Answer to Question #188078 in Electric Circuits for Subham Acharya

Question #188078

In the manufacture of copper wire a 5 m length of thick circular rod, which has a resistance of

 0.02 ohm is drawn out without change in volume until its new diameter is one-tenth of what it

 was. Calculate the length of the drawn conductor and its resistance. [Ans: 200 Q]


1
Expert's answer
2021-05-03T10:29:05-0400

Explanations & Calculations


  • If the diameter of the rod is "\\small d" and the new length of the drawn wire is "\\small L", then equalling the volume of both the products, the new length can be calculated.

"\\qquad\\qquad\n\\begin{aligned}\n\\small V&=\\small \\pi r^2l=\\pi(\\frac{d}{2})^2.l\\\\\n&\\propto d^2l\\\\\\\\\n\\small V_1&\\propto\\small d^2.5m\\cdots(1)\\\\\n\\small V_2 &\\propto\\small (\\frac{d}{10})^2.L\\cdots(2) \\\\\\\\\n\\small 5d^2&=\\small \\frac{d^2.L}{10^2}\\\\\n\\small L&=\\small 500\\,m\n\\end{aligned}"



  • Resistance of the rod/wire is given by

"\\qquad\\qquad\n\\begin{aligned}\n\\small R&=\\small \\frac{\\rho l}{A}=\\frac{\\rho l}{\\pi r^2}=\\frac{\\rho l}{\\pi\\frac{d^2}{4}}\\\\\n&\\propto\\small \\frac{l}{d^2}\n\\end{aligned}"

  • Then the resistance of the rod is

"\\qquad\\qquad\n\\begin{aligned}\n\\small 0.02\\Omega&\\propto\\frac{5}{d^2}\n\\end{aligned}"

  • That of the wire is,

"\\qquad\\qquad\n\\begin{aligned}\n\\small R&\\propto \\small \\frac{500}{(\\frac{d}{10})^2}\\propto \\frac{500\\times100}{d^2}\n\\end{aligned}"

  • On division of the two equations

"\\qquad\\qquad\n\\begin{aligned}\n\\small \\frac{R}{0.02}&=\\small 100\\times100\\\\\n\\small R&=\\small \\bold{200\\Omega}\n\\end{aligned}"


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