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4 VISCOSITY, 5 REYNOLDS NUMBER, 6 SURFACE TENSION, 7 SURFACE ENERGY, 8 CAPILLARY RISE, Let's understand Density.., Density: The density of material shows the denseness of that, material in a specific given area. The density of a material is, defined as the ratio of its mass per unit volume. Kisi bhi, object ka mass upon volume density hota hai. Is chapter ko, samjhne ke liye aapko water aur mercury ka density yaad rakhna, hoga., Density, , =, , m, v, , S.I unit: Kg / m, , 3, , Relative Density = (Density of the substance)/(Density of, reference substance), Example:Relative density of mercury with respect to water, = (Density of mercury)/(Density of water)

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3, , 13.6 × 10 Kg / m, =, , 3, 3, , Kg / m ), 1000, , = 13.6, , Density of water, , = 1000Kg / m, , Density of mercury, , 3, , 3, , = 13.6 × 10 Kg / m, , 3, , Average Density: The mean or average density is the total, mass of a region divided by the total volume of that region., = (Total mass)/(Total volume), Density scalar quantity hota hai. Iska sirf magnitude hota hai koi, direction nahi hota., Specific Density: The ratio of the density of a substance to the, density of a standard substance under specified conditions., = (density of material)/ (density of water), Density of water at 4 ∘ C, Density of air, , = 1.29kgm, , Density of Hydrogen, , 3, , = 1.00 × 10 kgm, , −3, , −3, , = 9 × 10, , −2, , kg / m, , 3, , Thrust by Fluid, Force applied by fluid perpendicular on the surface of the, container., , PRESSURE, PRESSURE, Pressure effect of force hota hai. Pressure is defined as the, force divided by the area perpendicular to the force over

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Note: A large force is experienced in a larger cross-section if a, smaller force 4cross is applied in smaller by the relation section., A column of height h of a liquid of density p exerts a pressure P, given P, , = hρg, , Niche diye figure me aap dekhenge jab pressure enclosed fluid, pr lagta hai to pressure sare directions me transmit ho jata hai.

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Hydraulic Press, Jab hum section 1 me force apply krte hai to section 2 me force, produce hota hai. By Pascal’s Law,, , F, A, , = constant, , F ∝ A, , If A < A , then F < F, 1, , 2, , 1, , 2, , Archimedes Principle and Buoyancy, The upward buoyant force that is exerted on a body, immersed in a fluid, whether partially or fully submerged, is, equal to the weight of the fluid that the body displaces and, acts in the upward direction at the center of mass of the, displaced fluid.

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Apparent weight of the body of density, , ρ, , when immersed in a, , liquid of density σ., Apparent weight = Actual weight - Upthrust, = ρV g − σV g, , Yaha, V volume of solid hai., , Floatation, Floatation ke teen cases hote hai, jab body partially immerse, hota hai, jab body fully immerse hota hai aur jab body sink krta, hai., , The principle of floatation states that when an object floats, on a liquid the buoyant force that acts on the object is equal, to the weight of the object. Floatation me niche diye points, ko aap yaad rakh sakte hai:, The displaced volume of fluid is equal to the volume of the, object which is immersed in the fluid., If the weight of the object is greater than the upthrust, then the, object will sink in the fluid., If the weight of the object is equal to the upthrust, then the, object is balanced making the object float.

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BERNOULLI’S, BERNOULLI’S THEOREM, THEOREM, Bernoulli’s principle Daniel Bernoulli ne formulate kiya tha. It, states that as the speed of a moving fluid increases (liquid or, gas), the pressure within the fluid decreases. Leonhard Euler, ne Bernoulli’s equation ko derive kiya tha year 1752 me., Bernoulli’s principle states that, The total mechanical energy of the moving fluid comprising, the gravitational potential energy of elevation, the energy, associated with the fluid pressure and the kinetic energy of, the fluid motion, remains constant., Bernoulli’s principle principle of conservation of energy se, derive ho sakta hai., , Bernoulli’s Principle Formula, Bernoulli’s equation formula pressure, kinetic energy, aur ek, container me fluid ka gravitational potential energy ke beech ka, relation batata hai., The formula for Bernoulli’s principle is given as:, 1, p =, , ρv, 2, , p, v, , 2, , + ρgh, , =constant, , = pressure exerted by the fluid, , = velocity of the fluid, , ρ, , = density of the fluid, , h, , = height of container, , Bernoulli’s Equation Derivation, Ek pipe consider krte hai jiska varying diameter aur height ho, aur jisse incompressible fluid flow kr raha ho. Areas of cross-

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1, dK =, 2, , m2 v, , 2, 2, , 1, −, 2, , m1 v, , 1, 2, , 1, =, , ρdV (v, 2, , 2, 2, , 1, , − v ), 2, , The change in potential energy is given as:, dU = mgy2 – mgy1 = ρdVg ( y, , 2, , – y1 ), , Therefore, the energy equation is given as:, dW = dK + dU, 1, (p1 – p2 )dV, , =, , ρdV (v, 2, , 1, (p1 – p2 ) =, , ρ(v, 2, , 2, 2, , 2, 2, , 1, , − v ) + ρdVg ( y, 2, , 2, , – y1 ), , 1, , − v ) + ρg(y2 – y1 ), 2, , Rearranging the above equation, we get;, 1, p1 +, , ρv, 2, , 1, 2, , 1, + ρgy1 = p2 +, , This is Bernoulli’s equation., , ρv, 2, , 2, 2, , + ρgy2, , POINT 1, Hydrstatics: The science of the mechanical properties of fluids is, called Hydrostatics., Bernoulli’s Principle: It states that the total energy of the water, always remains constant, therefore when the water flow in a system, increases, the pressure must decrease., Fluid Pressure: All liquids and gases are fluids. The force exerted, normally at a unit area of the surface of a fluid is called fluid, pressure., Critical Velocity: The critical velocity is that velocity of liquid flow,, upto which its flow is streamline and above which its flow becomes

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turbulent., Surface Tension: It is the property of the liquid by virtue of which, the free surface of liquid at rest tends to have minimum area and as, such it behaves as a stretched elastic membrane., Surface Energy: Energy possessed by the surface of the liquid is, called surface energy., , Example 1, What is the pressure on a swimmer 10m below the surface of lake?, g = 10ms, , −2, , , atmospheric pressure = 1.01 × 10, , 5, , Pa, , Watch Video, , Answer:, Here, h = 10m, , and ρ, , = 1000kg m, , −3, , . Take g, , = 10ms, , −2, , P = Pa + ρgh, 5, , = 1.01 × 10 P a + 1000kgm, , −3, , × 10ms, , −2, , × 10m, , 5, , = 2.01 × 10 P a, ≈ 2atm, , This is a 100% increase in pressure from surface level. At a depth of 1, km, the increase in pressure is 100 atmm! submarines are designed to, withstand such enormous pressures., , Example 2

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At a depth of 1000 m in an ocean (a) what is the absolute pressure?, (b) what is the gauge pressure? (c ) Find the force acting on the, window of area, , 20cm × 20cm, , of a submarine at this depth, the, , interior of which is maintained at sea-level atmospheric pressure., 3, , The density of sea water is, , 1.03 × 10 kgm, , Atmospheric pressure = 1.01 × 10, , 5, , Pa, , −3, , ,, , g = 10ms, , −2, , ., , ., , Watch Video, , Answer:, Here, , h = 100m, , 3, , and, , ρ = 1.03 × 10 kgm, , −3, , P = Pa + ρgh, , ltb rgt, , 5, , = 1.01 × 10 P a, 3, , + 1.03 × 10 kgm, , −3, , × 10ms, , −2, , × 1000m, , 5, , = 104.01 × 10 P a, ≈ 104atm, , (b). Gauge pressure is P, 3, , Pg = 1.03 × 10 kgm, , − P = ρgh = P, , −3, , × 10ms, , 2, , × 1000m, , 5, , = 103 × 10 P a, ≈ 103atm, , (c). The pressure outside the submarine is, pressure inside it is, is gauge pressure, A = 0.04m, , 2, , Pa, , P = Pa + ρgh, , and the, , . Hence, the net pressure acting on the window, , Pg = ρgh, , . Since the area of the window is, , , the force acting on it is, 5, , F = Pg A = 103 × 10 P a × 0.04m, , 2, , 5, , = 4.12 × 10 N, , Example 3, Two syringes of different cross-sections (without needles) filled, with water are connected with a tightly fitted rubber tube filled, with water. Diameters of the smaller piston and larger piston are, 1.0cm and 3.0 cm respectively.

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(a) Find the force on the larger piston when a force of 10N is, applied to the smaller piston., (b) The smaller piston is paused in through 6.0 cm, much does the, larger piston move out?, Watch Video, , Answer:, (a). Since pressure is transmitted undiminished throughout the fluid,, F2 =, , A2, A1, , π(3 / 2 × 10, , −2, , 2, , m), , F1 =, π(1 / 2 × 10, , −2, , 2, , × 10N, , m), , = 90N, , (b). Water is considered to be perfectly incompressible. Volume, covered by the movement of smaller piston inwards is equal to volume, moved outwards due to the larger piston, L 1 A1 = L 2 A2, , L2 =, , A1, A2, , π(1 / 2 × 10, , −2, , 2, , m), , L1 =, , ≅ 0.67 × 10, , π(3 / 2 × 10, −2, , −2, , 2, , × 6 × 10, , −2, , m, , m), , m = 0.67cm, , Note, atmospheric pressure is common to both pistons and has been, ignored., , PRINCIPLE, PRINCIPLE OF, OF CONTINUITY, CONTINUITY, Continuity equation kuchh quantities jaise ki fluid aur gas ke, transport ko describe krta hai. Ye equation explain krta hai ki

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Assuming A2< We get, v, , 2, , = √2gh, , Hence, the velocity of efflux is √2gh, , Bernoulli’s Principle Use, Bernoulli’s principle unsteady potential flow ko study krne me, use krte hai. Unsteady potential flow ko theory of ocean surface, waves aur acoustics me use kiya jata hai. It is also used for, approximation of parameters like pressure and speed of the, fluid., The other applications of Bernoulli’s principle are:, Venturi meter: It is a device that is based on Bernoulli’s, theorem and is used for measuring the rate of flow of liquid, through the pipes. Using Bernoulli’s theorem, Venturi meter, formula is given as:, 2hg, V, , = a1 a2 √, a, , 2, 1, , − a, , 2, 2

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Note: Working of an aeroplane: The shape of the wings is such, that the air passes at a higher speed over the upper surface, than the lower surface. The difference in airspeed is calculated, using Bernoulli’s principle to create a pressure difference., When we are standing at a railway station and a train comes we, tend to fall towards the train. This can be explained using, Bernoulli’s principle as the train goes past, the velocity of air, between the train and us increases. Hence, from the equation,, we can say that the pressure decreases so the pressure from, behind pushes us towards the train. This is based on Bernoulli’s, effect., , Relation between Conservation, Energy and Bernoulli’s Equation, , of, , Conservation of energy fluid ke flow pr apply kiya jata hai, Bernoulli’s equation ko produce krne ke liye. The net work, done is the result of a change in fluid’s kinetic energy and, gravitational potential energy. Bernoulli’s equation can be, modified depending on the form of energy that is involved., , Bernoulli’s Equation at Constant Depth, When the fluid moves at a constant depth that is when h1 = h2,, then Bernoulli’s equation is given as:, P1 +, , 1, 2, , ρv, , 2, 1, , = P2 +, , 1, 2, , ρv, , 2, 2, , Bernoulli’s Equation for Static Fluids, When the fluid is static, then v1 = v2 = 0, then Bernoulli’s, equation is given as:

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Example 4, A fully loaded Boeing aircraft has a mass of, , 5, , 3.3 × 10 kg, , . Its total, , 2, , wing area is 500m . It is in level flight with a speed of 960km / h., (a) Estimate the pressure difference between the lower and upper, surfaces of the wings, (b) Estimate the fractional increases in the speed of the air on the, upper surfaces of the wing relative to the lower surface. The, 3, , density of air is 1.2kg / m ., Watch Video, , Answer:, (a). The weight of the boeing aircraft is balanced by the upward force, due to the pressure difference, 5, , ΔP × A = 3.3 × 10 kg × 9.8, 5, , ΔP = (3.3 × 10 kg × 9.8ms, 3, , 6.5 × 10 N m, , −2, , ) / 500m, , 2, , −2, , (b). We ignore the small height difference between the top and bottom, sides in eQ. The pressure difference between them is then, ρ, ΔP =, , (v, 2, , 2, 2, , 2, , − v ), 1, , When v is the speed of air over the upper surface and, 2, , under the bottom surface., 2ΔP, (v2 − v1 ) =, , ρ(v2 + v1 ), , v1, , is the speed

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Taking the average speed, vav = (v2 + v1 ) / 2 = 960km / h = 267ms, , −1, , We have, v2 − v1, , ΔP, =, , vav, , ≈ 0.08, , 2, , ρvav, , the speed above the wing needs to be only 8% higher than that below., , Example 5, A metal plate of area, , 0.10m, , 2, , is connected to a, , 0.01kg, , mass via a, , string that passes over an ideal pulley (considered to be frictionless). A liquid with a film thickness of 3.0mm is placed between the, plate and the table. When released the plate moves to the right, with a constant speed of, , 0.085ms, , −1, , . Find the coefficient of, , viscosity of the liquid., Watch Video, , Answer:, The metal block moves to the right because of the tension in the string., The tension T is equal in magnitude to the weight of the suspended, mass m. Thus, the shear force F is, F = T = mg = 0.010kg × 9.8ms, , Shear stress on the fluid, straight rate, , v, , = 9.8 × 10, , 9.8 × 10, = F /A =, , 0.10, , 0.085, , =, , =, l, , 0.30 × 10, , Stress, η =, , −2, , s, , −3, , −1, , Strain rate, 3, , = ((9.8 × 10, , −2, , 0.30 × 10 m, )N ), (0.085ms, , 3, , = 3.46 × 10 P a, , s, , −1, , 2, , )(0.10m ), , −2, , N, , −2, , N /m, , 2

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Example 6, In a car lift, compressed air exerts a force, , F1, , on a small piston, , having a radius of 0.5cm. This pressure is transmitted to a second, piston of radius 10.0 cm . If the mass of the car to be lifted is 1350, kg. calculate F . What is the pressure necessary to accomplish this, 1, , task?, Watch Video, , Answer:, Since pressure is transmitted undiminished throughout the fluid,, F1 =, , π(5 × 10, , A1, , −2, , 2, , m), , = F2 =, , A2, , π(15 × 10, , −2, , 2, , (1350kg × 9.8ms, , −2, , ), , m), , = 1470N, 3, , ≈ 1.5 × 10 N, , The air pressure that will produce this force is, 3, , F1, P =, , 1.5 × 10 N, =, , A1, , π(5 × 10, , −2, , 2, , 5, , = 1.9 × 10 P a, , ) m, , This is almost double the atmospheric pressure. Hydraulic brakes in, automobiles also work on the same principle. when we apply a little, force on the pedal with our foot the master piston moves inside the, master cylider. and the pressure caused is transmitted through the, brake oil to act on a piston of larger area. A large force acts on the, piston and is pushed down expanding the brake shoes against brake, lining. In this way, a small force on the pedal produces a large, retarding force on the wheel. An important advantage of the system is, that the pressure set up by pressing pedal is transmitted equally to all, cylinders attached to the four wheels so that the braking effort is equal, on all wheels.

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Example 7, The terminal velocity of a copper ball of radius, through a tank of oil at 20, oil at 20, , ∘, , C, , ∘, , C, , 3, , 3, , falling, , is 6.5cm / s. Find the viscosity of the, , . Density of oil is 1.5 × 10, , 8.9 × 10 Kg / m, , 2mm, , 3, , Kg / m, , 3, , , density of copper is, , ., , Watch Video, , Answer:, We have v, , t, , g = 9.8ms, , = 6.5 × 10, −2, , (2 × 10, , −3, , , a = 2 × 10, , −3, , 2, , ) m, , 6.5 × 10, −1, , kgm, , −3, , m, , ., , −3, , . From Eq., , ×, 9, , = 9.9 × 10, , −1, , , ρ = 8.9 × 10 kgm, , σ = 1.5 × 10 kgm, , η =, , ms, , 3, , 3, , 2, , −2, , −1, , s, , 2, , × 9.8ms, , −2, , −2, 3, , × 7.4 × 10 kgm, ms, , −3, , −1, , −1, , INTRODUCTION, INTRODUCTION TO, TO VISCOSITY, VISCOSITY, Jyadatar fluids motion me kuchh resistance offer krte hai, is, resistance ko hum viscosity kehte hai. Viscosity arise hota hai, jab fluid ke layers ke beech relative motion hota hai. It, measures resistance to flow arising due to the internal, friction between the fluid layers as they slip past one, another when fluid flows. Viscosity can also be thought of as a, measure of a fluid’s thickness or its resistance to objects passing, through it.

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Use ke time, upper bulb suction se liquid draw krta hai aur phir, liquid niche capillary tube se flow krte hue lower bulb me jata, hai. Two marks (one above and one below the upper bulb), indicate a known volume. The time taken for the liquid to pass, between these marks is proportional to the kinematic viscosity., Most commercial units are provided with a conversion factor.

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Unhone ye bhi observe kiya ki flow ke type me suddenly, transition ho sakta hai, laminar se turbulent., Example 8, (a) Pressure decreases as one ascends the atmosphere. If the, density of air is, , ρ,, , What is the change in pressure dp over a, , differential height dh ? (b) Considering the pressure p to be, proportional to the density, find the pressure p at a height h if the, pressure on the sureface of the earth is, 5, , p0 = 1.03 × 10 N m, , −2, , , ρ0 = 1.29kgm, , −3, , what height will the pressure drop to, , p0 ., , and g = 9.8ms, (1 / 10), , (c ) If, −2, , ,, , at, , the value at the, , surface of the earth ? (d) This model of the atmosphere works for, relatively small distance. Identify the underlying assumption that, limits the model., Watch Video, , Answer:, Consider a horizontal parcel of air with cross-section A and height dh., , Let the pressure on the top surface and bottom surface be p and p+ dp., If the parcel is in equilibrium , then the net upward force must be, balanced by the weight.

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1.013 × 10, =, , 5, , 1.22 × 9.8, , = 16 × 10, , 3, , × 2.303 = 0.16 × 10, , 5, , m, , m, , (d) We know that, , p ∝ ρ, , (when T=constant i.e., isothermal, , pressure), Temperature (T) remains constant only near the surface of the earth ,, not at greater heights., , Example 9, A vessel filled with water is kept on a weighing pan and the scale, adjusted of zero. A block of mass M and density ρ is suspended by, a masseless spring of spring constant k. This block is submerged, inside into the water in the vessel. What is the reading of the scale /, Watch Video, , Answer:, Consider the diagram, The scale is adjusted to zero, therefore, when the block suspended to a, spring in immersed in water, then the reading of the scale will be equal, to the thrust on the block due to water . Thrust =weight of water, displaced, = V ρw g, , (where V is volume of the block and ρ is density of water), , ρ, , w, , ρw, , m, ρw g = (, , )mg, ρ

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(, , ∵, , Density of the block ρ, , mass, =, , m, =, , volume, , V, , ), , POINT 2, Angle of Contact: The angle which the tangent to the liquid surface, at the point of contact makes with the solid surface inside the liquid, is called angle of contact., Principle of Continuity: Property of a transition between two states, of matter, as between gas and liquid, during which there are no, abrupt changes in physical properties.

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Example 10, The free surface of oil in a tanker, at rest, is horizontal if the, tanker starts accelebrating the free surface will be titled by an, angle θ. If the acceleration is a ms, , −2, , what will be the slope of the, , free surface ?, Watch Video, , Answer:, Consider the diagram where a tanker is accelerating with acceleration, a., , Consider an elementary particle of the fluid of mass dm., The acting forces on the particle with respect to the tanker are shown, above., Now, balancing forces (as the particle is in equilibrium along the, inclined direction component of weight=component of pseudo force, dmg sin θ = dma cos θ, , (we have assumed that the surface is inclined, , at an angle θ) where, dma is pseudo force

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⇒ g sin θ = a cos θ, ⇒ a = g tan θ, a, ⇒ tan θ =, , slope, , =, , g, , Example 11, Iceberg floats in water with part of it submerged. What is the, fraction of the volume of iceberg submerged if the density of ice is, ρi = 0.917gcm, , −3, , ?, , Watch Video, , Answer:, Given , density of ice (ρ, Density of water (ρ, , w), , ice ), , = 0.917g / cm, , = 1g / cm, , 3, , 3, , Let V be te total volume of the iceberg and V' of its volume be, submerged in water. In floating condition., Weight of the iceberg=Weight of the water displaced by the submerged, part by ice, V ρice g = V ' ρw g, , or, , ρice, , V', =, V, , 0.917, =, , ρw, , = 0.917, 1, , ( ∵, , Weight =mg=vρg), , SURFACE, SURFACE TENSION, TENSION, Surface tension is the tendency of fluid surfaces to shrink into, the minimum surface area possible. Kya aapne kabhi notice kiya

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Examples of Surface Tension, 1. Water striders joki chhote insects hote hai wo pani me chal, sakte hai kyunki unka weight water ke surface ko penetrate krne, ke liye km hota hai. Aise hi nature me surface tension ke bahut, examples hai. Kuchh niche diye hai:, 2. Insects walking on water, 3. Floating a needle on the surface of the water., 4. Rainproof tent materials where the surface tension of water, will bridge the pores in the tent material, 5. Clinical test for jaundice, 6. Surface tension disinfectants (disinfectants are solutions of, low surface tension)., 7. Cleaning of clothes by soaps and detergents which lowers, the surface tension of the water, 8. Washing with cold water, 9. Round bubbles where the surface tension of water provides, the wall tension for the formation of water bubbles., This phenomenon is also responsible for the shape of liquid, droplets., , SURFACE, SURFACE ENERGY, ENERGY, Surface energy quantifies the disruption of intermolecular, bonds that occurs when a surface is created. It is also called, surface free energy or interfacial free energy. Surface energy

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can be defined as the work per unit area done by the force that, creates the new surface., , What is Surface Energy?, Have you ever tried drowning an ant? Doesn’t work, does it?, They seem to be body sliding across the water’s surface till they, safely reach solid ground. But that doesn’t make sense because, their breathing mechanism is kind of located in the underbelly!, They aren’t body sliding. They are just walking across just like, how they would do on land except with a different manoeuvring, technique. They are able to do this because of the, phenomenon known as surface tension. Because of cohesive, forces across liquid molecules, surface tension causes the, existence of a thin film (layer) across the surface. Surface energy, is the work done per unit area to produce this new surface. And, if the insect is light enough, to not damage (cause brittle failure, to) the layer, it can use the thin layer as support to walk across., , Surface energy ka S.I Unit N / m hai aur Dimension[M T, , −2, , ], , hai.

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= (, , 2S, , )cos θ(i), , a, , Thus the pressure of the water inside the tube, just at the, meniscus (air-water interface) is less than the atmospheric, pressure., Consider The two points A and B. They must be at the same, pressure,, P0 + hρg = Pi = PA, , where ρ is the density of water,and h is called the capillary, 2S cosθ, hρg = (Pi – P0 ) =, , a, , Therefore capillary rise surface tension ke karan hota hai., Chhote radius ke liye ye jyada hota hai., , Figure (a) Schematic picture of a narrow tube immersed in, water. (b) Enlarged picture near interface.

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DROPS, DROPS AND, AND BUBBLES, BUBBLES, Why are water and bubbles drops?, Whenever liquid is left to itself it tends to acquire the least, possible surface area so that it has least surface energy so it has, most stability., Therefore for more stability they acquire the shape of a sphere,, as the sphere has the least possible area.

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Example 12, The surface tension and vapour pressure of water at, 7.28 × 10, , −2, , Nm, , −1, , and, , 3, , 2.33 × 10 P a, , 20, , ∘, , C, , is, , , respectively. What is the, , radius of the smallest spherical water droplet which can form, without evaporating at 20, , ∘, , C, , ?, , Watch Video, , Answer:, Given, surface tension of water, (S) = 7.28 × 10, , Vapour pressure (p), , −2, , N /m, , = 2.33 × 10, , 3, , Pa

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The drop will evaporate, if the water pressure is greater than the, vapour pressure., Let a water droplet or radius R can be formed without evaporating., Vapour pressure=Excess pressure in drop., 2S, ∴ p =, R, , or R, , 2S, =, , p, , 2 × 7.28 × 10, =, , = 6.25 × 10, , 2.33 × 10, −5, , −2, , 3, , m, , Example 13, A hot air balloon is a sphere of radius 8 m . The air inside is at a, temperature of 60, , ∘, , C., , How large a mass can the balloon lift when, , the outside temperature is, R = 8.314J mole, , −1, , K, , −1, , 20, , ∘, , , 1atm., , membrane tension is 5N m, , −1, , (Assume air is an ideal gas, , C?, , 5, , = 1.013 × 10 P a, , ,, , the, , ), , Watch Video, , Answer:, Let the pressure inside the balloon be p and the outside pressure be p, i, , , then excess pressure is p, , 2S, i, , Where , S=Surface tension, r=radius of balloon, , − p∘ =, , r, , ., , ∘

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Considering the air to be in ideal gas, volume of the air displaced and, , n∘, , pi V, , = ni RTi, , where, V is the, , is the number of moles displaced, , and T is the temperature outside., ∘, , So, n, , Pi V, i, , =, , Mi, =, , RTi, , MA, , Where, M is the mass of air inside and M is the molar mass of air, i, , and n, , ∘, , Where,, , =, , A, , P∘ V, , =, , RT ∘, M∘, , M∘, MA, , is the mass of air outside that has been displaced . If w is, , the load it can raise, then w + M g, i, , = M∘ g, , ⇒ w = M ∘ g − Mi g, , As in atmosphere 21% O and 79 % N -is present, 2, , ∴, , 2, , Molar mass of air, , Mi = 0.21 × 32 + 0.79 × 28 = 28.84g, ∴, , Weight raised by the balloon, , w = (M ∘ − Mi )g, ⇒ w =, , MA V, R, , (, , P∘, T∘, , −, , Pi, Ti, , )g

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0.02884 ×, =, , 4, 3, , π × 8, , 3, , × 9.8, (, , 8.314, 0.02884, , =, , 4, 3, , π × 8, , 1.013 × 10, , 5, , 1.013 × 10, −, , 293, , 5, , 2 × 5, −, , 333, , 8 × 313, , 3, , 1, , 5, , 1, , × 1.013 × 10 (, , 8.314, , −, 293, , ) × 9.8, 333, , = 3044.2N, , ∴, , Mass lifted by the balloon, , ≈ 305kg, , 3044.2, , w, =, , g, , =, , 10, , ≈ 304.42, , kg ., , ., , Example 14, A hydraulic automobile lift is designed to lift cars with a maximum, mass of 300 kg. the area of cross-section of the piston carrying the, load is, , 425cm, , 3, , . What maximum pressure would smaller piston, , have to bear?, Watch Video, , Answer:, Pressure on the piston due to car, Weight of car, =, Area of piston, 3000 × 9.8, P =, 425 × 10, , −4, , Nm, , −2, , 5, , = 6.92 × 10 P a, , This is also the maximum pressure that the smaller piston would have, to bear., , Example 15, If a drop of liquid breaks into smaller droplets, it result in, lowering of temperature of the droplets. Let a drop of radius R,, breaks into N small droplets each of radius r. Estimate the drop in, temperature.

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Watch Video, , Answer:, When a big drop of radius R, breaks into n droplets each of radius r,, the volume remains constant., ∴, , Volume of big drop, , 4, 3, , πR, , or R, , 3, , or N, , 3, , 4, = N ×, , = Nr, R, =, r, , πr, , 3, , 3, , 3, , 3, , 3, , Now, change in surface area 4πR, = 4π(R, , 2, , Energy, , Volume of each small drop, , = N ×, , 2, , − N 4πr, , 2, , 2, , − Nr ), , released, , = T − ΔA = S4π(R, , 2, , 2, , − Nr ), , [T=Surface, , tension], Due to releasing of this energy, the temperature is lowered., If ρ is the density and s is specific heat of liquid and its temperature is, lowered by Δθ , then energy released, Δθ, , 2, , 4, , 2, , − Nr ) = (, , T × 4π(R, ⇒ Δθ =, , 4, 3, , 3T, ρs, , R, [, R, , 3T, =, , [s=specific heat, , =change in temperature], , T × 4π(R, , =, , = msΔθ, , ρs, , 3, , [, , ρs, , Nr, R, (R, , 3, , 3, , ], 3, , /r ) × r, R, , 1, −, , R, , ], r, , − nr ), , 2, , −, , 1, , 2, , 3, , R, , [, , 3, , πR ρ × s, , −, , 1, , 3T, =, , 2, , 2, , × R, , 3, , 2, , ], , 3, , 4, × ρ)sΔθ, , [ ∴ m = vρ =, , 3, , 3, , πR ρ]

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SUMMARY, 1. The basic property of a fluid is that it can flow. The fluid does not, have any resistance to change its shape. Thus, the shape of a fluid is, governed by the shape of its container., 2. A liquid is incompressible and has a free surface of its own. A gas, is compressible and it expands to occupy all the space available to it., 3. If, , F, , is the normal force exerted by a fluid on an area A then the, , average pressure, , Pa v, , is defined as the ratio of the force to area, , F, Pa v =, , ., A, , 4. The unit of the pressure is the pascal (Pa). It is the same as N m, . Other common units of pressure are 1 atm, 1 bar, 1 torr, , = 10, , 5, , = 1.01 × 10, , 5, , −2, , Pa, , Pa, , = 133P a = 0.133kP a1mmof H g, , = 1 torr, , = 133P a, , 5. Pascal’s law states that: Pressure in a fluid at rest is same at all, points which are at the same height. A change in pressure applied to, an enclosed fluid is transmitted undiminished to every point of the, fluid and the walls of the containing vessel., 6. The pressure in a fluid varies with depth h according to the, expression, , P = P a + ρgh, , where ρ is the density of the fluid,, , assumed uniform., 7. The volume of an incompressible fluid passing any point every, second in a pipe of non uniform crossection is the same in the steady

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flow. v A = constant (, , v, , is the velocity and A is the area of, , crossection) The equation is due to mass conservation in, incompressible fluid flow., 8. Bernoulli’s principle states that as we move along a streamline, the, sum of the pressure (P ), the kinetic energy per unit volume (ρv, and the potential energy per unit volume, 2, P + ρv, , + ρgy, 2, , (ρgy), , 2, ), 2, , remains a constant., , = constant. The equation is basically the, , conservation of energy applied to non viscuss fluid motion in steady, state. There is no fluid which have zero viscosity, so the above, statement is true only approximately. The viscosity is like friction, and converts the kinetic energy to heat energy., 9. Though shear strain in a fluid does not require shear stress, when a, shear stress is applied to a fluid, the motion is generated which, causes a shear strain growing with time. The ratio of the shear stress, to the time rate of shearing strain is known as coefficient of viscosity,, η, , . where symbols have their usual meaning and are defined in the, , text., 10. Stokes’ law states that the viscous drag force, , F, , on a sphere of, , radius a moving with velocity v through a fluid of viscosity is,, F = – 6πηav, , ., , 11. The onset of turbulence in a fluid is determined by a, dimensionless parameter is called the Reynolds number given by, d, Re = ρv, η, , Where, , d, , is a typical geometrical length associated with, , the fluid flow and the other symbols have their usual meaning.

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12. Surface tension is a force per unit length (or surface energy per, unit area) acting in the plane of interface between the liquid and the, bounding surface. It is the extra energy that the molecules at the, interface have as compared to the interior.