Question

If there is an error of $\pm 0.04\, cm$ in the measurement of the diameter of a sphere, then the approximate percentage error in its volume, when the radius is $10\, cm$, 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{d U}{d r}=\frac{4}{3} \pi \times 3 r^{2}=4 \pi r^{2} $
$\therefore \Delta V=\frac{d U}{d r} \Delta r=4 \pi r^{2} \,\Delta r$
$\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 $
$=\pm \frac{6}{10}=\pm 0.6$