Question

If the percentage error in measuring the surface area of a sphere is $\alpha$ %, then the error in its volume is

• A. $\frac{3}{2}\alpha \%$
• B. $\frac{2}{3}\alpha \%$
• C. $3\alpha \%$
• D. none of these

Answer 1

The correct option is A $\frac{3}{2}\alpha \%$

Volume of sphere $V=\frac{4}{3}\pi r^3$

$\frac{dV}{dr}=4\pi r^2$

Surface area of sphere $S=4\pi r^2$

$\frac{dS}{dr}=8\pi r$

$\Rightarrow \frac{dV}{dS}=\frac{r}{2}$

Also, given percentage error in measuring S is $\alpha$%

$\Rightarrow \frac{\Delta S}{S}=\frac{\alpha }{100}$

$\Rightarrow \Delta S=\frac{\alpha S}{100}=\frac{4\pi r^2\alpha }{100}$

Approximate error in V is $=dV=\left(\frac{dV}{dS}\right)\Delta S$

$=\frac{\alpha }{100}\left(2\pi r^3\right)$

Percentage error in V=$\frac{dV}{V}$

$=\frac{3}{2}\alpha$%