What is the resistance of a wire of length 5 | vedclass

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(D) The resistance $R$ of a conductor is given by $R = \rho \frac{l}{A}$.Here,the cross-sectional area $A$ is the area of the hollow part: $A = \pi (r

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(D) The resistance $R$ of a conductor is given by $R = ho \frac{l}{A}$.Here,the cross-sectional area $A$ is the area of the hollow part: $A = \pi (r_2^2 - r_1^2)$.Given: $r_2 = 10 \, cm = 0.1 \, m$,thickness $t = 5 \, mm = 0.005 \, m$.So,inner radius $r_1 = r_2 - t = 10 \, cm - 0.5 \, cm = 9.5 \, cm = 0.095 \, m$.$A = \pi [(0.1)^2 - (0.095)^2] = \pi [0.01 - 0.009025] = \pi [0.000975] \, m^2$.Now,$R = (1.7 	imes 10^{-8}) 	imes \frac{5}{\pi 	imes 0.000975} \approx 2.77 	imes 10^{-5} \, \Omega$.

What is the resistance of a wire of length $5 \, m$ and resistivity $1.7 \times 10^{-8} \, \Omega \cdot m$,given the cross-sectional area is a hollow cylinder with outer radius $r_2 = 10 \, cm$ and thickness $5 \, mm$?

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