Q28PE - College Physics (Urone)

Question

A car engine moves a piston with a circular cross-section of \({\bf{7}}.{\bf{500}}{\rm{ }} \pm {\rm{ }}{\bf{0}}.{\bf{002}}{\rm{ }}{\bf{cm}}\) diameter a distance of \({\bf{3}}.{\bf{250}}{\rm{ }} \pm {\rm{ }}{\bf{0}}.{\bf{001}}{\rm{ }}{\bf{cm}}\) to compress the gas in the cylinder. (a) By what amount is the gas decreased in volume in cubic centimeters. (b) Find the uncertainty in this volume.

Step-by-Step Solution

Answer

1. The volume change in cubic centimeters is \(143.5806{\rm{ }}c{m^3}\).
2. The uncertainty in this volume is \(0.083{\rm{ }}c{m^3}\).

Step 1: Definition of percentage uncertainty and volume:

Uncertainty as a percentage is just relative uncertainty multiplied by \({\rm{100}}\). The percent uncertainty likewise lacks units since it is a ratio of comparable values.

 

The volume V of a cylinder with radius r is the area of the base times the height.

Given data:

Consider the given data as below.

Diameter=\(d = 7.5{\rm{ }}cm\)

Height (distance)= \(h = 3.250{\rm{ }}cm\)

Step 2: (a) Define the volume:

The volume of the cylindrical block is,

\(\begin{array}{c}V = \pi {r^2}h\\ = \pi {\left( {\frac{d}{2}} \right)^2}h\end{array}\)

Substitute \(7.5{\rm{ }}cm\) for \(d\) and \(3.250{\rm{ }}cm\) for \(h\) in the above equation.

\(\begin{array}{c}V = 3.13 \times {\left( {\frac{{7.5}}{2}} \right)^2} \times 3.250\\ = 143.5806{\rm{ }}c{m^3}\end{array}\)

 

Hence, the volume change in cubic centimeters is \(143.5806{\rm{ }}c{m^3}\).

Step 3: (b) Uncertainty percentage:

The Percentage uncertainty in the length of the cylinder is 

\(\begin{array}{c}H = \frac{{\Delta h}}{h} \times 100\% \\ = \frac{{0.001}}{{3.250}} \times 100\% \\ = 0.031\% \end{array}\)

 

The percentage uncertainty in the diameter of the cylinder is

\(\begin{array}{c}D = \frac{{\Delta d}}{d} \times 100\% \\ = \frac{{0.002}}{{7.5}} \times 100\% \\ = 0.027\% \end{array}\)

Step 4: Determine the uncertainty in this volume:

Define the uncertainty of the volume as below.

\(\% uncertainty = \frac{{\delta V}}{V} \times 100\% \)

\(\begin{array}{c}\delta V = \frac{{\% uncertainty \times V}}{{100}}\\ = \frac{{0.058 \times 143.5806}}{{100}}\\ = 0.083{\rm{ }}c{m^3}\end{array}\)

Hence, the uncertainty in this volume is \(0.083{\rm{ }}c{m^3}\).