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 $4 .$