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.

Short 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-by-step solution

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

(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 \(143.5806{\rm{ }}c{m^3}\).

(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

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

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 \(0.083{\rm{ }}c{m^3}\).