Posted by **Mathslover Please help** on Wednesday, May 8, 2013 at 1:36am.

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.

- Physics Help -
**bobpursley**, Wednesday, May 8, 2013 at 2:03am
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

- Physics Help -
**Mathslover Please help**, Thursday, May 9, 2013 at 12:24am
Sir but 46/9.6 is 4.79 tell the right answer

- Physics Help -
**levina**, Thursday, May 9, 2013 at 7:30am
1000*9.8*0.2 = 0.072/r

1960*r = 0.072

r = 3,673 x 10^(-5) m

so r = 36,73 μm

- Physics Help -
**levina**, Thursday, May 9, 2013 at 7:33am
sorry i mean

r = 3.673 x 10^(-5) m

so r = 36.73 μm

haha still confused with . and , :)

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