Current Electricity 1 Question 12
12. 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 $2.7 \times 10^{-8} \Omega m$ and $1.0 \times 10^{-7} \Omega m$, respectively. The electrical resistance between the two faces $P$ and $Q$ of the composite bar is
(2015 Adv.)
(a) $\frac{2475}{64} \mu \Omega$
(b) $\frac{1875}{64} \mu \Omega$
(c) $\frac{1875}{49} \mu \Omega$
(d) $\frac{2475}{132} \mu \Omega$
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Solution:
- $\frac{1}{R}=\frac{1}{R _{Al}}+\frac{1}{R _{Fe}}=\frac{A _{Al}}{\rho _{Al}}+\frac{A _{Fe}}{\rho _{Fe}} \frac{1}{\ell}$
$$ =\frac{\left(7^{2}-2^{2}\right)}{2.7}+\frac{2^{2}}{10} \frac{10^{-6}}{10^{-8}} \times \frac{1}{50 \times 10^{-3}} $$
Solving we get,
$$ R=\frac{1875}{64} \times 10^{-6} \Omega=\frac{1875}{64} \mu \Omega $$
