Question

A wire of resistance 4$\Omega $ is stretched to double its original length. The resistance of the stretched wire would be
$
  {\text{A}}{\text{. 2}}\Omega \\
  {\text{B}}{\text{. 4}}\Omega \\
  {\text{C}}{\text{. 8}}\Omega \\
  {\text{D}}{\text{. 16}}\Omega \\
 $

Answer 1

Hint: We know that the resistance of a wire is equal to the product of resistivity and length of wire divided by the cross-sectional area. By calculating the change in the dimensions of the wire, we can calculate the resistance of the stretched wire and obtain the required answer.

Formula used:
The resistance of a wire of length l and cross-sectional area A is given as
$R = \dfrac{{\rho l}}{A}$

Complete step-by-step answer:
We are given a wire whose resistance is given as
$R = 4\Omega $
Let its length be l and its cross-sectional area be A. As we know that the resistance of a wire of length l and cross-sectional area A is given as
$R = \dfrac{{\rho l}}{A}$
Therefore we can insert the value of resistance and write the following expression.
$\dfrac{{\rho l}}{A} = 4$
Now it is given that the wire is stretched to twice its original length l. Let the new length be l’ which can be written as
$l' = 2l$
Now on stretching the wire, the cross-sectional area of the wire decreases. Let the new cross-sectional area be A’. In both the cases, the volume of the wire will remain same and its equal to the product of length of the wire and its cross-sectional A. Therefore, we can write for the two cases that
$
  lA = l'A' \\
  lA = 2lA' \\
   \Rightarrow A' = \dfrac{A}{2} \\
 $
The resistance for the stretched wire can be written as
$R' = \dfrac{{\rho l'}}{{A'}}$
Inserting the known values, we get
$R' = \dfrac{{\rho \times 2l}}{{\dfrac{A}{2}}} = 4 \times \dfrac{{\rho l}}{A} = 4 \times 4 = 16\Omega $
This is the required value of resistance. Hence, the correct answer is option D.

So, the correct answer is “Option D”.

Note: It should be noted that the resistivity of a wire independent of the dimensions of a wire. It depends on the material of the wire. Since we are dealing with the wire of the same material, the resistivity remains constant throughout our calculations.