   Chapter 9, Problem 10P

Chapter
Section
Textbook Problem

Mercury is poured into a U-tube as shown in Figure P9.10a. The left arm of the cube has cross-sectional area A1 of 10.0 cm2, and the right arm has a cross-sectional area A2 of 5.00 cm2. One hundred grams of water are then poured into the right arm as shown in Figure P9.10b. (a) Determine the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm? Figure P9.10

(a)

To determine
The length of the water column in the right arm of the U-tube.

Explanation
The volume of the water that poured into the right arm of the U-tube is given by Vw=mw/ρw and now the height at which the column has to be filled calculated as Vw=mw/ρw=A2hwhw=mw/(ρwA2) .

Given info: The right arm cross-sectional area of the U-tube is 5.00cm2 , mass of water is 100g , and the density of water is 1.00g/cm3 .

The formula for the length of the water column in the right arm of the U-tube is,

hw=mwρwA2

• mw is mass of the water

(b)

To determine
The distance h rises in the left arm

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