Two wires A and B of same mass and material are taken. Diameter of wire A is half of wire B. If resistance of wire A is `24Omega` find the resistance of wire B.

Explanation

Let `r_A and r_B` be radii and `l_A , l_B ` be lengths of wire A and B respectively . As mass and density of the wires is the same , we have `pir_A^2 l_A d = pir_B^2 l_B d`
Resistance of wires A and B are
`R_A = rho(l_A)/(pir_A^2) and R_B = rho (l_B)/(pi r_B^2)`
`therefore (R_A)/(R_B) = (l_A)/(l_B) xx (r_B^2)/(r_A^2) = (r_B^2)/(r_A^2) xx(r_B^2)/(r_A^2) = ((r_B)/(r_A))^4`
`implies r_A = (r_B)/(2) implies (R_A)/(R_B) = (2)^4 = 16`
`therefore R_B = (R_A)/(16) =(24)/(16) = 1.5 ` ohm.