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A uniform wire of length \(10 \mathrm{~m}\) and diameter \(0.6 \mathrm{~mm}\) is stretched by \(6 \mathrm{~mm}\) with certain force. If the Poisson's ratio of the material of the wire is 0.3 , then the change in diameter of the wire is
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Verified Answer
The correct answer is:
\(10.8 \times 10^{-8} \mathrm{~m}\)
Given, length of the wire, \(L=10 \mathrm{~m}\), diameter, \(D=0.6 \times 10^{-3} \mathrm{~m}\), poisson's ratio, \(\sigma=0.3\) and change in wire length, \(\Delta L=6 \times 10^{-3} \mathrm{~m}\)
As poisson's ratio, \(\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)
\(\sigma=\frac{\frac{\Delta D}{D}}{\frac{\Delta L}{L}}\)
\(\begin{array}{rlrl}
\therefore & \sigma =\frac{\Delta D L}{D \Delta L} \\
\Rightarrow & \Delta D =\frac{\sigma D \Delta L}{L} \\
& =\frac{0.3 \times 0.6 \times 10^{-3} \times 6 \times 10^{-3}}{10} \\
\Rightarrow \Delta D & =10.8 \times 10^{-8} \mathrm{~m}
\end{array}\)
Hence, the change in diameter of wire is \(10.8 \times 10^{-8} \mathrm{~m}\). So, the correct option is (c).
As poisson's ratio, \(\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)
\(\sigma=\frac{\frac{\Delta D}{D}}{\frac{\Delta L}{L}}\)
\(\begin{array}{rlrl}
\therefore & \sigma =\frac{\Delta D L}{D \Delta L} \\
\Rightarrow & \Delta D =\frac{\sigma D \Delta L}{L} \\
& =\frac{0.3 \times 0.6 \times 10^{-3} \times 6 \times 10^{-3}}{10} \\
\Rightarrow \Delta D & =10.8 \times 10^{-8} \mathrm{~m}
\end{array}\)
Hence, the change in diameter of wire is \(10.8 \times 10^{-8} \mathrm{~m}\). So, the correct option is (c).
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