The external diameter of a 5 , m long hollow | vedclass

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(A) Given: Length, $l = 5 \, m$.External diameter, $d_1 = 0.1 \, m$.External radius, $r_1 = \frac{d_1}{2} = \frac{0.1}{2} = 0.05 \, m$.Thickness, $t =

Vedclass · Accepted Answer

$5.7 	imes 10^{-5} \, \Omega$

Vedclass · Answer

(A) Given: Length, $l = 5 \, m$.External diameter, $d_1 = 0.1 \, m$.External radius, $r_1 = \frac{d_1}{2} = \frac{0.1}{2} = 0.05 \, m$.Thickness, $t = 0.005 \, m$.Internal radius, $r_2 = r_1 - t = 0.05 - 0.005 = 0.045 \, m$.Area of cross-section of the hollow tube, $A = \pi(r_1^2 - r_2^2)$.$A = 3.14 	imes [(0.05)^2 - (0.045)^2] = 3.14 	imes (0.0025 - 0.002025) = 3.14 	imes 0.000475 = 1.4915 	imes 10^{-3} \, m^2$.Resistance, $R = ho \cdot \frac{l}{A} = 1.7 	imes 10^{-8} 	imes \frac{5}{1.4915 	imes 10^{-3}}$.$R \approx 5.7 	imes 10^{-5} \, \Omega$.

The external diameter of a $5 \, m$ long hollow tube is $0.1 \, m$ and the thickness of its wall is $0.005 \, m$. If the resistivity $\rho = 1.7 \times 10^{-8} \, \Omega \cdot m$, then its resistance will be:

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