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Physics Challenge for<br />

Teachers and Students<br />

A Pipe Dream<br />

Challenge: An open glass tube has a length of 0.500 m.<br />

The tube is held vertically above a water-filled container so<br />

that the bottom end of the tube slightly touches the surface<br />

of the water. Under these conditions, the height of the<br />

water column in the tube is 20.0 mm.<br />

What would be the height of the water column in the<br />

tube if the top end of the tube is sealed before touching the<br />

water? The atmospheric pressure is 1.01 x 10 5 Pa.<br />

Solution: When the top end of the tube (of<br />

inner radius r) is open, there is atmospheric<br />

pressure at both the top and bottom surfaces<br />

of the water column of height h = 20 mm.<br />

Therefore its weight ρπr 2 hg (where ρ =<br />

1000 kg/m 3 is the density of water) is balanced<br />

by a capillary force F cap due to an adhesive<br />

attraction between the glass and the water<br />

around the circumference of its top surface,<br />

F cap = ρπr 2 hg. (1)<br />

(There is also an adhesive attraction between<br />

the curved side of the water column and the<br />

glass, but this produces no net force on the<br />

water.)<br />

If the top end of the tube is sealed before its<br />

bottom end is immersed, we trap an air column<br />

of initial length L = 500 mm that we can<br />

model as an isothermal ideal gas. Suppose this<br />

time the water column rises to height h' inside<br />

the tube. This compresses the air column to a<br />

gauge pressure of P, where<br />

PV i i = PfVf ⇒ PatmLπr = ( P + P)( L− h′ ) πr<br />

atm<br />

⇒ P= P<br />

Solutions to <strong>February</strong> 2006 Challenge<br />

2<br />

2<br />

h′<br />

P<br />

L− h′<br />

≈<br />

atm atm<br />

h′<br />

,<br />

L<br />

(2)<br />

with P atm = 101 kPa atmospheric pressure.<br />

Now we have four forces acting on the water<br />

column: the difference between atmospheric<br />

pressure upward on its bottom and the air column<br />

pressure downward on its top multiplied<br />

by the inner cross-sectional area of the tube,<br />

the same capillary force upward on its top circumference<br />

as before, and its weight downward.<br />

The force balance thus becomes<br />

ρπr 2 h' g + Pπr 2 = F cap . (3)<br />

Substituting Eqs. (1) and (2) and rearranging<br />

gives<br />

⎡ ⎤<br />

′ = ⎢ + ⎥ (4)<br />

⎣ ⎦<br />

=<br />

−1<br />

Patm<br />

h h 1 0. 925 mm.<br />

ρgL<br />

Note that h'


lege, Cleveland, OH)<br />

Art Hovey (Milford, CT)<br />

José Costa Leme (High School Lanheses, Viana<br />

do Castelo, Portugal)<br />

Stephen McAndrew (Trinity Grammar School,<br />

Summer Hill, NSW, Australia)<br />

Leo H. van den Raadt (Heemstede, The Netherlands)<br />

Many thanks to all contributors and we hope to<br />

hear from you in the future!<br />

Please send correspondence to:<br />

Boris Korsunsky<br />

444 Wellesley St.<br />

Weston, MA 02493-2631<br />

korsunbo@post.harvard.edu<br />

THE PHYSICS TEACHER ◆ Vol. 44, May 2006 29

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