- #1
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.
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.
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