A tube with a radius of 4.2 cm is holding oil that has a density of 0.92 g/cm³. The gauge pressure in the oil at a depth of 64 cm from the top of the surface is

Question

Quizplus · Accepted Answer

The gauge pressure in a liquid at a depth h can be found using the formula:

P = ρgh

where P is the gauge pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth.

Substituting the given values:

ρ = 0.92 g/cm³ 
h = 64 cm 
g = 9.8 m/s²

First, we need to convert the radius from cm to m:

r = 4.2 cm = 0.042 m

The volume of the oil in the tube is:

V = πr²h

Substituting the given values:

V = π(0.042m)²(64cm)

V = 0.054 m³

The mass of the oil in the tube is:

m = ρV

Substituting the given values:

m = (0.92 g/cm³)(0.054 m³)

m = 0.050 kg

The weight of the oil in the tube is:

W = mg

W = (0.050 kg)(9.8 m/s²)

W = 0.49 N

The gauge pressure in the oil at the bottom of the tube is:

P = gauge pressure at the bottom

P = atmospheric pressure + ρgh

Since the pressure at the top of the tube is atmospheric pressure, we can assume that atmospheric pressure is equal to 0 gauge pressure.

P = 0 + (0.92 g/cm³)(9.8 m/s²)(64 cm)

P = 5.8 x 10⁴ Pa

Therefore, the best choice is B, which is 5.8 x 10⁴ Pa in scientific notation.

A Tube with a Radius of 4