As shown in the figure,forces of 10^5,N each | vedclass

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(B) For the same material,the modulus of rigidity $\eta$ is constant,where $\eta = \frac{	ext{Shear Stress}}{	ext{Shear Strain}}$.Shear stress is gi

Vedclass · Accepted Answer

$0.25$

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(B) For the same material,the modulus of rigidity $\eta$ is constant,where $\eta = \frac{	ext{Shear Stress}}{	ext{Shear Strain}}$.Shear stress is given by $\sigma = \frac{F}{A}$,where $A = L^2$ is the area of the face.Shear strain is given by $\gamma = \frac{\Delta x}{L}$,where $\Delta x$ is the displacement and $L$ is the side length.Thus,$\eta = \frac{F/L^2}{\Delta x/L} = \frac{F}{L \cdot \Delta x}$.For the first cube: $L_1 = 10\,cm = 0.1\,m$,$\Delta x_1 = 0.5\,cm = 0.005\,m$,$F = 10^5\,N$.For the second cube: $L_2 = 20\,cm = 0.2\,m$,$\Delta x_2 = x$,$F = 10^5\,N$.Since the material is the same,$\eta_1 = \eta_2$:$\frac{F}{L_1 \cdot \Delta x_1} = \frac{F}{L_2 \cdot \Delta x_2}$$L_1 \cdot \Delta x_1 = L_2 \cdot \Delta x_2$$10\,cm 	imes 0.5\,cm = 20\,cm 	imes x$$5 = 20x$$x = \frac{5}{20} = 0.25\,cm$.

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