A vertical off-shore structure is built to withstand a maximum stress of `10^(9) Pa.` Is the structure suitable for putting up on top of an oil well in Bombay High? Take the depth of the son to be roughly 3 km, and ignore ocean currents.

Explanation

To determine if the vertical offshore structure can withstand the pressure at the depth of 3 km in the ocean, we need to follow these steps:

### Step 1: Understand the given data
- Maximum stress the structure can withstand: \( \sigma_{max} = 10^9 \, \text{Pa} \)
- Depth of the ocean: \( h = 3 \, \text{km} = 3000 \, \text{m} \)
- Density of seawater: \( \rho = 1000 \, \text{kg/m}^3 \) (approximately)
- Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate the pressure at the depth of 3 km
The pressure \( P \) at a certain depth in a fluid is given by the formula:
\[ P = \rho g h \]
Substituting the values we have:
\[ P = 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 3000 \, \text{m} \]

### Step 3: Perform the multiplication
Calculating the pressure:
\[ P = 1000 \times 9.8 \times 3000 \]
First, calculate \( 9.8 \times 3000 \):
\[ 9.8 \times 3000 = 29400 \, \text{m}^2/\text{s}^2 \]
Now, multiply by 1000:
\[ P = 1000 \times 29400 = 29400000 \, \text{Pa} = 2.94 \times 10^7 \, \text{Pa} \]

### Step 4: Compare the calculated pressure with the maximum stress
Now we compare the pressure at the depth with the maximum stress the structure can withstand:
\[ P = 2.94 \times 10^7 \, \text{Pa} \quad \text{and} \quad \sigma_{max} = 10^9 \, \text{Pa} \]
Since:
\[ 2.94 \times 10^7 \, \text{Pa} < 10^9 \, \text{Pa} \]
The pressure at the depth is much less than the maximum stress the structure can withstand.

### Step 5: Conclusion
Since the pressure at the depth of 3 km is significantly lower than the maximum stress the structure can handle, the structure is suitable for putting up on top of an oil well in Bombay High.

### Final Answer:
Yes, the structure is suitable for putting up on top of an oil well in Bombay High.

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