Measured along the surface of the water , a rectangular swimming pool has a length of 15 m . Along this length , the flat bottom of the pool slopes downward at an angle of `11^(@)` below the horizontal , from one end to the other . By how much does the pressure at the bottom of the deep end exceed the pressure at the bottom of the shallow end ?

Options

A. `2.9 xx 10^(4) Pa`
B. `6.1 xx 10^(4) Pa`
C. `4.3 xx 10^(4) Pa`
D. `9.7 xx 10^(4) Pa`

Answer

To find the difference in pressure between the deep end and the shallow end of the swimming pool, we can follow these steps:

### Step 1: Understand the Problem
We have a rectangular swimming pool that is 15 m long, and the bottom slopes downward at an angle of 11 degrees below the horizontal. We need to calculate the difference in pressure between the deep end and the shallow end of the pool.

### Step 2: Identify the Relevant Formula
The pressure difference due to a height difference in a fluid is given by the formula:
\[ \Delta P = \rho g h \]
where:
- \(\Delta P\) is the pressure difference,
- \(\rho\) is the density of the fluid (for water, \(\rho \approx 1000 \, \text{kg/m}^3\)),
- \(g\) is the acceleration due to gravity (\(g \approx 9.8 \, \text{m/s}^2\)),
- \(h\) is the height difference between the two points.

### Step 3: Calculate the Height Difference
To find the height difference \(h\) between the shallow end and the deep end, we can use trigonometry. The height difference can be calculated using the sine function:
\[ h = L \sin(\theta) \]
where:
- \(L\) is the length of the pool (15 m),
- \(\theta\) is the angle of the slope (11 degrees).

Calculating \(h\):
\[ h = 15 \sin(11^\circ) \]

Using a calculator, we find:
\[ \sin(11^\circ) \approx 0.1908 \]
Thus,
\[ h \approx 15 \times 0.1908 \approx 2.862 \, \text{m} \]

### Step 4: Calculate the Pressure Difference
Now we can substitute \(h\) into the pressure difference formula:
\[ \Delta P = \rho g h \]
Substituting the values:
\[ \Delta P = 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 2.862 \, \text{m} \]
Calculating this gives:
\[ \Delta P \approx 1000 \times 9.8 \times 2.862 \approx 28055.76 \, \text{Pa} \]
or approximately:
\[ \Delta P \approx 2.81 \times 10^4 \, \text{Pa} \]

### Step 5: Conclusion
The pressure at the bottom of the deep end exceeds the pressure at the bottom of the shallow end by approximately \(2.81 \times 10^4 \, \text{Pa}\).

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