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(B) The shear modulus $\eta$ is defined as $\eta = \frac{	ext{Shear Stress}}{	ext{Shear Strain}} = \frac{F/A}{\Delta x/L}$,where $A$ is the area of

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$x_{1}/2$

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(B) The shear modulus $\eta$ is defined as $\eta = \frac{	ext{Shear Stress}}{	ext{Shear Strain}} = \frac{F/A}{\Delta x/L}$,where $A$ is the area of the face,$\Delta x$ is the displacement,and $L$ is the side length.For the first cube: $\eta = \frac{F/L^2}{x_{1}/L} = \frac{F}{L x_{1}}$.Thus,$x_{1} = \frac{F}{\eta L}$.For the second cube of side $2L$: The area $A' = (2L)^2 = 4L^2$. The shear modulus $\eta$ remains the same as the material is the same.Let the new displacement be $x_{2}$. Then $\eta = \frac{F/A'}{x_{2}/(2L)} = \frac{F/(4L^2)}{x_{2}/(2L)} = \frac{F}{2L x_{2}}$.Equating the two expressions for $\eta$: $\frac{F}{L x_{1}} = \frac{F}{2L x_{2}}$.Solving for $x_{2}$: $x_{2} = \frac{x_{1}}{2}$.

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