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(D) The elongation $\Delta \ell$ of a rod of length $L$,cross-sectional area $A$,and Young's modulus $Y$ due to its own weight $W$ is given by the for

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$5$

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(D) The elongation $\Delta \ell$ of a rod of length $L$,cross-sectional area $A$,and Young's modulus $Y$ due to its own weight $W$ is given by the formula:$\Delta \ell = \frac{WL}{2AY}$Given values:Weight $W = 10 \, 	ext{N}$Length $L = 20 \, 	ext{cm} = 0.2 \, 	ext{m}$Area $A = 100 \, 	ext{cm}^2 = 100 	imes 10^{-4} \, 	ext{m}^2 = 10^{-2} \, 	ext{m}^2$Young's modulus $Y = 2 	imes 10^{11} \, 	ext{N/m}^2$Substituting these values into the formula:$\Delta \ell = \frac{10 	imes 0.2}{2 	imes 10^{-2} 	imes 2 	imes 10^{11}}$$\Delta \ell = \frac{2}{4 	imes 10^9} = 0.5 	imes 10^{-9} \, 	ext{m}$$\Delta \ell = 5 	imes 10^{-10} \, 	ext{m}$Thus,the elongation is $5 	imes 10^{-10} \, 	ext{m}$. The value in $	imes 10^{-10} \, 	ext{m}$ is $5$.

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