In an aluminium (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are `2.7 xx (10^-8) Omega m and 1.0 xx (10^-7) Omegam,` respectively. The electrical resistance between the two faces P and Q of the composite bar is

Options

A. `(2475)/(64)mu Omega`
B. `(1875)/(64)mu Omega`
C. `(1875)/(49)mu Omega`
D. `(2475)/(132)mu Omega`

Answer

b

Explanation

For Aluminimum
`R(A1)=(rho_(A1^(l)))/(A)=(2.7 xx10^(-8) xx50 xx10^(-3))/((49 -4)xx10^(-6))`
`= 3 xx 10^(-5) Omega`
For iron
`R_(Fe) = (10 xx 10^(-7) xx 50 xx 10^(-3))/(4 xx 10^(-4)) = 125 xx 10^(-5) Omega`
Both will act as two resistance joined in parallel
`:. R_(eq)=(R_(A1)xxR_(Fe))/(R_(A1)+R_(Fe))=(3xx10^(-5)xx125 xx10^(-5))/(3xx10^(-5)+125 xx10^(-5))`
`=(1875 xx 10^(-6) Omega)/(64) = (1875)/(64) mu Omega`