Forces of 10⁵N each are applied in opposite direction on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of the cube were 20 cm, the displacement would be

Correct Option: 0.25 cm

To find the displacement when the side of the cube is changed, we can use the concept of shear modulus (modulus of rigidity), which is a property of the material of the cube.

The shear modulus ($\eta$) of a material is defined as the ratio of shear stress to shear strain:
$\eta =\frac{\text{Shear Stress}}{\text{Shear Strain}}$

Let $F$ be the tangential force applied, $A$ be the area of the face, $L$ be the side length of the cube, and $\Delta x$ be the lateral displacement (shifting of the upper face).
The area of the face of the cube is given by:
$A=L^2$

Thus, the formulas for shear stress and shear strain are:
$\text{Shear Stress}=\frac{F}{A}=\frac{F}{L^2}$
$\text{Shear Strain}=\frac{\Delta x}{L}$

Substituting these into the expression for the shear modulus:
$\eta =\frac{F/L^2}{\Delta x/L}=\frac{F}{L·\Delta x}$

Since the material of the cube remains the same, the shear modulus $\eta$ is constant. The applied force $F$ is also the same in both cases.
Therefore, the product of the side length ($L$) and the displacement ($\Delta x$) must be constant:
$L_1·\Delta x_1=L_2·\Delta x_2$

Given:
Initial side length of the cube, $L_1=10\text{cm}$
Initial displacement, $\Delta x_1=0.5\text{cm}$
New side length of the cube, $L_2=20\text{cm}$

Substitute these values into the equation:
$10·0.5=20·\Delta x_2$
$5=20·\Delta x_2$
$\Delta x_2=\frac{5}{20}=0.25\text{cm}$

Thus, the displacement of the upper face if the side of the cube were 20 cm would be 0.25 cm.