Length of a hollow tube is $5\,m$, it’s outer diameter is $10\, cm$ and thickness of it’s wall is $5\, mm$. If resistivity of the material of the tube is $1.7 \times 10^{-8} \,\Omega m$ then resistance of tube will be
Medium
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(a) By using $R = \rho .\frac{l}{A};$ here $A = \pi (r_2^2 - r_1^2)$
Outer radius $r_2 = 5\,cm$
Inner radius $r_1 = 5 -0.5 = 4.5\, cm$
So $R = 1.7 \times {10^{ - 8}} \times \frac{5}{{\pi \{ {{(5 \times {{10}^{ - 2}})}^2} - {{(4.5 \times {{10}^{ - 2}})}^2}\} }}$
$ = 5.6 \times {10^{ - 5}}\,\Omega $
Outer radius $r_2 = 5\,cm$
Inner radius $r_1 = 5 -0.5 = 4.5\, cm$
So $R = 1.7 \times {10^{ - 8}} \times \frac{5}{{\pi \{ {{(5 \times {{10}^{ - 2}})}^2} - {{(4.5 \times {{10}^{ - 2}})}^2}\} }}$
$ = 5.6 \times {10^{ - 5}}\,\Omega $
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