Question 5: Units , Dimensions and Error Analysis

Question

The density $\rho$ of a uniform cylinder is determined by measuring its mass $m$, length $l$ and diameter $d$. The measured values of $m, l$ and $d$ are $97.42 \pm 0.02 \mathrm{~g}$, $8.35 \pm 0.05 \mathrm{~mm}$ and $20.20 \pm 0.02 \mathrm{~mm}$, respectively. Calculated percentage fractional error in $\rho$ is $\_\_\_\_$ .

Accepted Answer

The density $(ho)$ of a body is defined as the ratio of its mass (m) to its volume (V).
$$ ho=\frac{m}{V} $$
For a uniform cylinder of length l and diameter D , the volume is,
$$ V=\pi r^2 l=\pi\left(\frac{D}{2}ight)^2 l=\frac{\pi D^2 l}{4} $$
Substituting the volume into the density formula:
$$ ho=\frac{\mathrm{m}}{\mathrm{~V}}=\frac{4 \mathrm{~m}}{\pi \mathrm{D}^2 \mathrm{l}} $$
To find the fractional error, we take the natural logarithm $(\ln )$ on both sides:
$$ \ln (ho)=\ln (4)+\ln (m)-\ln (\pi)-2 \ln (D)-\ln (l) $$
Differentiating both sides to find the relative change
$$ \frac{\mathrm{d} ho}{ho}=\frac{\mathrm{dm}}{\mathrm{~m}}-\frac{2 \mathrm{dD}}{\mathrm{D}}-\frac{\mathrm{dl}}{\mathrm{l}} $$
The maximum possible fractional error is the sum of the absolute values of the individual relative errors:
$$ \frac{\Delta ho}{ho}=\frac{\Delta \mathrm{m}}{\mathrm{~m}}+\frac{2 \Delta \mathrm{D}}{\mathrm{D}}+\frac{\Delta \mathrm{l}}{\mathrm{l}} $$
For the percentage fractional error,
$$ \% 	ext { Error in } ho=\left(\frac{\Delta \mathrm{m}}{\mathrm{~m}}+\frac{2 \Delta \mathrm{D}}{\mathrm{D}}+\frac{\Delta \mathrm{l}}{\mathrm{l}}ight) 	imes 100 $$
The measured value of mass is $97.42 \pm 0.02 \mathrm{~g} \Rightarrow \mathrm{~m}=97.42 \mathrm{~g}$ and the absolute error is $\Delta \mathrm{m}=0.02 \mathrm{~g}$ So, relative error in mass is,
$$ \frac{\Delta \mathrm{m}}{\mathrm{~m}}=\frac{0.02}{97.42} \approx 0.000205 $$
The measured value of length is $8.35 \pm 0.05 \mathrm{~mm} \Rightarrow \mathrm{l}=8.35 \mathrm{~mm}$ and the absolute error is $\Delta \mathrm{l}=0.05 \mathrm{~mm}$
So, relative error in length is,
$$ \frac{\Delta \mathrm{l}}{\mathrm{l}}=\frac{0.05}{8.35} \approx 0.005988 $$
The measured value of diameter is $20.20 \pm 0.02 \mathrm{~mm} \Rightarrow \mathrm{D}=20.20 \mathrm{~mm}$ and the absolute error is $\Delta \mathrm{D}=$ 0.02 mm
So, relative error in diameter is,
$$ \frac{2 \Delta \mathrm{D}}{\mathrm{D}}=\frac{2 	imes 0.02}{20.20}=\frac{0.04}{20.20} \approx 0.001980 $$
Putting the values in the expression for the percentage error in density,
$$ \frac{\Delta ho}{ho} 	imes 100 \%=(0.000205+0.005988+0.001980) 	imes 100 \%=0.008173 	imes 100 \% $$
$$\Rightarrow $$ $$ \frac{\Delta ho}{ho} 	imes 100 \%=0.8173 \% \approx 0.82 \% $$
Therefore, the calculated percentage fraction error in density is approximately $0.82 \%$.
Hence, the correct option is (B).

Answer

0.63%

Answer

0.82%

Answer

0.72%

Answer

0.25%

Units , Dimensions and Error Analysis

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