Question:
If the percentage error in the measurement of radius is \(2\%\), then the error in measurement of volume of a sphere is
If the percentage error in the measurement of radius is \(2\%\), then the error in measurement of volume of a sphere is
Show Hint
For quantities of the form
\[
y=kx^n,
\]
the percentage error in \(y\) is approximately
\[
n\times
(\text{percentage error in }x).
\]
Here \(V\propto r^3\), so the error gets multiplied by \(3\).
Updated On: Jul 29, 2026
- \(6\%\)
- \(8\%\)
- \(4\%\)
- \(10\%\)
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The Correct Option is A
Solution and Explanation
Concept:
If
\[
y=x^n,
\]
then the percentage error in \(y\) is approximately
\[
n\times
(\text{percentage error in }x).
\]
Step 1: Write the formula for the volume of a sphere. \[ V=\frac{4}{3}\pi r^3. \]
Step 2: Differentiate logarithmically. Taking logarithms, \[ \log V = \log\left(\frac43\pi\right) + 3\log r. \] Differentiating, \[ \frac{dV}{V} = 3\frac{dr}{r}. \] Hence, \[ \frac{\Delta V}{V} = 3\frac{\Delta r}{r}. \]
Step 3: Convert to percentage error. Given percentage error in radius \[ = 2\%. \] Therefore, \[ \text{Percentage error in volume} = 3\times2\%. \] \[ =6\%. \] Therefore, \[ \boxed{6\%} \] \[ \boxed{\text{Answer = (A)}} \]
Step 1: Write the formula for the volume of a sphere. \[ V=\frac{4}{3}\pi r^3. \]
Step 2: Differentiate logarithmically. Taking logarithms, \[ \log V = \log\left(\frac43\pi\right) + 3\log r. \] Differentiating, \[ \frac{dV}{V} = 3\frac{dr}{r}. \] Hence, \[ \frac{\Delta V}{V} = 3\frac{\Delta r}{r}. \]
Step 3: Convert to percentage error. Given percentage error in radius \[ = 2\%. \] Therefore, \[ \text{Percentage error in volume} = 3\times2\%. \] \[ =6\%. \] Therefore, \[ \boxed{6\%} \] \[ \boxed{\text{Answer = (A)}} \]
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