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A wooden cubical block of mass, $m=20 \mathrm{~kg}$ is measured within an error of $10 \mathrm{~g}$. Its side length, $l=100 \mathrm{~cm}$ is measured within an error of $1 \mathrm{~mm}$. Then, the relative error in the measurement of its density is
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Verified Answer
The correct answer is:
$3.5 \times 10^{-3}$
Density $=\frac{\text { Mass }}{\text { Volume }} \Rightarrow \rho=\frac{m}{l^3}$
So, relative error in density is
$\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{3 \Delta l}{l}$
Here, $m=20 \mathrm{~kg}, \Delta m=10 \mathrm{~g}=10 \times 10^{-3} \mathrm{~kg}$, $l=100 \mathrm{~cm}, \Delta l=1 \mathrm{~mm}=0.1 \mathrm{~cm}$.
So,
$\begin{aligned}
\frac{\Delta \rho}{\rho} & =\frac{10 \times 10^{-3}}{20}+\frac{3 \times 01}{100} \\
& =0.5 \times 10^{-3}+3 \times 10^{-3}=3.5 \times 10^{-3}
\end{aligned}$
So, relative error in density is
$\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{3 \Delta l}{l}$
Here, $m=20 \mathrm{~kg}, \Delta m=10 \mathrm{~g}=10 \times 10^{-3} \mathrm{~kg}$, $l=100 \mathrm{~cm}, \Delta l=1 \mathrm{~mm}=0.1 \mathrm{~cm}$.
So,
$\begin{aligned}
\frac{\Delta \rho}{\rho} & =\frac{10 \times 10^{-3}}{20}+\frac{3 \times 01}{100} \\
& =0.5 \times 10^{-3}+3 \times 10^{-3}=3.5 \times 10^{-3}
\end{aligned}$
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