Question

A wire of a given material having length $l$ and area of cross-sectional $A$ has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length $\frac{l}{2}$ and area of cross-section $2A$?

Answer 1

We know that, $R=\frac{\rho l}{A}$
For first wire:

$R=\frac{\rho l}{A}=4\Omega$

Now, For the second wire:
Now, $R^{\prime }=\frac{\rho l^{\prime }}{A^{\prime }}$ ......(1)

Here
$l^{\prime }=\frac{l}{2}$ and
$A^{\prime }=2A$
Putting the values in equation (1).
$R^{\prime }=\frac{\rho l^{\prime }}{A^{\prime }}=\frac{\frac{\rho l}{2}}{2A}=\frac{\rho l}{4A}=\frac{4}{4}=1\Omega$


So, the resistance of the new wire is 1Ω.