A `5 kg` rod of square cross section `5 cm` on a side and `1m` long is pulled along a smooth horizontal surface by a force applied at one end. The rod has a constant acceleration of `2 ms^(-12)`. Determine the elongation in the rod. (Young's modulus of the material of the rod is `5 xx 10^(3) N//m^(9)`).

Options

A. Zero, as for elongation to be there, equal and opposite forces must act on the rod
B. Non-zero but cannot be determine from the give, situation
C. `0.4 mum`
D. `16 mum`

Explanation

To determine the elongation in the rod, we will follow these steps:

### Step 1: Identify the given data
- Mass of the rod, \( m = 5 \, \text{kg} \)
- Cross-sectional area, \( A = (5 \, \text{cm})^2 = (0.05 \, \text{m})^2 = 0.0025 \, \text{m}^2 \)
- Length of the rod, \( L = 1 \, \text{m} \)
- Acceleration, \( a = 2 \, \text{m/s}^2 \)
- Young's modulus, \( Y = 5 \times 10^3 \, \text{N/m}^2 \)

### Step 2: Calculate the force acting on the rod
Using Newton's second law, the force \( F \) acting on the rod can be calculated as:
\[ F = m \cdot a = 5 \, \text{kg} \cdot 2 \, \text{m/s}^2 = 10 \, \text{N} \]

### Step 3: Use the formula for elongation
The elongation \( \Delta L \) in the rod can be calculated using the formula:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
Substituting the values we have:
\[ \Delta L = \frac{10 \, \text{N} \cdot 1 \, \text{m}}{0.0025 \, \text{m}^2 \cdot 5 \times 10^3 \, \text{N/m}^2} \]

### Step 4: Calculate the denominator
Calculating the denominator:
\[ 0.0025 \, \text{m}^2 \cdot 5 \times 10^3 \, \text{N/m}^2 = 12.5 \, \text{N} \]

### Step 5: Calculate the elongation
Now substituting back into the elongation formula:
\[ \Delta L = \frac{10 \, \text{N}}{12.5 \, \text{N}} = 0.8 \, \text{m} \]

### Step 6: Convert to micrometers
To express the elongation in micrometers:
\[ \Delta L = 0.8 \, \text{m} = 0.8 \times 10^6 \, \mu m = 800000 \, \mu m \]

### Final Result
The elongation in the rod is \( \Delta L = 800000 \, \mu m \).

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