Q134 / 160 · AP EAMCET 2024 (23 May Shift 1) · Physics

The work done on a wire of volume of $2 \mathrm{~cm}^3$ is $16 \times 10^2 \mathrm{~J}$. If the young's modulus of the material of the wire is $4 \times 10^{12} \mathrm{Nm}^{-2}$. Then the strain produced in the wire is

  • A
    0.03 m
  • B
    0.04 m
  • C
    0.01 m
  • Correct answer
    D
    0.02 m

Solution

$\begin{aligned} & \mathrm{W}=16 \times 10^2 \mathrm{~J}, \mathrm{~V}=2 \mathrm{~cm}^3=2 \times 10^{-6} \mathrm{~m}^2 \\ & \mathrm{Y}=4 \times 10^{12} \mathrm{Nm}^{-2} \\ & \therefore \quad \mathrm{~W}=\frac{1}{2} \mathrm{y} \varepsilon^2 \mathrm{~V} \Rightarrow \varepsilon=\sqrt{\frac{2 \mathrm{~W}}{\mathrm{Yv}}} \\ & \therefore \quad \varepsilon=\sqrt{\frac{2 \times 16 \times 10^2}{4 \times 10^{12} \times 2 \times 10^{-6}}}=0.02\end{aligned}$

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

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