Mechanical Properties of Solids

JEE Main level, numerical answer, medium

19 of 118

Question 19 of 118

Two persons pull a wire towards themselves. Each person exerts a force of 200 N200 \mathrm{~N} on the wire. Young's modulus of the material of wire is 1×1011 N m−21 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}. Original length of the wire is 2 m2 \mathrm{~m} and the area of cross section is 2 cm22 \mathrm{~cm}^2. The wire will extend in length by ______ μm\mu \mathrm{m}.

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Answer: 20

200 N⟷O200 \mathrm{~N} \longleftrightarrow O
FA=YΔℓℓ⇒Δℓ=FℓAYΔℓ=200×22×10−4×1011=2×10−5=20μm\begin{aligned} & \frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta \ell}{\ell} \Rightarrow \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \\ & \Delta \ell=\frac{200 \times 2}{2 \times 10^{-4} \times 10^{11}}=2 \times 10^{-5}=20 \mu \mathrm{m}\end{aligned}

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