Answer
a)The average volume of the cookie is 2.8 $cm^3$ and the uncertainty in the volume is $\pm0.3 cm^3$ i.e, volume of the cookie with uncertainty is $2.8\pm0.3 cm^3$
b)The ratio of the average diameter to average thickness is 170 and the uncertainty in this ratio is $\pm20$ i.e, $170\pm20$
Work Step by Step
Given: Diameter(d)=$8.50\pm0.02 cm$
Thickness(t)=$0.050\pm0.005 cm$
a)The average volume of the cookie,V=$\pi(\frac{d}{2})^{2}t$
V=$\pi\times(\frac{8.50}{2})^{2}\times0.050cm^3$
=2.8$cm^3$
We now use the extreme values of the given data to give us the maximum and minimum values of the volume from which we can find the uncertainty.
The maximum value for the volume is $V_{max}=\pi\times(\frac{8.52}{2})^2\times0.055 cm^3$ =3.1$cm^3$
The minimum value for the volume is $V_{min}=\pi\times(\frac{8.48}{2})^2\times0.045 cm^3$ =2.5$cm^3$
We find that the maximum volume is 0.3$cm^3$ larger than the average volume and the minimum volume is also 0.3$cm^3$ smaller than the average value .
Therefore ,the uncertainty in volume is $\pm0.3cm^3$ . We can express the volume as $V=2.8\pm0.3cm^3$
b)The ratio of the average diameter to the average thickness is $\frac{8.50}{0.050}=170$
We can obtain the uncertainty in this ratio as follows:
The maximum of this ratio is $\frac{8.52}{0.045}$=190
The minimum of this ratio is $\frac{8.48}{0.055}$=150
We find that the maximum value of the ratio is 20 larger than than the average ratio and the minimum value of the ratio is 20 smaller than the average ratio.
Therefore ,the uncertainty in the ratio is $\pm20$ . We can write the ratio as $170\pm20$
