The length,radius and resistivity of two wires are each in the ratio 1:3. The resistance of the comparatively thin wire is `20 Omega`. Determine the resistance of the other wire.

Explanation

`R=rho1/A=(rho1)/((pid^2)/4)=(4rho1)/(pid^2)`[d=diameter of thwe wire]
For the first wire `R_1=(4rho_1l_1)/(pid_2^2)`
For the second wire ,`R_2=(4rho_2l_2)/(pid_2^2)`
`therefore=R_1/R_2=rho_1/rho_2timesl_1/l_2times(d_2/d_1)^2=1/3times1/3times(3/1)^2=1`
or `R_1=R_2`
The diameter of the first wire is less. So it is thin.
`thereforeR_1=20Omega,soR_2=20Omega`