Fluid Mechanics - Chapter 2 (2023) PDF

Chapter 2: Force & motion in fluids (7%) 2023  Fluid: is a substance that adjusts or adapts to the boundaries of any container in which it is put in. Unlike solid, fluid can flow when external force is applied. Examples are water, honey, water vapor, ox...

Chapter 2: Force & motion in fluids (7%) 2023  Fluid: is a substance that adjusts or adapts to the boundaries of any container in which it is put in. Unlike solid, fluid can flow when external force is applied. Examples are water, honey, water vapor, oxygen etc. Acceleration due to gravity (g) Let earth be a homogenous sphere of radius ‘R’ and mass ‘Me’, then by Newton’s law of gravitation (inverse square law), the force ‘Fg’ felt by a body or weight ‘W’ of a small mass m on the surface of the earth is given by Me  m W = Fg = G 2.............. This equation shows that g is independent of mass of R the body and hence it is same for all the bodies. By Newton's 2nd law, Me F = ma  F = mg........... g =G 2 This equation also shows that g is nearly constant R since G and Me are constants and R varies only slightly near the earth’s surface. = Me  m mg = G R2 The acceleration due to gravity varies depending on where you are on the Earth's surface. It is generally greater at the poles than at the equator. This is because the Earth is not a perfect sphere, but rather an oblate spheroid, which means that it bulges at the equator and is flatter at the poles. Another factor that affects the acceleration due to gravity is altitude. The acceleration due to gravity decreases as you move further away from the Earth's surface. This is because the force of gravity weakens with distance. At sea level, the acceleration due to gravity is approximately 9.81 meters per second squared (m/s²). However, at an altitude of 10,000 meters (approximately 32,800 feet), the acceleration due to gravity is about 9.67 m/s², which is slightly lower than at sea level. This decrease in acceleration due to gravity with altitude has important implications for things like spacecraft and satellites, which operate at high altitudes and must take into account the weaker force of gravity.  Verification of value of g from calculation For Earth: G = 6.67 x 10 -11 Nm2 /kg2 , R = 6.38 x 10 6 m and Me = 5.98 x 10 24 kg Me 5.98 x 1024 g earth = G 2 = 6.67 x 10  -11 =9.78m/s 2 R ( 6.38 x 10 ) 6 2 For Moon: G = 6.67 x 10 -11 Nm2 /kg2, R = 1.74 x 10 6 m and Mm = 7.35 x 10 22 kg Mm 7.35 x 1022 g moon = G 2 = 6.67 x 10  -11 = 1.62m / s 2 (1.74 x 106 ) 2 R The acceleration due to gravity near the earth’s surface is 9.8 m/s2. On the moon, the acceleration due to gravity g = 1.67 m/s2. If we consider a body of mass 1 kg, its mass on the moon will remain the same as the earth but its weight will be 1.67 N, while on the earth its weight will be 9.8 N.  Motion of a body falling in a uniform gravitational field with fluid (air) resistance: Whenever a body of mass m is falling vertically in the presence of fluid (air), there are two forces acting in opposite direction on the body, gravitational force ‘Fg’ and resisting force ‘Fa’ of the fluid or viscous drag. Terminal velocity (vt): As the velocity ‘v’ keeps on increasing, the resisting force Fa also increases till it becomes equal to the weight of the body. At this point, the velocity becomes constant and maximum. This velocity is called terminal velocity. mg vt = k K-proportionality constant or drag constant (depends On dimension of the body and viscosity of the fluid)  Instantaneous velocity: velocity at any instant of time during the fall. − kt v = vt (1 − e m ) It shows that velocity increases exponentially. Case 1: at time t=0 Case 2: at time t=∞ − kt − kt v = vt (1 − e m )becomes v = vt (1 − e m ) becomes v = vt (1 − e0 ) 𝑣 = 𝑣𝑡 (1-𝑒 −∞ ) 1 v = vt (1 − 1)  v = 0 𝑣 = 𝑣𝑡 (1 − ∞ ) 𝑒 1 𝑣 = 𝑣𝑡 (1- ∞) 𝑣 = 𝑣𝑡 (1 − 0) 𝑣 = 𝑣𝑡 Graphical representation of instantaneous velocity:  Instantaneous acceleration: acceleration at any time instant during the fall: − kt a = ge m This equation shows that acceleration decreases exponentially. Case 2: when t = ∞ Case 1: when t=0 − kt a = ge m becomes − kt a = ge m becomes 𝑎 = 𝑔𝑒 −∞ 1 a=ge0  a = g 𝑎=𝑔 ∞ 𝑒 1 𝑎=𝑔 ∞ 𝑎 =𝑔×0=0 Graphical representation of instantaneous acceleration:  Displacement or position of the body with respect to time:  m − kt m  y = vt t − 1− e  k    This equation shows that displacement increases exponentially. Graphical representation of position: Case 1: when t=0 Case 2: when t=∞  m  y = vt t − 1 − e m   becomes − kt  m − kt m  y = vt t − 1 − e   k     k       m  y = vt 0 − 1 − e0    y = 0 becomes  k  𝑚 𝑦 = 𝑣𝑡 [∞ − [1 − 𝑒 −∞ ]] 𝑘 𝑦 = 𝑣𝑡 [∞] 𝑦=∞  Surface tension (s) Surface tension (s):is a property by virtue of which free surface of a liquid possesses a tendency to contract so as to acquire minimum surface area. This tendency makes the surface as a stretched membrane and this is due to the molecular nature of the liquid (cohesive force) and not due to gravity. For a given volume, spheres have the least surface area, therefore, raindrops, soap bubbles, mercury drops all take spherical shape. Surface tension is also defined as force per unit length in the plane of liquid surface acting at right angles (or perpendicularly) on either side of the imaginary line drawn on that surface. F Mathematically, S= SI unit is N/m. L Dimensional formula: F ma kgms −2 S= = = L L m S = kgs −2 S   M 1 L0T −2   Molecular theory for surface tension: Molecules A, B and C are called sphere of influence or sphere of molecular activity. The radius of these molecules is equal to the molecular range. The maximum distance up to which the attractive force between the molecules exists is called molecular range. For solid and liquid, the molecular range is of order of 10 -9 m. Molecule A experiences force of attraction equally in all directions due to the neighboring molecules. Therefore, the net force on A is zero. Molecule B also experiences attractive force but not equally in all directions and the net force on it is downwards. Molecule C experiences maximum downward pull because the sphere of influence is half out side the liquid and half inside. If molecule A is to be moved from the interior of the liquid to the surface, work has to be done on it against the inward force of attraction. This work done is stored as additional potential energy of the molecule on the surface but in stable equilibrium, a system must have minimum potential energy. Therefore, to have minimum potential energy, the surface film tends to have minimum surface area, so that the number of molecules in the surface film becomes minimum and behaves as a stretched membrane. This tendency is exhibited as surface tension.  Surface film Surface film: is a thin film of liquid near its surface and has thickness equal to molecular range of that liquid. This film is responsible for surface tension. All the molecules situated in this film experiences downward cohesive force. Factors affecting surface tension of a liquid: Surface tension is inversely proportional to temperature. That is why hot soup tastes better than cold one. When the temperature is high the soup covers a larger area of the tongue (which is actually due to reduced surface tension) giving better taste. Highly soluble substances increase the surface tension of the liquid whereas sparingly soluble substances reduce the surface tension of the liquid. Example: detergent when added to water reduces the surface tension thereby covering more area of the dirty cloths.  Surface energy Out of syllabus Surface energy : Free surface of a liquid always have the tendency and possess minimum surface area. If there is a need to increase the surface area then extra work has to be done. This work done is stored in the liquid surface as potential energy. This potential energy per unit area of the surface plane is called surface energy. When the surface area of the film increases, it loses more heat. To maintain the temperature, heat from the surrounding is taken up by the surface film. This additional energy gained by the surface film is also called surface energy. Mathematically, work done in increasing the area of the surface film surface energy= increase in the area of the surface film at a constant temperature W SE = A SI unit is J/m2 Dimensional formula: W F  dis ma  dis SE = = = A A A kgms −2  m = 2 = kgs −2 m =  M 1 L0T −2   Relation between surface tension and surface energy: Consider a liquid film to be formed on a rectangular surface ABCD. On forming a film, the surface film contracts (see arrows) due to surface tension i.e., AB is pulled inwards. To keep AB in its original position, a force F is applied in the outward direction, all along its length. If T is force due to surface tension (i.e. T=F/l). Then F = T x 2l. Since the film has two free surfaces, the RHS of the equation is multiplied by 2. If the wire AB is to be pulled by an additional distance dx, then the additional work done W in this case is, W = F  x W = T  2l  x ( 2l  x = A increase in area) W = T  A i.e. Work done in increasing the surface area by W unity, is equal to the surface tension T at T= =E constant temperature. Therefore, this work A done, is the surface energy of the film. If A = 1unit , then T =W Examples of Surface Tension: Applications of Surface Tension: Drops of Water Washing cloth more effectively with soap Beading of Raindrops Add flux in tin for soldering The spreading of antiseptics such as dettol, savlon, etc. Duck floating on water. In all these cases the surface tension is lowered. 5.A soap film is trapped between a frame and a wire of length 10 cm as shown. If the surface tension is given as 0.0049 N/m, what will be the value of mass m such that the wire remains in equilibrium? F = S  2l mg = S  2l S  2l 0.0049  2 10 10−2 m= = g 9.8 m = 110−4 kg Oil spreads over the surface of water whereas water does not spread over the surface of the oil, due to (a) Surface tension of water is very high (b) Surface tension of water is very low (c) Viscosity of oil is high (d) Viscosity of water is high Ans. (a) The paints and lubricating oils have low surface tension. Why? Answer. Paints and lubricating oils with low surface tension can spread over a large surface area. An oil drop on a hot cup of soup spreads over when the temperature of the soup falls. Why? Answer. Since hot water has a lower surface tension than oil, oil drops do not spread over it. Water’s surface tension increases as it cools. At low temperatures, the surface tension of water exceeds that of oil, and thus oil drops begin to spread over it. What shape does a liquid take when it weighs nothing? Give reason to support your answer. Answer. When a liquid has no weight, the only force acting on it is surface tension. The liquid surface tends to occupy the smallest amount of surface area due to surface tension. Because the surface area of a sphere is the smallest for a given volume, the liquid takes on a spherical shape. What is the effect of contamination and temperature on the surface tension of a liquid? Answer. Temperature Effect: As the temperature rises, the surface tension of the liquid decreases. It decreases linearly for small temperature differences. A liquid’s surface tension is zero at its boiling point. As a result, the boiling point is referred to as the critical temperature of that liquid. Contaminations effect: If there is dust, grease, oil, or other impurities on the liquid’s surface, the surface tension decreases.  It is the angle inside the liquid between the tangent to the solid surface and tangent to the liquid Angle of contact (θ): surface at the point of contact. θ determines whether a liquid will spread or form droplets on the surface of the solid. If θ is less than 900, the liquid will spread on the solid surface and if θ is more than 900, then the liquid will form droplets on the solid surface. θ is fixed for a given pair of solid and liquid. Value of θ for Pure water and glass =00 Ordinary water and glass = 80 Mercury and glass = 1350 Soap, detergents and dying substances are called wetting agents because when they are added to liquid, the angle of contact decreases, thus wetting the surface. Water proofing agents on the other hand are added to increase the angle of contact so that water does not stick to or wet the surface.  Factors affecting the value of θ: θ is inversely proportional to temperature. It depends on the nature of solid and liquid in contact and the surrounding medium. It is independent of angle of inclination between the solid and liquid. Two forces that act on liquid are: 1. Gravitational force 2. Force of surface tension Gravitational force tries to squash the droplets whereas force due to surface tension tends to make the droplets spherical. Gravitational force is more prominent on larger droplets whereas force of surface tension is more prominent in smaller droplets. Gravitational force depends on mass of the droplets whereas force of surface tension depends on the area of the droplets. Capillarity: The phenomenon of rise or fall of liquid in a capillary tube is called capillarity. Liquid whose angle of contact is less than 900 show capillary rise. Liquid whose angle of contact is more than 900 show capillary fall or depression. Liquid whose angle of contact is equal to 900 show neither rise nor fall in the capillary tube. Factors that affect the flow of liquid through porous media: 1. The movement of liquid depends on how hydrophilic a surface is. More hydrophilic, more absorption, hence more rise. 2. It depends on how small the pore is. Smaller the pore, more the absorption. Consider a liquid contained in a vessel. The liquid solid adhesive force is very weak as compared to the cohesive force in the liquid. The shape of the liquid surface near the solid shall be (a) Horizontal (b) Almost vertical (c) Concave (d) Convex (d) Convex Mercury does not wet glass, wood or iron because (a) Cohesive force is less than adhesive force (b) Cohesive force is greater than adhesive force (c) Angle of contact is less than 90o (d) Cohesive force is equal to adhesive force Ans. (b) https://www.examsegg.com/surface-tension-questions.html#gsc.tab=0  Some of the applications of capillarity in day-to-day life: Oil rises through a wick due to capillarity. Ink spreads on a blotting paper. Water rises to the tip of the plants through the tiny fibers in the plant. A towel gets soaked with water on account of capillary action. Ploughing of fields to preserve moisture in the soil. Ploughing of fields can help preserve moisture in the soil in several ways: 1.Breaking up soil crust: Ploughing helps break up the soil crust, which is formed when the surface of the soil dries out. The crust can prevent moisture from penetrating into the soil, and breaking it up allows water to infiltrate and reach the deeper layers of the soil. 2.Increasing soil porosity: Ploughing also increases the porosity of the soil, creating spaces for air and water to circulate. This helps prevent waterlogging and ensures that the soil can hold onto moisture for longer periods. 3.Improving soil structure: Ploughing can also improve the structure of the soil, allowing it to hold onto moisture more effectively. This is because ploughing can break up clods and create a more crumbly soil structure, which allows water to penetrate and be held in the soil. Overall, ploughing can be an effective way to preserve moisture in the soil, which is essential for promoting healthy plant growth and maintaining soil health. Rise of liquid in a capillary (Ascent formula) 2 S cos  h=......... Ascent formula adg where S is the surface tension d is the density of the liquid a is the circular cross-section radius of the tube ‘r’ be the radius of the meniscus.  2𝑆 𝑃𝑖 − 𝑃𝑜 = 𝑟 r- radius of meniscus Where 𝑃𝑖 = pressure on the top of the 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑃𝑖 − 𝑃𝑜 =pressure of liquid column of height h open vessel and 𝑃0 = pressure on the top of the capillary tube. 𝑃𝑖 − 𝑃𝑜 = hdg 2𝑆 ℎ= 𝑎 2𝑆 = hdg 𝑐𝑜𝑠𝜃 𝑑𝑔 𝑟 2𝑆 ℎ = 𝑟𝑑𝑔---------------- 2𝑆𝑐𝑜𝑠𝜃 ℎ= 𝑎𝑑𝑔 From the diagram, 𝑎 𝑐𝑜𝑠𝜃 = 𝑟 𝑎 𝑟= −−− − 𝑐𝑜𝑠𝜃 a- radius of the tube Ascent Formula Class 11 Physics Important Derivations - YouTube https://www.youtube.com/watch?v=2Rz0MwWllY8 4. Calculate the radius of a capillary tube that would raise sap 100 m to the top of a giant redwood, assuming that sap’s density is 1050 kg/𝑚3 , its contact angle is zero, and its surface tension is 0.0728 N/m. 1.41x10-7 m  In a surface tension experiment with a capillary tube, the water rises up to 0.1m. If the same experiment is repeated on an artificial satellite revolving around the Earth, the rise of water in the capillary tube will A. be 0. B. remain the same. C. be half-length of the capillary tube. D. be full length of the capillary tube. D. be full length of the capillary tube.  TYPES OF LIQUID FLOW Streamline flow: A flow in which every particle of the liquid follow the same path as that of the preceding particle and has the same velocity as that of the preceding particle while passing through that point. Tube of flow: Bundle of streamlines having same velocity of liquid particles over any cross-sectional area perpendicular to the direction of the flow of liquid. Laminar flow: If a liquid is flowing over a horizontal surface with a steady flow and moves in the form of liquid layers of different velocities which does not mix with each other, such a flow is called laminar flow.  Turbulent flow: when the motion of particles of liquid become disorderly due to increase in velocity (more than critical velocity), such a flow is called a turbulent flow. In laminar flow, molecules follow the same lane whereas in turbulent flow, molecules do not stay in the same lane. Critical velocity (vc): the velocity of liquid flow up to which its flow is streamlined and above which its flow becomes turbulent, is called critical velocity.  Re Mathematically, vc = where Re is Reynold's number r  − vis cos ity  − density r − radius of the pipe density  obstacle length  flow speed Re = vis cos ity As Re increases, the flow goes from viscous-dominated to inertia-dominated, that is, from laminar to turbulent. 1. How can Reynolds number help to predict the flow pattern in fluid mechanics? The Reynolds number (Re) helps predict flow patterns in different fluid flow situations. At low Reynolds numbers, flows tend to be dominated by laminar (sheet-like) flow, while at high Reynolds numbers flows tend to be turbulent. Viscosity of fluids The real fluid flowing in the pipe has the slowest velocity at the walls of the pipe due to friction-like resistive force (which is due to irregularities in the pipe walls). Other examples of irregularities are Rust on the interiors of the pipe. Accumulation of matter on the inner walls of the blood vessel. Each layer of the fluid stick to the molecules of the adjacent layer. This is called viscosity.  Firefighters use a firehose to adjust water pressure. If the firefighters have to reach the sixth floor of a building to extinguish a fire, what should be the flow of water in the pipe? A. streamlined B. laminar flow C. turbulent flow D. turbulent and streamlined A. laminar flow Force needed to move a layer of viscous fluid with a constant velocity or viscosity on quantitative form: Unit of viscosity Fh Nm = = 2 −1  Nm −2 s 1 Av m ms F  A, F  , F  v h Av F SI unit: h Av F = where  is a constant called coefficient of vis cos ity h Fh Ph  F  = =  = P  Fh Av v  A  = Av unit of P is pascal(Pa) CGS unit is poise. ( Pa ) m Relation between SI and CGS unit:  −1 = Pa s or poiseuille ms 1 poise = 0.1 Pa s  Coefficient of viscosity in terms of shear stress and strain: Definition of coefficient of viscosity: Fh F = = Av A v ( h) v dv = = velocity gradient (it represents the h dx Shear strain: is the change in configuration per unit original configuration. change in velocity for every unit change in height ) dx  shear strain, tan  =  = F h  = F A Fh = A dv( dx ) Stress  rate of change of strain v h Av dv shear strain dx dx If A = 1 unit , = 1 unit then, strain rate = = h= dx time dt hdt F =  = F dx v 1 1 = v  strain rate = dt h Therefore, coefficient of viscosity is the tangential force (F) required shear stress (or pressure) = to maintain a unit velocity gradient between two parallel layers of strain rate liquid, each of unit area. Dimensional formula: Fh (ma)h (kgms −2 )m = = = 2 Av Av m (ms −1 ) kgs −1  = kgm −1s −1 m   M 1 L−1T −1  Factors on which viscosity depends: 1. Nature of the fluid: Viscosity of liquid is much greater than that of gas. 2. Temperature: viscosity decreases with increase in the temperature for liquid whereas for gases viscosity increases with the increase in temperature. For ideal fluid viscosity is zero. 7. The shear stress at a point in a liquid is found to be 0.03 N/m2. The velocity gradient at the point is 0.15 s-1. What will be it’s viscosity (in Poise)? 2 poise a) There are two lubricants A and B, whereby A has co-efficient of viscosity slightly greater than that of B. Which lubricant would you prefer in summer and why? Q. “Car lubricants play a major role in controlling friction and wear to the engine. If the car is used at high altitudes, we have to use lubricants of low viscosity”. Do you agree with the statement given above? Support your answer with an appropriate reason. Yes, —------------------------------------------------------------------------------------------[0.5] Because the temperature is low at a high altitude, the viscosity of lubricants must be low so that particles can freely move which will support the smooth movement of the engine parts. ------------------------------------------------------------------------------------- As the temperature gets lower, the viscosity of lubricant will be higher.--------------[0.5] Q. Why do people not use blotting paper or newspaper to write with an ink pen? It is not possible to write directly on blotting paper or newspaper with an ink pen because capillarity absorbs the ink. ------------------------ OR Only capillarity-----  Equation of continuity (ideal fluid) Mass coming into the pipe = mass going out of the pipe (Law of Conservation of Mass) Δ𝑉 × 𝜌 = 𝐴1 × 𝑣1 × ∆𝑡1 × 𝜌 = 𝐴2 × 𝑣2 × ∆𝑡2 × 𝜌 = 𝑚𝑎𝑠𝑠 This equation represents the conservation of mass in case of moving fluid. In other words, mass is also conserved or remains constant. A1v1 = A 2v2..........This is the equation of continuity 𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 Av = constant( k ) Volume flow rate(Q ) Δ𝑉 = 𝐴 × 𝑣 × Δ𝑡 or Av = k V = A  v  t Δ𝑉 × 𝜌 = 𝐴 × 𝑣 × Δ𝑡 × 𝜌 k V 𝑚𝑎𝑠𝑠 = 𝐴 × 𝑣 × Δ𝑡 × 𝜌 A= = A v v t 𝑚𝑎𝑠𝑠 =𝐴×𝑣×𝜌 unit : m3 / s Δ𝑡 1 A 𝑢𝑛𝑖𝑡: 𝑘𝑔/𝑠 v Cross-sectional area A is inversely proportional to velocity v 8. A syringe has a diameter of 1.5 cm. Area of hole in needle is only 0.01 𝑚𝑚2. If the plunger of syringe is pushed at velocity of 1.0 cm/s, what speed must medicine come out of needle? The motion of real fluid is complicated. Therefore, to discuss about motion, ideal fluid is taken which is assumed to have the following characteristics: 1. Steady flow: The velocity of the moving fluid at any fixed point does not change with time. 2. Incompressible flow: Density of ideal fluid remains constant throughout. 3. Non-viscous flow: An object will move at constant speed in the ideal fluid (no viscous drag). 4. Irrotational flow: A body floating in an ideal fluid will not rotate about its axis through its own center of mass. Two streamlines can never intersect because if the fluid element reaches the intersection, it would have two different velocities at a point and this is impossible.  Pressure energy: is the energy possessed by a liquid by virtue of its pressure. Bernoulli’s principle According to Bernoulli’s principle, for a streamline flow of ideal liquid, the total energy (pressure energy + kinetic energy + potential energy) of liquid per unit mass remains constant at every cross-section throughout the liquid. Mathematically, pressure energy + kinetic energy + potential energy = constant mass 1 P+  v 2 +  gh = k (constant) 2 𝑊 𝐹×𝑑 𝐹 = =𝐴=𝑃 𝑣𝑜𝑙 𝐴×𝑑 𝐾𝐸 1Τ 𝑚𝑣 2 1 = 2 𝑚Τ =2 𝜌𝑣 2 𝑉𝑜𝑙 𝜌 𝑃𝐸 𝑚𝑔ℎ = = 𝜌𝑔ℎ 𝑣𝑜𝑙 𝑚Τ𝜌  Case 2: when ℎ1 =h2 = ℎ ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑛𝑑 𝑉2 > 𝑉1 1 1 P1 +  gh1 +  v12 = P2 +  gh 2 +  v2 2becomes 2 2 1 1 P1 +  v12 = P2 +  v2 2 1 1 2 2 P1 +  gh1 +  v12 = P2 +  gh 2 +  v2 2 = constant 2 2 1 P +  gh +  v 2 = constant → Bernoulli's equation Since v 2  v1 ( A  1 )  P2  P1 2 v Case 1: If v1 = 𝑣2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑛𝑑 ℎ1 < ℎ2 1 1  pressure at the bigger area is more. P1 +  gh1 +  v12 = P2 +  gh 2 +  v2 2becomes 2 2 P1 +  gh1 = P2 +  gh 2 Note: 1. High velocity: Low pressure since h1  h 2  P1  P2 2. Low velocity : High pressure  pressure at A (closer to the ground) is more CYU 2: Water flows smoothly through the pipe shown in the figure, descending in the process. Rank the four numbered sections of the pipe according to (a) the volume flow rate through them, (b) the flow speed v through them, and (c) the water pressure P within them, greatest first. Sol: (a) All same since volume flow rate is constant. (b) 1, 2 and 3 same, 4 because area is smallest for 1(wider means slower) (c) 4, 3, 2, 1 because lower means more pressure.  What will be the relation between H & L? A. H>L B. H L https://ophysics.com/fl2.html While watering her garden, Pema observed that the speed of water can be increased by pressing the mouth of the pipe with the thumb and the water also travels a longer distance. Justify the above statement with a diagram and an equation. Answer Diagram [labeling with A1, A2, v1 and v2] ---------------- [0.5] According to the equation of continuity, the area of the cross section is inversely proportional to the velocity of fluid. This implies that when the area of cross section is less, then the velocity of water will be more, therefore, the water flows faster and covers long distances.------[1.5] OR IF only formula, v∝1A (Where v is velocity of water and A is area of cross-section) --------------[0.5]  Factors that determine volume flow rate (Q) of liquid or Poiseuille’s law : Out of syllabus 1 Note: Q L Because of this fact, long pipe lines have pumping stations at various intervals to compensate for pressure drop. Q  ( P2 − P1 ) or P Q  R4 Poiseuille’s law states that a fluid whose viscosity is  , flowing through 1 a pipe of radius R and length L has a volume of flow rate Q given by Q L 1  R ( P2 − P1 ) 4 Q Q=  8 L R 4 ( P2 − P1 ) where P1 and P2 are the pressure at the ends of the pipe. Q L  R ( P2 − P1 ) 4 Q=..........this is Poiseuille ' s law 8 L Stokes’ law: According to Stokes’ law, when a small spherical body moves through a viscous medium, the backward dragging force F acting on the spherical body is directly proportional to the radius ‘R’ of the body,  coefficient of viscosity ‘ ’ of the fluid through which the body is moving and the velocity ‘v’ of the body. Out of syllabus FR F  F v  F  Rv F = 6 Rv........stokes ' law  Terminal velocity (vt): When a spherical body of radius ‘r’ falls freely in a viscous medium, gravitational force ‘Fg’, viscous drag ‘Fa’ and buoyant force ‘Fb’ (which is equal to weight of the liquid displaced) acts on the body. When these three forces balance each other, then the net force acting on the body is zero. In such a situation, the sphere moves with a constant velocity called terminal velocity. At terminal velocity, (  −  )2r 2 g mb is the mass of the body vt = mL is the mass of the liquid displaced 9 density of the body(  ) density of liquid ( )  𝐶𝑎𝑠𝑒 1: 𝐼𝑓𝜌 > 𝜎, (𝜌 − 𝜎)2𝑟 2 𝑔 𝑣𝑡 = 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒. 9𝜂 It means that the body will sink or fall downwards. Case 2 : If    , vt will be negative. It means that the body will rise up. Example : bubbles in fizzy drinks. Case 3 : If  =  , vt will be zero. It means that the body will float.  Questions for you 1. What is the speed of efflux (v) of a fluid through a spigot at the bottom of a tank with height d that is placed x meters above the ground and filled to a height y above the ground? Show your work. A. 2 g ( y − x) B.  g ( x + d ) C. 2 gy D.  gy 2. Blood takes about 1.0 seconds to pass through a 1.0 millimeter long capillary in the human circulatory system. If the diameter of the capillary is 8.00 μm and the pressure drop is 2.0 kPa, what is the viscosity of the blood, assuming laminar flow? 3. The rise in the level of a liquid in a tube is h. What will be the rise in the level if the same amount of liquid is poured into a tube of half the diameter.[2h]  Differentiate between turbulent and laminar flow. Mercury has an angle of contact equal to 140o with glass. A narrow tube of glass of radius 1mm is dipped in a trough of mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 N/m. Density of mercury = 13.6x103 kgm-3 and g=9.8m/s2. When paddling a canoe upstream, it is wisest to travel as near to the shore as possible. When canoeing downstream, it may be best to stay near the middle. Explain why? Imagine yourself falling vertically downwards. Represent graphically how your velocity changes as you fall in presence and in absence of drag force. A liquid does not wet the surface of a solid, if the angle of contact is A. Zero B. Acute angle C. Right angle D. Obtuse angle What happens to the acceleration of a sky diver after his parachute opens? A. Increases and becomes maximum C. increases B. Decreases and becomes zero D. decreases Oil spreads over the surface whereas water does not spread over the surface of the oil due to the reason that A. Surface tension of water is very high C. Viscosity of oil is high B. Surface tension of oil is very high D. Viscosity of water is high Water flows through a horizontal pipe. The area of cross section at one place A1 =10cm2, velocity of water flow is 1m/s and pressure is 2000 Pa. At another place area, A2 =5cm2.what is the pressure at A2 ? A capillary tube of diameter 2x 10-3m is dipped in a liquid of surface tension 0.072 N/m and density 999.73 kg/m3.Calculate the height to which the liquid will rise in the capillary tube. (given g = 10m/s2, angle of contact = 45o)  Blowing of roof of house or tin sheet is a common incident observed in windy weather. Identify the principle in the above situation. A group of Materials Science and Engineering (MSE) students from Bhutan is designing a new engine oil to be used by motor cars in Bhutan. Of the listed principle factors affecting viscosity, which one of the following factors should be the dependent variable in their experiment? A. Oil composition. B. Dissolved gas. C. Temperature. D. Pressure. Dorji was playing with the mercury during the chemistry experiment. He dropped the mercury down vertically and found that shape of the mercury on the floor is spherical. He increases the size of the mercury drop and later he noticed the shape, to his surprise the shape is an oval. Explain. One of the students of class VII from UA is excited to see a water strider walks easily on the normal water in the swimming pool but she is surprised to see that same water strider drown in contaminated water (soap water). Surprised student comes to seek your help. As a physics student, what answer would you give? Support your answer with proper justification. Karma and Sonam are watering the plants; Karma was surprised to see that Sonam can reach the greater distance to water the plant without moving from place to place. As a physics student Sonam might have used the important concept learned in the classroom. What could be the concept learnt. Support your answer with an equation. In one of the biology classes, teacher explained about absorption capacity of different papers. For demonstration, teacher took normal paper and desho (handmade paper), teacher dropped a drop of ink on each paper. After an hour students found that the ink spread more in desho. Considering yourself as one of the students, explain why ink spread more in desho than normal paper? i. If the angle of contact of a liquid and solid surface is ______________________than 𝟗𝟎𝟎 ,then the liquid forms a _______________________ on the surface of solid. The distilled water is not an ideal fluid. Justify with two reasons. i. Water is flowing in a horizontal pipe of non-uniform cross section. At the most contracted region of the pipe: A. Velocity of water will be maximum and pressure minimum B. Pressure of water will be maximum and velocity minimum C. Both pressure and velocity of water will be maximum D. Both pressure and velocity of water will be minimum. i) The angle of contact between pure water and glass, ordinary water and glass, and mercury and glass is 0o, 8o and 135o respectively. What can you conclude from these data? i) A large pipe 5 meter above the ground and a small pipe 1 meter above the ground. The velocity of the water in a large pipe is 36 km/h with a pressure of 9.1 x 105 Pa, while the pressure in the small pipe is 2x105 Pa. ii) What is the water velocity in the small pipe? Water density = 103 kg/m3. iii)Which law of conservation is used in the above calculation? [39.9]  i. Washing soap is used for cleaning the clothes because A. It absorbs dirt. B. It increase the surface tension. C. It reduces the surface tension. D. It increases the viscosity of the liquid. i. It is difficult to separate two papers glued together. Why? a) How is the understanding of Bernoulli’s principle helpful while bowling in cricket? a) Mr. Jamphel, a chemistry teacher of Chukha Central school was conducting an experiment on calorimeter. During the experiment he noticed that the ends of a glass tube become rounded on heating. Explain the science behind it? 3.Why is a jet attached to the head of a firefighter's water pipes? 4.Why does a fast-moving train drag anything in and nearby its path? 5.Deep water is practically motionless. What exactly does it clarify?