Question

As shown in the figure, forces of $10^5N$ each are applied in opposite directions, on the upper and lower faces of a cube of side 10cm, shifting the upper face parallel to itself by 0.5 cm. If the side of another cube of the same material is 20 cm, then under similar conditions as above, the displacement will be

• A. 0.10 cm
• B. 0.25 cm
• C. 0.50 cm
• D. 1.00 cm

Answer 1

Answer: (B)

0.25 cm

Answer 2

The deformation of the cube under the applied forces is due to shear stress. The relationship between shear stress ($\tau$), shear strain ($\gamma$), and shear modulus ($G$) is given by Hooke's law for shear: $\tau = G \gamma$ Shear stress is defined as the force ($F$) applied per unit area ($A$): $\tau = \frac{F}{A}$ Shear strain is defined as the ratio of the displacement ($\Delta x$) of the upper face to the height ($h$) of the object: $\gamma = \frac{\Delta x}{h}$ Substituting these into Hooke's law, we get: $\frac{F}{A} = G \frac{\Delta x}{h}$ Rearranging the formula to solve for displacement ($\Delta x$): $\Delta x = \frac{Fh}{GA}$ For a cube of side length $s$, the height $h$ is equal to $s$, and the area of the face $A$ is $s^2$. Thus, the formula becomes: $\Delta x = \frac{F \cdot s}{G \cdot s^2} = \frac{F}{G s}$ This shows that for a constant force $F$ and a constant shear modulus $G$ (since the material is the same), the displacement $\Delta x$ is inversely proportional to the side length $s$ of the cube: $\Delta x \propto \frac{1}{s}$ Let's denote the properties of the first cube with subscript 1 and the second cube with subscript 2. Given for the first cube: Side length $s_1 = 10$ cm Displacement $\Delta x_1 = 0.5$ cm Applied force $F_1 = 10^5$ N

Given for the second cube: Side length $s_2 = 20$ cm The material is the same, so $G_1 = G_2 = G$. The conditions are similar, which implies the applied force is the same: $F_2 = 10^5$ N.

Using the proportionality $\Delta x \propto \frac{1}{s}$, we can write the ratio of displacements: $\frac{\Delta x_2}{\Delta x_1} = \frac{F_2 / (G s_2)}{F_1 / (G s_1)}$ Since $F_1 = F_2$ and $G$ is the same for both cubes: $\frac{\Delta x_2}{\Delta x_1} = \frac{1/s_2}{1/s_1} = \frac{s_1}{s_2}$ Now, we can calculate $\Delta x_2$: $\Delta x_2 = \Delta x_1 \cdot \frac{s_1}{s_2}$ Substituting the given values: $\Delta x_2 = (0.5 \text{ cm}) \cdot \frac{10 \text{ cm}}{20 \text{ cm}}$ $\Delta x_2 = (0.5 \text{ cm}) \cdot \frac{1}{2}$ $\Delta x_2 = 0.25 \text{ cm}$