A cube of aluminium of side 6 cm is subjected to a tangential force such that the top face is shears through 0.012cm relative to the bottom face. The tangential force is `k xx 10^(10)` dyne. What is the value of k? [Shear modulus of the material is `eta= 2 xx 10^(11)` dyne `cm^(-2) `]

Answer

To solve the problem, we will use the relationship between shear modulus, force, area, and shear strain. Let's break it down step by step.

### Step 1: Understand the given data
- Side of the cube, \( A = 6 \, \text{cm} \)
- Shear displacement, \( \Delta L = 0.012 \, \text{cm} \)
- Shear modulus, \( \eta = 2 \times 10^{11} \, \text{dyne/cm}^2 \)
- Tangential force, \( F = k \times 10^{10} \, \text{dyne} \)

### Step 2: Calculate the area of the face of the cube
The area \( A \) of one face of the cube can be calculated as:
\[ A = \text{side}^2 = (6 \, \text{cm})^2 = 36 \, \text{cm}^2 \]

### Step 3: Calculate the shear strain
Shear strain \( \phi \) is defined as the ratio of the shear displacement to the original length:
\[ \phi = \frac{\Delta L}{L} = \frac{0.012 \, \text{cm}}{6 \, \text{cm}} = 0.002 \]

### Step 4: Use the formula for shear modulus
The shear modulus \( \eta \) is given by the formula:
\[ \eta = \frac{F/A}{\phi} \]
Rearranging this gives:
\[ F = \eta \cdot A \cdot \phi \]

### Step 5: Substitute the known values
Substituting the values we have:
\[ F = (2 \times 10^{11} \, \text{dyne/cm}^2) \cdot (36 \, \text{cm}^2) \cdot (0.002) \]

### Step 6: Calculate the force \( F \)
Calculating \( F \):
\[ F = 2 \times 10^{11} \cdot 36 \cdot 0.002 = 2 \times 10^{11} \cdot 0.072 = 1.44 \times 10^{10} \, \text{dyne} \]

### Step 7: Relate the force to \( k \)
Since \( F = k \times 10^{10} \, \text{dyne} \), we can set up the equation:
\[ 1.44 \times 10^{10} = k \times 10^{10} \]

### Step 8: Solve for \( k \)
Dividing both sides by \( 10^{10} \):
\[ k = 1.44 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{1.44} \]

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