A U-shaped tube is open at both ends. The tube initially has some water in it. A

Question

A U-shaped tube is open at both ends. The tube initially has some water in it. Anamount of a second liquid, which is immiscible with water, is poured into one arm of thetube. Once the arrangement comes to equilibrium, the top of the water in the tube stands16.6 cm above the interface between that liquid and the water (which is at level a in the diagram). The top of the water column in the other arm is 3.32 cm above the top of theunknown liquid.

(a) Which liquid is denser: the unknown, or water?

(b) What is the density of the unknown liquid? Hint: The pressures at points a and b are equal (why is this)?

Explanation / Answer

Here, The pressures at points a and b are equal

This is because,when two fluids are in contact with each other and are at rest,the pressure must be same at two sides of interface.

So, Pa=Pb

When a tube is filled with two different immiscible liquids of different densities ,if one liquid has a higher density than other, the liquid with lower density rises to a higher level.

Therefore, the unknown liquid is denser than water because water has risen to a higher level.

As, Pa=Pb

Pa=for unknown liquid pressure at point a

Pa=Patm+Punknwn

where,Patm=atmospheric pressure at surface

Punknwn=pressure due to liquid at point a with distance 16.6 cm

x=density of unknown liquid

Pb=for water pressure at point b

Pb=Patm+Pwater

where,Patm=atmospheric pressure at surface

Pwater=pressure due to water at point b with distance (16.6+3.32 cm).

w=density of water=1000 kg/m3

So, Patm + x *g* 16.6 = Patm + w *g* 19.92

x *16.6 = w *19.92

x = (1000 * 19.92)/16.6

x = 1200 kg/m3

Therefore, density of unknown liquid is 1200 kg/m3

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