Answer:
R_a/R_b=3
Explanation:
The resistance in terms of the area and the length of the wire is given by:
R=pL/A
if we have two wires, the first one is a solid wire with a diameter of d_A = 1 x 10^-3 m, and the second one is a hollow wire with inner diameter of d_B,i = 1 x 10^-3 m and outer diameter of d_B,σ= 2 x 10^-3 m, so the cross sectional area of the first wire is: Â
A_a=Ï€r^2_a
A_a=Ï€d^2_A/4
hence the resistance is: Â
R_a=(4*p*L_a)/Ï€*d^2_A Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (1)
the area of the second wire is: Â
A_b=π*r^2_B,σ-π*r^2_B,i
A_b=π/4(d^2_B,σ-d^2_B,i)
hence the resistance is: Â
R_b=(4*p*L_b)/π(d^2_B,σ-d^2_B,i)          (2)
To find the ratio between the resistances R_a/R_b, we divide (1) over (2) to get: Â
R_a/R_b=(d^2_B,σ-d^2_B,i)*L_a/(d^2_a*L_b)
but the wires have the same length, therefore: Â
R_a/R_b=(d^2_B,σ-d^2_B,i)/(d^2_a)
substitute with the given values to get:
R_a/R_b=3
