Question

If the error in the measurement of radius of a sphere is 2 % then the error in the determination of volume of the sphere will be -

• A. $8\%$
• B. $2\%$
• C. $4\%$
• D. $6\%$

Answer 1

The correct option is D $6\%$

$\Rightarrow$$V=\frac{4}{3}\pi r^3$

$\Rightarrow$$\Delta V=4\pi r^2\Delta r$

$\Rightarrow$$\frac{\Delta V}{V}\times 100=3\times (\frac{\Delta r}{r}\times 100)$

$\%$error in volume $=3\times \%$error in radius

$=3\times 2=6$%