Question

Mercury is poured into a U-tube as shown in Figure (a). The left arm of the tube has cross-sectional area of 10.2 cm2, and the right arm has a cross-sectional area A2 of 4.60 cm2. One hundred grams of water are then poured into the right arm as shown in Figure (b). Determine the length of the water column in the right arm of the U-tube. cm Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?

Answer 1

First you have to get water's density in order to find it's volume and the length of the water column in the right arm.

d = density of water
m= mass
v = volume

d = m/v

so

v = m/d = 100gr/(1g/cm^3) = 100cm^3

now you know that

a= area
h = length

v = a*h

so

h = v/a = 100cm^3/4.6cm^2 = 21.74 cm

that would be item a

now for b you know that the pressures in the contact area of water and mercury have to be the same, so

P1 = pressure of water

P2 = pressure of mercury

d1 = density of water

d2 = density of mercury

g = acceleration of gravity

h1 = height of water

h2 = height of mercury

P1 = P2

d1gh1=d2gh2

d1h1=d2h2

so

h2 = d1h1/d2

h2 = (1gm/cm^3)(21.74cm)/(13.6gm/cm^3)= 1.5985cm

so that would be the height of mercury from the point it touches the water.

Hope this helps.

Answer 2

First you have to get water's density in order to find it's volume and the length of the water column in the right arm.

d = density of water
m= mass
v = volume

d = m/v

so

v = m/d = 100gr/(1g/cm^3) = 100cm^3

now you know that

a= area
h = length

v = a*h

so

1) h = v/a = 100cm^3/4.6cm^2 = 21.74 cm

that would be item a

now for b you know that the pressures in the contact area of water and mercury have to be the same, so

P1 = pressure of water

P2 = pressure of mercury

d1 = density of water

d2 = density of mercury

g = acceleration of gravity

h1 = height of water

h2 = height of mercury

P1 = P2

d1gh1=d2gh2

d1h1=d2h2

so

2) h2 = d1h1/d2

h2 = (1gm/cm^3)(21.74cm)/(13.6gm/cm^3)= 1.5985cm

so that would be the height of mercury from the point it touches the water.

Hope this helps.

Answer 3

First you have to get water's density in order to find it's volume and the length of the water column in the right arm.

d = density of water
m= mass
v = volume

d = m/v

so

v = m/d = 100gr/(1g/cm^3) = 100cm^3

now you know that

a= area
h = length

v = a*h

so

1) h = v/a = (100cm^3)/(4.6cm^2) = 21.74 cm

that would be item a

now for b you know that the pressures in the contact area of water and mercury have to be the same, so

P1 = pressure of water

P2 = pressure of mercury

d1 = density of water

d2 = density of mercury

g = acceleration of gravity

h1 = height of water

h2 = height of mercury

P1 = P2

d1gh1=d2gh2

d1h1=d2h2

so

2) h2 = d1h1/d2

h2 = (1gm/cm^3)(21.74cm)/(13.6gm/cm^3)= 1.5985cm

so that would be the height of mercury from the point it touches the water.

Hope this helps

Answer 4

1) 21.74 cm
2) 0.4968 cm