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Math  /  Algebra

QuestionThe height a fluid will rise in a capillary tube is given as: H=2σρgrH=\frac{2 \sigma}{\rho g r}
The fluid being tested has a specific gravity of 0.87[−]0.87[-]. Use the graph shown to determine the surface tension ( σ\sigma ) of the fluid used in this experiment, in units of kilograms per second squared [kgs2]\left[\frac{\mathrm{kg}}{\mathrm{s}^{2}}\right].

Studdy Solution

STEP 1

What is this asking? We need to find the surface tension of a fluid, given its specific gravity and a graph relating the height it rises in a tube to the tube's radius. Watch out! Don't forget to convert specific gravity to density!
Also, make sure the units work out correctly.

STEP 2

1. Find the density
2. Relate the equations
3. Calculate surface tension

STEP 3

We're given the **specific gravity** of the fluid as 0.870.87.
Specific gravity is the ratio of a substance's density to the density of water.
The density of water is typically 1000 kg/m31000 \, \text{kg/m}^3.
So, to find the fluid's density, we **multiply** the specific gravity by the density of water.
Why? Because specific gravity is defined as ρfluid/ρwater\rho_{fluid}/\rho_{water}, and we want to find ρfluid\rho_{fluid}!

STEP 4

Let's **calculate** the density: ρ=0.87⋅1000 kg/m3=870 kg/m3 \rho = 0.87 \cdot 1000 \, \text{kg/m}^3 = \mathbf{870 \, \text{kg/m}^3} So, the **density** of our fluid is 870 kg/m3\mathbf{870 \, \text{kg/m}^3}.
Awesome!

STEP 5

We're given the equation for the height of the fluid in the capillary tube: H=2σρgr H = \frac{2 \sigma}{\rho g r} And the graph tells us: H=1.67×10−5 r−1 H = 1.67 \times 10^{-5} \, r^{-1} These equations both describe the same thing: how high the fluid goes!
So, they must be equal to each other.
This is a key insight!

STEP 6

Let's set them **equal** to each other: 2σρgr=1.67×10−5 r−1 \frac{2 \sigma}{\rho g r} = 1.67 \times 10^{-5} \, r^{-1} Now we have an equation that relates surface tension (σ\sigma) to the other quantities.
Perfect!

STEP 7

We want to **solve** for σ\sigma, the surface tension.
Let's **multiply** both sides of our equation by ρgr2\frac{\rho g r}{2} to isolate σ\sigma.
Remember, we're dividing to one and multiplying to one to move things around! σ=1.67×10−5 r−1⋅ρgr2 \sigma = \frac{1.67 \times 10^{-5} \, r^{-1} \cdot \rho g r}{2} Notice that r−1r^{-1} and rr divide to one, simplifying things nicely.

STEP 8

Now, let's **plug in** the values we know.
We found ρ=870 kg/m3\rho = 870 \, \text{kg/m}^3, gg (acceleration due to gravity) is approximately 9.81 m/s29.81 \, \text{m/s}^2, and we have that constant from the graph. σ=1.67×10−5⋅870 kg/m3⋅9.81 m/s22 \sigma = \frac{1.67 \times 10^{-5} \cdot 870 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2}{2}

STEP 9

Time to **calculate**! σ=0.1426 kg/s22 \sigma = \frac{0.1426 \, \text{kg/s}^2}{2} σ=0.0713 kg/s2 \sigma = \mathbf{0.0713 \, \text{kg/s}^2}

STEP 10

The surface tension of the fluid is 0.0713 kg/s2\mathbf{0.0713 \, \text{kg/s}^2}.

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