Question

A wire of 3 ohms resistance and 15 cm in length is stretched to 45 cm in length. Calculate the following:

i) New cross sectional area

ii) New resistance

iii) Resistivity.

Assume that wire has a uniform cross sectional area.

Answer 1

(i) Let the initial area be A

The volume will be the same in both conditions
$Volume=A\times Length$

$\therefore A\left(15\right)=A_2\left(45\right)$
---> new area $A_2=A\left(\frac{15}{45}\right)=\frac{A}{3}$

New cross-sectional area = 1/3 of the initial area

(ii)
$R=\frac{\rho L}{A}$

New length is 3 times the initial length

The new area is 1/3 of the initial area

So new resistance $R^{\prime }=\rho \left(\frac{3L}{\frac{A}{3}}\right)=9\left(\frac{\rho L}{A}\right)=9R=9\left(3\right)=27\Omega$

(iii)

$R=\rho \left(\frac{L}{A}\right)$

$\therefore 3=\rho \left(\frac{0.15}{A}\right)$

$\therefore \rho =\frac{3A}{0.15}$

$\therefore \rho =20A$

where A is the area of the conductor