Two parallel and opposite forces each 5000 N are applied tangentially to the upper and lower faces of a cubical metal block of side 25 cm. the angle of shear is (The shear modulus of the metal is 80 Gpa)

Options

A. `10^(-4) rad`
B. `10^(-5) rad`
C. `10^(-6) rad`
D. `10^(-7) rad`

Explanation

To find the angle of shear for the given cubical metal block, we can follow these steps:

### Step 1: Identify the Given Values
- Force applied (F) = 5000 N
- Side of the cube (L) = 25 cm = 0.25 m (convert to meters)
- Shear modulus (G) = 80 GPa = 80 × 10^9 Pa (convert to Pascals)

### Step 2: Calculate the Area of the Faces
The area (A) of one face of the cube can be calculated using the formula:
\[ A = L^2 \]
Substituting the value of L:
\[ A = (0.25 \, \text{m})^2 = 0.0625 \, \text{m}^2 \]

### Step 3: Calculate the Shear Stress
Shear stress (τ) is defined as the force (F) applied per unit area (A):
\[ \tau = \frac{F}{A} \]
Substituting the values:
\[ \tau = \frac{5000 \, \text{N}}{0.0625 \, \text{m}^2} = 80000 \, \text{Pa} \]

### Step 4: Relate Shear Stress to Shear Modulus and Angle of Shear
The relationship between shear stress (τ), shear modulus (G), and angle of shear (θ) is given by:
\[ \tau = G \cdot \theta \]
Rearranging this to find the angle of shear:
\[ \theta = \frac{\tau}{G} \]

### Step 5: Substitute the Values to Find Angle of Shear
Substituting the values of τ and G:
\[ \theta = \frac{80000 \, \text{Pa}}{80 \times 10^9 \, \text{Pa}} \]
Calculating this gives:
\[ \theta = 1 \times 10^{-6} \, \text{radians} \]

### Final Answer
The angle of shear is:
\[ \theta = 1 \times 10^{-6} \, \text{radians} \]