Question

A. A wire is 1m long 0.2mm in diameter and has resistance of 10 ohm. Calculate the
resistivity.​

Answer 1

Explanation:

explained above

ans is 3.14 × 10^-3

Answer 2

Given :

• Length of wire = 1m
• Diameter = 0.2mm
• Resistance = 10 ohm

To find :

• resistivity

Formula used :

$\bf \large \: R = \rho \dfrac{l}{a}$

$\sf \: Where :- \\ \sf \: \: \: \: \: R = resistance \\ \sf \: \: \: \: \: \rho=resistivity \\ \sf \: \: \: \: \: \: l = length \: of \: wire \\ \sf \: \: \: \: \: \: \: \: a = area \: of \: cross \: section$

$\sf \: radius = \dfrac{diameter}{2}$

$\sf1mm = {10}^{ - 3} m$

Solution :

Length of wire = 1m

Diameter = 0.2mm

We know that :-

$\sf \: radius = \dfrac{diameter}{2}$

$\sf \therefore radius = \dfrac{0.2}{2} = 0.1mm$

Now , we know that :-

$\sf1mm = {10}^{ - 3} m$

$\therefore\sf0.1mm = 1 \times {10}^{ - 4} m$

Resistance = 10 ohm

$\implies \sf\: R = \rho \dfrac{l}{a}$

$\implies \sf \: 10 = \rho \times \dfrac{1}{\pi {(1 \times {10}^{ - 4})}^{2} }$

$\implies \sf \: 10 = \rho \times \dfrac{1}{\pi { {10}^{ - 8}} }$

$\implies \sf \: 10 \times\pi \times { {10}^{ - 8}} = \rho$

put π = 3.14

$\implies \sf \: 10 \times3.14 \times { {10}^{ - 8}} = \rho$

$\sf \implies \:3.14\times { {10}^{ - 7}} Ohms-meter= \rho$

Answer :

$\bf\:3.14\times { {10}^{ - 7}} Ohms-meter$