A cast iron column has internal diameter of 200 mm. What sholud be the minimum external diameter so that it may carry a load of `1.6 MN` without the stress exceeding ` 90 N//mm^(2)`?

Explanation

To solve the problem, we need to determine the minimum external diameter of a cast iron column that can carry a load of 1.6 MN without exceeding a stress of 90 N/mm².

### Step-by-Step Solution:

1. **Identify Given Values:**
- Internal diameter (d_i) = 200 mm
- Load (F) = 1.6 MN = 1.6 × 10^6 N
- Maximum allowable stress (σ) = 90 N/mm²

2. **Convert Internal Diameter to Radius:**
- Internal radius (r_i) = d_i / 2 = 200 mm / 2 = 100 mm = 0.1 m

3. **Use the Formula for Stress:**
- Stress (σ) is defined as the force (F) divided by the cross-sectional area (A).
- For a hollow circular column, the area A can be expressed as:
\[ A = \pi (r_e^2 - r_i^2) \]
where \( r_e \) is the external radius.

4. **Set Up the Stress Equation:**
- Rearranging the stress formula gives:
\[ \sigma = \frac{F}{A} \implies A = \frac{F}{\sigma} \]
- Substituting the area expression:
\[ \pi (r_e^2 - r_i^2) = \frac{F}{\sigma} \]

5. **Substitute Known Values:**
- Plug in the values:
\[ \pi (r_e^2 - (0.1)^2) = \frac{1.6 \times 10^6}{90} \]
- Calculate the right side:
\[ \frac{1.6 \times 10^6}{90} \approx 17777.78 \text{ mm}^2 \]

6. **Solve for External Radius (r_e):**
- Rearranging gives:
\[ r_e^2 - (0.1)^2 = \frac{17777.78}{\pi} \]
- Calculate \( \frac{17777.78}{\pi} \):
\[ r_e^2 - 0.01 = 5647.24 \]
- Therefore:
\[ r_e^2 = 5647.24 + 0.01 \approx 5647.25 \]
- Taking the square root:
\[ r_e \approx \sqrt{5647.25} \approx 75.2 \text{ mm} \]

7. **Calculate External Diameter:**
- The external diameter (d_e) is given by:
\[ d_e = 2r_e \approx 2 \times 75.2 \approx 150.4 \text{ mm} \]

### Final Answer:
The minimum external diameter should be approximately **150.4 mm**.