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

Option: 1

1.5 \times 10^{-4}


Option: 2

1.0 \times 10^{-4}


Option: 3

0.5 \times 10^{-4}


Option: 4

0.25 \times 10^{-4}


Answers (1)

best_answer

\mathrm{ \begin{aligned} & \frac{\Delta L}{L}=\frac{\text { Stress }}{Y}=\frac{5 \times 10^7}{2 \times 10^{11}}=2.5 \times 10^{-4} \text {. Given } \frac{\Delta V}{V} \\\\ & =\frac{0.02}{100}=2 \times 10^{-4} \end{aligned} }

Now, \mathrm{\quad V=\pi r^2 L.}  Therefore,

\mathrm{ \begin{aligned} & \frac{\Delta V}{V}=\frac{\Delta\left(\pi r^2 L\right)}{\pi r^2 L}=\frac{2 \pi r \Delta r L+\pi r^2 L \Delta L}{\pi r^2 L} \\\\ & =2 \frac{\Delta r}{r}+\frac{\Delta L}{L} \\\\ & \text { or } \quad \frac{2 \Delta r}{r}=\frac{\Delta V}{V}-\frac{\Delta L}{L}=2 \times 10^{-4}-2.5 \times 10^{-4} \\\\ & =-0.5 \times 10^{-4} \\ & \text { or } \quad \frac{\Delta r}{r}=-0.25 \times 10^{-4} \\\\ & \end{aligned} }
\therefore Fractional decrease in r=0.25 \times 10^{-4}

which is choice (d).

Posted by

Divya Prakash Singh

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