Answer:
Explanation:
We know that the volume V for a sphere of radius r is
[tex]V(r) = \frac{4}{3} \ \pi \ r^3[/tex]
If we got an uncertainty [tex]\Delta r[/tex] the formula for the uncertainty of V is:
[tex]\Delta V(r) = \sqrt{ (\frac{dV}{dr} \Delta r)^2 }[/tex]
We can calculate this uncertainty, first we obtain the derivative:
[tex]\frac{dV}{dr} = 3 * \frac{4}{3} \ \pi \ r^2[/tex]
[tex]\frac{dV}{dr} = 4 \ \pi \ r^2[/tex]
And using it in the formula:
[tex]\Delta V(r) = \sqrt{ (4 \ \pi \ r^2\Delta r)^2 }[/tex]
[tex]\Delta V(r) = \sqrt{ 4^2 \ \pi^2 \ r^4 \Delta r^2 }[/tex]
[tex]\Delta V(r) = 4 \ \pi \ r^2 \Delta r [/tex]
The relative uncertainty is:
[tex]\frac{\Delta V(r)}{V(r)}[/tex]
[tex]\frac{ 4 \ \pi \ r^2 \Delta r }{ \frac{4}{3} \ \pi \ r^3}[/tex]
[tex]\frac{ 3 \Delta r }{ r}[/tex]
Using the values for the problem:
[tex]\frac{ 3 * 0.09 m }{ 5.66 m} = 0.0477[/tex]
This is, a percent uncertainty of 4.77 %
