A wire of length 1.0, m and diameter 0.6, cm | vedclass

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(B) The resistance of a conductor is given by $R = ho \frac{l}{A}$.For the wire: $l_1 = 1.0\, m$,$d_1 = 0.6\, cm = 0.6 	imes 10^{-2}\, m$,$r_1 = 0.

Vedclass · Accepted Answer

$2.7 	imes 10^{-7}\, \Omega$

Vedclass · Answer

(B) The resistance of a conductor is given by $R = ho \frac{l}{A}$.For the wire: $l_1 = 1.0\, m$,$d_1 = 0.6\, cm = 0.6 	imes 10^{-2}\, m$,$r_1 = 0.3 	imes 10^{-2}\, m$,$R_1 = 3.0 	imes 10^{-3}\, \Omega$.$R_1 = ho \frac{l_1}{\pi r_1^2} \Rightarrow ho = \frac{R_1 \pi r_1^2}{l_1}$.For the disc: $l_2 = 1.0\, mm = 1.0 	imes 10^{-3}\, m$,$d_2 = 2.0\, cm = 2.0 	imes 10^{-2}\, m$,$r_2 = 1.0 	imes 10^{-2}\, m$.$R_2 = ho \frac{l_2}{\pi r_2^2} = \left( \frac{R_1 \pi r_1^2}{l_1} ight) \frac{l_2}{\pi r_2^2} = R_1 \left( \frac{l_2}{l_1} ight) \left( \frac{r_1}{r_2} ight)^2$.Substituting the values: $R_2 = (3.0 	imes 10^{-3}) 	imes \left( \frac{1.0 	imes 10^{-3}}{1.0} ight) 	imes \left( \frac{0.3 	imes 10^{-2}}{1.0 	imes 10^{-2}} ight)^2$.$R_2 = (3.0 	imes 10^{-6}) 	imes (0.3)^2 = (3.0 	imes 10^{-6}) 	imes 0.09 = 0.27 	imes 10^{-6} = 2.7 	imes 10^{-7}\, \Omega$.

$A$ wire of length $1.0\, m$ and diameter $0.6\, cm$ has a resistance of $3.0 \times 10^{-3}\, \Omega$. What will be the resistance between the faces of a disc made of the same material with diameter $2.0\, cm$ and thickness $1.0\, mm$?

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