Question

A wire of length 2L and r is stretched between A and B without the application of any tension. If Y is the Young's modulus of the wire and it is stretched like ACB (d<<L) ,
then what will be the tension of the wire ?

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Answer 1

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Correct question :-

A wire of length 2L and radius 'r' is stretched between A and B without the application of any tension. If Y is the Young's modulus of the wire and it is stretched like ACB (d<<L),  then what will be the tension of the wire ?

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Solution :-

The length of the wire is 2L.

When the wire is stretched, then the change in length can be calculated as :

$\displaystyle{\frac{\delta l}{l}=\frac{l_f-l_i}{l_i} }$

$\displaystyle{AC = \sqrt{l^2+d^2} }\\\\\text{(By using Pythagoras Theorem.)}$

$\text{The total length becomes }2\sqrt{l^2+d^2}.$

As the length of BC is equal to the length of AC, and the values are added.(AB =AC+BC)

◘ Let the initial length $(l_i)$ be 2L, and the final length $(l_f)$ be $2\sqrt{l^2+d^2}$.

$\displaystyle{\frac{\delta l}{l}=\frac{2\sqrt{l^2+d^2} }{2l}} \\\\\displaystyle{\implies \frac{\delta l}{l} =\frac{l\sqrt{l+\frac{d^2}{l^2}-l } }{l} }\\\\\displaystyle{\implies \frac{\delta l}{l}=\sqrt{l+\frac{d^2}{l^2} }-l }\\\\\displaystyle{\implies \frac{\delta l}{l}=(l+\frac{d^2}{l^2}^\frac{1}{2} ) -1}\\\\\displaystyle{\implies \frac{\delta l}{l}=1+\frac{1}{2}*\frac{d^2}{l^2}-1 }\\\\\displaystyle{\implies \frac{\delta l}{l}=\frac{d^2}{2l^2} }$

Now, we have,

$\displaystyle{y=\frac{Fl}{A \delta l} }\\\\\displaystyle{\implies F=yA\frac{\delta l}{l} }$

Area of the wire = πr²

$\huge\boxed{\displaystyle{\implies F=y*\pi r^2*\frac{d^2}{2l^2} }}$

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*F is the tension required.

Thank you☺.

Answer 2

Correct question:

A wire of length 2l, and radius r is stretched between A and B without the application of any tension. If Y is the Young's modulus of the wire and it is stretched like ACB, then what will be the tension in the wire?

Theory :

Young modulus of elasticity (Y):

It is defined as the ratio of normal stress to the longitudinal strain within elastic limit.

Y = stress/strain

Solution :

Given : L = 2l

radius =r

Mean length =d

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$\sf \triangle l = (ac + bc) - 2l$

Now apply Pythagoras theorem in ∆AOC

and in ∆BOC

We get AC =BC = √l²+d²

$\sf \implies \triangle \: l = 2 \sqrt{l {}^{2} + d {}^{2} } - 2l$

$\sf\:Stress=\dfrac{Force}{Area}$

$\sf strain = \dfrac{ \triangle l}{l}$

$\sf = \frac{ \cancel{2} (\sqrt{l {}^{2} + d {}^{2} } - l) }{ \cancel{2}l}$

$\sf Y = \dfrac{T}{\pi r {}^{2} } \times \dfrac{l}{ \sqrt{l {}^{2} + d {}^{2} } }$

$\sf T= \dfrac{Y\pi r {}^{2}( \sqrt{l {}^{2} + d {}^{2} } - l)}{l}$

How ,use binomial expansion

$\sf \sqrt{l {}^{2} +d {}^{2} }= l(1 + \dfrac{d {}^{2} }{2l {}^{2} } )$

$\sf\:\sqrt{1+x}=(1+\dfrac{x}{2})$

$\bf \implies T= Y\pi r {}^{2} l(\dfrac{d {}^{2} }{2l {}^{2} } )$