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
B
2l
C
zero
D
l/2

Solution

## The correct option is A $$1$$According to hook's lawModules 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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