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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
