Question

A cylindrical wire has a mass $(0.3\pm 0.003)g$, radius $(0.5\pm 0.005)mm$ and length $(6\pm 0.06)cm$. The maximum percentage error in the measurement of its density is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

$\rho =\frac{m}{V}=\frac{m}{\pi r^{2}l}$
or,$\frac{\Delta \rho }{\rho }=\frac{\Delta m}{m}+2\frac{\Delta r}{r}+\frac{\Delta l}{l}$
or, $\frac{\Delta \rho }{\rho }\times 100=\left(\frac{0.003}{0.3}+2\frac{0.005}{0.5}+\frac{0.06}{6}\right)\times 100$
or, $\frac{\Delta \rho }{\rho }\times 100=(0.01+0.02+0.01)\times 100$
or, $\frac{\Delta \rho }{\rho }\times 100=0.04\times 100$
$(\Delta \rho /\rho )\times 100=4\%$
Therefore, the percentage error in the measurement of density is $4.$