A uniform heavy rod of weight 10 kg `ms^(-2,)` cross-sectional area 100 `cm^(2)` and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is `2 xx 10^(11) Nm^(-2)` Neglecting the lateral contraction, find the elongation of rod due to its own weight:

Options

A. `2xx10^(-9)` m
B. `5xx10^(-10)` m
C. `4xx10^(-8)` m
D. `8xx10^(-8)` m

Answer

To find the elongation of the rod due to its own weight, we will use the concepts of stress, strain, and Young's modulus. Here’s a step-by-step solution:

### Step 1: Identify Given Data
- Weight of the rod (W) = 10 kg × 9.8 m/s² = 98 N (using g = 9.8 m/s² for weight calculation)
- Cross-sectional area (A) = 100 cm² = 100 × 10⁻⁴ m² = 10⁻² m²
- Length of the rod (L) = 20 cm = 0.2 m
- Young's modulus (Y) = 2 × 10¹¹ N/m²

### Step 2: Calculate the Stress at a Distance x from the Top
The stress at a distance x from the top of the rod is given by the weight of the portion of the rod below that point. The weight of the rod below a distance x is:
\[ W_x = \frac{W}{L} \cdot x = \frac{98}{0.2} \cdot x = 490x \text{ N} \]

The stress (σ) at that point is:
\[ \sigma = \frac{W_x}{A} = \frac{490x}{10^{-2}} = 49000x \text{ N/m²} \]

### Step 3: Relate Stress to Strain
The strain (ε) is related to stress by Young's modulus:
\[ Y = \frac{\sigma}{\epsilon} \]
Thus,
\[ \epsilon = \frac{\sigma}{Y} = \frac{49000x}{2 \times 10^{11}} \]

### Step 4: Express Strain in Terms of Change in Length
Strain is also defined as:
\[ \epsilon = \frac{dL}{dx} \]
Therefore, we can write:
\[ \frac{dL}{dx} = \frac{49000x}{2 \times 10^{11}} \]

### Step 5: Integrate to Find Total Change in Length
To find the total change in length (ΔL), we integrate from 0 to L:
\[ \Delta L = \int_0^L \frac{49000x}{2 \times 10^{11}} \, dx \]
\[ \Delta L = \frac{49000}{2 \times 10^{11}} \int_0^L x \, dx \]
\[ \Delta L = \frac{49000}{2 \times 10^{11}} \cdot \left[ \frac{x^2}{2} \right]_0^L \]
\[ \Delta L = \frac{49000}{2 \times 10^{11}} \cdot \frac{L^2}{2} \]

### Step 6: Substitute Values and Calculate ΔL
Substituting L = 0.2 m:
\[ \Delta L = \frac{49000}{2 \times 10^{11}} \cdot \frac{(0.2)^2}{2} \]
\[ \Delta L = \frac{49000}{2 \times 10^{11}} \cdot \frac{0.04}{2} \]
\[ \Delta L = \frac{49000 \cdot 0.02}{2 \times 10^{11}} \]
\[ \Delta L = \frac{980}{2 \times 10^{11}} \]
\[ \Delta L = \frac{490}{10^{11}} \]
\[ \Delta L = 4.9 \times 10^{-9} \text{ m} \]

### Final Answer
The elongation of the rod due to its own weight is:
\[ \Delta L = 4.9 \times 10^{-9} \text{ m} \]

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