1.

If an increase in length of copper wire is 0.5% due to stretching, the percentage increase in its resistance will be

Answer»

0.001
0.002
0.01
0.02

Solution :GIVEN, increase in LENGTH of wire, i.e., `(DELTA L)/(L)=0.5 %`
We know, resistance `(R )= rho(l)/(A)= rho (l)/(A)xx(A)/(A)= rho (V)/(A^(2)) ""` (`because`VOLUME, i.e., `V=l xx A`)
Percentage increase in resistance `(= rho (V)/(l^(4)))`
(area, i.e., `A = l^(2)`, i.e., l is length of a wire)
i.e., `(Delta R)/(R )xx100=4(Delta l)/(l)xx100=4xx0.5%=2%`


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