Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2261 Upvotes
Verified Answer
The correct answer is:
4
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 $\rho$,
$\begin{aligned}
&\frac{\Delta \rho}{\rho} \times 100=\frac{\Delta m}{m} \times 100+\frac{2 \Delta r}{r} \times 100+\frac{\Delta l}{l} \times 100 \\
&=\frac{0.003}{0.3} \times 100+2 \times \frac{0.005}{0.5} \times 100+\frac{0.06}{6} \times 100 \\
&=1+2(1)+1=4 \%
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.