8.00Ω
The resistance (R) of a wire is related to the length (L) of the wire as follows;
R = ρL/ A ----------------(i)
Where;
ρ = resistivity of the wire
A = crossectional area of the wire.
Taking the resistivity and area constant, equation (i) can be re-written as;
R = k L -------------(ii)
where;
k = constant = ρ/L
From equation (ii);
The resistance is directly proportional to length. i.e when the resistance increases, the length will also increase and vice-versa. This can be written as follows;
[tex]\frac{R}{L}[/tex] = k
This implies that;
[tex]\frac{R_{1} }{L_{1}}[/tex] = [tex]\frac{R_{2} }{L_{2}}[/tex] -------------------(iii)
Where;
[tex]R_{1}[/tex] and [tex]L_{1}[/tex] are the initial values of the resistance and length respectively
[tex]R_{2}[/tex] and [tex]L_{2}[/tex] are the final values of the resistance and lenght respectively.
From the question;
[tex]R_{1}[/tex] = 4.00Ω
[tex]L_{1}[/tex] = 8.00m
[tex]L_{2}[/tex] = 16.00m
Substitute these values into equation(iii) as follows;
[tex]\frac{4.00}{8.00}[/tex] = [tex]\frac{R_{2} }{16.00}[/tex]
Cross multiply;
4.00 x 16.00 = [tex]R_{2}[/tex] x 8.00
64.00 = 8.00[tex]R_{2}[/tex]
Solve for [tex]R_{2}[/tex];
[tex]R_{2}[/tex] = 64.00/ 8.00
[tex]R_{2}[/tex] = 8.00Ω
Therefore the resistance of the wire after it has been stretched is 8.00Ω
