Question 92734
{{{R[1] = k[1]L[1]/d[1]^2}}} ___________  (1)
where {{{R[1]}}} (in Ohm) = resistance of the first electrical wire of length = {{{L[1]}}} (in metre) and diameter = {{{d[1]}}} (in centimetre) and {{{k[1]}}} = constant of variation with appropriate units


Similarly, {{{R[2] = k[2]L[2]/d[2]^2}}} __________ (2)
where {{{R[2]}}} (in Ohm) = resistance of the second electrical wire of length = {{{L[2]}}} (in metre) and diameter = {{{d[2]}}} (in centimetre) and {{{k[2]}}} = constant of variation with appropriate units

Dividing equation (1) by equation (2)

{{{R[2]/R[1] = (k[2]L[2]d[1]^2)/(k[1]L[1]d[2]^2)}}}
As both the wires are of same material {{{k[1] = k[2]}}}
So, {{{R[2]/R[1] = (L[2]d[1]^2)/(L[1]d[2]^2)}}} _________ (3)


Here {{{L[1] = 20}}}, {{{d[1] = 0.6}}}, {{{R[1] = 36}}}, {{{L[2] = 60}}} and {{{d[2] = 1.2}}}


Then, from substituting these values in equation (3)
{{{R[2]/36 = (60*0.6^2)/(20*1.2^2)}}}
{{{R[2]/36 = 3/4}}}
{{{R[2] = (3*36)/4}}}
{{{R[2] = 27}}}