Mechanical Properties of Solids

JEE Main level, single correct, medium

79 of 118

Question 79 of 118

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:

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Answer: (C) 0.25×10−40.25 \times 10^{-4}

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 \times 10^7 \mathrm{Nm}^{-2} \\ & \Delta \mathrm{V}=0.02 \%=2 \times 10^{-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}
From eqns (i) and (ii) putting the value of Δℓ,ℓ0\Delta \ell, \ell_0 and ΔV\Delta \mathrm{V} and solving we get
Δrr=0.25×10−4\frac{\Delta \mathrm{r}}{\mathrm{r}}=0.25 \times 10^{-4}

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