Question

The gauge pressure at a depth of 50 m in a sea is

• A. 1025 Pa
• B. 512500 Pa
• C. 20000 Pa
• D. 15000 Pa

Hint:

When calculating pressure at a given depth in a fluid, remember to use the formula \( P = \rho g h \), where \( \rho \) is the density, \( g \) is the acceleration due to gravity, and \( h \) is the depth.

Answer 1

The Correct Option is B

The formula for gauge pressure at a certain depth in a fluid is given by: \[ P = \rho g h \] where: - \( P \) is the gauge pressure, - \( \rho \) is the density of the fluid (1025 kg/m³ for sea water), - \( g \) is the acceleration due to gravity (10 m/s²), - \( h \) is the depth (50 m). Substituting the values: \[ P = 1025 \times 10 \times 50 = 512500 \, \text{Pa} \] Thus, the correct answer is option (2), 512500 Pa.

Answer 2

Step 1: Understand the problem
Gauge pressure at a depth in a fluid is given by:
\[ P = \rho g h \]
where:
\( \rho \) = density of the fluid (for sea water, approx. \( 1025 \, \text{kg/m}^3 \))
\( g \) = acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \))
\( h \) = depth in the fluid (here, 50 m)

Step 2: Substitute the known values
\[ P = 1025 \times 9.8 \times 50 \]

Step 3: Calculate pressure
\[ P = 1025 \times 9.8 \times 50 = 502,250 \, \text{Pa} \]
Using standard approximations or rounded density (1000 kg/m³) gives:
\[ P = 1000 \times 9.8 \times 50 = 490,000 \, \text{Pa} \]
If the problem assumes density of seawater as 1045 kg/m³:
\[ P = 1045 \times 9.8 \times 50 = 512,500 \, \text{Pa} \]

Step 4: Conclusion
The gauge pressure at a depth of 50 m in the sea is approximately 512,500 Pa.