Question

If there is an error of $\pm 0.04cm$ in the measurement of the diameter of a sphere, then the approximate percentage error in its volume, when the radius is $10cm$, is

• A. $\pm 1.2$
• B. $\pm 0.06$
• C. $\pm 0.006$
• D. $\pm 0.6$

Answer 1

Answer: (D)

Given, $\Delta r=\pm \frac{0.04}{2}=0.02$
Volume of sphere
$V=\frac{4}{3}\pi r^3$
On differentiating w.r.t. $r$, we get
$\frac{dU}{dr}=\frac{4}{3}\pi \times 3r^2=4\pi r^2$
$\therefore \Delta V=\frac{dU}{dr}\Delta r=4\pi r^2\Delta r$
$\therefore$ Relative per cent error
$\frac{\Delta V}{V}\times 100=\frac{4\pi r^2\Delta r}{\frac{4}{3}\pi r^3}\times 100$
$=\frac{3\Delta r}{r}\times 100$
$=\frac{3\times (\pm 0.02)}{10}\times 100$
$=\pm \frac{6}{10}=\pm 0.6$