PRESSURE OF LIQUID : aroragaurav.wordpress.com 1 PRESSURE OF LIQUID The normal force (or thrust) exerted by a liquid at rest per unit area of the surface incontact with it is called pressure of liquid or hydrostatic pressure. If F be the normal force acting on a surface area A of the object incontact with liquid. The pressure exerted by liquid on this surface of object is : P = F/A UNIT of Pressure. The SI unit of pressure is Nm–2 or Pascal (Pa) and the cgs unit is dyne cm–2. The dimensional formula for pressure is [ML–1T–2]. Pressure is a scalar quantity, because at one level inside the liquid, the pressure due to liquid is exerted equally in all directions. It shows that a definite direction is not associated with the pressure due to liquid.
Slide 2 : aroragaurav.wordpress.com 2 DENSITY AND RELATIVE DENSITY Density of a substance is defined as the mass per unit volume of the substance, i.e. r = The S.I. unit of density is kg m–3 Density is a positive scalar quantity. Relative density of a substance is defined as the ratio of its density to the density of water at 4°C, i.e., The density of water at 4°C (= 277 K) is 1.0 × 103 kg m–3. Relative density =
Slide 3 : aroragaurav.wordpress.com 3 VARIATION OF PRESSURE WITH DEPTH Consider a liquid of density r contained in a vessel in equilibrium of rest. Let C and D be two points inside the liquid at a vertical distance h. Imagine a cylinder of liquid with axis CD, cross-sectional area A and length h, such that the points C and D lie on the flat faces of the cylinder. Mass of the liquid in the imaginary cylinder will be
M = volume × density
= Ah × r = Ahr
Slide 4 : aroragaurav.wordpress.com 4 Let P1 and P2 be the pressures of liquid at points C and D respectively. The liquid cylinder is under the action of following vertical forces. Force F1 = P1A, acting vertically downwards Force F2 = P2A, acting vertically upwards Weight, Mg = A h r g of the liquid cylinder acting vertically downwards. As the liquid is in equilibrium of rest, hence imaginary cylinder of liquid is also
in equilibrium state of rest. Therefore, the net force on it must be zero i.e. F1 + Mg – F2 = 0 P1A + A h r g – P2A = 0 P2– P1 = h r g
Slide 5 : aroragaurav.wordpress.com 5 If points C and D lie at the same level in liquid, then h = 0.
P2 – P1 = 0 or P1 = P2 the pressure is same at all points inside the liquid lying at the same horizontal plane. Gauge pressure at a point in a liquid is the difference of total pressure at that point and atmospheric pressure.
Slide 6 : aroragaurav.wordpress.com 6 HYDROSTATIC PARADOX Each pressure metre will record the same pressure even though the quantity of liquid in different vessels is different. It means the liquid pressure at a point is independent of the quantity of liquid but depends upon the depth of point below the liquid surface. This is known as Hydrostatic paradox.
Slide 7 : aroragaurav.wordpress.com 7 PASCAL’S LAW It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest is same. This law also accounts for the principle of transmission of pressure in liquids or gases. Pascal’s law states that the increase in pressure at one point of the enclosed liquid in equilibrium of rest is transmitted equally to all other points of the liquid and also to the walls of the container.
Slide 8 : aroragaurav.wordpress.com 8 ATMOSPHERIC PRESSURE The gaseous envelope surrounding the earth is called earth’s atmosphere. The atmospheric pressure at any point is equal to the weight of a column of air of unit cross-sectional area extending from that point to the top of the earth’s atmosphere. The value of pressure on the surface of earth at sea level called one atmosphere (1 atm.) is nearly 1.013 × 105 Nm–2 or pascal in S.I. and 1.013 × 106 dyne/cm2 in cgs system.
Slide 9 : aroragaurav.wordpress.com 9 TORRICELLI’S EXPERIMENT Take a glass tube 1m in length and of uniform area of cross-section, closed at one end. Fill the whole tube with pure and dry mercury. Close the open end of tube tightly with your thumb and invert the tube. Take this inverted end of the tube along with your thumb inside the pure and dry mercury contained in a trough. We find that the mercury in the tube will fall down at first and then will be stopped at a particular position. The height of level of mercury in the tube is nearly 76 cm above the free level of mercury in the trough. If the given tube is inclined or raised up or lowered in mercury trough, the vertical height of mercury level in tube is always found to be constant.
Slide 10 : aroragaurav.wordpress.com 10 Torricelli explained that the mercury column is supported by the atmospheric pressure acting on the free surface of mercury in the trough. Hydrostatic pressure exerted by the vertical mercury column in the tube above the free surface of mercury in the trough measures the atmospheric pressure. The density of mercury,
r = 13.6 × 103 kg m–3
And g = 9.8 ms–2
Atmospheric pressure,
Pa = h r g = 0.76 × (13.6 × 103) × 9.8
= 1.013 × 105 Nm–2 or Pa
Slide 11 : aroragaurav.wordpress.com 11 VARIOUS UNITS FOR ATMOSPHERIC PRESSURE S.I. unit of atmospheric pressure is Nm–2 or pascal (denoted by Pa) and cgs unit is dyne/cm2. Atmospheric pressure is also measured in mm or in cm of mercury column Torr : 1 torr = one mm of mercury column. bar : 1 bar = 105 Pa and 1 milli bar = 10–3 bar = 10–3 × 105 Pa = 100 Pa.
Slide 12 : aroragaurav.wordpress.com 12 BUOYANCY This upward force acting on the body immersed in a fluid is called upward thrust or buoyant force or simply buoyancy. The buoyant force acts at the centre of buoyancy which is the centre of gravity of the liquid displaced by the body when immersed in the liquid.
Slide 13 : aroragaurav.wordpress.com 13 ARCHIMEDE’S PRINCIPLE It states that when a body is immersed wholly or partly in a liquid at rest, it loses some of its weight. The loss of weight of the body in the liquid is equal to the weight of the liquid displaced by the immersed part of the body Proof : Consider a rectangular body of height h and mass M, Let x be the depth of top face of the body below the free surface of liquid. Therefore the depth of bottom face of the body = x + h. Liquid pressure on the top face of the body, P1 = x r g
Vertical downward thrust
F1 = P1a = x r g a Liquid pressure on the bottom face of the body, P2 = (x + h) r g
Vertical upward thrust on the bottom face of body,
F2 = P2a = (x + h) r g a Let a be the cross-sectional area of the top or bottom face of the body. The net upward thrust F = F2 – F1 = (x + h) r g a – x r g a = h r g a = (h a) r g = mg = weight of liquid displaced.
Slide 14 : aroragaurav.wordpress.com 14 EQUILIBRIUM OF FLOATING BODIES There will be equilibrium of floating bodies if the following conditions are fulfilled. A body can float if the weight of the liquid displaced by the immersed part of body must be equal to the weight of the body. A body can be in equilibrium if the centre of gravity of the body and centre of buoyancy must be along the same vertical line. The body will be in stable equilibrium if centre of gravity of body lies vertically below the centre of buoyancy and in the unstable equilibrium if centre of gravity lies vertically above the centre of buoyancy.