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 :
Answer & explanation
Correct answer: option 3
Volume of sphere = \(\frac{4}{3}\)\(\pi\)R3
\(\frac{\Delta V}{V} = \frac{3 \Delta R}{R} \)
Percentage error:
(\(\frac{\Delta V}{V}×100) = 3(\frac{\Delta R}{R} ×100)\)
=\(3 ×2%\)
= 6%