Question

If the radius of a sphere is found to be $3cm\pm 0.01cm,$ find the percentage error in the determination of its volume.

Answer 1

The volume of sphere is given as,

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

$=\frac{4}{3}\pi {\left(3\right)}^3$

$=113.097c{m}^3$

The error equation is given as,

$\frac{\Delta V}{V}=3\left(\frac{\Delta r}{r}\right)$

$\Delta V=113.097\times 3\times \left(\frac{0.01}{3}\right)$

$\Delta V=\pm 1.1309c{m}^3$

$\frac{\Delta V}{V}\times 100=\frac{1.13097}{113.097}\times 100$

$\frac{\Delta V}{V}\times 100=1\%$

Thus, the percentage error in volume is $1\%$.