Q131 / 160 · TS EAMCET 2023 (12 May Shift 1) · Physics

A wire of length $40 \mathrm{~cm}$ is stretched by $0.1 \mathrm{~cm}$. The strain on the wire is

  • Correct answer
    A
    $25 \times 10^{-4}$
  • B
    $40 \times 10^{-4}$
  • C
    $10 \times 10^{-4}$
  • D
    $12.5 \times 10^{-4}$

Solution

$\begin{aligned} & \text {Strain }=\frac{\Delta \ell}{\ell} \\ & =\frac{0.1}{40}=25 \times 10^{-4} \mathrm{~cm}\end{aligned}$

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

Relevant Questions