Two metal rods are joined end to end, as shown in the figure . Both have same cross sectional area of `10^(-6)m^2` and each rod is 1m long. The rod A has electrical resistivity of `2xx10^(-6) Omega` cm and B has electrical resistivity of `10^(-5) Omega cm` . The voltage V required of produce current of 2A in the rods is

Options

A. 0.12volt
B. 0.24volt
C. 0.20volt
D. 0.36volt

Answer

B

Explanation

`V=iR " and " R=rhoL/A`
`R_A = 2xx10^(-8) xx(1)/(10^(-6)) = 2xx10^(-2) Omega`
`R_B = 10^(-7) xx(1)/(10^(-6)) =10^(-1) Omega`
`R_(eq) = R_A + R_B = .02+0.1=0.12Omega`
`V=iR_(eq)=2xx0.12`
=0.24V