Uneven U-Tube

## **Problem**

Mercury is poured into an uneven u-shaped tube. The cross-sectional area of the left column is

10.8
c

m
2

{\displaystyle 10.8 cm^2 }

, while the cross-sectional area of the right column is

5.60
c

m
2

{\displaystyle 5.60 cm^2 }

. Then 300 grams of water is poured into the right column. Given that the density of water is

1.00
g

/

c

m
3

{\displaystyle 1.00 g/cm^3 }

 and the density of mercury is

13.6
g

/

c

m
3

{\displaystyle 13.6 g/cm^3 }

, calculate the displaced height of the mercury in the left column.

## **Solution**

Let

P

{\displaystyle P}

 be the pressure at the base of the water column. Since both sides of the tube are open to the atmosphere,

# P

P
0

ρ

H
g

(
g
)

(

h

H
g

)

=

P
0

ρ

H
2

O

(
g
)

(

h

H
2

O

)

{\displaystyle P = P\_0 + {\rho}\_{Hg}(g)\left({h}\_{Hg}\right) = P\_0 + {\rho}\_{H\_2 O}(g)\left({h}\_{H\_2 O}\right) }

.
Subtract the atmospheric pressure on both sides. Then divide the gravitational acceleration on both sides. Thus,

ρ

H
g

(

h

H
g

)

=

ρ

H
2

O

(

h

H
2

O

)

{\displaystyle {\rho}\_{Hg}\left({h}\_{Hg}\right) = {\rho}\_{H\_2 O}\left({h}\_{H\_2 O}\right) }

.
It is known that

ρ

H
g

=
13.6
g

/

c

m
3

{\displaystyle {\rho}\_{Hg} = 13.6 g/cm^3 }

ρ

H
2

O

=
1.00
g

/

c

m
3

{\displaystyle {\rho}\_{H\_2 O} = 1.00 g/cm^3 }

but both heights are not known. We cannot solve for

h

H
g

{\displaystyle {h}\_{Hg} }

, however there is enough information to solve for

h

H
2

O

{\displaystyle {h}\_{H\_2 O} }

. Using the mass of the added water, we can exploit the fact

m

H
2

O

=

ρ

H
2

O

V

H
2

O

=

ρ

H
2

O

A
2

h

H
2

O

{\displaystyle {m}\_{H\_2 O} = {\rho}\_{H\_2 O} {V}\_{H\_2 O} = {\rho}\_{H\_2 O} A\_2 {h}\_{H\_2 O} }

to solve for the height of the water column.

300
# g
(
1
g

/

c

m
3

)
(
5.6
c

m
2

)

(

h

H
2

O

)

{\displaystyle 300 g = (1 g/cm^3) (5.6 cm^2)\left({h}\_{H\_2 O}\right) }

So

h

H
2

O

=
53.57
c
m

{\displaystyle {h}\_{H\_2 O} = 53.57 cm }

.
Since

h

H
g

=

ρ

H
2

O

ρ

H
g

h

H
2

O

{\displaystyle {h}\_{Hg} = \frac{{\rho}\_{H\_2 O}}{{\rho}\_{Hg}} {h}\_{H\_2 O} }

,

h

H
g

=

1.00
g

/

c

m
3

13.6
g

/

c

m
3

(
53.57
c
m
# )
3.939
c
m

{\displaystyle {h}\_{Hg} = \frac{1.00 g/cm^3}{13.6 g/cm^3}(53.57 cm) = 3.939 cm }

.
Let

h

{\displaystyle h}

 be the displaced height of the mercury in the left column, and

h
2

{\displaystyle h\_2}

 be the height from the dotted line to the base of the water column. It can be deduced that

h

H
g

=
h

h
2

{\displaystyle {h}\_{Hg} = h + h\_2 }

.
So far it appears we have reached a dead end. There is only one equation, but there are two unknowns. However, we should note the volume of water below the dotted line is the same volume of the displaced mercury above the dotted line. Hence, if

V

{\displaystyle V}

 represents the volume of mercury displaced above the dotted line, then

# V

A
1

# h

A
2

h
2

{\displaystyle V = A\_1 h = A\_2 h\_2 }

.
This shows

h
2

=

10.8
c

m
2

5.6
c

m
2

# h
1.9286
h

{\displaystyle h\_2 = \frac{10.8 cm^2}{5.6 cm^2} h = 1.9286h }

Now we can solve for

h

{\displaystyle h}

.

3.939
c
# m
h
1.9286
# h
2.9286
h

{\displaystyle 3.939 cm = h + 1.9286h = 2.9286h }

Thus, the displaced height of the mercury in the left column is

# h
1.345
c
m

{\displaystyle h = 1.345 cm }

.