Question

a wire of resistance R and of radius r is drawn into another wire of radius 2r, the new resistance of the wire will be-

Answer 1

Step 1: Given that:

The resistance of the wire $=R$

The radius of the wire $=r$

The new radius of the wire $r^{\prime }=2r$

Step 2: Formula used:

As the same wire of radius r is converted into the wire of radius $2r$.

Therefore the volume of the wire in both cases will be equal. That is

$Volumeofthewire=Al=constant$

Thus,

The volume of the wire in the original case= The volume of the wire in the new case

Now,

Resistance of a conducting wire is given by

$R=\frac{\text{ρ}l}{A}\times \frac{A}{A}$

$Since,Al=Volume\left(V\right)$

$R=\frac{\text{ρ}V}{A^2}$

Step 3: Calculation of new resistance of the wire:

According to the question, Originally

$R=\frac{\text{ρ}V}{A^2}$

$R=\frac{\text{ρ}V}{{\left(\pi r^2\right)}^2}$

$R=\frac{\text{ρ}V}{{\pi }^2{r}^4}$

In new case if the resistance of the wire is R' then

$R^{\prime }=\frac{\text{ρ}V}{{A^{\prime }}^2}$

$R^{\prime }=\frac{\text{ρ}V}{{\left(\pi r^{\prime 2}\right)}^2}$

$R^{\prime }=\frac{\text{ρ}V}{{\left(\pi {\left(2r\right)}^2\right)}^2}$

$R^{\prime }=\frac{\text{ρ}V}{16{\pi }^2{r}^4}$

$But,R=\frac{\text{ρ}V}{{\pi }^2{r}^4}$

$Therefore,$

$R^{\prime }=\frac{R}{16}$

Thus, the new resistance of the wire will be $\frac{R}{16}$ .