To understand what happens to the resistance of a copper wire if its length is reduced by half, we should refer to the formula that relates resistance to the physical properties of a material:
[tex]\[ R = \rho \frac{L}{A} \][/tex]
where:
- [tex]\( R \)[/tex] is the resistance of the wire,
- [tex]\( \rho \)[/tex] (Greek letter rho) is the resistivity of the copper wire (a material constant),
- [tex]\( L \)[/tex] is the length of the wire,
- [tex]\( A \)[/tex] is the cross-sectional area of the wire.
Resistance ([tex]\( R \)[/tex]) is directly proportional to the length ([tex]\( L \)[/tex]) of the wire: if you change the length of the wire, the resistance will change by the same factor. The resistivity ([tex]\( \rho \)[/tex]) is a property of the material and does not change unless the material's temperature or intrinsic properties change. The cross-sectional area ([tex]\( A \)[/tex]) also remains constant unless the wire is physically deformed.
Now, let's consider what happens when the length of the wire is reduced by half. Let's say the original length is [tex]\( L \)[/tex], and thus the resistance is [tex]\( R \)[/tex]. If the length of the wire is now reduced by half, the new length is [tex]\( \frac{L}{2} \)[/tex].
Substituting this new length into the formula, the new resistance [tex]\( R' \)[/tex] will be:
[tex]\[ R' = \rho \frac{L/2}{A} \][/tex]
We can also express this in terms of the original resistance [tex]\( R \)[/tex]:
[tex]\[ R = \rho \frac{L}{A} \][/tex]
If we divide every term in this equation by [tex]\( 2 \)[/tex], we get:
[tex]\[ \frac{R}{2} = \rho \frac{L}{2A} \][/tex]
Since [tex]\( \rho \frac{L}{2A} \)[/tex] is the new resistance [tex]\( R' \)[/tex], we deduce that:
[tex]\[ R' = \frac{R}{2} \][/tex]
Therefore, when the length of the wire is reduced by half, the resistance of the wire is halved.
So, the correct answer to the question is:
O It is halved.
