A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^{−8}Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

#### Solution

Area of cross-section of the wire, A =π (d/2) 2

Diameter= 0.5 mm = 0.0005 m

Resistance, R = 10 Ω

We know that

radius of wire r = 0.5/2 mm = 0.25 mm = 0.00025 m

`A=pir^2=3.14 xx (0.00025)^2=0.000000019625 m^2`

`l=(RA)/rho=(10xx0.000000019625)/(1.6xx10^-8)=122.72 m`

If the diameter (radius) is doubled, the new radius r = 0.5 mm = 0.0005 m

`A=pir^2=3.14 xx (0.0005)^2=0.000000785m^2`

So, the new resistance will be

`R'=(rho l)/A=(1.6 xx 10^(-8)xx 122.72)/(0.000000785)=2.5 Omega`

Now

(R')/R=2.5/10=1/4

R'=1/4 R

Hence, the new resistance will become 1⁄4 times the original resistance.