Q. 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) $6\%$
Explanation
As $v = \frac{4}{3}\ \pi r^{3}$ $\frac{dv}{v} = 3\left( \frac{dr}{r} \right)$ $\therefore$ Percentage error in determination of volume $= 3$ (Percentage error in measurement of radius) $= 3(2\%) = 6\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.