A rubber eraser 3 `cmxx1cmxx8cm` is clamped at one end with 8 cm edge as vertical. A horizontal force of 2.1N is applied on the free surface. Shear modulus of rubber `=1.4xx10^(5)Nm^(-2)`. The horizontal displacement (in mm) of the top face is x. Find x.

Explanation

To find the horizontal displacement \( x \) of the top face of the rubber eraser when a horizontal force is applied, we can follow these steps:

### Step 1: Calculate the Area of the Rubber Eraser
The dimensions of the rubber eraser are given as \( 3 \, \text{cm} \times 1 \, \text{cm} \times 8 \, \text{cm} \). The area \( A \) on which the force is applied is:
\[ A = \text{length} \times \text{width} = 3 \, \text{cm} \times 1 \, \text{cm} = 3 \, \text{cm}^2 \]

### Step 2: Calculate the Shear Stress
Shear stress \( \sigma \) is defined as the force \( F \) applied divided by the area \( A \):
\[ \sigma = \frac{F}{A} \]
Given \( F = 2.1 \, \text{N} \) and \( A = 3 \, \text{cm}^2 \):
\[ \sigma = \frac{2.1 \, \text{N}}{3 \, \text{cm}^2} = \frac{2.1 \, \text{N}}{3 \times 10^{-4} \, \text{m}^2} = 7000 \, \text{N/m}^2 \]

### Step 3: Relate Shear Strain to Shear Stress
Shear strain \( \epsilon \) is defined as the displacement \( x \) divided by the height \( h \):
\[ \epsilon = \frac{x}{h} \]
Here, \( h = 8 \, \text{cm} = 0.08 \, \text{m} \).

### Step 4: Use Shear Modulus to Relate Shear Stress and Shear Strain
The shear modulus \( G \) is defined as:
\[ G = \frac{\sigma}{\epsilon} \]
Rearranging gives:
\[ \epsilon = \frac{\sigma}{G} \]

### Step 5: Substitute Values to Find Displacement \( x \)
Substituting the values we have:
\[ \epsilon = \frac{7000 \, \text{N/m}^2}{1.4 \times 10^5 \, \text{N/m}^2} \]
Calculating \( \epsilon \):
\[ \epsilon = \frac{7000}{1.4 \times 10^5} = 0.050 \, \text{(dimensionless)} \]

Now, substituting back to find \( x \):
\[ x = \epsilon \times h = 0.050 \times 0.08 \, \text{m} = 0.004 \, \text{m} = 4 \, \text{mm} \]

### Final Answer
The horizontal displacement \( x \) of the top face is:
\[ \boxed{4 \, \text{mm}} \]