Question

If there is an error of $\pm 0.04cm$ in the measurement of the diameter of a sphere, then the approximate percentage error in its volume, when the radius is $10cm$, is

• A. $\pm 1.2$
• B. $\pm 0.06$
• C. $\pm 0.006$
• D. $\pm 0.6$

Answer 1

The correct option is D

$\pm 0.6$

Step 1: Given

Error in diameter: $\pm ∆d=\pm 0.04cm$

Radius: $r=10cm$

Step 2: Formula Used

The volume of a sphere: $V=\frac{4}{3}\pi r^3$

Step 3: Find the percentage error in volume

Calculate the error in radius by dividing the error in diameter by 2.

$\begin{array}{rcl}\pm ∆r& =& \frac{\pm ∆d}{2}\\ & =& \frac{\pm 0.04}{2}cm\\ & =& \pm 0.02cm\end{array}$

Take the logarithm on both sides of the formula and then differentiate it on both sides.

$$\begin{gathered} \log V=\log \left(\frac{4}{3}\pi r^3\right) \\ \Rightarrow \frac{d}{dV}\log V=\frac{d}{dV}\log \left(\frac{4}{3}\pi r^3\right) \\ \Rightarrow \frac{1}{V}∆V=3\times \frac{1}{r}∆r \end{gathered}$$

Substitute the values and multiply both sides by 100 to find the percentage

$\begin{array}{rcl}\frac{∆V}{V}\times 100& =& 3\frac{∆r}{r}\times 100\\ & =& 3\frac{\pm 0.02}{10}\times 100\\ & =& \pm 0.06\times 10\\ & =& \pm 0.6\%\end{array}$

Hence, the percentage error in volume is $\begin{array}{rcl}& & \pm 0.6\%\end{array}$.

Therefore, option (D) is the correct answer.