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

`I_(Al) = I_(Fe) = 50 xx 10^(-3) m, A_(Al) = (7^(2)-2^(2)) xx 10^(-6) m^(2)`
`A_(Fe) = 2^(2) xx 10^(-6) m^(2)`
Here, two resistances are parallel
`R = (R_(Al) xx R_(Fe))/(R_(Al) + R_(Fe)) = (I_(Al)rho_(Al) rho_(Fe))/(rho_(Al) A_(Fe) + rho_(Fe)A_(Al))`
`= (50 xx 10^(-3) xx 2.7 xx 10^(-8) xx 1 xx 10^(-7))/(2.7 xx 10^(-8) xx 4 xx 10^(-6) +10^(-7) xx 45 xx 10^(-6))`
`= 1875/64 xx 10^(-6) Omega`