# Question #222af

Dec 18, 2016

Given

$d \to \text{Diameter of the wire} = 0.46 m m = 46 \times {10}^{-} 5 m$

$l \to \text{Length of the wire} = 730 c m = 7.3 m$

$\rho \to \cdot \text{Resistivity of the wire} = 110 \times {10}^{-} 8 o h m m$

So $A \to \text{Area of cross section} = \frac{\pi {d}^{2}}{4}$

We know resustance $R$ of the wire is rlated with above noted physical quantities as follows.

$R = \rho \frac{l}{A} = \frac{4 \cdot \rho \cdot l}{\pi \cdot {d}^{2}}$

$= \frac{4 \cdot 110 \times {10}^{-} 8 \times 7.3}{3.14 \cdot {\left(46 \times {10}^{-} 5\right)}^{2}} \approx 48.3 \Omega$