Question

If an error of $k\%$ is made in measuring the radius of a sphere, then percentage error in its volume is

• A. $k\%$
• B. $3k\%$
• C. $2k\%$
• D. $\frac{2}{3}k\%$

Answer 1

The correct option is B $3k\%$

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

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

Given, percentage error in measuring radius =k%

$\Rightarrow \frac{\Delta r}{r}=\frac{k}{100}$

$\Rightarrow \Delta r=\frac{kr}{100}$

Now, approximate error in measuring V$=dV=\left(\frac{dV}{dr}\right)\Delta r$

$=\frac{k}{100}4\pi r^3=\frac{3k}{100}V=3k\%$ of V

Percentage error in measuring $V=3k\%$