Calculating Liquid Pressure: Water and Mercury

[guan721]

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

Notes:
Water density = 1000 kg/m3
Mercury density = 13600 kg/m3
g = 10 N/kg

Hi all, please help on this questions. The attempt from me
a)
(0.7m)(1000kg/m3)(g) = (h)(13600kg/m3)(g)
h = 0.0514m
h = 5.14cm

Assume h = 5.14cm is based on same surface area between both side of tube, however, for different surface area of tube, the h = 5.14cm have to be adjusted

Let d = diameter of wider tube
Surface area of narrower tube = (1/8 x d) ^2 = 1/64d2
h = 5.14cm, based on (1/64)(d^2)

Surface area of wider tube = (1/2 x d) ^ 2 = 1/4d2
to get the same volume, h of wider tube have to be lesser

h1d1 = h2d2
(5.14cm)(1/64)(d^2) = (h2)(1/4)(d^2)
h2 = 0.32cmb) By refer to the a) calculation, the level dropped from mercury in narrower tube = 5.14cm

c) I have no clue about this question, as I can't confirm the concept I used to solve a) and b) whether correct or wrong.

Kindly help, many thanks.

 

[haruspex]

h = 5.14cm

In terms of the initial level of the mercury, its final level in the narrow tube, its final level in the wide tube, and the final top surface of the water, what exactly does this calculated h represent?