Question

A drawn wire of resistance 5 ohm is further drawn so that is diameter becomes one fifth of th original what is its resistance with volume remaining the same

Answer 1

Answer:The Resistance of the finally drawn wire will be $3125 \Omega$.

Explanation;

Initial diameter of the wire = d

Initial length of the wire = l

Final diameter of the wire = d'

Final length of the wire = l'

Initial volume of the wire = Final volume of the wire = V

$\pi\frac{d^2}{4}l=\pi\frac{d'^2}{4}l'$

Given that :$d'=d\times \frac{1}{5}$..(1)

So,$l'=25 l$...(2)

Initial Resistance of the wire = R

$5 \Omega=\rho\times \frac{l}{A}=\rho \times \frac{l}{\pi \frac{d^2}{4}}$

$R'=\rho\times \frac{l'}{A'}=\rho \times \frac{l'}{\pi \frac{d'^2}{4}}$

By using (1) and (2) we get:

$R'=R\times 25\times 25=5\Omega \times 25\times 25 =3125 \Omega$

The Resistance of the finally drawn wire will be $3125 \Omega$.