MHT CET · Physics · Mechanical Properties of Solids
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 \(\ell^{4}\)
- C \(\ell^{2}\)
- D \(\ell^{3}\)
Answer & Solution
Correct Answer
(C) \(\ell^{2}\)
Step-by-step Solution
Detailed explanation
(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
\(\begin{array}{l}
W=\frac{1}{2} \times Y \times(\operatorname{strain})^{2} \times A l \\
=\frac{1}{2} \times Y\left[\frac{y}{1}\right]^{2} \times A l=\frac{1}{2}\left[\frac{Y A}{1}\right] y^{2} \Rightarrow W \propto y^{2}
\end{array}\)
\(\mathrm{V}\) 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
\(\begin{array}{l}
W=\frac{1}{2} \times Y \times(\operatorname{strain})^{2} \times A l \\
=\frac{1}{2} \times Y\left[\frac{y}{1}\right]^{2} \times A l=\frac{1}{2}\left[\frac{Y A}{1}\right] y^{2} \Rightarrow W \propto y^{2}
\end{array}\)
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