The length of a rod is 20 , cm and the area o | vedclass

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(A) Given: Length $L = 20 \, cm = 0.2 \, m$,Area $A = 2 \, cm^2 = 2 	imes 10^{-4} \, m^2$,Young's modulus $Y = 1.4 	imes 10^{11} \, N/m^2$,Force $F

Vedclass · Accepted Answer

$8.57 	imes 10^{-6}$

Vedclass · Answer

(A) Given: Length $L = 20 \, cm = 0.2 \, m$,Area $A = 2 \, cm^2 = 2 	imes 10^{-4} \, m^2$,Young's modulus $Y = 1.4 	imes 10^{11} \, N/m^2$,Force $F = 5 \, kg-wt = 5 	imes 9.8 \, N = 49 \, N$ (taking $g = 9.8 \, m/s^2$).The elastic potential energy $U$ stored in a rod is given by the formula:$U = \frac{1}{2} 	imes \frac{F^2 L}{AY}$Substituting the values:$U = \frac{1}{2} 	imes \frac{(49)^2 	imes 0.2}{2 	imes 10^{-4} 	imes 1.4 	imes 10^{11}}$$U = \frac{1}{2} 	imes \frac{2401 	imes 0.2}{2.8 	imes 10^7}$$U = \frac{480.2}{5.6 	imes 10^7} \approx 8.57 	imes 10^{-6} \, J$.

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