Question

If the error in measurement of the surface area of a sphere is 4%, then the error in measurement of its volume will be

• A. 4%
• B. 6%
• C. 8%
• D. 16%

Answer 1

The correct option is B 6%

$\mathrm{Su}rfaceare\mathrm{a,}s=4\pi r^2,where4\pi iscons\tan t,\log \left(s\right)=\log \left(4\pi \right)+2\log \left(r\right)Therefore,\frac{\delta s}{s}=2\frac{\delta r}{r}=4\%\frac{\delta r}{r}=2\%$
Volume, $v=\frac{4}{3}{\mathrm{\pi r}}^3$
$\log v=\log \left(4{\mathrm{\pi r}}^3\right)-\log \left(3\right)\log v=\log \left(4\pi \right)+3\log \left(r\right)-\log \left(3\right)\frac{\delta v}{v}=3\frac{\delta r}{r}=3\times 2=6\%$