Question

An 8.00-m-long length of wire has a resistance of 4.00 Ω. The wire is uniformly stretched to a length of 16.0 m. Find the resistance of the wire after it has been stretched.

Answer 1

Answer:

8.00Ω

Explanation:

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;

$\frac{R}{L}$ = k

This implies that;

$\frac{R_{1} }{L_{1}}$ = $\frac{R_{2} }{L_{2}}$         -------------------(iii)

Where;

$R_{1}$ and $L_{1}$ are the initial values of the resistance and length respectively

$R_{2}$ and $L_{2}$ are the final values of the resistance and lenght respectively.

From the question;

$R_{1}$ = 4.00Ω

$L_{1}$ = 8.00m

$L_{2}$ = 16.00m

Substitute these values into equation(iii) as follows;

$\frac{4.00}{8.00}$ = $\frac{R_{2} }{16.00}$

Cross multiply;

4.00 x 16.00 = $R_{2}$ x 8.00

64.00 = 8.00$R_{2}$

Solve for $R_{2}$;

$R_{2}$ = 64.00/ 8.00

$R_{2}$ = 8.00Ω

Therefore the resistance of the wire after it has been stretched is 8.00Ω