A brass rod of cross-sectional area 1,c{m^2} | vedclass

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Question

(b) $U = \frac{1}{2} 	imes \frac{{{{{m{(stress)}}}^{m{2}}}}}{Y} 	imes {m{volume}}$$=$ $\frac{1}{2} 	imes \frac{{{F^2} 	imes A 	imes L}}{{{A

Vedclass · Accepted Answer

$2.5 	imes {10^{ - 5}}\,J$

Vedclass · Answer

(b) $U = \frac{1}{2} 	imes \frac{{{{{m{(stress)}}}^{m{2}}}}}{Y} 	imes {m{volume}}$$=$ $\frac{1}{2} 	imes \frac{{{F^2} 	imes A 	imes L}}{{{A^2} 	imes Y}}$

$=$$\frac{1}{2} 	imes \frac{{{F^2}L}}{{AY}} = \frac{1}{2} 	imes \frac{{{{(50)}^2} 	imes 0.2}}{{1 	imes {{10}^{ - 4}} 	imes 1 	imes {{10}^{11}}}}$$=$ $2.5 	imes {10^{ - 5}}J$

A brass rod of cross-sectional area $1\,c{m^2}$ and length $0.2\, m$ is compressed lengthwise by a weight of $5\, kg$. If Young's modulus of elasticity of brass is $1 \times {10^{11}}\,N/{m^2}$ and $g = 10\,m/{\sec ^2}$, then increase in the energy of the rod will be

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