Question

A swimming pool has depth of 3 meters. The pressure at the bottom of the pool due to water alone is (Density of water is $1000$ kg/m$^3$, $g = 10$ m/s$^2$)

• A. $10^4$ Pa
• B. $3 \times 10^3$ Pa
• C. $29 \times 10^3$ Pa
• D. $30 \times 10^3$ Pa

Hint:

Pressure in Fluids:

• $P = \rho g h$ gives gauge pressure due to fluid column.
• SI units: $\rho$ in kg/m$^3$, $g$ in m/s$^2$, $h$ in m, $P$ in Pascal.
• 1 Pa = 1 N/m$^2$ = 1 kg·m$^-1$·s$^-2$.

Answer 1

The Correct Option is D

Hydrostatic pressure at depth $h$ is $P = \rho g h$.
Substitute: $\rho = 1000$, $g = 10$, $h = 3$ m. So, $P = 1000 \cdot 10 \cdot 3 = 30000$ Pa.
That is $30 \times 10^3$ Pa.
This is pressure due to water alone (gauge pressure). Atmospheric pressure is not included in this.

Answer 2

Step 1: Understand the problem
We need to find the pressure at the bottom of a swimming pool due to the water column above.

Step 2: Given data
- Depth of water, h = 3 m
- Density of water, ρ = 1000 kg/m³
- Acceleration due to gravity, g = 10 m/s²

Step 3: Formula for pressure due to a liquid column
Pressure, P = ρgh
where P is the pressure exerted by the water column at depth h.

Step 4: Calculate the pressure
P = 1000 × 10 × 3 = 30,000 Pa = 30 × 10³ Pa

Step 5: Conclusion
The pressure at the bottom of the pool due to water alone is 30 × 10³ Pascal.