KCET · Physics · Mathematics in Physics
A cylindrical wire has a mass \((0.3 \pm 0.003) g\), radius \((0.5 \pm 0.005) \mathrm{mm}\) and length \((6 \pm 0.06) \mathrm{cm}\). The maximum percentage error in the measurement of its density is
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation
Given, mass, \(m=(0.3 \pm 0.003) \mathrm{g}\)
Radius, \(r=(0.5 \pm 0.005) \mathrm{mm}\)
Length, \(l=(6 \pm 0.06) \mathrm{cm}\)
Density of cylinder, \(\rho=\frac{\text { Mass }}{\text { Volume }}=\frac{m}{\pi r^{2} l}\)
Fraction error in \(\rho\),
\(\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{2 \Delta r}{r}+\frac{\Delta l}{l}\)
Percentage error in \(\rho\),
\(\begin{aligned}
&\frac{\Delta \rho}{\rho} \times 100=\frac{\Delta m}{m} \times 100+\frac{2 \Delta r}{r} \times 100+\frac{\Delta l}{l} \times 100 \\
&=\frac{0.003}{0.3} \times 100+2 \times \frac{0.005}{0.5} \times 100+\frac{0.06}{6} \times 100 \\
&=1+2(1)+1=4 \%
\end{aligned}\)
Radius, \(r=(0.5 \pm 0.005) \mathrm{mm}\)
Length, \(l=(6 \pm 0.06) \mathrm{cm}\)
Density of cylinder, \(\rho=\frac{\text { Mass }}{\text { Volume }}=\frac{m}{\pi r^{2} l}\)
Fraction error in \(\rho\),
\(\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{2 \Delta r}{r}+\frac{\Delta l}{l}\)
Percentage error in \(\rho\),
\(\begin{aligned}
&\frac{\Delta \rho}{\rho} \times 100=\frac{\Delta m}{m} \times 100+\frac{2 \Delta r}{r} \times 100+\frac{\Delta l}{l} \times 100 \\
&=\frac{0.003}{0.3} \times 100+2 \times \frac{0.005}{0.5} \times 100+\frac{0.06}{6} \times 100 \\
&=1+2(1)+1=4 \%
\end{aligned}\)
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