Question:
There is an error of $\pm 0.04$ cm in the measurement of the diameter of a sphere. When the radius is 10 cm, the percentage error in the volume of the sphere is
There is an error of $\pm 0.04$ cm in the measurement of the diameter of a sphere. When the radius is 10 cm, the percentage error in the volume of the sphere is
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For $y = x^n$, the percentage error in $y$ is $n \times$ (percentage error in $x$).
Updated On: Apr 10, 2026
- $\pm 1.2$
- $\pm 1.0$
- $\pm 0.8$
- $\pm 0.6$
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The Correct Option is D
Solution and Explanation
Step 1: Given Data
Error in diameter $dD = \pm 0.04$ cm $\implies$ Error in radius $dr = \pm 0.02$ cm. Radius $r = 10$ cm.
Step 2: Formula for Percentage Error
Volume $V = \frac{4}{3} \pi r^3$. $\frac{dV}{V} \times 100 = 3 \frac{dr}{r} \times 100$.
Step 3: Calculation
Percentage Error $= 3 \cdot \frac{\pm 0.02}{10} \cdot 100 = 3 \cdot (\pm 0.2) = \pm 0.6$.
Final Answer: (d)
Error in diameter $dD = \pm 0.04$ cm $\implies$ Error in radius $dr = \pm 0.02$ cm. Radius $r = 10$ cm.
Step 2: Formula for Percentage Error
Volume $V = \frac{4}{3} \pi r^3$. $\frac{dV}{V} \times 100 = 3 \frac{dr}{r} \times 100$.
Step 3: Calculation
Percentage Error $= 3 \cdot \frac{\pm 0.02}{10} \cdot 100 = 3 \cdot (\pm 0.2) = \pm 0.6$.
Final Answer: (d)
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