The cross-sectional area of all wires is 10^{ | vedclass

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(B) The displacement of point $C$ is the sum of the elongations of wire $AB$ and wire $BC$.Elongation of wire $AB$ $(\Delta L_{AB})$ = $\frac{Mg L_{AB

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$9 	imes 10^{-6} \, m$

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(B) The displacement of point $C$ is the sum of the elongations of wire $AB$ and wire $BC$.Elongation of wire $AB$ $(\Delta L_{AB})$ = $\frac{Mg L_{AB}}{A Y_{AB}} = \frac{10 	imes 10 	imes 0.1}{10^{-4} 	imes 2.5 	imes 10^{10}} = \frac{1}{2.5} 	imes 10^{-5} = 0.4 	imes 10^{-5} = 4 	imes 10^{-6} \, m$.Elongation of wire $BC$ $(\Delta L_{BC})$ = $\frac{Mg L_{BC}}{A Y_{BC}} = \frac{10 	imes 10 	imes 0.2}{10^{-4} 	imes 4 	imes 10^{10}} = \frac{2}{4} 	imes 10^{-5} = 0.5 	imes 10^{-5} = 5 	imes 10^{-6} \, m$.Total displacement of point $C$ = $\Delta L_{AB} + \Delta L_{BC} = 4 	imes 10^{-6} + 5 	imes 10^{-6} = 9 	imes 10^{-6} \, m$.

The cross-sectional area of all wires is $10^{-4} \, m^2$. What is the displacement of point $C$?

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