Fluid Statics

Matter exists in three states solid, liquid and gas. Liquids and gases are referred to as fluids. Any state of matter that can flow is a fluid. Study of a fluid at rest is called fluid statics or hydrostatics and the study of fluid in motion is called fluid dynamics of hydrodynamics. Both combined are called fluid mechanics. The intermolecular force in liquids are comparatively weaker than in solids. Therefore, their shape can be changed easily. Thus liquids assume the shape of the container. But their volume (or density) cannot be changed so easily. Liquids are incompressible and have free surface of their own. The intermolecular forces are weakest in gases, so their shape and size can be changed easily. Gases are compressible and occupy all the space of the container. Density () Mass per unit volume is defined as density. So density at a point of a fluid is represented as   Lim V 0 SI Unit : kg/m3 m dm  V dV Density is a positive scalar quantity. CGS Unit : g/cc Dimensions : [ML–3] Relative Density It is defined as the ratio of the density of the given fluid to the density of pure water at 4°C. Relative density (R.D.) = density of given liquid density of pure water at 4°C Relative density or specific gravity is a unitless and dimensionless positive scalar physical quantity. Being a dimensionless/unitless quantity R.D. of a substance is same in SI and CGS system. Specific Gravity It is defined as the ratio of the specific weight of the given fluid to the specific weight of pure water at 4°C. Specific gravity =   g l specific weight of given liquid = R.D. of the liquid  l  3 specific weight of pure water at 4 °C (9 .8 1 kN m ) w  g w Thus specific gravity of a liquid is numerically equal to the relative density of that liquid and for calculation purposes they are used interchangeably. Ex. A hollow metallic sphere has inner and outer radii, respectively, as 5 cm and 10 cm. If the mass of the sphere is 2.5 kg, find (a) density of the material, (b) relative density of the material of the sphere. Sol. The volume of the material of the sphere is  10  3  5  3  4 4 4 V     r23  r13   3 .1 4      1 00    3  3.1 4   0.0 01  0 .0 00 12 5  3 3  10 0    =  4 × 3.14 × 0.000875 m3 = 0.00367 m3 3 (a) Therefore, density of the material of the sphere is  = (b) Relative density of the material of the sphere r  M 2 .5 = kg/m3 = 681.2 kg/m3 V 0 .0 0 36 7 68 1.2 = 0.6812 1 00 0 MECHANICAL PROPERTIES OF MATTER Density of a Mixture of substance in the proportion of mass Let a number of substances of masses M1, M2 , M3 etc., and densities 1 , 2 , 3 etc. respectively are mixed together. The total mass of the mixture = M1 + M2 + M3 +.... The total volume = M1 M2 M M1  M2  M3 .... + + 3 + .... therefore, the density of the mixture is   M1 M2 M3 3 1 2    .... 1 2 3 For two substances the density of the mixture   Ex. Sol. 1 2  M1  M2  1 M2  2 M1 Two immiscible of densities 2.5 g/cm3 and 0.8 g/cm3 are taken in the ratio of their masses as 2:3 respectively. Find the average density of the liquid combination.  2 m 3m   Let masses be 2m g & 3m g, then V = V1 + V2 =  cm3 2 .5 0 .8  Total mass = 2m + 3m = 5m g. Therefore, the average density  av = 5m 5m 10 5 = = = gm/cm3 = 1.09 gm/cm3 2m 3m 2 3 V 9 .1   2.5 0.8 2.5 0.8 Density of a mixture of substance in the proportion of volume Suppose that a number of substances of volume V1, V2, V3 etc. and densities 1 , 2 , 3 etc. respectively are mixed. The total mass of the mixture is = 1 V1  2 V2  3 V3  .... The total volume of the mixture is = V1 + V2 + V3+......... Therefore, the density of the mixture is  Therefore, for two substances we can write   Ex. Sol. 1 V1  2 V2  3 V3 V1  V2  V3  .... 1 V1  2 V2 V1  V2 Two miscible liquids of densities 1.2 gm/cc and 1.4 gm/cc are mixed with a proportion ratio of their volumes equal to 3:5: What is the density of resulting liquid? V 1 1  2 1 V1  2 V2 1.2  3 / 5   1 .4 3 .6  7 V2 1   =   = =   = 1.325 V1  V2 8 V1 / V2  1 3 / 5   1 PRESSURE The pressure P is defined as the magnitude of the normal force acting on a unit surface area. P= F here F= normal force on a surface area A A The pressure is a scalar quantity. This is because hydrostatic pressure is transmitted equally in all directions when force is applied, which shows that a definite direction is not associated with pressure. CONSEQUENCES OF PRESSURE (i) Railway tracks are laid on large sized wooden or iron sleepers. This is because the weight (force) of the train is spread over a large area of the sleeper. This reduces the pressure acting on the ground and hence prevents the yielding of ground under the weight of the train. (ii) A sharp knife is more effective in cutting the objects than a blunt knife. The pressure exerted = Force / area. The sharp knife transmits force over a small area as compared to the blunt knife. Hence the pressure exerted in case of sharp knife is more than in case of blunt knife. (iii) A camel walks easily on sand but a man cannot inspite of the fact that a camel is much heavier than man. This is because the area of camel's feet is large as compared to man's feet. So the pressure exerted by camel on the sand is very small as compared to the pressure exerted by man. Due to large pressure, sand under the feet of man yields and hence he cannot walk easily on sand. TYPES OF PRESSURE There are three types of pressure (i) Atmospheric pressure (P0) 1. (ii) Gauge pressure (Pgauge) (iii) Absolute pressure (Pabs.) Atmospheric pressure : Force exerted by air column on unit cross–section area of sea level called up to top of atmosphere atmospheric pressure (Po) Po = F =101.3 kN/m2 A  Po = 1.013 × 105 N/m2 sea level Barometer is used to measure atmospheric pressure. air column Which was discovered by Torricelli. Atmospheric pressure varies from place to place and at a particular place from time to time. 2. Gauge Pressure : Excess Pressure ( P– Patm) measured with the help of pressure measuring instrument called Gauge pressure. Patm manometer g as Pabsolute h Pgauge = hg or Pgauge h Gauge pressure is always measured with help of "manometer" 3. Absolute Pressure : Sum of atmospheric and Gauge pressure is called absolute pressure. Pabs = Patm + Pgauge Pabs = Po + hg The pressure which we measure in our automobile tyres is gauge pressure. 2 area=1m MECHANICAL PROPERTIES OF MATTER P1 VARIATION OF PRESSURE WITH DEPTH (i) y1 Let pressure at L is P1 and pressure at M is P2 Then, P2A = P1 A + gA (y2 – y1)  P2 = P1 + g(y2 – y1) Here pressure gradient (ii) L A M A y2 dP = g dy P2 Pressure is same at two points in the same horizontal level. As body is in equilibrium, P1A = P2A  P1 = P2 P1 Note : Pressure P is independent of shape of container L M A A P2 Ex. Assuming that the atmosphere has a uniform density of (1.3 kg/m3) and an effective height of 10 km, find the force exerted on an area of dimensions 100 m × 80 m at the bottom of the atmosphere . Sol. F = PA = gh A = (1.3) (9.8) (104) (100 × 80) = 12.74 × 8× 107 N = 1.0192 × 109 N PRESSURE EXERTED BY A LIQUID (EFFECT OF GRAVITY) : Consider a vessel containing liquid. As the liquid is in equilibrium, so every volume element of the fluid is also in equilibrium. Consider one volume element in the form of a cylindrical column of liquid of height h and of area of cross section A. The various forces acting on the cylindrical column of liquid are : (i) Force, F1 = P1 A acting vertically downward on the top face of the column. P1 is the pressure of the liquid on the top face of the column and is known as atmospheric pressure. (ii) Force, F2 = P2 A acting vertically upward at the bottom face of the cylindrical column. P2 is the pressure of the liquid on the bottom face of the column. (iii) Weight, W = mg of the cylindrical column of the liquid acting vertically downward. Since the cylindrical column of the liquid is in equilibrium, so the net force acting on the column is zero. i.e. F1 + W – F2 =0  P1A +mg – P2A = 0 P1A + mg = P2 A  P2 = P1 + mg A ...(i) Now, mass of the cylindrical column of the liquid is, m = volume × density of the liquid = Area of cross section × height × density = Ah  equation (i) becomes P2 =P1 + Ah g , P2 = P1 + hg A ...(ii) P2 is the absolute pressure at depth h below the free surface of the liquid. Equation (ii), shows that the absolute pressure at depth h is greater than the atmospheric pressure (P1) by an amount equal to hg. Equation (ii) can also be written as (P2–P1) = hg which is the difference of pressure between two points separated by a depth h. PRESSURE IN CASE OF ACCELERATING FLUID (i) Liquid placed in elevator : When elevator accelerates upward with acceleration a0 then pressure in the fluid, at depth 'h' may be given by, P  h  g  a 0  a0 h (ii) Free surface of liquid in case of horizontal acceleration : tan   ma 0 a 0 = mg g If P1 and P2 are pressures at point 1 & 2 then P1–P2 = g (h1– h2) = gtan = a0 (iii) Rotating Vessel Consider a cylindrical vessel, rotating at constant angular velocity about its axis. If it contains fluid then after an initial irregular shape, it will rotate with the tank as a rigid body. The acceleration of fluid particles located at a distance r from the axis of rotation will be equal to 2r, and the direction of the acceleration is toward the axis of rotation as shown in the figure. The fluid particles will be undergoing circular motion. Lets consider a small horizontal cylinder of length dr and cross-sectional area A located y below the free surface of the fluid and r from the axis. This cylinder is accelerating in ground frame with acceleration 2r towards the axis hence the net horizontal force acting on it should be equal to the product of mass (dm) and acceleration. dm = Adr P2A – P!A = (Adr)2r If we say that the left face of the cylinder is y below the free surface of the fluid then the right surface is y + dy below the surface of liquid. Thus P2 – P1 = gdy Thus solving we get dy r 2  dr g and, therefore, the equation for surfaces of constant pressure is y  2 r 2 + constant 2g This equation means that these surface of constant pressure are parabolic as shown in figure. The pressure varies with the distance from the axis of rotation, but at a fixed radius, the pressure varies hydrostatically in the vertical direction as shown in figure. Ex. An open water tanker moving on a horizontal straight road has a cubical block of cork floating over its surface. If the tanker has an acceleration of 'a' as shown, the acceleration of the cork w.r.t. container is (ignore viscosity) a \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ MECHANICAL PROPERTIES OF MATTER Sol. marel = mgsin – macos but for water surface tan  = a/g  arel = 0 Ex. An open rectangular tank 1.5 m wide, 2 m deep and 2m long is half filled with water. It is accelerated horizontally at 3.27 m/s2 in the direction of its length. Determine the depth of water at each end of tank. [g = 9.81 m/s2] Sol. Here tan   a 1  , depth at corner 'A' = 1– 1.5 tan  = 0.5 m g 3 depth at corner 'B' = 1 + 1.5 tan  = 1.5 m MEASUREMENT OF ATMOSPHERIC PRESSURE 1. Mercury Barometer : To measure the atmospheric pressure experimentally, Torricelli invented a mercury barometer in 1643. pa =hg The pressure exerted by a mercury column of 1mm high is called 1 Torr. 1 Torr = 1 mm of mercury column Torricelli vaccum h A Mercury Trough OPEN TUBE MANOMETER : Open-tube manometer is used to measure the pressure gauge. When equilibrium is reached, the pressure at the bottom of left limb is equal to the pressure at the bottom of right limb. i.e. p + y1 g = pa + y2 g p – pa = g (y2 – y1) = gy p – pa = g (y2 – y1) = gy p = absolute pressure, p – pa = gauge pressure. Thus, knowing y and  (density of liquid), we can measure the gauge pressure. Ex. The manometer shown below is used to measure the difference in water level between the two tanks. Calculate this difference for the conditions indicated Liquid (sp gravity = 0.9) 40cm Water Sol. pa + h1 g – 401g + 40g = pa + h2 g h2 g – h1 g = 40 g – 40 1g as 1 = 0.9 (h2 – h1) g = 40g – 36g h2 – h1 = 4 cm Water WATER BAROMETER 1 .013  1 0 5 g The height of the water column in the tube will be 10.3 m. Such a long tube cannot be managed easily, thus water barometer is not feasible. Let us suppose water is used in the barometer instead of mercury. hg = 1.013 × 105 or h = Ex. In a given U-tube (open at one-end) find out relation between p and pa. Sol. y=25cm d2 Given d2 = 2 × 13.6 gm/cm d1 = 13.6 gm/cm 3 3 p x=26cm Pressure in a liquid at same level is same i.e. at A – A–, d1 p a  d 2 yg  xd1 g  p In C.G.S. pa + 13.6 × 2 × 25 × g + 13.6 × 26 × g = p pa + 13.6 × g [50 + 26] = p Ex. Sol. 2pa = p as pa = 13.6 × g × 76 Find out pressure at points A and B. Also find angle ‘’. Pressure at A – PA = Patm – 1 g sin  Pressure at B PB = Patm + 2 gh  But PB is also equal to PB = PA + 3 g sin  Hence - Patm + 2 gh = PA + 3 g sin  Patm + 2 gh = Patm – 1 g sin  + 3 g sin sin   Ex. 2 h (3  1 )l . In the given figure, the container slides down with acceleration ‘a’ on an incline of angle ‘’. Liquid is stationary with respect to container. Find out (i) Angle made by surface of liquid with horizontal plane. (ii) Angle if a = g sin . Sol. Consider a fluid particle on surface. The forces acting on it are shown in figure. = Resultant force acting on liquid surface, will always normal to it tan  = Thus angle of liquid surface with the horizontal is equal to  = tan–1 ma cos  a cos  = mg  ma sin  (g  a sin ) a cos  (g  a sin ) MECHANICAL PROPERTIES OF MATTER  a cos   (ii) If a = g sin , then  = tan–1  g  g sin 2   = tan–1 g sin  cos  g cos2  = tan–1 (tan )  =  Ex. An L shaped glass tube is kept inside a bus that is moving with constant acceleration. During the motion, the level of the liquid in the left arm is at 12 cm whereas in the right arm, it is at 8 cm when the orientation of the tube is as shown. Assuming that the diameter of the tube is much smaller than levels of the liquid and neglecting effect of surface tension, acceleration of the bus find the (g = 10 m/s2). Sol. tan   h 2  h1 a 4 cm 2   g h 2 tan 45   h 1 tan 4 5  20 cm  a = 2 m/s FORCE ON SIDE WALL OF VESSEL Force on the side wall of the vessel can not be directly determined as a different depths pressures are different. To find this we consider a strip of width dx at a depth x from the surface of the liquid as shown in figure, and on this strip the force due to the liquid is given as : dF = xg × bdx x dx h dF b a This force is acting in the direction normal to the side wall. Net force can be evaluated by integrating equation h F   dF   xgbdx F 0 gbh 2 2 Average pressure on side wall The absolute pressure on the side wall cannot be evaluated because at different depths on this wall pressure is different. The average pressure on the wall can be given as : P av  F 1 gbh 2 1   gh bh 2 bh 2 Equation shows that the average pressure on side vertical wall is half of the net pressure at the bottom of the vessel. TORQUE ON THE SIDE WALL DUE TO FLUID PRESSURE As shown in figure, due to the force dF, the side wall experiences a torque about the bottom edge of the side which is given as d  dF   h  x   xgb dx  h  x  h 2 This net force is    d   gb  hx  x  dx 0   h3 h3  1  gb     gbh 3 3  6 2 Ex. Water and liquid is filled up behind a square wall of side . Find out (a) Pressures at A, B and C (b) Forces in part AB and BC (c) Total force and point of application of force. (Neglect atmosphere pressure in every calculation) Sol. (a) As there is no liquid above ‘A’ So pressure at A, pA = 0 Pressure at B, pB = gh1 Pressure at C, pC = gh1 + 2gh2 (b) Force at A = 0 Take a strip of width ‘dx’ at a depth ‘x’ in part AB. Pressure is equal to gx. Force on strip = pressure × area dF = gx dx h1 gxl h 12 1 00 0  1 0  10  5  5   1 .25  10 6 N 2 2 0 In part BC for force take a elementary strip of width dx in portion BC. Pressure is equal to =gh1 + 2g(x – h1) Total force upto B :F  gxl dx  Force on elementary strip = pressure × area dF = [gh1 + 2g(x – h1)]  dx Total force on part BC F  l h1 [gh 1  2 g (x  h 1 )]l dx A l   x2  =  gh 1 x  2 g  2  h 1 x   l    h1  h1 B  l 2  h 12   h 1 l  h 12  = gh1h2  + 2g  2   = gh2 [h1 + h2] = gh22 = 1000 × 10 × 5 × 10 × 10 = 5 × 106 N (c) Total force = 5 × 106 + 1.25 × 106 = 6.25 × 106 N Taking torque about A Total torque of force in AB =  dF  x  h1 dx C 2 g l = gh1h2 + [2 + h12 – 2h1] = gh1h2 +g ( – h1)2 2 h1  gxl dx.x 0  gl x 3  gl h 13 1 00 0  10  10  1 25 1 .2 5  10 7    Nm   3 3 3  3 0 x MECHANICAL PROPERTIES OF MATTER Total torque of force in BC =  dF  x On solving we get = gh1h2[h1 + h2 2 h2 ] + gh22[h1 + ] 2 3 = 1000 × 10 × 5 × 5 × 10 [5 + 2.5] + 1000 × 10 × 25 × 10 [5 + = 2.5 × 7.5 × 106 + 10 ] 3 118.75 6 2 .5 × 106 = × 106 3 3 1 1 .8 7 5  1 0 7 1 .2 5  1 0 7 1 3 .1 2 5  1 0 7 + = 3 3 3 Total torque = total force × distance of point of application of force from top = F.xp Total torque = 6.25 × 106 xp = 1 3 .1 2 5  1 0 7 3 xp = 7m PASCAL's LAW If the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude. Pascal's law is stated in following ways – (i) The pressure in a fluid at rest is same at all the points if gravity is ignored. (ii) A liquid exerts equal pressures in all directions. (iii) If the pressure in an enclosed fluid is changed at a particular point, the change is transmitted to every point of the fluid and to the walls of the container without being diminished in magnitude. Applications of pascal's law hydraulic jacks, lifts, presses, brakes, etc For the hydraulic lift Pressure applied = F1 A1 F1 F2  Pressure transmitted = A2  Upward force on A2 is F2 = Ex. F2 F1  = A2 A1 P A1 Q A2 F1 × A2 = × F1 A1 A1 A vertical U–tube of uniform cross–section contains mercury in both arms. A glycerine (relative density = 1.3) column of length 10 cm is introduced into one of the arms. Oil of density 800 kg m–3 is poured into the other arm until the upper surface of the oil and glycerine are at the same horizontal level. Find 10cm the length of the column. Density of mercury is 13.6 × 103 kgm–3 Sol. A2 C h 0.1 h Pressure at A and B must be same Pressure at A = P0 + 0.1 × (1.3 × 1000) × g Pressure at B = P0 + h × 800 × g + (0.1 – h) × 13.6 × 1000 g 0.1 × 1300 = 800 h + (0.1 – h) × 13600 h = 0.096 m = 9.6 cm A B BUOYANCY AND ARCHIMEDE'S PRINCIPLE (i) Buoyant force : If a body is partially or wholly immersed in a fluid, it experiences an upward force due to the fluid surrounding it. This phenomenon of force exerted by fluid on the body is called buoyancy and force is called buoyant force or upthrust. (ii) Archimede's principle : It states that the buoyant force on a body that is partially or totally immersed in a liquid is equal to the weight of the fluid displaced by it. Now consider a body immersed in a liquid of density . Top surface of the body experiences a downward force F1 = AP1 = A[h1g + P0] ...(i) Lower face of the body will experiences a upward force F2 = AP2 = A[h2g + P0] As h2 > h1. F1 L ...(ii) h1 h2 F2 So F2 is greater than F1 So net upward force : F = F2 – F1 = Ag[h2 – h1]  F = AgL = Vg [ V = AL] FLOATATION When a body of density () and volume (V) is completely immersed in a liquid of density (), the forces acting on the body are : (i) Weight of the body W = Mg = Vg (directed vertically downwards through C.G. of the body). (ii) Buoyant force or Upthrust Th = Vg (directed vertically upwards through C.B.). The apparent weight WApp is equal to W – Th. The following three cases are possible : Case I Density of the body () is greater than that of liquid ) In this case if>thenW > Th So the body will sink to the bottom of the liquid. WApp = W – Th = Vg – Vg = Vg (1 – /) = W (1 – /). Case II Density of the body is equal to the density of liquid (= ) In this case if =then W = Th So the body will float fully submerged in the liquid. It will be in neutral equilibrium. WApp = W – Th = 0 Case III Density of the body is lesser than that of liquid (< ) In this case if<then W < Th So the body will float partially submerged in the liquid. In this case the body will move up and the volume of liquid displaced by the body (Vin) will be less than the volume of body (V). So as to make Th equal to W WApp = W – Th = 0 The above three cases constitute the law of flotation which states that a body will float in a liquid if weight of the liquid displaced by the immersed part of the body is at least equal to the weight of the body. Rotatory – Equilibrium in Floatation : When a floating body is slightly tilted from equilibrium position, the centre of buoyancy B shifts. The vertical line passing through the new centre of buoyancy B' and initial vertical line meet at a point M called meta – centre. If the metacentre M is above the centre of gravity the couple due to forces at G (weight of body W) and at B' (upthrust) tends to bring the body back to its original position (figure) . So for rotational equilibrium of floating body the meta–centre must always be higher than the centre of gravity of the body. MECHANICAL PROPERTIES OF MATTER However, if meta–centre goes below centre of gravity, the couple due to forces at G and B' tends to topple the floating body. This is why a wooden log cannot be made to float vertical in water or a boat is likely to capsize if the sitting passengers stand on it. In these situations centre of gravity becomes higher than meta centre and so the body will topple if slightly tilted. Ex. A cubical block of wood of edge 3 cm floats in water. The lower surface of the cube just touches the free end of a vertical spring fixed at the bottom of the pot. Find the maximum weight that can be put on the block without wetting it. Density of wood = 800 kg/m3 and spring constant of the spring = 50 N/m. Take g = 10 m/s2 Sol. The specific gravity of the block = 0.8. Hence the height inside water = 3 cm × 0.8 = 2.4 cm. The height outside water = 3 cm – 2.4 = 0.6 cm. Suppose the maximum weight that can be put without wetting it is W. The block in this case is completely immersed in the water. The volume of the displaced water= volume of the block = 27 × 10–6 m3. Hence, the force of buoyancy = (27 × 10–6 m3) × 1(1000 kg/m3) × (10 m/s2) = 0.27 N. The spring is compressed by 0.6 cm and hence the upward force exerted by the spring = 50 N/m × 0.6 cm = 0.3 N. The force of buoyancy and the spring force taken together balance the weight of the block plus the weight W put on the block. The weight of the block is W= (27 × 10–6 m) × (800 kg/m3) × (10 m/s2) = 0.22 N. Thus, W = 0.27 N + 0.3 N – 0.22 N = 0.35 N. Ex. A wooden plank of length 1 m and uniform cross-section is hinged at one end to the bottom of a tank as shown in figure. Sol. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. Find the angle  that the plank makes with the vertical in the equilibrium position. (Exclude the case  = 0). The forces acting on the plank are shown in the figure. The height of water level is  = 0.5 m. The length of the plank is 1.0 m = 2. The weight of the plank acts through the centre B of the plank. We have OB = . The buoyant force F acts through the point A which is the middle point of the dipped part OC of the plank. l OC We haveOA = = . 2 cos  2 Let the mass per unit length of the plank be . Its weight mg = 2g.  l  The mass of the part OC of the plank =  .  cos   1 l 2l  The mass of water displaced = = . 0 .5 cos  cos  2 l g The buoyant force F is, therefore, F = . cos  Now, for equilibrium, the torque of mg about O should balance the torque of F about O. 1  2l    l  1 2 So, mg (OB) sin = F(OA) sin or, (2) =  or, , cos = or, cos = , or,, = 45°.    cos    2 cos   2 2 (i)  (ii)  Buoyant force act vertically upward through the centre of gravity (C.G.) of the displaced fluid. This point is called centre of buoyancy (C.B.). Thus centre of buoyancy is the point through which the force of buoyancy is supposed to act. Buoyant force or upthrust does not depend upon the characteristics of the body such as its mass, size, density, etc. But it depends upon the volume of the body inside the liquid. (iii) It depends upon the nature of the fluid as it is proportional to the density of the fluid. This is the reason that upthrust on a fully submerged body is more in sea water than in pure water (iv) It depends upon the effective acceleration. If a lift is accelerated downwards with acceleration a (a < g) then Th = Vin (g – a) If a lift is accelerated downwards with a = g then Th = Vin (g – a) = 0 If a lift is accelerated upward with acceleration a then Th = Vin (g + a)  (v) If a body is weighed in air (WA), in water (WW) and in a oil (WO), then Specific gravity of oil = WA  WO loss of weight in oil = WA  WW loss of weight in water Ex. A body weighs 160 g in air, 130 g in water and 136 g in oil. What is the specific gravity of oil? Sol. Specific gravity of oil = Ex. An iceberg is floating partially immersed in sea–water. The density of sea–water is 1.03 gm/cm3 and that of ice is 0.92 gm/cm3. What is the fraction of the total volume of the iceberg above the level of sea–water ? Sol. In case of flotation weight = upthrust i.e. loss of weight in oil 16 0 – 136 24 8 = = = = 0.8 loss of weight in water 16 0 – 1 30 30 1 0 mg = Ving Vg = Ving  fout   Vin =  V    So Vout = V – Vin = V 1     Vout   0.92  0.11  1   1    1.03  0.10 6 fout = 10.6 % V  1.03    Ex. A rubber ball of mass 10 gm and volume 15 cm3 is dipped in water to a depth of 10m. Assuming density of water uniform throughout the depth if it is released from rest. Find (take g = 980 cm/s2) (a) the acceleration of the ball, and (b) the time taken by it to reach the surface. Sol. The maximum buoyant force on the ball is FB = Vw g  15 × 1× 980 dyne = 14700 dyne. The weight of the ball is mg = 10 × g = 10 × 980 = 9800 dyne The net upward force, F = (15 × 980 – 10 × 980) dyne = 5 × 980 dyne = 4900 dyne (a) Therefore, acceleration of the ball upward a  (b) Time taken by it reach the surface is t 2h a F 5  980 = 490 cm/s2 = 4.9 m/s2  m 10 2 10 4.9