Question

What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?

Answer 1

Answer:

• 4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

$V(r) = \frac{4}{3} \ \pi \ r^3$

If we got an uncertainty $\Delta r$ the formula for the uncertainty of V is:

$\Delta V(r) = \sqrt{ (\frac{dV}{dr} \Delta r)^2 }$

We can calculate this uncertainty, first we obtain the derivative:

$\frac{dV}{dr} = 3 * \frac{4}{3} \ \pi \ r^2$

$\frac{dV}{dr} = 4 \ \pi \ r^2$

And using it in the formula:

$\Delta V(r) = \sqrt{ (4 \ \pi \ r^2\Delta r)^2 }$

$\Delta V(r) = \sqrt{ 4^2 \ \pi^2 \ r^4 \Delta r^2 }$

$\Delta V(r) = 4 \ \pi \ r^2 \Delta r$

The relative uncertainty is:

$\frac{\Delta V(r)}{V(r)}$

$\frac{ 4 \ \pi \ r^2 \Delta r }{ \frac{4}{3} \ \pi \ r^3}$

$\frac{ 3 \Delta r }{ r}$

Using the values for the problem:

$\frac{ 3 * 0.09 m }{ 5.66 m} = 0.0477$

This is, a percent uncertainty of 4.77 %