Question

A wire has a mass (0.1 $\pm$ 0.001)g radius (0.5 $\pm$ 0.005) mm and length (10 $\pm$ 0.1) cm.
The maximum percentage error in the measurement of its density is.

• A. 1 %
• B. 2 %
• C. 3 %
• D. 4 %

Answer 1

Answer: (D)

Here, mass of wire, $m=(0.1\pm 0.001)g$
Radius, $r=(0.5\pm 0.005)mm$ and length,
$L=(10\pm 0.1)cm$
As density of a wire,
$\rho =\frac{mass}{volume}=\frac{mass}{area\times length}$
$\Rightarrow \rho =\frac{m}{\pi R^2\times L}$
So relative error in density,
$\pm \frac{\Delta \rho }{\rho }=\pm \left[\frac{\Delta m}{m}+2\frac{\Delta R}{R}+\frac{\Delta L}{L}\right]$
$\Rightarrow \frac{\Delta \rho }{\rho }=\frac{0.001\times 10^{-3}}{0.1\times 10^{-3}}+2\times \frac{0.005\times 10^{-3}}{0.5\times 10^{-3}}+\frac{0.1\times 10^{-2}}{10\times 10^{-2}}$
$\Rightarrow \frac{\Delta \rho }{\rho }=0.01+0.02+0.01$
$\Rightarrow \frac{\Delta \rho }{\rho }=0.04$
Maximum percentage error
$=\frac{\Delta \rho }{\rho }\times 100=0.04\times 100=4\%$
Hence, the maximum percentage error in the measurement of density is $4\%$.