Q133 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Physics

What is the work done in stretching a uniform metal wire of length from 2 m to 2.004 m with an area of cross section $10^{-6} \mathrm{~m}^2 ?$ [Young's modulus of the wire = 2 × 1011 N/m2]

  • A
    1.6 J
  • Correct answer
    B
    0.8 J
  • C
    8 J
  • D
    16 J

Solution

$\begin{aligned} & \text { Work done }=\frac{1}{2} \mathrm{Kx}^2 \\ & =\frac{1}{2} \frac{\mathrm{yA}}{\ell} \times(\Delta \ell)^2=\frac{1}{2} \times \frac{2 \times 10^{11} \times 10^{-6}}{2} \times(0.004)^2 \\ & =0.8 \mathrm{~J}\end{aligned}$

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

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