Question

A wire of resistance $4\Omega$ is stretched to twice its original length. The resistance of stretched wire would be
(A) $8\Omega$
(B) $16\Omega$
(C) $2\Omega$
(D) $4\Omega$

Answer 1

The resistance of a wire or conductor is directly related to the length of the wire. Since there’s no mention of a change in any other properties such as area, it is safe to assume they remain constant.

Formula used: In this solution we will be using the following formulae;
$R = \rho \dfrac{l}{A}$ where $R$ is the resistance of a conductor or wire, $\rho$ is the resistivity of the material, $l$ is the length of the wire, and $A$ is the cross section.

Complete Step-by-Step solution
To solve the question above, we shall recall that the resistance provided by a wire can be given by
$R = \rho \dfrac{l}{A}$ where $\rho$ is the resistivity of the material, $l$ is the length of the wire, and $A$ is the cross section.
In the question the length was doubled and nothing was said about the area, hence we assume constancy.
Hence,
$R = kl$
$\Rightarrow {R_1} = k{l_1}$ and ${R_2} = k{l_2}$ .
But according to question, ${l_2} = 2{l_1}$ then inserting into above expression, we get
${R_2} = k\left( {2{l_1}} \right) = 2k{l_1}$
$\Rightarrow {R_2} = 2{R_1}$
But according to the question, we have ${R_1} = 4\Omega$
Hence, by inserting into expression ${R_2} = 2{R_1}$
${R_2} = 2\left( 4 \right) = 8\Omega$
Hence, the correct option is A.

Note
For clarity, in actuality, when the length of a substance stretches the area reduces also due to what can be called the Poisson ratio. The Poisson ratio is the ratio of the decrease in width (called lateral length; is the length perpendicular to the length stretched) to the increase in length of the substance in the direction that is stretched. This hence causes a reduction in the cross sectional area. The Poisson ratio is a constant of the material of the substance.