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

Show Answer

Solution:

  1. $\frac{1}{R}=\frac{1}{R _{Al}}+\frac{1}{R _{Fe}}=\frac{A _{Al}}{\rho _{Al}}+\frac{A _{Fe}}{\rho _{Fe}} \frac{1}{\ell}$

$$ =\frac{\left(7^{2}-2^{2}\right)}{2.7}+\frac{2^{2}}{10} \frac{10^{-6}}{10^{-8}} \times \frac{1}{50 \times 10^{-3}} $$

Solving we get,

$$ R=\frac{1875}{64} \times 10^{-6} \Omega=\frac{1875}{64} \mu \Omega $$



NCERT Chapter Video Solution

Dual Pane