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What length of wire (Specific resistance 48\times10^{-8}\mathrm{\Omega m} is needed to make a resistance of \mathrm{4.2\Omega } ? \mathrm{(Diameter =0.4mm) }

Option: 1

\mathrm{3.1m}


Option: 2

\mathrm{4.1m}


Option: 3

\mathrm{2.1m}


Option: 4

\mathrm{1.1m}


Answers (1)

best_answer

Given, \rho=48 \times 10^{-8} \Omega \mathrm{m}

\mathrm{ R=4.2 \Omega }

\mathrm{ D=0.4 \mathrm{~mm}}

use,\mathrm{R=\rho \frac{l}{A} }

\mathrm{ l=\frac{R A}{\rho}=\frac{4.2 \times 22 \times\left(0.2 \times 10^{-3}\right)^2}{7 \times 48 \times 10^{-8}} }

\mathrm{ =\frac{0.6 \times 22 \times 0.04 \times 10^{-6}}{48 \times 10^{-8}} }

\mathrm{ l=\frac{13.2 \times 0.04 \times 10^2}{48} }

\mathrm{ l=1.1 \mathrm{~m} }

Hence option 4 is correct.






 

Posted by

Sanket Gandhi

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