A copper wire has diameter 0.5 mm and resistivity of `1.6 xx 10^(-8)Omega` m. What will be the length of this wire to make its resistance `10Omega`. How much does the resistance change if the diameter is doubled?

Explanation

(i) Diameter, d = 0.5 mm
Radius, `r= (0.5)/(2) mm = 0.25 mm = (0.25)/(1000)m= 0.25 xx 10^(-3)m`
Area of cross-section, `A= pir^(2)= (22)/(7) xx (0.25 xx 10^(-3))^(2)= 0.1964 xx 10^(-6)m^(2)`
Resistance, `R =10Omega`, Resistivity, `rho =1.6xx 10^(-8)Omegam`, Length, I = ?
We know that `R= (rhol)/(A)`
`l= (RA)/(rho)= (10 xx 0.1964 xx 10^(-6))/(1.6xx 10^(-8))= (1964)/(16)= 122.75m`
(ii) The area of cross section `A_(1)= pir^(2)= pi((d)/(2))^(2)= (pirl^(2))/(4)`
Its resistance, `R_(1)= (rhol)/(A_(1))= (4rhol)/(pid^(2))` ...(1)
When the diameter is doubled area of cross section,
`A_(2)= pi((2d)/(2))^(2)= pid^(2)`
Its resistance, `R_(2)= (rhol)/(A_(2))= (rhol)/(pid^(2))` ...(2)
From (1) and (2) we have
`(R_(2))/(R_(1))= (rhol)/(pid^(2)) xx (pid^(2))/(4rhol)= (1)/(4)`
`R_(2)= (1)/(4) R_(1)`
Thus on doubling the diameter the area of cross section becomes 4 times and the resistance becomes one fourth.