Q#40 (Past Exam Paper – November 2017 Paper 12 Q15)
Air is trapped inside a glass bulb which is immersed in water and attached to a U-tube containing mercury. The densities of water and mercury are ρw and ρm respectively. The surface of the water is open to the atmosphere where atmospheric pressure is P.
The acceleration of free fall is g.
What is the pressure of the air in the glass bulb?
A P + gρwhw + gρmhm
B P + gρwhw – gρmhm
C gρwhw + gρmhm
D gρwhw – gρmhm
Solution:
Answer: B.
The pressure is the same at the same level in a liquid.
Pressure is always due to what is above the level being considered.
Consider the level of mercury on the left-hand side of the U-tube. It is at a depth of hw below the surface of water. Let this level be O.
At this level,
Pressure on left-had side = Pressure on right-hand side
Again, remember that pressure is always due to what is ABOVE that level.
On the left-hand side, we have water of depth hw above level O. And above the water surface, we have the atmospheric pressure P acting on the level O.
Pressure on left-hand side = due to water of depth hw + due to atmospheric pressure
Pressure on left-hand side = hw ρw g + P
On the right-hand side, we have a height hm of mercury of level O. In addition to the mercury, the air inside the glass bulb also exerts a pressure (Pair).
Pressure on right-hand side = due to mercury of height hm + due to air
Pressure on right-hand side = hm ρm g + Pair
At level O,
Pressure on left-hand side = Pressure on right-hand side
hw ρw g + P = hm ρm g + Pair
Pair = hw ρw g + P – hm ρm g
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