Please Answer This ( BOth part a and Part B ) And please Explain your Answer in
ID: 1603210 • Letter: P
Question
Please Answer This ( BOth part a and Part B ) And please Explain your Answer in Detail and any equation that you are using.
Thank You
Q1: S&J; Chapter 13, Q16 Mercury is poured into a U-tube as shown in the figure 13.16a. The left arm of the tube has a cross-sectional area A1 of 10.0cm and the right arm has a cross sectional area A2 of 5.00cm2. One hundred grams of water is then poured into the right arm as shown in figure 13.16b Water a) Determine the length of the Al A2 Al A2 water column in the right arm of the U-tube. Mercury Figure P13.16 b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?Explanation / Answer
Density of water 1 g /cm^3
100 g of water has 100 cm^3
Volume = area *height
100 = 5* h
h = 20 cm
20 cm of water will be balanced by h *13.6 cm of mercury
h = 20 /13.6 = 1.47 cm.
But this is the height from the common interface of water and mercury in the right arm.
In the right arm the mercury has fallen through a distance x from its original level and has risen in the left arm by the same amount.
1.4 cm of mercury has a volume 1.4*10 = 14 cm ^3 of mercury.
If x is the distance through which the mercury falls in the right tube, or rises in the left tube
x ( A1 + A2 ) = 14
x ( 15 ) = 14
x = 0.93 cm
Subtracting this distance from 1.4 we get 1.4 – 0.93 = 0.47 cm
Thus the mercury level rises in the left arm by 0.47 cm from its original level and in the right arm it falls through 0.47 cm from its original level.
( Note that the from the new common level the mercury is 0.93 + 0.47 = 1.4 cm )