I will assume

• Instead of a battery you use an ideal voltage source to provide the potential difference across the wire.

• The connections from the source terminals to the ends of the 1-meter wire are very short and/or made of much larger-diameter wire than the wire being tested.

• The voltage produced by the voltage source is \$V\$

Then your voltmeter will read \$V/2\$.

Because the resistance of a section of wire is given by $$R=\rho\frac{l}{A}$$ where \$\rho\$ is the resistivity of the wire, \$l\$ is the length of the wire, and \$A\$ is the cross-sectional area of the wire.

By measuring dividing your wire into two segments of length \$x/2\$ you create a voltage divider with two equal resistors, and the voltage across either of them is then half the voltage across the series combination.

I will assume

• Instead of a battery you use an ideal voltage source to provide the potential difference across the wire.

• The connections from the source terminals to the ends of the 1-meter wire are very short and/or made of much larger-diameter wire than the wire being tested.

• The voltage produced by the voltage source is \$V\$

Then your voltmeter will read \$V/2\$.

Because the resistance of a section of wire is given by $$R=\rho\frac{l}{A}$$ where \$\rho\$ is the resistivity of the wire, \$l\$ is the length of the wire, and \$A\$ is the cross-sectional area of the wire.

By measuring between one end and the midpoint of the wire, you are effectively dividing your wire into two segments of length \$x/2\$. Thus you create a voltage divider with two equal resistors, and the voltage across either of them is then half the voltage across the series combination.

Actually you could measure the voltage across any section (not just one starting at the end of the wire) of the wire of length \$x/2\$ and you would get \$V/2\$, it would just take a little more math to show why.