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JEE Mains · Physics · STD 11- Mechanical properties of solids
  • JEE Mains 2023
  • Easy
  • Numerical

An aluminium rod with Young's modulus \(Y =7.0\) \(\times 10^{10} N / m ^2\) undergoes elastic strain of \(0.04 \%\). The energy per unit volume stored in the rod in SI unit is:

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Answer & Solution

Correct Answer

(A) \(5600\)

Step-by-step Solution

Detailed explanation

\(Y =7 \times 10^{10}\,N / m ^2\)

\(\text { Strain }=\frac{0.04}{100}\)

\(\text { Energy }=\frac{1}{2}\left(\frac{ YA }{l}\right) \Delta x ^2\)

\(\text { Energy }=\frac{1}{2} YA \left(\frac{\Delta x }{l}\right)^2 \times l\)

\(\frac{ E }{ V }=\frac{1}{2} \times Y \times \operatorname{strain}{ }^2\)

\(=\frac{1}{2} \times 7 \times 10^{10} \times \frac{0.04 \times 0.04}{10^4}=56 \times 10^2\)

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