Question

A cube has side length \(5 \text{ cm}\) and modulus of rigidity \(10^5 \text{ N/m}^2\). The displacement produced by a force of \(10 \text{ N}\) in the upper face of the cube is ____ \text{mm}.

Answer 1

Correct Answer: 2

Step 1: Understanding the Question:
The topic of this question is Elasticity, specifically concerning the Shear Modulus or Modulus of Rigidity.
We are asked to calculate the lateral displacement of the top face of a cube when a tangential force (shear force) is applied to it.
Step 2: Key Formula or Approach:
The Modulus of Rigidity (\(\eta\)) is defined as the ratio of shear stress to shear strain:
\[ \eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F/A}{\Delta x / L} \]
To find the displacement (\(\Delta x\)), we rearrange this formula:
\[ \Delta x = \frac{F \cdot L}{A \cdot \eta} \]
Step 3: Detailed Explanation:
First, we list the given values and convert them to SI units:
Force, \(F = 10 \text{ N}\).
Modulus of Rigidity, \(\eta = 10^5 \text{ N/m}^2\).
Side length, \(L = 5 \text{ cm} = 0.05 \text{ m}\).
Area of the face, \(A = L^2 = (0.05 \text{ m})^2 = 0.0025 \text{ m}^2\).
Now, we substitute these values into the formula for displacement:
\[ \Delta x = \frac{(10 \text{ N}) \times (0.05 \text{ m})}{(0.0025 \text{ m}^2) \times (10^5 \text{ N/m}^2)} \]
\[ \Delta x = \frac{0.5}{2500 \times 10^{-4} \times 10^5} = \frac{0.5}{250} \text{ m} \]
\[ \Delta x = 0.002 \text{ m} \]
Finally, we convert the displacement from meters to millimeters:
\[ \Delta x = 0.002 \times 1000 \text{ mm} = 2 \text{ mm} \]
Step 4: Final Answer:
The displacement produced is \(2 \text{ mm}\).