Question

If the percentage error in the surface area of sphere is $k$, then the percentage error in its volume is

• A. $\frac{3k}{2}$
• B. $\frac{2k}{3}$
• C. $\frac{k}{3}$
• D. $\frac{4k}{3}$

Answer 1

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

$\text{Given - Percentage error in Surface Area =}k$

$\begin{array}{c}\text{We Know that Surface area of sphere}\\ SA=4\pi r^2\end{array}$

$\begin{array}{c}\text{By error analysis}\\ \text{Taking logarithm both Sides}\\ \log \left(SA\right)=2\cdot \log r+\log \left(4\pi \right)\end{array}$

$\begin{array}{c}\text{Differentiating Both Sides-}\\ \begin{array}{cc}\frac{d\left(SA\right)}{SA}& =2\cdot \frac{dr}{r}=k\text{(given)}\\ \Rightarrow \frac{dr}{r}& =\frac{k}{2}\end{array}\end{array}$

$\begin{array}{c}\text{Now we know that Volume of Sphere}\\ V=\frac{4}{3}\pi r^3\\ \text{Similarly by error analysis we get-}\\ \frac{dV}{V}=3\frac{dr}{r}=\frac{3}{2}k\\ \Rightarrow \text{Percentage error in volume is}\frac{3}{2}k\end{array}$