Consider a glass with full of water of mass density ρ=1,000 kg/m3 and height h=20 cm. There's a circular hole in the bottom of the glass of radius r. The maximum pressure that pushes the water back into the hole is roughly (on the order of) p=σ/r, where σ=0.072 N/m is the water's surface tension. This extra pressure comes from the curvature of the water surface, and it tends to flatten out the surface.
Estimate the largest possible radius of the hole in μm such that water doesn't drip out of the glass.
Details and assumptions
The gravitational acceleration is g=−9.8 m/s2 and the glass is placed vertically.
Neglect any other effects that can influence the pressure from other external sources.
net pressure on hole=weight-pushback
= densitywater*g*height-.072/radius
when net pressure=zero, then
radius=.072/(g*.2*densitywater) meters
= .46/9.6 millimeters=46.9 micrometers.
check that
Sir but 46/9.6 is 4.79 tell the right answer
1000*9.8*0.2 = 0.072/r
1960*r = 0.072
r = 3,673 x 10^(-5) m
so r = 36,73 μm
sorry i mean
r = 3.673 x 10^(-5) m
so r = 36.73 μm
haha still confused with . and , :)
To estimate the largest possible radius of the hole in micrometers (μm) such that water doesn't drip out of the glass, we need to consider the equilibrium between the gravitational force pulling the water down and the pressure force pushing the water back up due to surface tension.
Let's first calculate the pressure force pushing the water back into the hole. According to the given information, the maximum pressure pushing the water back into the hole is roughly on the order of p = σ/r, where σ = 0.072 N/m is the water's surface tension and r is the radius of the hole.
Next, we need to calculate the gravitational force on the water inside the hole. The gravitational force is given by F = mg, where m is the mass of water in the hole and g is the gravitational acceleration.
First, let's calculate the mass of the water in the hole. The volume of the water in the hole is equal to the volume of a cylindrical shape, given by V = πr^2h, where r is the radius of the hole and h is the height of the water column.
Since the height h is given as 20 cm, we need to convert it to meters by dividing by 100:
h = 20 cm / 100 = 0.2 m
Now, we can calculate the mass of the water in the hole. The mass is given by m = ρV, where ρ = 1000 kg/m^3 is the mass density of water.
Substituting the values, we have:
m = ρπr^2h
Now, let's equate the gravitational force and the pressure force to find the maximum radius of the hole without water dripping out:
mg = σ/r
Substituting the expressions for mass and gravitational force, we have:
ρπr^2h * g = σ/r
Simplifying the equation, we get:
r^3 = σ/(ρπgh)
Taking the cube root of both sides, we have:
r = (σ/(ρπgh))^(1/3)
Now, we can substitute the given values:
ρ = 1000 kg/m^3
σ = 0.072 N/m
g = -9.8 m/s^2 (negative sign indicates downward direction)
h = 0.2 m
Calculate r:
r = (0.072/((1000)(π)(-9.8)(0.2)))^(1/3)
r ≈ 4.0399 x 10^(-5) m
Finally, to convert the radius to micrometers (μm), we multiply by 10^6:
r ≈ 4.0399 x 10^(-5) m * 10^6 = 40.399 μm
Therefore, the largest possible radius of the hole such that water doesn't drip out of the glass is approximately 40.399 micrometers (μm).