Q114 / 150 · MHT CET 2020 (19 Oct Shift 2) · Physics

A metal rod has length, cross-sectional area and Young's modulus as ' $\mathrm{L}^{\prime},{ }^{\prime} \mathrm{A}^{\prime}$ and $\mathrm{Y}^{\prime}$ respectively. If the elongation in the rod produced is ' $\ell^{\prime}$ then work done is proportional to

  • A
    $\ell$
  • B
    $\ell^{4}$
  • Correct answer
    C
    $\ell^{2}$
  • D
    $\ell^{3}$

Solution

(C) $\mathrm{V}$ olume $=\mathrm{A} \times \mathrm{L}$ or $\mathrm{V}=\mathrm{Al}$ strain $=\frac{\text { Elongation }}{\text { Original length }}=\frac{\mathrm{Y}}{\mathrm{l}}$ Young's modulus $=\frac{\text { stress }}{\text { strain }}$ Work done, $W=\frac{1}{2} \times$ stress $\times$ strain $\times$ volume $\begin{array}{l} W=\frac{1}{2} \times Y \times(\operatorname{strain})^{2} \times A l \\ =\frac{1}{2} \times Y\left[\frac{y}{1}\right]^{2} \times A l=\frac{1}{2}\left[\frac{Y A}{1}\right] y^{2} \Rightarrow W \propto y^{2} \end{array}$

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

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