Mechanical Properties of Solids

JEE Main level, single correct, medium

30 of 118

Question 30 of 118

A wire of length LL and radius rr is clamped at one end. If its other end is pulled by a force FF, its length increases by ll. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:

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Answer: (D) 22 times

The formula to calculate the Young's modulus of the material of the wire, in the first case, can be written as
Y=Fπr2lL...(1)Y=\dfrac{\dfrac{F}{\pi {r}^{2}}}{\dfrac{l}{L}}...\left(1\right)
From equation (1), it follows that
F=Yπr2×lL...(2)F=Y\pi {r}^{2}\times \dfrac{l}{L}...\left(2\right)
For the second case, the Young's modulus is given by
Y=F/2πr2/4l′L...(3)Y=\dfrac{\dfrac{F/2}{\pi {r}^{2}/4}}{\dfrac{l'}{L}}...\left(3\right)
From equation (3), it follows that
F=Yl′L×2×πr24...(4)F=Y\dfrac{l'}{L}\times 2\times \dfrac{\pi {r}^{2}}{4}...\left(4\right)
From equations (2) and (4), it implies that
Yπr2×lL=Yl′L×2×πr24⇒l′=2lY\pi {r}^{2}\times \dfrac{l}{L}=Y\dfrac{l'}{L}\times 2\times \dfrac{\pi {r}^{2}}{4} \Rightarrow l'=2l

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