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Question: When a tension \(F\) is applied, the elongation produced in uniform wire of length \(l\) , radius \(...

When a tension FF is applied, the elongation produced in uniform wire of length ll , radius rr is ee, when tension 2F2F is applied, the elongation produced in another uniform wire of length 2l2l and radius 2r2r made of same material is:
A) 0.5e0.5\,e
B) 1.0e1.0\,e
C) 1.5e1.5\,e
D) 2.0e2.0\,e

Explanation

Solution

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=FAelY = \dfrac{{\dfrac{F}{A}}}{{\dfrac{e}{l}}}
Where YY is the young’s modulus of the wire, FF is the force applied on the wire, AA is the surface area of the wire, ee is the elongation produced in the wire and ll is the length of the wire.

Complete step by step solution:
In the given data
In a wire of the length ll and the radius rr , the tension produced is ee
Let us consider the xx is the elongation produced in the wire when the force of the 2F2F is produced in the wire of the radius 2r2r and the length 2l2l .
It is known that the wire of the same material has the same young’s modulus is also same.
Y=YY = {Y’}
Substituting the values of the known parameters in the above formula,
\Rightarrow FAel=FAx2l\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 1Ae1=1Ax2\dfrac{{\dfrac{1}{A}}}{{\dfrac{e}{1}}} = \dfrac{{\dfrac{1}{{A'}}}}{{\dfrac{x}{2}}}
By cross multiplying the terms in the above equation,
\Rightarrow 1A×x2=1A×e1\dfrac{1}{A} \times \dfrac{x}{2} = \dfrac{1}{{A'}} \times \dfrac{e}{1}
By further simplification of the above equation,
\Rightarrow x2A=eA\dfrac{x}{{2A}} = \dfrac{e}{{A'}}
By substituting the areas in the above equation,
\Rightarrow 1πr2e=44πr2x\dfrac{1}{{\pi {r^2}e}} = \dfrac{4}{{4\pi {r^2}x}}
\Rightarrow x=1.0ex = 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.