A copper wire has diameter 0.5 mm and resistivity of 1.6 xx 10^(-6) 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?

A copper wire has diameter 0.5 mm and resistivity of 1.6 xx 10^(-6) Om

1.	A copper wire has diameter 0.5 mm and resistivity of 1.6 xx 10^(-6) 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?
Answer» Solution :We are given, the DIAMETER of the wire `d =0.5 mm =0.5 xx 10^(-3)m = 5 xx 10^(-4) m` Resistivity of copper `rho =1.6 xx 10^(-8) OMEGA m` Required RESISTANCE R = `10 Omega` Length l =? `AsR = (rho l)/A`, `l = (RA)/rho` `=(R(pir^2))/rho ( :. A = pi r^2)` `= (R (pid^2))/rho ( :. r = d/2)` =`(pi R d^2)/(4rho)` `=(3.14 xx 10 xx (5 xx 10^(-4))^2)/(4 xx 1.6 xx 10^(-8))` `= (31.4 xx 25 xx 10^(-8))/(6.4 xx 10^(-8)) = 785/6.4 = 122.7 m` Since resistance `R = (rho l)/A` `= (rho l)/(pi r^2) =(rho l)/(pi(d/2)^2) =(4rho l)/(pid^2)` `R prop 1/d^2` (If thereis no change in `rho and l`.) So. when diameter d is doubled. then resistance R BECOMES one-fourth of its original value.