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 $4\%$.