Question

A U-shaped tube, open to air on both ends, contains mercury. Water is poured into the left arm until the water in the column is 10 cm deep. How far upward from its initial position does the mercury in the right arm rise? Answer is .003 meters or 3 mm, but I need help on why

Answer 1

Is the cross-sectional area of both ends of the tubes the same?

Answer 2

I assume it is. This was all the problem said

Answer 3

I haven't done fluid mechanics/dynamics in a while so I may or may not be able to help; I need to think.

Answer 4

the pressure from the water will be
pressure = rhow g h,
rhow=density, g = gravity, h = height of water in that arm.

The difference in the mercury levels, d, will equal this

pressure = rhom g d

so

rhow g h = rhom g d

d = (rhow/rhom) h

and the right arm moves d/2 up as the left moves d/2 down

Answer 5

@douglaswinslowcooper According to your explanation: \(\large d={\rho_w\over\rho_{Hg}}h\) such that
d=(1/13.6)(10 cm) but that is not the correct answer according to @Prominence

Answer 6

recall I wrote that d/2 is what the right arm will do

10/2(13.6) = 3.7 mm

Answer 7

I do not know how they got 3mm

Answer 8

convert 10 top .1 to change it to meters and you get that

Answer 9

it is not the mm I question it is 3 vs 3.7 that puzzles me.

Answer 10

The entire answer is 3.68mm

Answer 11

ok that makes more sense

Answer 12

@Prominence It kinda helps to give the full answer to avoid confusion :)
@douglaswinslowcooper Thanks!

Answer 13

Yes. d/2 = 10cm/2(13.6) right.

Answer 14

thank you

Answer 15

You are welcome. For a minute there....