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

Option: 1

1.0\times 10^{-4}
 


Option: 2

1.5\times 10^{-4}


Option: 3

0.25\times 10^{-4}

 


Option: 4

5\times 10^{-4}


Answers (1)

best_answer

\begin{aligned} &Y=\frac{\text {Stress}}{\Delta L / L}\\ &\frac{\Delta L}{L}=\frac{\text {stress}}{Y}=\frac{5 \times 10^{7}}{2 \times 10^{11}}=2.5 \times 10^{-4}\\ &\text { Now } V=\pi r^{2} L\\ &\frac{\Delta V}{V}=\frac{\pi \Delta\left(r^{2} L\right)}{\pi r^{2} L}=\frac{r^{2} \Delta L+L \times 2 r \Delta r}{r^{2} L} \end{aligned}

\begin{aligned} &\frac{\Delta L}{L}+2 \frac{\Delta r}{r}\\ &2 \frac{\Delta r}{r}=\frac{\Delta V}{V}-\frac{\Delta L}{L}=\frac{0.02}{100}-2.5 \times 10^{-4}\\ &\frac{\Delta r}{r}=1 \times 10^{-4}-\frac{2.5}{2} \times 10^{-4}=-0.25 \times 10^{-4} \end{aligned}

Posted by

himanshu.meshram

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