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.

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;

= k

This implies that;

=         -------------------(iii)

Where;

and are the initial values of the resistance and length respectively

and are the final values of the resistance and lenght respectively.

From the question;

= 4.00Ω

= 8.00m

= 16.00m

Substitute these values into equation(iii) as follows;

=

Cross multiply;

4.00 x 16.00 = x 8.00

64.00 = 8.00

Solve for ;

= 64.00/ 8.00

= 8.00Ω

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

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and 1and 1