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If a copper wire is stretched to make it a cross-sectional radius 0.1% thinner than what is the percentage increase the resistance
Asked by 2238750243078948 | 31 Aug, 2018, 18:02: PM
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Resistance R = begin mathsize 12px style rho L over A space equals rho fraction numerator L over denominator pi r squared end fraction equals rho L over pi 1 over r squared equals C over r squared end style...............(1)
where ρ is resistivity of the material of the wire, L is length, A is crosssection area and r is radius.
C is constant because length and resistivity are not changed due to stretching of wire
 
By taking logarithm on both sides of (1), we get,   ln R = ln C - 2 ln r ............(2)
 
by diferentiating (2), we get,  begin mathsize 12px style fraction numerator partial differential R over denominator R end fraction space equals space minus 2 fraction numerator partial differential r over denominator r end fraction end style
 
begin mathsize 12px style fraction numerator partial differential R over denominator R end fraction cross times 100 space equals space minus 2 fraction numerator partial differential r over denominator r end fraction cross times 100 space equals space minus 2 cross times 0.1 percent sign space equals space minus 0.2 percent sign end style
Hence resistance decrese by 0.2%
Answered by Thiyagarajan K | 02 Sep, 2018, 06:58: AM
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