Question

A hollow copper tube of length $5\text{m}$ has got external diameter equal to $10\text{cm}$ and the walls are $5\text{mm}$ thick. If the specific resistance of copper is $1.7\times {10}^{-8}\Omega -\text{m}$, the resistance of the tube across its ends will be:

• A. $5.77\times {10}^5\Omega$
• B. $5.77\times {10}^{-5}\Omega$
• C. $7.77\times {10}^7\Omega$
• D. $5.90\times {10}^5\Omega$

Answer 1

The correct option is B $5.77\times {10}^{-5}\Omega$


The cross-sectional area of copper tube for flow of current is

$A=\pi (R^2-r^2)$

Where $R$ & $r$ are outer & inner radii.

Specific resistance of tube i.e.

$\rho =1.7\times {10}^{-8}\Omega -\text{m}$

External radius, $R=\frac{10}{2}=5\text{cm}$

So, inner radius will be

$r=\frac{D-2t}{2}=\frac{10-(2\times 0.5)}{2}=4.5\text{cm}$

$A=\pi \left[\right(5{)}^2-\left(4.5{)}^2\right]\times {10}^{-4}{\text{m}}^2$

$A=\pi [25-20.25]\times {10}^{-4}=14.92\times {10}^{-4}{\text{m}}^2$

Using relation, $R=\frac{\rho l}{A}$

$\Rightarrow R=\frac{1.7\times {10}^{-8}\times 5}{14.92\times {10}^{-4}}=0.577\times {10}^{-4}$

$\therefore R=5.77\times {10}^{-5}\Omega$

Hence, option $\left(b\right)$ is correct.