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(A) In the first case,the tension in the wire is $T_1 = Mg$. The extension $\Delta \ell_1$ is given by $\Delta \ell_1 = \frac{MgL}{AY}$,where $L$ is t

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(A) In the first case,the tension in the wire is $T_1 = Mg$. The extension $\Delta \ell_1$ is given by $\Delta \ell_1 = \frac{MgL}{AY}$,where $L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Given $\Delta \ell_1 = 4.0 \ mm$.When the load is immersed in a liquid,a buoyant force $B$ acts upwards. The new tension is $T_2 = Mg - B$.The buoyant force $B = V ho_{liquid} g$,where $V$ is the volume of the load.The weight of the load is $Mg = V ho_{load} g$.Thus,$B = Mg \left( \frac{ho_{liquid}}{ho_{load}} ight) = Mg \left( \frac{2}{8} ight) = \frac{1}{4} Mg$.The new tension $T_2 = Mg - \frac{1}{4} Mg = \frac{3}{4} Mg$.The new extension $\Delta \ell_2 = \frac{T_2 L}{AY} = \frac{3}{4} \left( \frac{MgL}{AY} ight) = \frac{3}{4} \Delta \ell_1$.Substituting $\Delta \ell_1 = 4.0 \ mm$,we get $\Delta \ell_2 = \frac{3}{4} 	imes 4.0 \ mm = 3.0 \ mm$.

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