An aluminium rod with Young's modulus Y =7.0 × 10^{10} N / m ^2 undergoes elastic strain of 0.0

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8.Mechanical Properties of Solids

easy

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:

A

$5600$

B

$8400$

C

$2800$

D

$11200$

(JEE MAIN-2023)

Solution

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

$\text { Strain }=\frac{0.04}{100}$

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

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

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

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

Standard 11

Physics

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