Question

A thick wire is stretched, so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to the initial resistance of wire?

• A. 2:01
• B. 4:01
• C. 3:01
• D. 1:04

Answer 1

Answer: (C)

3:01

Answer 2

Resistance of the wire is given by
$R = \rho \frac {l}{A} = \rho \frac {l^2}{(Al)} = \frac {\rho l^2}{V} \, \, \, \, \, (\therefore Al = V)$
So, $R ? l ^2$
(If density remains same)
or $\, \, \, \, \, \, \, \, \, \, \frac {R'}{R} = \frac {(2l)^2}{(l)^2} = 4$
$\, \, \, \, \, \, \, \, \, \, R' = 4R$
Hence, change in resistance
Therefore, $\frac {Change \, in \, resistance }{Original \, resistance} = \frac {3R}{R} = 3 : 1$