Question

The external diameter of a $314m$ long copper tube is $1.2cm$ and the internal diameter is $1cm$. Calculate its resistance if the specific resistance of copper is $2.2\times {10}^{-8}\Omega -m$

• A. $5.0\times {10}^{-2}\Omega$
• B. $4.4\times {10}^{-2}\Omega$
• C. $3.14\times {10}^{-2}\Omega$
• D. $0.2\Omega$

Answer 1

Answer: (D)

Given, length $l=314m$
Specific resistance $p=2.2\times {10}^{-8}\Omega -m$
External diameter $D_1=1.2\times {10}^{-2}m$
Internal diameter $D_2=1\times {10}^{-2}m$
As resistance of tube
$R=\frac{\rho l}{A}$
Or $R=\frac{\rho l}{\pi [D_1^2-D_2^2]}$
$[\because A=\pi r^2]$
Or $R=\frac{4\times 2.2\times {10}^{-8}\times 314}{\pi [{(1.2)}^2-{(1.0)}^2]\times {10}^{-4}}$
$=0.2\Omega$