Mercury is poured into a U-tube as shown in Figure a. The left arm of the tube h
ID: 1289955 • Letter: M
Question
Mercury is poured into a U-tube as shown in Figure a. The left arm of the tube has cross-sectional area A1 of 10.7 cm2, and the right arm has a cross-sectional area A2 of 4.90 cm2. Two hundred grams of water are then poured into the right arm as shown in Figure b.
(a) Determine the length of the water column in the right arm of the U-tube.
200g /4.9 cm2= 40.81633 cm
(b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?
= ?
Please help explain how to get part B. Thanks!
Mercury is poured into a U-tube as shown in Figure a. The left arm of the tube has cross-sectional area A1 of 10.7 cm^2, and the right arm has a cross-sectional area A2 of 4.90 cm^2. Two hundred grams of water are then poured into the right arm as shown in Figure b. Mercury is poured into a U-tube as shown in Figure (a) Determine the length of the water column in the right arm of the U-tube. 200g /4.9 cm^2= 40.81633 cm (b) Given that the density of mercury is 13.6 g/cm^3, what distance h does the mercury rise in the left arm? = ? Please help explain how to get part B. Thanks!Explanation / Answer
Let h2 be the depth of mercury level from the initial level in right figure.
Volume of mercury transferred from right limb to left limb =
h A1 = h2A2
h = (4.9/10.7) h2 = 0.457 h2
h2 = 2.18h ------------------1
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If L is the total water column in the right figure,
L cm of water balances (h +h2) cm of mercury.
(h +h2) ? = L*1 (since density of water is 1gm/cm^3)
(h +h2)*13.6 = L----------------------------2
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But L = mass of water /( area *density of water)
L = 200/4.9 =40.8163cm ------------3
Substituting for L, and h2 in equation 2
(h +2.18h)*13.6 = 40.8163
h = 0.94cm