Question 3: Units , Dimensions and Error Analysis

Question

The percentage error in the calculated volume of a sphere, if there is $2 \%$ error in its diameter measurement, is $\_\_\_\_$ .

Accepted Answer

The volume $V$ of a sphere is given by the formula $V=\frac{4}{3} \pi r^3$, where $r$ is the radius of the sphere.
Since the error is given in terms of diameter (D), we express the radius as $r=\frac{D}{2}$.
Substituting this into the volume formula:
$$ \mathrm{V}=\frac{4}{3} \pi\left(\frac{\mathrm{D}}{2}ight)^3 $$
$$\Rightarrow $$ $\mathrm{V}=\frac{4}{3} \pi\left(\frac{\mathrm{D}^3}{8}ight)$
$$\Rightarrow $$ $$ V=\frac{1}{6} \pi D^3 $$
To find the relative error, we take the natural logarithm $(\ln )$ on both sides of the equation :
$$ \ln (V)=\ln \left(\frac{\pi}{6}ight)+\ln \left(D^3ight) $$
$$\Rightarrow $$ $$ \ln (V)=\ln \left(\frac{\pi}{6}ight)+3 \ln (D) $$
Now, on differentiating both sides.
$$ \frac{\mathrm{dV}}{\mathrm{~V}}=0+3\left(\frac{\mathrm{dD}}{\mathrm{D}}ight) $$
$$\Rightarrow $$ $$ \frac{\mathrm{dV}}{\mathrm{~V}}=3 \frac{\mathrm{dD}}{\mathrm{D}} $$
To express this as a percentage error, we multiply both sides by 100 :
$$ \left(\frac{\mathrm{dV}}{\mathrm{~V}} 	imes 100ight)=3 	imes\left(\frac{\mathrm{dD}}{\mathrm{D}} 	imes 100ight) $$
It is given that the percentage error in the diameter is $2 \%$.
$$ \frac{d D}{D} 	imes 100=2 \% $$
So, percentage error in Volume $=3 	imes 2 \%$
Percentage error in Volume $=6 \%$
Therefore, the correct option is (C).

Answer

1

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2

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6

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8

Units , Dimensions and Error Analysis

The percentage error in the calculated volume of a sphere, if there is 2 \% error in its diameter measurement, is \_\_\_\_ .

Choose the correct answer: