SSC JE Electrical Engineering 2017 Set 3 Paper

Resistance of a strip and copper of rectangular cross section is 2Ω.
If other material of twice resistivity is coated on its upper surface to a thickness equal to that of copper strip.
Then resistance of other material = 4Ω
Resistivity is 2 times then resistance is = 2 × 2 = 4Ω
both are connected in parallel.
∴ ${\mathit{R}}_{\mathrm{eq}}=\frac{2\times 4}{2+4}=\frac{4}{3}\mathit{\Omega }$