The length of a hollow tube is 5,m,its outer | vedclass

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(A) The resistance $R$ of a conductor is given by $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectio

Vedclass · Accepted Answer

$5.6 	imes 10^{-5} \,\Omega$

Vedclass · Answer

(A) The resistance $R$ of a conductor is given by $R = ho \frac{l}{A}$,where $ho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.For a hollow tube,the cross-sectional area $A$ is given by $A = \pi (r_2^2 - r_1^2)$,where $r_2$ is the outer radius and $r_1$ is the inner radius.Given:Length $l = 5\,m$Outer diameter $D_2 = 10\,cm$,so outer radius $r_2 = 5\,cm = 0.05\,m$.Wall thickness $t = 5\,mm = 0.5\,cm = 0.005\,m$.Inner radius $r_1 = r_2 - t = 5\,cm - 0.5\,cm = 4.5\,cm = 0.045\,m$.Resistivity $ho = 1.7 	imes 10^{-8} \,\Omega m$.Calculating the area $A$:$A = \pi [(0.05)^2 - (0.045)^2] = \pi [0.0025 - 0.002025] = \pi [0.000475] \,m^2$.Now,calculating the resistance $R$:$R = (1.7 	imes 10^{-8}) 	imes \frac{5}{\pi 	imes 0.000475} \approx \frac{8.5 	imes 10^{-8}}{0.001492} \approx 5.69 	imes 10^{-5} \,\Omega$.Rounding to the nearest provided option,the resistance is approximately $5.6 	imes 10^{-5} \,\Omega$.

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

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