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JEE Main 201322 AprilOnline
Exam
JEE Main
Year
2013
Date
22 April
Mode
Online
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2013 · 22 April · Online · Physics · Mechanical Properties of Solids

A uniform wire (Young's modulus 2 × 10^ 11 Nm^ -2 ) is subjected to...

A uniform wire (Young's modulus 2×1011Nm−22 \times 10^{11} \mathrm{Nm}^{-2} ) is subjected to longitudinal tensile stress of 5×107Nm−25 \times 10^7 \mathrm{Nm}^{-2}. If the overall volume change in the wire is 0.02%0.02 \%, the fractional decrease in the radius of the wire is close to:

Options

  1. A1.0×10−41.0 \times 10^{-4}
  2. B1.5×10−41.5 \times 10^{-4}
  3. C0.25×10−40.25 \times 10^{-4}
  4. D5×10−45 \times 10^{-4}

Correct answer

C. 0.25×10−40.25 \times 10^{-4}

Step-by-step solution

Given, y=2×1011Nm−2y=2 \times 10^{11} \mathrm{Nm}^{-2} Stress (FA)=5×107Nm−2ΔV=0.02%=2×10−4 m3Δrr=?\begin{aligned}& \text{ Stress }\left(\frac{\mathrm{F}}{\mathrm{A}}\right)=5 \times10^7 \mathrm{Nm}^{-2}\\ & \Delta \mathrm{V}=0.02 \%=2 \times10^{-4} \mathrm{~m}^3 \\ & \frac{\Delta \mathrm{r}}{\mathrm{r}}=? \end{aligned}γ= stress  strain ⇒strain⁡(Δℓℓ0)=γ stress ΔV=2πrℓ0Δr−πr2Δℓ\begin{aligned}& \gamma=\frac{\text{ stress }}{\text{ strain }} \Rightarrow \operatorname{strain}\left(\frac{\Delta \ell}{\ell_0}\right)=\frac{\gamma}{\text{ stress }}\\ & \Delta \mathrm{V}=2 \pi \mathrm{r} \ell_0 \Delta \mathrm{r}-\pi \mathrm{r}^2 \Delta \ell \end{aligned}Fromeqns(i)and(ii)puttingthevalueofFrom eqns (i) and (ii) putting the value of\Delta \ell, \ell_0 and \Delta \mathrm{V} and solving we get Δrr=\frac{\Delta \mathrm{r}}{\mathrm{r}}=0.25 ×\times10^{-4}
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