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

Hint: 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.