A cube of aluminium of sides 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be

Options

A. 0.02
B. 0.1
C. 0.005
D. 0.002

Explanation

To find the shearing strain in the given problem, we will follow these steps:

### Step 1: Understand the definition of shearing strain
Shearing strain (γ) is defined as the change in displacement (or deformation) per unit length of the material. It can be mathematically expressed as:
\[ \text{Shearing Strain} (\gamma) = \frac{\text{Displacement}}{\text{Original Length}} \]

### Step 2: Identify the given values
From the problem, we have:
- Displacement (Δx) = 0.02 cm = 0.02 × 10^{-2} m = 0.0002 m
- Original length (L) = 0.1 m

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the formula for shearing strain:
\[ \gamma = \frac{0.0002 \, \text{m}}{0.1 \, \text{m}} \]

### Step 4: Calculate the shearing strain
Now, we perform the calculation:
\[ \gamma = \frac{0.0002}{0.1} = 0.002 \]

### Step 5: Final result
Thus, the shearing strain is:
\[ \gamma = 0.002 \]

### Summary of the solution
The shearing strain for the cube of aluminum subjected to a shearing force is 0.002.

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