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| Pressure in a Liquid |
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| We can investigate pressure in a liquid by using the apparatus shown in figure below. |
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| Pressure due to a liquid |
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| It is a simple pressure gauge and it measures differences in pressure exerted at the two ends of the apparatus. It is called manometer. The mouth of a thistle funnel is tightly covered with a thin plastic sheet. The thistle funnel is connected to a U-tube manometer containing water, by a rubber tubing. If you press the plastic sheet lightly with a finger, the air inside the manometer gets compressed and it exerts greater pressure at that end than the atmospheric pressure exerted at the open end of the U-tube. As a result, the liquid level alters and goes up in the other open arm of the U-tube. When the pressure exerted on the plastic sheet is greater, the difference in the levels (h) is also greater. This difference is the measure of the difference in the pressures at the two ends of the manometer. |
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| Now, lower the mouth of the funnel into a glass vessel containing water. You will notice that the deeper it goes, greater is the difference in the levels of the water in the manometer. This indicates that the pressure in a liquid increases with depth. |
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| Repeat the experiment by turning the thistle funnel in different directions keeping the depth constant. You will observe that as long as the depth remains the same there is no change in the level of the water in the manometer. Thus, the pressure exerted by a liquid at a given depth is the same in all directions. |
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| Pressure at a given depth |
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| Now, lower the thistle funnel to the same depth in a number of liquids having different densities. You will notice that in liquids having greater density the pressure at the same depth is greater. This indicates that greater the density of the liquid, greater is the pressure at the same depth. |
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| Suppose a horizontal area A (m2) is supporting a column of liquid of height h (m) and density d (kgm-3). |
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| Pressure = hdg |
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| The pressure on A is due to the weight of the liquid acting on it. |
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| Mass of the liquid = Volume x Density |
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| = A x h x d |
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| Weight in newtons = A x h x d x g (which is force) |
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| The above expression also applies to any depth in the liquid. Hence, the pressure at any point in a liquid at rest depends only on the depth and on the density of the liquid. It is independent of the cross-sectional area. |
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| Liquids find their own level |
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| Figure above demonstrates this. Notice how the water fills each of the tubes to the same level. This is what we mean when we say that a liquid always finds its own level. Liquids will flow to equalise any pressure differences. Therefore the pressure at the top of each tube must be equal and likewise those at the bottom of each tube. This confirms the independence of pressure and cross-sectional area for a liquid. |
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| Figure below shows a hydraulic weight bridge which works on the principle of Pascal's law. |
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| (i) What is the pressure at B? |
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| (ii) What is the pressure at A? |
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| (iii) What is the weight of the vegetable on the large piston A if the weight bridge is in equilibrium? |
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(i)  |
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| = 5 Ncm-2 |
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| (ii) Pressure at A = Pressure at B = 5 Ncm-2 (By Pascal's law) |
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(iii)  |
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| The water tank in figure below is 8 m above the tap. What pressure forces the water out from the tap? (Density of water = 1000 kgm-3) |
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| Pressure at the tap is due to the water in the pipe and tank above it. |
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| Pressure = hpg |
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| = 8 x 1000 x 10 |
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| = 80,000 Pa |
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| A regularly shaped object is immersed in water of density 1000 kgm-3. |
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| (i) Calculate the water pressure at the top and the bottom of the object. |
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| (ii) What is the resultant pressure on the object? |
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| (i) Pressure exerted by water at the top surface of the object |
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| = h1pg |
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| = 0.1 x 1000 x 10 = 1000 Pa |
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| Pressure exerted by water at the bottom surface of the object |
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| = h2pg |
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| = 0.15 x 1000 x 10 = 1500 Pa |
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| (ii) Resultant pressure on the object |
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| = (1500 - 1000) Pa |
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| = 500 Pa (i.e., the object experiences an upward force) |
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| Calculate the pressure due to water column of height 100 m (Take g = 10 m s-2 and density of water = 103 kg m-3). What height of mercury column will exert the same pressure? (density of mercury = 13.6 x 103 kg m-3) |
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| P = hdg = 100 x 103 x 10 = 106 Pa |
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| P2 due to mercury column = P1 due to water column |
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| h2 d2 g = h1 d1 g |
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h2 =  |
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| The pressure of water on the ground floor is 40,000 Pa and at first floor is 10,000 Pa. Find the height of the first floor (density of water = 1000 Kg m-3, g = 10 ms-2) |
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| P1 on ground floor = h1 dg = 40,000 |
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| P2 on first floor = h2 dg = 10,000 |
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| Height of first floor = (h1 - h2) |
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| (h1 - h2) dg = 40,000 - 10,000 |
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| = 3 m. |
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