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. $2.2\times {10}^{-2}\Omega$
• E. $0.2\Omega$

Answer 1

Answer: (E)

$0.2\Omega$

Given, $lengthl=314m$
Specific resistance $\rho l=2.2\times {10}^{-8}\Omega -m$
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 $R=\frac{4\times 2.2\times {10}^{-8}\times 314}{\pi \left[\right(1.2{)}^2-\left(1.0{)}^2\right]\times {10}^{-4}}$
$=0.2\Omega$.