Fluid Pressure & Buoyancy Fluid Pressure & Buoyancy A fluid can spend its energy in two main accounts: pressure and motion. When flow speeds up, the spending shifts to kinetic energy and the sideways push on walls drops. That trade-off is the heart of Bernoulli’s idea and it explains everyday sights like a fast jet of water that pinches inward and a Venturi tube where manometer levels fall at the throat. But in many situations fluids are at rest or close to rest. There, pressure is the amount of normal push per unit area, and it grows with depth because deeper layers must support the weight of layers above. This simple gradient creates buoyancy: the bottom of a submerged object faces higher pressure than its top, giving a net upward force equal to the weight of the displaced fluid. Whether a body sinks, floats, or is neutrally buoyant is then decided by densities and displaced volume, not by the object’s shape alone. In clinics and labs you meet these ideas as blood pressure readings, IV drips, sphygmomanometers and manometers, syringes and hydraulic lifts, and the reading of barometers in weather reports. To master questions, think in three connected steps. First, identify the state - static or flowing - and choose the right model. For static fluids, use P = P 0 + g h and Archimedes’ principle. For enclosed fluids, use Pascal’s law to transmit pressure to remote pistons. For modest steady flows, continuity A v = constant and Bernoulli P + 1 2 v 2 + g h = constant predict which region has higher or lower pressure. Second, track pressures as absolute or gauge, and keep a clear sign for level differences in manometers. Third, check limits: if an object is fully submerged in an incompressible fluid, the buoyant force becomes independent of depth; if a tube is wide or the flow is viscous and unsteady, Bernoulli and simple continuity can fail. Build intuition first - where is the push stronger and why - then let the equations do the precise accounting. remember Crowded hallway model: in a wide hallway students move slowly and bump sideways a lot (high pressure). At a narrow doorway they sprint in line (high speed, low sideways bumping, low pressure). Core ideas that everything else leans on Pressure is normal force per area. Absolute pressure counts everything including atmospheric pressure above the fluid; gauge pressure measures only the excess above atmosphere. Pascal’s law says any externally applied pressure change to a confined incompressible fluid is passed undiminished to every point and to the walls. Manometers convert pressure differences into height differences of a heavy fluid column. Buoyancy is the net upward force arising from the pressure gradient in a fluid. Archimedes’ principle packages it neatly: the buoyant force equals the weight of displaced fluid. For floating bodies, the fraction submerged equals the ratio of body density to fluid density. These basics also support simple flow results: conservation of mass through a pipe gives A 1 v 1 = A 2 v 2 , and the energy balance along a streamline gives Bernoulli’s equation. Even if a problem looks dynamic, most NEET questions reduce to these few statements once you identify what is known and what is sought. Fluid pressure Normal force per unit area exerted by a fluid on a surface. P = F /A . Absolute pressure Total pressure measured relative to perfect vacuum. P abs = P atm + P gauge . Excess pressure above atmospheric pressure. In static fluids: P gauge = g h . Gauge pressure Mass per unit volume, = m/V . Density Relative density (specific gravity) Ratio of a substance’s density to water’s density at 4 C . Pascal’s law A pressure change applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and the walls. Device using Pascal’s law: equal pressure on unequal areas gives force multiplication. Hydraulic lift Buoyant force Resultant upward force on an immersed body due to pressure variation with depth. Archimedes’ principle Buoyant force equals the weight of the fluid displaced by the body. Law of flotation A floating body displaces its own weight of fluid; fraction submerged equals body / fluid . Manometer U-tube (or differential) device that measures pressure difference via height difference of a heavy fluid column. Barometer Instrument using a mercury column to measure atmospheric pressure. Point about which a tilted floating body oscillates; higher metacentre relative to centre of gravity indicates stable equilibrium. Metacentre (qualitative) Definition of pressure Normal force per unit area. SI unit: Pa = N m -2 . Atmospheric pressure conversions Use 1 mm Hg 133.3 Pa for quick conversions. Linear increase with depth for incompressible fluids. Pressure with depth in a static fluid This relationship shows that pressure increases linearly with depth in a stationary fluid, a key factor in understanding buoyancy. Equal pressure on both pistons: P = F 1/A 1 = F 2/A 2 . Pascal transmission in hydraulic lift Pressure applied to an enclosed, incompressible fluid is transmitted undiminished to every point within the fluid. Archimedes’ principle Use displaced volume V of fluid, not necessarily the full object volume when floating. The formula P = P 0 + g h assumes constant density with depth. For liquids over small to moderate depths this is excellent. For gases, density varies with height, so a barometric formula is needed and the change is not strictly linear. Pascal’s law needs the fluid to be enclosed and essentially incompressible. Real hydraulic lifts use oil, seals, and one-way valves; while friction reduces efficiency, the ideal pressure multiplication remains a good first estimate. In buoyancy, always identify the displaced volume: for fully submerged bodies it is the whole volume; for floating bodies it is the submerged part only. Pressure increases with depth h in static fluid. Consider a horizontal pair of small opposite faces of area A separated by height difference h. Net upward force on that slab due to pressure difference. Volume of the slab between the two faces. This net force equals the buoyant force on that displaced volume. Fluid is incompressible and at rest Uniform gravitational field Object displaces volume V of the fluid F b = V g Archimedes’ Principle: F b = V g Mass flow rate through any cross-section. Conservation of mass between two sections. Incompressibility implies constant density. Cancel density to obtain continuity. Continuity for incompressible flow: A 1 v 1 = A 2 v 2 Steady flow Incompressible fluid No sources or sinks A 1 v 1 = A 2 v 2 Steady, non-viscous flow Incompressible fluid Along a streamline P + 1 2 v 2 + g h = constant Bernoulli’s equation: P + 1 2 v 2 + g h = constant Work by pressure forces on a fluid element of volume V. Work by gravity as element moves from 1 to 2. Change in kinetic energy of the element. Work-energy theorem for the fluid element. Divide by V and rearrange. Bernoulli’s form along a streamline. h = 2 g r Static equilibrium Cylindrical narrow tube Constant contact angle Capillary rise: h = 2 g r Vertical component of surface tension along the contact line. Weight of liquid column of height h. Force balance at equilibrium. Solve for h. Fractional change in volume. Pressure change produces uniform compression. Define bulk modulus with a minus sign so B is positive. Final relation. B = -V , P V Small uniform compression Elastic response Bulk modulus: B = -V , P V Manometers translate pressure differences into measurable height differences of a dense fluid like mercury. Connect one limb to the unknown pressure and the other to a reference (often atmosphere). If the limb on the unknown side is lower by h , then for a simple open-end manometer with mercury the gauge pressure is P = Hg g h . For differential manometers, write hydrostatic balance moving along a path through connected fluids: pressures at the same horizontal level in the same fluid are equal. Each step adds or subtracts g h depending on moving down or up. Consistent sign handling of level differences is the most common place students lose marks, so always sketch and label heights clearly. Buoyant force equals weight of displaced fluid: F b = V g . Use submerged volume only; independent of object density for a given displaced volume. For steady incompressible flow, area-speed product is constant: A 1 v 1 = A 2 v 2 . Narrower section implies higher speed. Along a streamline in ideal flow, P + 1 2 v 2 + g h is constant. Faster flow often means lower static pressure if height remains similar. Resistance to uniform compression: B = -V , P/ V . Large B means small compressibility. Static rise or fall in a narrow tube: h = 2 g r . Rise for wetting liquids, fall for non-wetting. Floating and stability: a body floats if its average density is less than the fluid’s. At equilibrium the weight W equals F b . If you gently push a floating body down, extra volume submerges and buoyant force rises. If this restoring effect acts through a point above the centre of gravity, the couple returns the body upright. Wide-bottomed boats keep the metacentre high; loading must keep the centre of gravity low. In NEET problems, stability criteria are rarely computed numerically, but you should be able to reason which configuration is more stable by comparing where the metacentre lies qualitatively. Find gauge and absolute pressure at a depth of 5 m in water. Take = 1000 kg/m 3 , g = 9.8 m/s 2 , P atm = 1.013 10 5 Pa . Use P gauge = g h and P abs = P atm + P gauge . h = 5 m rho = 1000 kg/ m 3 g = 9.8 m/ s 2 Patm = 1.013e5 Pa easy Gauge pressure and absolute pressure at depth Pa V = 2.0e-3 m 3 rho fluid = 800 kg/m 3 g = 9.8 m/ s 2 easy Apply Archimedes: F b = V g . A solid object of volume 2.0 10 -3 m 3 is fully submerged in kerosene of density 800 kg/m 3 . Find the buoyant force. Buoyant force Minimum input force on the small piston (ideal) In ideal lift, F 1/A 1 = F 2/A 2 and F 2 = mg . A hydraulic lift has a small piston of area A 1 = 0.005 m 2 and a large piston of area A 2 = 0.50 m 2 . What least input force is needed to lift a 1200 kg car? A1 = 0.005 m 2 A2 = 0.50 m 2 m car = 1200 kg g = 9.8 m/ s 2 medium Absolute pressure of the gas Pa An open-end mercury manometer measures the pressure of a gas. The mercury level on the gas side is lower by 20 cm than the open atmospheric limb. Find the gas absolute pressure. For open-end manometer, P gas = P atm + Hg g h if the gas side is lower. hard rho Hg = 13600 kg/m 3 h = 0.20 m g = 9.8 m/ s 2 Patm = 1.013e5 Pa Fraction submerged = 0.75 rho water = 1000 kg/m 3 medium A wooden block floats in water with 75% of its volume submerged. Find the density of the wood. For floating: body / fluid = fraction submerged. kg/ m 3 Density of the wood Venturi meter glass pipe with labeled manometer tubes showing pressure variation. Venturi tube with manometer taps: highest fluid level at wide inlet and outlet (higher pressure), lowest at the narrow throat (lower pressure). Blue streamlines bunch at the throat. Aerofoil illustration: faster red streamlines over the curved top indicate lower pressure; denser blue particles below indicate higher pressure, creating lift. Airplane wing diagram showing high and low pressure regions and net lift arrow. Garden hose nozzle: as the jet speed increases at the narrow tip, the static pressure drops compared to the thick hose behind it. Close-up of water jet from a narrow nozzle with arrows indicating pressure and kinetic energy changes. custom Pressure P Pa Depth h control rho g P0 control P0 Surface pressure P0 + rho g h Linear growth with depth Pressure increases uniformly with depth in a static liquid. Straight line: P = P0 + rho g h for an incompressible liquid. Reference constants for pressure and buoyancy questions. Water density at 4 °C 1000 kg m -3 Use for most NEET numericals unless stated otherwise Mercury density ρ Hg 13600 kg m -3 Preferred manometer fluid due to high density Atmospheric pressure P atm 1.013× 10 5 Pa ≈760 mm Hg at sea level Acceleration due to gravity 9.8 m s -2 Use 9.8 unless question specifies 10 Quantity Symbol Typical value Notes Absolute Measured from vacuum P abs = P atm + P gauge Barometer readings, gas laws Gauge Above atmospheric P gauge = ρ g h (static column) Tyre pressure, manometers Vacuum (partial) Below atmospheric P vac = P atm - P abs Suction systems Type of pressure Definition Formula When used Pressure at the same horizontal level in a connected region of the same static fluid is equal. Use this to step through manometer limbs. tip neet-alert In Archimedes’ principle, use displaced fluid volume, not the object’s total volume when it is only partially submerged. Many students wrongly plug V object and overestimate F b . neet-alert Bernoulli applies along a streamline for steady, non-viscous flow. Do not apply it across a pump, turbine, or where viscosity and turbulence dominate. tip Hydraulic lift advantage trades force for distance. If force is multiplied by 100, the large piston must move 100 times less distance for the same volume transfer. Float test: Light in density, Large in fraction above water. If body < fluid , it Floats. If equal, it is neutrally buoyant; if greater, it sinks. For a given displaced volume in a given fluid, F b depends on the fluid’s density, not the object’s. Object density only decides how much volume must be submerged to balance weight. Buoyant force depends on the density of the object. A fully submerged body experiences greater buoyant force if taken deeper. In an incompressible liquid, once fully submerged, the displaced volume is fixed and the buoyant force stays the same with depth. In a narrower pipe water slows down because there is less space. For steady incompressible flow, continuity requires A v = constant . Narrower area implies higher speed, not lower. Worked method for manometer questions: 1) Draw the U-tube and mark fluid interfaces. 2) Choose a reference horizontal line through the lower interface in the heavy fluid. 3) Write pressure equality at the reference in the same fluid. 4) March along each limb to the measurement points, adding g h when moving downward and subtracting when moving upward. 5) Solve for the unknown pressure. If two immiscible fluids fill the limbs, include both densities correctly. Keep track of units - convert cm to m before multiplication by g . Applications you can recognize quickly: 1) Intravenous drip height determines hydrostatic head feeding the vein - the higher the bag, the larger g h and the faster the flow if resistance is small. 2) Car jacks and dentists’ chairs rely on Pascal’s law; if leaks are negligible, small hand forces balance large loads. 3) Submarines and divers sense pressure increasing roughly 1 atm every 10 m in seawater. 4) Hot-air balloon lift arises from gaseous buoyancy; warm air inside reduces density, so displaced cooler air weighs more and provides net upward buoyant force. 5) Venturi meters and Pitot tubes infer speed from pressure differences via Bernoulli; for NEET, the algebra often reduces to equating heads and solving for v . Checklist before writing equations Static or flowing? If static, use hydrostatics and Archimedes. If flowing, check continuity and Bernoulli conditions. Absolute or gauge? Convert before adding or comparing. Known densities and heights? Keep cm to m conversion in mind. Is the fluid enclosed (Pascal) or open to atmosphere (barometer/manometer)? For floating bodies, identify the submerged fraction first. Edge cases and limits to remember: At h 0 , P P 0 at the free surface. As h in a real ocean, compressibility of water and variation of g and make P = P 0 + g h only approximate. For gases, density varies with height, so barometric formula replaces the linear rule. For very narrow capillaries the meniscus curvature is spherical only if the tube is sufficiently thin; for thick tubes the formula for h breaks down. Continuity and Bernoulli ignore viscosity; if Reynolds number is high and flow turns turbulent, the simple energy sum is not conserved due to dissipation. Law of flotation in three crisp lines At equilibrium: W = F b . So body V g = fluid V sub g . Hence V sub /V = body / fluid . Dimensional checks help avoid mistakes. Pressure has dimension [M L -1 T -2 ] . Buoyant force is a force, so also [M L T -2 ] . In F b = V g , has [M L -3 ] , V has [L 3 ] , and g has [L T -2 ] , giving the correct result. In the continuity equation A v , the dimension is [L 3 T -1 ] which is volume flow rate. Such quick checks catch errors like using total area instead of cross-sectional area or mixing up v and Q . Worked thought experiment: two identical cubes made of different materials are fully submerged in water, one aluminum and one plastic, both attached to spring scales. The heavier aluminum cube has a larger weight reading in air but experiences exactly the same buoyant force as the plastic one if their volumes are equal, because both displace the same amount of water. The spring scales in water read different apparent weights W app = W - F b , differing only due to their different real weights. This reinforces that buoyant force depends on displaced fluid, not the material itself. Barometer logic: a long tube filled with mercury and inverted into a mercury reservoir leaves a vacuum at the top. The column height adjusts so that P atm = Hg g h baro . Daily weather changes shift h baro by a few millimeters. In contrast, a water barometer would need a column roughly 10.3 m tall, which is why mercury is chosen despite its toxicity. For exam calculations, treat mercury density as 13.6 10 3 kg/m 3 and 1 mm Hg = 133.3 Pa . Pascal’s law and syringes: A nurse pressing on a syringe plunger applies a force over a small area, creating a pressure rise that pushes liquid through the needle. If the needle is narrow, viscous losses dominate and a larger force is needed to maintain flow, but the initial pressure transmission still follows Pascal. In hydraulic brakes, small pedal movements create large braking forces at the wheels via large-area pistons, but safety valves limit pressure to avoid lock or failure. Qualitative buoyancy in gases: Helium balloons float because helium’s density is around 0.18 kg/m 3 while air near the surface is about 1.2 kg/m 3 . The net upward force is F b - W = ( air - He ) V g - W skin . Heating air in a hot-air balloon lowers its density, increasing lift until drag and fuel limits cap the rise. NEET rarely demands exact gas-buoyancy calculations, but understanding the density contrast explains the behavior succinctly. Common unit traps: 1) Heights given in cm must be converted to m before using with g . 2) Densities may appear as relative densities without kg/m 3 . Multiply by 1000 for water-based liquids. 3) Atmospheric pressure used as 1.013 10 5 Pa ; do not round to 10 5 Pa if a question mixes mm Hg and Pa, because the small difference can affect a digit in final answers. Distance in the nth second confusion has an analog here: do not mistake total depth h for the height difference between manometer limbs. Use the difference in levels, not the absolute height from the bench. neet-alert Worked visual link to the Venturi image: The manometer level is highest at the wide inlet and outlet because static pressure is higher there. At the throat, the cross-section is smallest, continuity forces speed to rise, and Bernoulli says static pressure must drop to keep total pressure constant along a streamline at similar height. The difference in manometer readings is therefore directly connected to the difference in kinetic heads 1 2 v 2 . Stability of floating bodies in words: When a boat tilts, the shape of the submerged volume changes and the centre of buoyancy shifts. If the new line of action of buoyant force passes to the same side of the centre of gravity as the tilt, it creates a restoring couple. The point where the buoyant force line crosses the original vertical when the tilt is infinitesimal is the metacentre. A higher metacentre than the centre of gravity means stability. Increasing width or lowering mass high up improves this. Pressure on curved surfaces: For hemispherical or curved gates, the resultant fluid force does not act at the geometric center. Hydrostatic pressure increases with depth, so the line of action lies below the centroid of the area. NEET usually avoids full calculus here, but be aware that engineers compute the center of pressure using the second moment of area. For flat vertical surfaces, the center of pressure is at a depth h + I G/(A h ) where h is the depth of centroid, a detail beyond most NEET questions. Quick proportionalities: F b for a fixed displaced volume, and F b V for a fixed fluid. Hydraulic advantage F 2/F 1 = A 2/A 1 . In capillarity, h 1/r , so halving capillary radius doubles rise if other parameters are unchanged. In Bernoulli flows at the same level, an increase in v by factor k lowers P by 1 2 (k 2-1) v 1 2 compared to the original state if total head is fixed. Practice blueprint: Start by writing what is known - densities, areas, heights. Mark absolute or gauge. Sketch every problem: sections of pipes, pistons, U-tubes, or bodies in water. Identify the governing equation and list the assumptions. Solve symbolically first for clarity, then substitute numbers late to reduce errors. Finally, check reasonableness: does a lighter object float more, does a smaller capillary give larger h , do manometer levels differ in the correct direction? Pressure (P) Normal force per area. SI unit: Pa. Absolute vs Gauge Absolute counts from vacuum; gauge adds only the excess over atmosphere. Pascal’s law Applied pressure change transmits undiminished in enclosed incompressible fluids. Uses Pascal’s law to multiply force: F 2/F 1 = A 2/A 1 . Hydraulic lift Buoyant force equals weight of displaced fluid, F b = V g . Archimedes’ principle V sub /V = body / fluid for floats. Law of flotation Measures pressure difference via height difference of a column fluid. Manometer Continuity For steady incompressible flow, A v is constant. Total head constant along a streamline for ideal flow. Bernoulli Measure of compressibility: B = -V , P/ V . Bulk modulus Capillary rise Height due to surface tension: h = 2 /( g r) . End-of-lesson recap