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Question

A wire elongates by $$l$$ $$mm$$ when a load $$W$$ is hanged from it. If the wire goes over a pulley and two weights $$W$$ each are hung at the two ends, the elongation of the wire will be (in $$mm$$):


A
1
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B
2l
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C
zero
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D
l/2
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Solution

The correct option is A $$1$$
According to hook's law
Modules of elasticity $$, E = \dfrac{W}{A} \times \dfrac{l}{1}$$
where$$,$$ $$l= $$ original length of the wire 
$$A=$$ cross sectional area of the wire 
$$\therefore \,Elongation\,,\,\Delta l = \dfrac{{Wl}}{E}$$
on either side of the wire$$,$$ tension is $$W$$ and length is $$\frac{1}{2}$$
$$\Delta l = \dfrac{{Wl}}{{\dfrac{2}{{AE}}}} = \dfrac{{Wl}}{{2AE}} = \dfrac{1}{2}$$
$$\therefore $$ Total elongation in the wire $$\dfrac{1}{2} + \dfrac{1}{2} = 1$$ 
Hence,
option $$(A)$$ is correct answer.

Physics

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