Question

Length of a hollow tube is $5m$, it's outer diameter is $10cm$ and thickness of it's wall $5mm$. If resistivity of the material of the tube is $1.7\times {10}^{-8}\Omega m$ then resistance of tube accross its ends will be

• A. $5.6\times {10}^{-5}\Omega$
• B. $2\times {10}^{-5}\Omega$
• C. $4\times {10}^{-5}\Omega$
• D. $Noneofthese$

Answer 1

The correct option is A $5.6\times {10}^{-5}\Omega$

$\text{Step 1 : Resistance of the tube}$

From figure,

Inner radius, $r_1=$ Outer radius $-$ Thickness $=5cm-0.5cm=4.5cm$,

$\rho =1.7\times {10}^{-8}\Omega m$, $l=5m$

We know Resistance can be given as :

$R=\rho \frac{l}{A}$
Since we have to find its resistance across the right and left ends, so we will consider the length between the ends as $5m$ and Area will be the area of cross section of the pipe, i.e. $A=\pi (r_2^2-r_1^2)$.

$\Rightarrow R=1.7\times {10}^{-8}\times \frac{5}{\pi \left\{\right(5\times {10}^{-2}{)}^2-(4.5\times {10}^{-2}{)}^2\}}$
$=5.6\times {10}^{-5}\Omega$

Hence option(A) is correct.

$\text{Note 1:}$

If the Resistance was asked across the inner and outer ends of the pipe, then in the formula of resistance$\left(\rho l/A\right)$, length $l$ will be the thickness$\left(5mm\right)$, and area $A$ will be the curved surface area of the pipe$\left(2\pi rl\right)$.

$\text{Note 2:}$

When the thickness of the pipe is small as compared to its radius, then we can asume the cross section as a thin rectangular strip of length $2\pi r$, therefore its Area can be directly written as $2\pi r\times$ Thickness