Q133 / 160 · TS EAMCET 2007 · Physics

When a wire of length $10 \mathrm{~m}$ is subjected to a force of $100 \mathrm{~N}$ along its length, the lateral strain produced is $0.01 \times 10^{-3} \mathrm{~m}$. The Poisson's ratio was found to be 0.4 . If the area of cross-section of wire is $0.025 \mathrm{~m}^2$, its Young's modulus is

  • Correct answer
    A
    $1.6 \times 10^8 \mathrm{~N} / \mathrm{m}^2$
  • B
    $2.5 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$
  • C
    $1.25 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
  • D
    $16 \times 10^9 \mathrm{~N} / \mathrm{m}^2$

Solution

Poisson's ratio $=\frac{\text { lateral strain }}{\text { longitudinal strain }}$ $\begin{aligned} \text { ie, } & 0.4 & =\frac{0.01 \times 10^{-3}}{\Delta L / L} \\ \text { or } & \frac{L}{\Delta L} & =\frac{0.4}{0.01 \times 10^{-3}}=4 \times 10^4\end{aligned}$ Young's modulus $\begin{aligned} Y & =\frac{F L}{A \Delta l} \\ & =\frac{100}{0.025} \times 4 \times 10^4 \\ & =1.6 \times 10^8 \mathrm{~N} / \mathrm{m}^2\end{aligned}$

Hence, the correct answer is option: 'A'.

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