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

Answer 1

Answer: (B)

Volume of a sphere $=\frac{4}{3}\pi$ (radius)${}^3$

or $V=\frac{4}{3}\pi R^3$

Taking logarithm on both sides, we have

log $V=$ log $\frac{4}{3}\pi +3$ log $R$

Differentiating, we get $\frac{\Delta V}{V}=0+\frac{3\Delta R}{R}$

Accordingly, $\frac{\Delta R}{R}=2\%$

Thus, $\frac{\Delta V}{V}=3\times 2\%=6\%$