Question

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

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

Answer 1

The correct option is C $3k\%$

Given that $\frac{\Delta r}{r}\times 100=k\Rightarrow \Delta r=0.01kr$
Volume of the sphere,
$V=\frac{4}{3}\pi r^3$
Taking loagarithm on both sides
$\Rightarrow \frac{dV}{dr}=\frac{3}{4}\pi \left(3r^2\right)=4\pi r^2$
$\therefore \Delta V\approx \left(\frac{dV}{dr}\right)\times \Delta r$
$\Rightarrow \Delta V\approx 4\pi r^2\times 0.01kr$
$\Rightarrow \Delta V\approx 0.04k\pi r^3$
Percentage error in its period is :
$\frac{\Delta V}{V}\times 100\%\approx \frac{0.04k\pi r^3}{\frac{4}{3}\pi r^3}\times 100\%\approx 3k\%$
So, percentage error in its period is $3k\%$
Hence, Option $\left(B\right)$ is correct.