A brass rod of cross-sectional area 1 , cm^2 | vedclass

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(B) The elastic potential energy stored in a rod under compression is given by the formula: $U = \frac{1}{2} 	imes \frac{	ext{stress}^2}{Y} 	imes \

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The elastic potential energy stored in a rod under compression is given by the formula: $U = \frac{1}{2} 	imes \frac{	ext{stress}^2}{Y} 	imes 	ext{volume}$.Substituting $	ext{stress} = \frac{F}{A}$ and $	ext{volume} = A 	imes L$,we get:$U = \frac{1}{2} 	imes \frac{F^2}{A^2 Y} 	imes (A 	imes L) = \frac{F^2 L}{2AY}$.Given values:$F = m 	imes g = 5 \, kg 	imes 10 \, m/s^2 = 50 \, N$.$L = 0.2 \, m$.$A = 1 \, cm^2 = 1 	imes 10^{-4} \, m^2$.$Y = 1 	imes 10^{11} \, N/m^2$.Substituting these values into the formula:$U = \frac{(50)^2 	imes 0.2}{2 	imes (1 	imes 10^{-4}) 	imes (1 	imes 10^{11})} = \frac{2500 	imes 0.2}{2 	imes 10^7} = \frac{500}{2 	imes 10^7} = 250 	imes 10^{-7} = 2.5 	imes 10^{-5} \, J$.

$A$ brass rod of cross-sectional area $1 \, cm^2$ 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 $g = 10 \, m/s^2$,then the increase in the energy of the rod will be:

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