How much a hollow cylindrical piller of height 6m and external diameter 28cm and internal diameter 22cm will contract under a load of 60,000 kg? Given Young's modulus of the material to be `21 xx 10^(10) N //m^(2)`.

Explanation

To solve the problem of how much a hollow cylindrical pillar will contract under a given load, we can follow these steps:

### Step 1: Identify the given data
- Height of the pillar (L) = 6 m
- External diameter (D) = 28 cm = 0.28 m
- Internal diameter (d) = 22 cm = 0.22 m
- Load (W) = 60,000 kg
- Young's modulus (Y) = 21 x 10^10 N/m²

### Step 2: Convert the load from mass to weight
To find the weight (W) in Newtons, we use the formula:
\[ W = m \times g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).
\[ W = 60000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 588600 \, \text{N} \]

### Step 3: Calculate the cross-sectional area (A) of the hollow cylinder
The cross-sectional area of the hollow cylinder can be calculated using:
\[ A = \frac{\pi}{4} (D^2 - d^2) \]
Substituting the values:
\[ A = \frac{\pi}{4} ((0.28)^2 - (0.22)^2) \]
\[ A = \frac{\pi}{4} (0.0784 - 0.0484) = \frac{\pi}{4} (0.03) \]
\[ A \approx \frac{0.03 \times 3.14}{4} \approx 0.0236 \, \text{m}^2 \]

### Step 4: Calculate the stress (σ)
Stress is defined as:
\[ \sigma = \frac{W}{A} \]
Substituting the values:
\[ \sigma = \frac{588600 \, \text{N}}{0.0236 \, \text{m}^2} \approx 24900000 \, \text{N/m}^2 \]

### Step 5: Calculate the strain (ε)
Using the relationship between stress and strain:
\[ \epsilon = \frac{\sigma}{Y} \]
Substituting the values:
\[ \epsilon = \frac{24900000}{21 \times 10^{10}} \approx 0.00118 \]

### Step 6: Calculate the change in length (ΔL)
The change in length can be calculated using:
\[ \Delta L = \epsilon \times L \]
Substituting the values:
\[ \Delta L = 0.00118 \times 6 \approx 0.00708 \, \text{m} \]

### Step 7: Convert the change in length to millimeters
To convert meters to millimeters:
\[ \Delta L \approx 0.00708 \, \text{m} \times 1000 \approx 7.08 \, \text{mm} \]

### Final Answer
The hollow cylindrical pillar will contract approximately **7.08 mm** under the load of 60,000 kg.

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