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In this problem we have two questions which are all related to some uncertainties. And let's get

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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 \( 0.82 \mathrm{~mm}^{3} \) 14

          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 \( 0.82 \mathrm{~mm}^{3} \)
14

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 ± 0.01 mm. Here's an example of measuring three ball bearings and estimating their volumes:

Ball Bearing 1:
Diameter measurement: 5.12 mm± 0.01 mm (using the caliper)

Radius =( diameter 5)/(2)=2.56 mm± 0.005 mm=r+Δ r
Volume =(4)/(3)π r^3=(4)/(3)π(r+Δ r)^3

∴Δ r is very small as compared to r so therefore square and higher power will be neglected

Volume =(4)/(3)π r^3± 4 π r^2Δ r
.V ±Δ V=(4)/(3)(3.14)(2.56)^3± 4(3.14)(2.56)^2(0.005))

=(70.24 ± 0.41) mm^3

Volume of ball bearing 1 is found to be 70.24 mm^3 with uncertainty of 0.41 mm^3
Ball Bearing 2:

Diameter measurement: 3.78 mm± 0.01 mm
Radius = diameter / 2=1.89 mm± 0.005 mm

V ±Δ V=(4)/(3)π r^3± 4 π r^2Δ r

V ±Δ V=(28.26 ± 0.22) mm^3

Volume of ball bearing 2 is found to be 28.26 mm^3 with uncertainty of 0.22 mm^3
Ball Bearing 3:

Diameter measurement: 7.25 mm± 0.01 mm
Radius = diameter / 2=3.62 mm± 0.005 mm

V ±Δ V=(4)/(3)π r^3± 4 π r^2Δ r

V ±Δ V=(198.60 ± 0.82) mm^3

Volume of ball bearing 3 is found to be 198.60 mm^3 with uncertainty of 0.82 mm^3
14

Added by Farman H.

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