A copper wire is stretched to make its radius decreased by `0.1` % .Find the percentage increase in resistance .

Explanation

Now , volume will remain constant
`rArr pir^(2)l = pir^(2)l` [ l is its new length ]
`rArr pir^(2)l = pi (0.99)^(2)l`
`rArr " " l l/((0.99)^(2))`
If R is new area ,then
`A/A = (pir^(2))/(pi(0.99r)^(2)) =1/(0.99)^(2)rArr A = (0.99)^(2)A`
Percentage increase in resistance
`= (R-R)/R xx 100 = =(rhol/A-(rhol)/A)/(rho l/A) xx 100`
` = (1.041 -1)/1 xx 100 = 4.1 %`