Question

A wire of 3 ohm 15cm in length is stretched to 45cm length. Calculate :-
(1)new cross section​ area
(2) new resistance
(3) resistivity

Answer 1

Hey

--> Assuming the Wire was stretched and additional wire was not used..

$R = \frac{\rho \ l }{A^2} \\ \\ R_0 = \frac{\rho \ 0.15 }{A_0^2} = 3 ohm \\ \\ =\ \textgreater \ 20 A_0^2 = \rho \\ \\ Volume \ of \ wire = 0.15 * A_0 = 0.45 * A_1 \\ ( a \ constant \ since \ the \ wire \ is \ stretched ) \\ \\ R_1 = \frac{\rho \ 0.45 }{A_1^2} = \frac{ 20 A_0^2 \ 0.45 }{A_1^2} =9 ( \frac{A_0}{A_1} )^2 = 9 * 3^2 = 81 ohm$

[tex]Hence, we \ find \\ (1) New \ Area = (0.45/0.15) = (1/3) \ of \ the \ Initial \ Area \\ (2) New \ Resistance = 81 ohm \\ (3) Resistivity = 20 A_0^2 = 180 A_1^2 = \rho \ ( \ constant \ )[/tex]

If only we were given any clue about the area, we might easily figure out the rest ...

Hope it helps ^_^