Question

If there is an error of 3% in the volume of a sphere then the percentage error in its radius is

• A. 1%
• B. 2%
• C. 3%
• D. 3/10%

Hint:

For $y = x^{n}$, the percentage error in $y$ is $n$ times the percentage error in $x$. Since volume is $r^{3}$, the factor is 3.

Answer 1

The Correct Option is A

Step 1: Concept
Use the formula for the volume of a sphere, $V = \frac{4}{3}\pi r^{3}$, and the relationship between relative errors.

Step 2: Meaning
The percentage error in volume is given by $\frac{\Delta V}{V} \times 100$, and we need to find $\frac{\Delta r}{r} \times 100$.

Step 3: Analysis
Differentiating $V$ with respect to $r$ gives $dV = 4\pi r^{2} dr$. Dividing by $V$ yields $\frac{dV}{V} = \frac{4\pi r^{2} dr}{\frac{4}{3}\pi r^{3}} = 3 \frac{dr}{r}$. Thus, the percentage error in volume is 3 times the percentage error in radius.

Step 4: Conclusion
Given $3\% = 3 \times (\text{error in radius})$, the percentage error in radius is $3\% / 3 = 1\%$. Final Answer: (A)