A uniform heavy rod of mass 20,kg,cross-secti | vedclass

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(B) The elongation $\Delta \ell$ of a rod of length $\ell$,mass $m$,and cross-sectional area $A$ due to its own weight is given by the formula:$\Delta

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$25$

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(B) The elongation $\Delta \ell$ of a rod of length $\ell$,mass $m$,and cross-sectional area $A$ due to its own weight is given by the formula:$\Delta \ell = \frac{mg\ell}{2AY}$Given values:$m = 20\,kg$$A = 0.4\,m^{2}$$\ell = 20\,m$$Y = 2 	imes 10^{11}\,N/m^{2}$$g = 10\,m/s^{2}$Substituting these values into the formula:$\Delta \ell = \frac{20 	imes 10 	imes 20}{2 	imes 0.4 	imes 2 	imes 10^{11}}$$\Delta \ell = \frac{4000}{1.6 	imes 10^{11}}$$\Delta \ell = 2500 	imes 10^{-11}\,m$$\Delta \ell = 25 	imes 10^{-9}\,m$Comparing this with $x 	imes 10^{-9}\,m$,we get $x = 25$.

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