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**Hint:**If the wires are made of the same material, then the young’s modulus of the wires are the same. Compare the young’s modulus of the wire and apply the formula and substitute the known values to find the elongation produced in the wire.

**Formula used:**

The formula of the young’s modulus is

$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{e}{l}}}$

Where $Y$ is the young’s modulus of the wire, $F$ is the force applied on the wire, $A$ is the surface area of the wire, $e$ is the elongation produced in the wire and $l$ is the length of the wire.

**Complete step by step solution:**

In the given data

In a wire of the length $l$ and the radius $r$ , the tension produced is $e$

Let us consider the $x$ is the elongation produced in the wire when the force of the $2F$ is produced in the wire of the radius $2r$ and the length $2l$ .

It is known that the wire of the same material has the same young’s modulus is also same.

$Y = {Y’}$

Substituting the values of the known parameters in the above formula,

$\Rightarrow$ $\dfrac{{\dfrac{F}{A}}}{{\dfrac{e}{l}}} = \dfrac{{\dfrac{F}{{A'}}}}{{\dfrac{x}{{2l}}}}$

By cancelling the similar terms in the above formula, we get

$\Rightarrow$ $\dfrac{{\dfrac{1}{A}}}{{\dfrac{e}{1}}} = \dfrac{{\dfrac{1}{{A'}}}}{{\dfrac{x}{2}}}$

By cross multiplying the terms in the above equation,

$\Rightarrow$ $\dfrac{1}{A} \times \dfrac{x}{2} = \dfrac{1}{{A'}} \times \dfrac{e}{1}$

By further simplification of the above equation,

$\Rightarrow$ $\dfrac{x}{{2A}} = \dfrac{e}{{A'}}$

By substituting the areas in the above equation,

$\Rightarrow$ $\dfrac{1}{{\pi {r^2}e}} = \dfrac{4}{{4\pi {r^2}x}}$

$\Rightarrow$ $x = 1.0\,e$

**Thus the option (B) is correct.**

**Note:**The young modulus is the modulus of the elasticity which is used in the elastic deformation that the object may deform when the force is applied and regains its original position after the force is released. This depends on the kind of the material.

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