Sure, let's solve this step-by-step.
### Step 1: Understanding the factors that affect resistance
The resistance [tex]\( R \)[/tex] of a wire can be expressed by the formula:
[tex]\[ R = \rho \frac{L}{A} \][/tex]
where:
- [tex]\( \rho \)[/tex] is the resistivity of the material,
- [tex]\( L \)[/tex] is the length of the wire,
- [tex]\( A \)[/tex] is the cross-sectional area of the wire.
### Step 2: Effect of doubling the length on the cross-sectional area
When the length of the wire is doubled, the volume of the wire remains constant because the material of the wire is conserved. The volume [tex]\( V \)[/tex] of the wire can be given by:
[tex]\[ V = L \times A \][/tex]
If the length is doubled from [tex]\( L \)[/tex] to [tex]\( 2L \)[/tex], to keep the volume [tex]\( V \)[/tex] constant, the cross-sectional area [tex]\( A \)[/tex] must adjust. The new volume equation when the length is [tex]\( 2L \)[/tex] is:
[tex]\[ V = 2L \times A' \][/tex]
Since the volume must remain constant, we have:
[tex]\[ L \times A = 2L \times A' \][/tex]
[tex]\[ A' = \frac{A}{2} \][/tex]
So, the new cross-sectional area [tex]\( A' \)[/tex] becomes half of the original area [tex]\( A \)[/tex].
### Step 3: Calculating the new resistance
The new resistance [tex]\( R' \)[/tex] with the new length [tex]\( 2L \)[/tex] and new cross-sectional area [tex]\( A' = \frac{A}{2} \)[/tex] can be calculated using the resistance formula:
[tex]\[ R' = \rho \frac{2L}{A'} \][/tex]
Since [tex]\( A' = \frac{A}{2} \)[/tex], we substitute this into the equation:
[tex]\[ R' = \rho \frac{2L}{\frac{A}{2}} \][/tex]
[tex]\[ R' = \rho \frac{2L \times 2}{A} \][/tex]
[tex]\[ R' = 4 \left( \rho \frac{L}{A} \right) \][/tex]
We know that the original resistance [tex]\( R = \rho \frac{L}{A} \)[/tex], so:
[tex]\[ R' = 4R \][/tex]
Hence, the new resistance of the wire after doubling its length is [tex]\( 4R \)[/tex].
