A `50 kg` motor rests on four cylindrical rubber blocks. Each block has a height of `4 cm` and a cross-sectional area of `16 cm^(2)`. The shear modulus of rubber is `2 xx 10^(6) N//m^(2)`. A sideways force of `500 N` is applied to the motor. The distance that the motor moves sideways is

Options

A. `0.156 cm`
B. `1.156 cm`
C. `0.312 cm`
D. `0.204 cm`

Explanation

To solve the problem, we need to calculate the sideways displacement of the motor when a force is applied to it. We will use the relationship between shear stress, shear strain, and the shear modulus.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the motor (m) = 50 kg
- Height of each rubber block (h) = 4 cm = 0.04 m (convert to meters)
- Cross-sectional area of each block (A) = 16 cm² = 16 × 10⁻⁴ m² (convert to square meters)
- Shear modulus of rubber (η) = 2 × 10⁶ N/m²
- Applied sideways force (F) = 500 N

2. **Calculate the Total Cross-Sectional Area:**
Since there are 4 blocks, the total cross-sectional area (A_total) is:
\[ A_{total} = 4 \times A = 4 \times 16 \times 10^{-4} \, \text{m}^2 = 64 \times 10^{-4} \, \text{m}^2 = 6.4 \times 10^{-3} \, \text{m}^2 \]

3. **Calculate the Shear Stress (τ):**
Shear stress is given by the formula:
\[ \tau = \frac{F}{A_{total}} \]
Substituting the values:
\[ \tau = \frac{500 \, \text{N}}{6.4 \times 10^{-3} \, \text{m}^2} = 78125 \, \text{N/m}^2 \]

4. **Relate Shear Stress to Shear Strain:**
The relationship between shear stress and shear strain is given by:
\[ \tau = \eta \cdot \frac{X}{h} \]
Where:
- \(X\) is the displacement (shear strain),
- \(h\) is the height of the rubber block.

5. **Rearranging for X:**
Rearranging the equation to solve for \(X\):
\[ X = \frac{\tau \cdot h}{\eta} \]
Substituting the values:
\[ X = \frac{78125 \, \text{N/m}^2 \cdot 0.04 \, \text{m}}{2 \times 10^6 \, \text{N/m}^2} \]

6. **Calculating X:**
\[ X = \frac{3125}{2000000} = 0.0015625 \, \text{m} = 0.15625 \, \text{cm} \]

7. **Final Result:**
The distance that the motor moves sideways is approximately:
\[ X \approx 0.156 \, \text{cm} \]

### Summary:
The sideways displacement of the motor when a force of 500 N is applied is approximately **0.156 cm**.