Question

What length of German silver wire, diameter $0.050cm$, is needed to make a $28\Omega$ resistor, if the resistivity of German silver is $2.2\times {10}^{-7}\Omega m$?

• A. $20m$
• B. $25m$
• C. $30m$
• D. $35m$

Answer 1

The correct option is B $25m$

Given :

Resistance R = 28$\Omega$

Using formula, $R=\rho \frac{l}{A}$,

where

$l$ is length of wire

$A$ is the cross-section area and,

$\rho$ is resistivity of the wire.

To find length $l$:

$A=\pi r^2$

Radius $r=\frac{d}{2}=\frac{0.050}{2}cm=0.00025m$

Area $A=\pi {\left(0.00025\right)}^2{m}^2=625\pi \times {10}^{-10}{m}^2$

$28=2.2\times {10}^{-7}\frac{\left(l\right)}{A}$

So,

$l=\frac{28A}{2.2\times {10}^{-7}}$

$l=\frac{28\times 625\pi \times {10}^{-3}}{2.2}=24.98m\approx 25m$