Question

Two metal wires have the resistivity of 1.62 × 10^-18 Ωm and 4.86 × 10^-18 Ωm. Compare the resistance of 2 wires if A is twice as long as B and half of its diameter.

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Answer 1

$\textsf{Factors which control the Resistance in a Wire are :}\\\\\bigstar\;\; \textsf{The Material with which the wire is made up of\; (\rho -\sf{Resistivity}) }\\\\\bigstar\;\; \textsf{The Length of the Wire\;(L)}\\\\\bigstar\;\; \textsf{The Area of Cross section of the Wire\;(A)}}$

$\textsf{All Materials are assigned a value of Resistivity which indicates how well that} \\\textsf{material will conduct an electric current. Hence this Property Resistivity}\\\textsf{depends on how the material is made by the Nature. We cannot decide the}\\\textsf{Resistivity of a Material.It is a constant and is denoted by\; {\rho} }$

$\sf{We\;know\;that : As\;the\;Length\;of\;the\;Wire\;Increases \implies Resistance \;increases}\\\\\textsf{Because, The longer the length of the Wire, The Electrons need to travel a}\\\textsf{longer distance with obstructions in between. So, We can conclude that}\\\textsf{Resistance is directly proportional to Length of the Wire}$

$\sf{We\;know\;that : As\;the\;Area\;of\;the\;Cross\;section\;of\;the\;Wire\;Increases}\\\implies {Resistance\;Decreases}\\\\\textsf{Because, The larger the Area of cross section of the Wire, The Wider the} \\\textsf{Path for the electric current to flow through which results in less obstructions}\\ \textsf{So, We can conclude that Resistance is inversely proportional to the}\\ \textsf{Area of cross section of the Wire.}$

$\textsf{Now based on the above Explanation, We can say that :}\\\\\sf{\bigstar\;\;R \propto L}\\\\\sf{\bigstar\;\;R \propto\dfrac{1}{A}}\\\\\textsf{Combining both we get :}\\\\\sf{\implies R \propto \dfrac{L}{A}}\\\\\textsf{Removing the Proportionality, We get a Proportionality constant - Resistivity}\\\\\implies R = \dfrac{\rho\;L}{A}$

$\textsf{\underline{Problem} : Let the Resistance of the Wire A be R_a }$

$\textsf{Let the Resistance of Wire B be : R_b }$

$\sf{Given : The\;Resistivity\;(\rho_a)\;of\;Wire\;A\;is\;1.62 \times 10^-^1^8}$

$\sf{Given : The\;Resistivity\;(\rho_b)\;of\;Wire\;B\;is\;4.86 \times 10^-^1^8}$

$\textsf{Given : Length of Wire A is Twice the Length of Wire B}\;\sf{\implies L_a = 2L_b}$

$\textsf{Given : Diameter of Wire A is Half of the Diameter of Wire B}\;\sf{\implies 2D_a = D_b}$

$\textsf{We know that,\;Area of Cross Section (circular) is given by} : \sf{\dfrac{\pi D^2}{4}}$

$\sf{\implies \dfrac{R_a}{R_b} = \dfrac{(\rho_a)(L_a)(A_b)}{(\rho_b)(L_b)(A_a)}}\\\\\\\sf{\implies \dfrac{R_a}{R_b} = \dfrac{(1.62 \times 10^-^1^8)(2L_B)\big(\dfrac{\pi (D_b)^2}{4}\big)}{(4.86 \times 10^-^1^8)(L_B)\big(\dfrac{\pi (D_a)^2}{4}\big)}$

$\sf{\implies \dfrac{R_a}{R_b} = \dfrac{(1.62)(2){(D_b)^2}}{(4.86){(D_a)^2}}}\\\\\\\sf{\implies \dfrac{R_a}{R_b} = \dfrac{(1.62)(2){(2D_a)^2}}{(4.86){(D_a)^2}}}\\\\\\\sf{\implies \dfrac{R_a}{R_b} = \dfrac{(1.62)(2)(4){(D_a)^2}}{(4.86){(D_a)^2}}}\\\\\\\sf{\implies \dfrac{R_a}{R_b} = \dfrac{(1.62)(8)}{(4.86)}}\\\\\\\sf{\implies \dfrac{R_a}{R_b} = \dfrac{(12.96)}{(4.86)}}\\\\\\\sf{\implies \dfrac{R_a}{R_b} = 2.67}$

Answer 2

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