Question

A silver wire of resistivity 1.7 x 10-8 2 m has a diameter of 0.2 mm. What will be the length of this wire to offer a resistance of 2 Q? What will be the resistance if the diameter is increased four times? ​

Answer 1

Step 1: The resistance of wire is expressed as

R= AρL ....(1)

Given, R=1Ω,

The area of cross section, A=πr

2 =3.14×0.001

2=3.14×10^−6

.

Resistivity ρ=1.7×10^−8

ohm-meter

Step 2: Putting above values in equation (1)

We get

1= 1.7×10^−8×L ÷ 3.14×10^−6

⇒L=1×3.14×10^−6 ÷ 1.7×10^−8

= 1.7×10^−8 ÷ 3.14×10^−6

=1.847×10^2 m.

Therefore, the length of the wire is 184.7 meters.

hope this helps you :)

Answer 2

Explanation:The formula used in above answer is incorrect

R=pL/A

Area is πr² = π*(10^-4)² converted mm in m and d in r

2π×10^-8/1.7×10^-8 = L

So L is 3.69m

Diameter is increased 4 times so r will be 2 times and r² will be 4 times

But R is inversely proportional to A

So Resistance will be 1/4th