A certain fluid has a density of 1045 kg/m3 and is observed to rise to a height of 2.4 cm in a 1.0-mm-diameter tube. The contact angle between the wall and the fluid is zero. Calculate the surface tension of the fluid.
Answer in N/m
2 answers
I guess we'll never know
Here's how to solve the problem:
First, we need to calculate the weight of the fluid that rises up in the tube. We can use the equation:
weight = density x volume x gravity
where density = 1045 kg/m3 (given), volume = πr2h (where r = 0.5 mm = 0.0005 m and h = 2.4 cm = 0.024 m), and gravity = 9.81 m/s2.
weight = 1045 x π x (0.0005)2 x 0.024 x 9.81 = 0.000903 N
Next, we need to calculate the force due to the surface tension, which acts around the circumference of the tube. We can use the equation:
force = 2πr x surface tension x cosθ
where r = 0.5 mm = 0.0005 m, θ = 0 (given), and force = weight = 0.000903 N.
Solving for surface tension, we get:
surface tension = force / (2πr x cosθ)
surface tension = 0.000903 / (2π x 0.0005 x 1) = 0.573 N/m
Therefore, the surface tension of the fluid is 0.573 N/m.
First, we need to calculate the weight of the fluid that rises up in the tube. We can use the equation:
weight = density x volume x gravity
where density = 1045 kg/m3 (given), volume = πr2h (where r = 0.5 mm = 0.0005 m and h = 2.4 cm = 0.024 m), and gravity = 9.81 m/s2.
weight = 1045 x π x (0.0005)2 x 0.024 x 9.81 = 0.000903 N
Next, we need to calculate the force due to the surface tension, which acts around the circumference of the tube. We can use the equation:
force = 2πr x surface tension x cosθ
where r = 0.5 mm = 0.0005 m, θ = 0 (given), and force = weight = 0.000903 N.
Solving for surface tension, we get:
surface tension = force / (2πr x cosθ)
surface tension = 0.000903 / (2π x 0.0005 x 1) = 0.573 N/m
Therefore, the surface tension of the fluid is 0.573 N/m.