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In an aluminum bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown is the figures. The electrical resistivities of Al and Fe are 2.7 \times 10^{-8} \Omega \mathrm{m} and 1 \times 10^{-7} \Omega \mathrm{m} respectively. The electrical resistance between the two faces P and Q of the composite bar is

Option: 1

\frac{2475}{132} \mu\Omega


Option: 2

\frac{2475}{64} \mu \Omega


Option: 3

\frac{1875}{49} \mu \Omega


Option: 4

\frac{1875}{64} \pi \Omega


Answers (1)

R_{F e}=\frac{\rho_{\text {Fe }} \times 50 \times 10^{-3}}{\left(2 \times 10^{-3}\right)^2} =\frac{2.7 \times 10^{-8} \times 50 \times 10^{-3}}{4 \times 10^{-6}} \\

                                             =1250 \mu \Omega

 

R_{Al} =\frac{P_{Al} \times 50 \times 10^{-3}}{(49-4) \times 10^{-6}}=30 \mu \Omega \\

R_{eq} =\frac{1250 \times 30}{1280}=\frac{1875}{64} \mu \Omega \\

Posted by

Ramraj Saini

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