Question

Dealt with the flow of saline solution in an IV system. Verify that a pressure of (1.62 X 10⁴ N/m²) is created at a depth of 1.61 m in a saline solution, assuming its density to be that of seawater.

A) (1.00 X 10⁴ N/m²)
B) (1.62 X 10⁴ N/m²)
C) (2.00 X 10⁴ N/m²)
D) (2.50 X 10⁴ N/m²)

Answer 1

Using the formula P = ρgh with the given depth of 1.61 m and the density of seawater, the calculated pressure is indeed 1.62 × 10⁴ N/m², which confirms option B) as the correct answer.

Step-by-step explanation:

To verify that a pressure of 1.62 × 10⁴ N/m² is created at a depth of 1.61 m in saline solution with the same density as sea water, we can use the formula for the pressure due to the weight of a fluid column:

P = ρgh

Where:

• P is the pressure,
• ρ (rho) is the density of the fluid,
• g is the acceleration due to gravity (9.81 m/s²), and
• h is the depth of the fluid.

The density of seawater is approximately 1025 kg/m³. Plugging in the values, we get:

P = (1025 kg/m³)(9.81 m/s²)(1.61 m)

P = 1.62 × 10⁴ N/m²

Therefore, option B) is correct, verifying that the pressure at the given depth matches the value in the question.