A resistance wire made from German silver has a resistance of `4.25 Omega`. Calculate the resistance of another wire, made from the same material, such that its length increases by `4` times and area of cross-section decrease by `3` times.

Answer

`51 ohm`

Explanation

`R = (rho l)/(A), R'= (rho(4 l))/((A//3)) = 12(rho l//A) =12 R = 12 (4.25 Omega) = 51 Omega`.