An aluminium rod with Young's modulus Y = 7.0 | vedclass

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(A) The energy per unit volume $(u)$ stored in a stretched wire is given by the formula:$u = \frac{1}{2} 	imes 	ext{Young's modulus} 	imes (	ext{s

Vedclass · Accepted Answer

$5600$

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(A) The energy per unit volume $(u)$ stored in a stretched wire is given by the formula:$u = \frac{1}{2} 	imes 	ext{Young's modulus} 	imes (	ext{strain})^2$Given:$Y = 7.0 	imes 10^{10} \ N/m^2$$	ext{Strain} = 0.04 \% = \frac{0.04}{100} = 4 	imes 10^{-4}$Substituting the values into the formula:$u = \frac{1}{2} 	imes (7.0 	imes 10^{10}) 	imes (4 	imes 10^{-4})^2$$u = \frac{1}{2} 	imes 7.0 	imes 10^{10} 	imes 16 	imes 10^{-8}$$u = 3.5 	imes 16 	imes 10^2$$u = 56 	imes 10^2 = 5600 \ J/m^3$Thus,the energy per unit volume is $5600 \ J/m^3$.

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 $SI$ units is:

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