \mathrm{~mm} \\
V \pm \Delta V=\frac{4}{3} \pi r^{3} \pm 4 \pi r^{2} \Delta r \\
V \pm \Delta V=(28.26 \pm 0.22) \mathrm{mm}^{3}
\end{array}
\]
Volume of ball bearing 2 is found to be \( 28.26 \mathrm{~mm}^{3} \) with uncertainty of \( 0.22 \mathrm{~mm}^{3} \)
Ball Bearing 3:
\[
\begin{array}{c}
\text { Diameter measurement: } 7.25 \mathrm{~mm} \pm 0.01 \mathrm{~mm} \\
\text { Radius }=\text { diameter } / 2=3.62 \mathrm{~mm} \pm 0.005 \mathrm{~mm} \\
V \pm \Delta V=\frac{4}{3} \pi r^{3} \pm 4 \pi r^{2} \Delta r \\
V \pm \Delta V=(198.60 \pm 0.82) \mathrm{mm}^{3}
\end{array}
\]
Volume of ball bearing 3 is found to be \( 198.60 \mathrm{~mm}^{3} \) with uncertainty of \( 0.82 \mathrm{~mm}^{3} \)
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