A cube of side 40 cm has its upper face displaced by 0.1 mm by a tangential force of 8 Kn. The shearing modulus of cube is :-

Options

A. `2xx10^(9)N-m^(-2)`
B. `4xx10^(9)N-m^(-2)`
C. `2xx10^(8)Nm^(-2)`
D. `4xx10^(8)N-m^(-2)`

Answer

To find the shearing modulus of the cube, we will follow these steps:

### Step 1: Convert all measurements to consistent units
Given:
- Side of the cube = 40 cm = 400 mm
- Displacement of the upper face (ΔL) = 0.1 mm

### Step 2: Calculate shear strain
Shear strain (γ) is defined as the ratio of the displacement (ΔL) to the original length (L):
\[ \text{Shear Strain} (\gamma) = \frac{\Delta L}{L} = \frac{0.1 \text{ mm}}{400 \text{ mm}} = \frac{0.1}{400} = 0.00025 \]

### Step 3: Calculate shear stress
Shear stress (τ) is defined as the force (F) divided by the area (A). The force given is 8 kN, which we convert to Newtons:
\[ F = 8 \text{ kN} = 8000 \text{ N} \]
The base area (A) of the cube is:
\[ A = \text{side}^2 = (400 \text{ mm})^2 = 160000 \text{ mm}^2 = 160000 \times 10^{-6} \text{ m}^2 = 0.16 \text{ m}^2 \]
Now, we can calculate shear stress:
\[ \text{Shear Stress} (\tau) = \frac{F}{A} = \frac{8000 \text{ N}}{0.16 \text{ m}^2} = 50000 \text{ N/m}^2 = 5 \times 10^5 \text{ N/m}^2 \]

### Step 4: Calculate shearing modulus
Shearing modulus (G) is defined as the ratio of shear stress to shear strain:
\[ G = \frac{\tau}{\gamma} = \frac{5 \times 10^5 \text{ N/m}^2}{0.00025} = 2 \times 10^9 \text{ N/m}^2 \]

### Final Answer
The shearing modulus of the cube is \( 2 \times 10^9 \text{ N/m}^2 \).

---