{"id":10344,"date":"2019-12-07T16:37:18","date_gmt":"2019-12-07T11:07:18","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10344"},"modified":"2021-07-02T15:43:21","modified_gmt":"2021-07-02T10:13:21","slug":"1st-puc-physics-question-bank-chapter-10","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-question-bank-chapter-10\/","title":{"rendered":"1st PUC Physics Question Bank Chapter 10 Mechanical Properties Of Fluids"},"content":{"rendered":"

You can Download Chapter 10 Mechanical Properties Of Fluids Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Physics Question Bank Chapter 10 Mechanical Properties Of Fluids<\/h2>\n

1st PUC Physics Mechanical Properties Of Fluids Textbook Questions and Answers<\/h3>\n

Question 1.
\nExplain why
\n1. The blood pressure In humans is greater at the feet than at the brain.
\nAnswer:
\nThe height of the blood column in the human body is more at the feet than at the brain. As the pressure increases with the height of the column, blood exerts\u00a0 more pressure at the feet than at the brain (P = \u03c1gh)<\/p>\n

2. Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km.
\nAnswer:
\nThe density of air is maximum near the surface of the earth (at sea level) and it decreases rapidly with height. The density of air at a height of 6 km reduces to almost half it\u2019s value at sea level. Hence the atmospheric pressure at a height of 6 km decreases to nearly half of its value at sea level even though the height of the atmosphere is more than 100km.<\/p>\n

3. Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
\nAnswer:
\nThe pressure created while applying a force on a liquid is transmitted equally in all direction inside the liquid. That is why there is no fixed direction for the pressure transmitted. Hence we can say that hydrostatic pressure is a scalar quantity.<\/p>\n

Question 2.
\nExplain why
\n1. The angle of contact of mercury with glass Is obtuse, while that of water with glass is acute.
\nAnswer:
\nWhen a small quantity of liquid is poured on solid, three interfaces, namely liquid – air, solid – air, and solid-liquid are formed. The surface tension corresponding to theses three layers are SLA<\/sub>, SSA, <\/sub>and SSL<\/sub> respectively. Let \u03b8 be the angle of contact between the liquid and solid.<\/p>\n

The molecules in the region, where the three interfaces meet is in equilibrium. It means that net force acting on them is zero. For the molecule at K to be in equilibrium, we have
\nSSL<\/sub> + SLA<\/sub> cos \u03b8 = SSA <\/sub>or cos \u03b8 = \\(\\frac{\\mathrm{S}_{\\mathrm{SA}}-\\mathrm{S}_{\\mathrm{SL}}}{\\mathrm{S}_{\\mathrm{LA}}}\\) In case of mercury-glass, SSA<\/sub> < SSL<\/sub>, therefore cos \u03b8 is negative or \u03b8 > 90\u00b0 i.e. (obtuse). In case of water-glass SSA<\/sub> > SSL<\/sub>, therefore cos \u03b8 is positive or \u03b8 < 90\u00b0 (acute).
\n\"1st<\/p>\n

2. Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops.
\nAnswer:
\nIn the case of mercury the mercury glass, angle of contact is obtuse. To obtain this obtuse value of angle of contact, mercury tends to form a drop. But in the case of water, the water-glass angle of contact is acute. Thus the water tends to spread out to achieve this acute angle of contact.<\/p>\n

3. The surface tension of a liquid is independent of the area of the surface.
\nAnswer:
\nThe surface tension of a liquid is the force acting per unit length on a line drawn tangentially to the liquid surface which is at rest. As this force is independent of the area of the liquid surface, the surface tension is also independent of the area of the liquid surface.<\/p>\n

4. Water with detergent dissolved in ‘ it should have small angles of contact.
\nAnswer:
\nThe cloth has narrow spaces in the form of capillares. The rise of liquid in a capillary tube is directly proportional to cos \u03b8. If \u03b8 is small cos \u03b8 will be large. Hence capillary rise will be more so that the detergent will penetrate more in cloth and hence gets dissolved.<\/p>\n

5. A drop of liquid under no external forces is always spherical in shape.
\nAnswer:
\nIs the absence of external forces, the surface of the liquid drop tends to acquire the minimum surface area due to surface tension. For a given volume, the surface area of the sphere is least and hence the liquid drop takes the spherical shapes.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFill in the blanks using the word(s) from the list appended with each statement:
\n1. Surface tension of liquids generally …….with temperatures (increases \/decreases)
\nAnswer:
\ndecreases<\/p>\n

2. Viscosity of gases ……with temperature, whereas Viscosity of liquids …….with temperature (increases\/decreases).
\nAnswer:
\nincreases, decreases<\/p>\n

3. For solids with elastic modulus of rigidity, the shearing force is proportional to……, while for fluids it is proportional to …..(shear strain \/rate of shear strain).
\nAnswer:
\nShear strain, rate of shear strain.<\/p>\n

4. For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass \/ Bernoulli\u2019s principle).
\nAnswer:
\nBoth conservation of mass and Bernoulli\u2019s principle.<\/p>\n

5. For the model of a plane in a wind tunnel, turbulence occurs at a ….. speed for turbulence for an actual plane (greater \/ smaller).
\nAnswer:
\nGreater<\/p>\n

Question 4.
\nExplain why.
\n1. To keep a piece of paper horizontal, you should blow over, not under, it.
\nAnswer:
\nIf we blow over a piece, velocity of air above the paper becomes more than that below the paper. This reduces the pressure of the air above the paper in accordance with Bernoulli\u2019s theorem. But the pressure of air below the paper is still atmospheric and is higher than the pressure above. Henc.e the paper remains horizontal and does not fall.<\/p>\n

2. When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
\nAnswer:
\nWhen we try to close the water top with our fingers, the area of outlet of water eject gets reduced. Therefore in accordance with the principle of Continuity (AV = constant), the velocity of the water will increase creating fast jets of water.<\/p>\n

3. The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
\nAnswer:
\nThe size of the needle controls the velocity of flow and the thumb pressure controls the pressure. According to Bernoulli\u2019s equation:
\nP + \u03c1gh+ \\(\\frac{1}{2} \\rho\\) v2<\/sup> = constant
\nwe can note that the equation varies linearly with pressure, P but varies as the square of the velocity. Hence the contribution of velocity of flow is greater, due to which the needle has a better control over the flow.<\/p>\n

4. A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
\nAnswer:
\nWhen a fluid flows out of a small hole in a vessel, it acquires a large velocity and hence has a large momentum. Since no external force is acting on the system, a backward velocity must be acquired by the vessel (By-law of conservation of momentum). As a result backward thrust is experienced by the vessel.<\/p>\n

5. A spinning cricket ball In air does not follow a parabolic trajectory.
\nAnswer:
\nA spinning ball displaces air. The ball moves formed and relative to it, the air moves backward. Hence, the velocity of air above the ball relative to it is larger and below it is smaller. This difference in the velocities of air results in a pressure difference between the lower and upper faces leading to a net upward force on the ball. Therefore the ball does not follow a parabolic trajectory.<\/p>\n

Question 5.
\nA 50 kg girl wearing high heel shoes balances on a single heel. The heel Is circular with a diameter of 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
\nAnswer:
\nForce exerted by the heel due to the weight of the girl, F = mg = 50 \u00d7 10 = 500N
\nDiameter of the heel\u2019s circular are, d = 1 cm
\n= 10-2<\/sup>m
\n\u2234 Area of the heel \\(\\frac{\\pi \\mathrm{d}^{2}}{4}\\)
\n= \\(\\frac{\\pi \\times\\left(10^{-2}\\right)^{2}}{4}=7.85 \\times 10^{-5} \\mathrm{m}^{2}\\)
\n\u2234 Pressure exerted by the heel = \\(\\frac{F}{A}\\)
\n\\(=\\frac{500}{7.85 \\times 10^{-5}}\\)
\n= 6.366 \u00d7 106<\/sup> Pa.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nToricelli\u2019s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m-3<\/sup>. Determine the height of the wine column for normal atmospheric pressure. (Assume g = 9.8 m\/s2<\/sup>)
\nAnswer:
\nNormal atmospheric pressure = 1 atm
\n\u21d2 gauge pressure of wine = 1 atm
\n\u03c1gh = 1.013 \u00d7 105<\/sup> Pa (Since 1 atm = 1.013 \u00d7 105<\/sup>pa)
\n\u21d2 h = \\(\\frac{1.013 \\times 10^{5}}{984 \\times 9.8}\\)
\n\u21d2 h = 10.5 m.<\/p>\n

Question 7.
\nA vertical off-shore structure is built to withstand a maximum stress of 109<\/sup> Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.
\nAnswer:
\nMaximum stress that can be taken by the off-share structure, Pmax<\/sub> = 109<\/sup>Pa.
\ndensity of water = 103<\/sup> kg \/m3<\/sup> and g = 9.8 m\/s2<\/sup> and h = 3km (depth of the sea)
\n\u2234 Pressure exerted by seawater, P = \u03c1gh
\n= 103<\/sup> \u00d7 9.8 \u00d7 3 \u00d7 103<\/sup>
\nP = 2.94 \u00d7 107<\/sup> Pa
\nNote that, P < Pmax<\/sub>
\nSince the pressure exerted by the seawater is less than the maximum pressure the structure can withstand, the structure is suitable for putting on the top of the oil well.<\/p>\n

Question 8.
\nA hydraulic automobile lift is designed to lift cars with a maximum mass of 3000kg. The area of cross\u00acsection of the piston carrying the load is 425 cm2<\/sup>. What maximum pressure would the smaller piston have to bear.
\nAnswer:
\nLet Fmax<\/sub> be the maximum force experienced by the Bigger piston.
\nFmax<\/sub> = mg = 3000 \u00d7 9.8 = 29400 N
\nArea of the bigger piston = 425 cm2<\/sup>
\n= 425 x 10-4<\/sup>m2<\/sup>
\nMaximum pressure on the bigger piston
\npmax<\/sub> = \\(\\frac{F_{\\max }}{A}\\)
\n\\(P=\\frac{29400}{425 \\times 10^{-4}}\\) = 6.92 \u00d7 105<\/sup> Pa
\nSince pressure is transmitted uniformly throughout the liquid the smaller piston will also bear a pressure of 6.92 \u00d7 105<\/sup>Pa.<\/p>\n

Question 9.
\nA U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are In level with 10.0 cm of water in one arm and 12.5 cm of spirit In the other. What is the specific gravity of spirit?
\nAnswer:
\n\"1st
\n\\(\\rho_{1}\\) \u2192 density of water
\n\\(\\rho_{2}\\) \u2192 density of spirit
\nh1<\/sub> \u2192 height of water column
\nh2<\/sub> \u2192 height of spirit column
\nLet Po<\/sub> be the atmospheric pressure
\n\u21d2 PA<\/sub> = Po<\/sub>\u00a0+ \\(\\rho_{1}\\) gh1<\/sub> ….(1)
\nPB<\/sub> = Po<\/sub>\u00a0+ \\(\\rho_{2}\\) gh2<\/sub> …..(2)
\nSince the mercury levels are same at point A and B we can say that,
\nPA<\/sub> = PB<\/sub>
\n\u21d2 Po<\/sub>\u00a0+ \\(\\rho_{1}\\) gh1<\/sub> = Po<\/sub>\u00a0+ \\(\\rho_{2}\\) gh2<\/sub>
\nfrom (1) and (2)
\n\u21d2 \\(\\rho_{1}\\) gh1<\/sub> = \\(\\rho_{2}\\) gh2<\/sub>
\n\u21d2 \\(\\frac{\\rho_{2}}{\\rho_{1}}=\\frac{h_{1}}{h_{2}}=\\frac{10}{12.5}=0.8\\)
\n\u2234 specific gravity of spirit
\n\\(=\\frac{\\text { density of spirit }}{\\text { density of water }}=\\frac{\\rho_{2}}{\\rho_{1}}=0.8\\)<\/p>\n

Question 10.
\nIn the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6)
\nAnswer:
\nSince we are putting equal amount of water and spirit and specific gravity of spirit is < 1 we can assume that the mercury level will rise on the spirit side,
\n\\(\\rho_{1}\\) \u2192 density of water,
\n\\(\\rho_{2}\\) \u2192 density of spirit,
\n\\(\\rho_{3}\\) \u2192 density of mecury,
\n\"1st
\nSince A and B are at the same height,
\nPA<\/sub> = PB<\/sub>
\nwe know that, PA<\/sub> = \\(\\rho_{1}\\)gh1<\/sub>
\nand PB<\/sub> = \\(\\rho_{2}\\) gh2<\/sub> + \\(\\rho_{3}\\) gx<\/sub>
\n\u21d2 \\(\\rho_{1}\\) gh1<\/sub> = \\(\\rho_{2}\\) gh2<\/sub> + \\(\\rho_{3}\\) gx<\/sub>
\n\u21d2 x = \\(\\frac{\\rho_{1} h_{1}-\\rho_{2} h_{2}}{\\rho_{3}}\\)
\nx = \\(\\frac{\\mathrm{h}_{1}-\\frac{\\rho_{2}}{\\rho_{1}} \\mathrm{h}_{2}}{\\frac{\\rho_{3}}{\\rho_{1}}}\\)
\nwhere \\(\\frac{\\rho_{2}}{\\rho_{1}}\\) \u21d2 specific gravity of spirit and \\(\\frac{\\rho_{3}}{\\rho_{1}}\\) \u21d2 specific gravity of mercury.
\nThus x = \\(\\frac{25-0.8 \\times 27.5}{13.6}\\) = 0.221 cm
\nTherefore the difference in the mercury level in the arms is 0.221 cm.<\/p>\n

Question 11.
\nCan Bernoulli\u2019s equation be used to describe the flow of water through a rapid in a river? Explain.
\nAnswer:
\nNo. The flow of water through a rapid in a river is turbulent and hence Bernoulli\u2019s principle cannot be applied.<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nDoes it matter if one uses gauge Instead of absolute pressures in applying Bernoulli\u2019s equation? Explain.
\nAnswer:
\nNo, unless the atmospheric pressure at the two points where Bernoulli\u2019s equation is applied are significantly different.<\/p>\n

Question 13.
\nGlycerine flows steadily through a horizontal tube of length 1.5 m and radius 1cm. If the amount of glycerine collected per second at one end is 4.0 x 10-3<\/sup> kg s-1<\/sup>, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 x 103<\/sup> kg m-3<\/sup> and viscosity of glycerine = 0.83 Pas). [You may also like to check if the assumption of laminar flow in the tube is correct].
\nAnswer:
\nThe volume of liquid floawing out per second is given by, \\(Q=\\frac{\\pi R^{4}\\left(p_{1}-p_{2}\\right)}{8 h l}\\)
\n\u21d2 pressure difference at tube ends, \\(\\mathrm{P}_{1}-\\mathrm{P}_{2}=\\frac{8 \\mathrm{Q} \\eta}{\\pi \\mathrm{r}^{4}}\\)
\nwhere Q is the volume of liquid flowing per second = \\(\\frac{\\text { mass of liquid flowing per second }}{\\text { density of liquid }}\\)
\n\\(=\\frac{4 \\times 10^{-3} \\mathrm{kg} \/ \\mathrm{s}}{1.3 \\times 10^{3} \\mathrm{kg} \/ \\mathrm{m}^{3}}\\)
\n= 3.077 x 10-6<\/sup> m3<\/sup>\/s
\nTherefore,
\nP1<\/sub> – P2<\/sub> = \\(=\\frac{8 \\times 0.83 \\times 1.5 \\times 3.077 \\times 10^{-6}}{3.14 \\times(1 \\times 10)^{3}}\\)
\n= 9.76 x 102<\/sup> N\/m2<\/sup>
\nA pressure difference of 9.76x 102<\/sup> N\/m2<\/sup> is obtained.
\nReynolds Number,
\nR = \\(\\frac{4 \\rho V}{\\pi d n}\\) = \\(\\frac{4 \\times 1.3 \\times 10^{3} \\times 3.08 \\times 10^{-6}}{\\pi \\times(0.02) \\times 0.83}\\) = 0.3
\nSince the Reynolds number is about 0.3, the flaw is laminar.<\/p>\n

Question 14.
\nIn a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s-1<\/sup> and 63 m s-1<\/sup> respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3 kg m-3<\/sup>.
\nAnswer:
\nP1<\/sub> = pressure at the lower surface of wings
\nP2<\/sub> = pressure at the lower surface of wings
\n\u03c1 = Density of air
\nV1<\/sub> = Speed of wind at the lower surface
\n= 63 m\/s.
\nV2<\/sub> = speed of wind at upper surface
\n= 70 m\/s
\nAccording to Bermolli\u2019s theorem,
\nP1<\/sub> + \\(\\frac{1}{2} \\rho V_{1}^{2}\\) = P2<\/sub> + \\(\\frac{1}{2} \\rho V_{2}^{2}\\)
\nor P1<\/sub> – P2<\/sub> = \\(\\frac{1}{2} \\rho\\left(V_{1}^{2}-V_{2}^{2}\\right)\\)
\nP1<\/sub> – p2<\/sub> = \\(\\frac{1}{2}\\) \u00d7 1.3 \u00d7 (702<\/sup> – 632<\/sup>) = 605.15 pa
\nForce on wings = (P1<\/sub> – P2<\/sub>) \u00d7 area
\n= 605.15 \u00d7 2.5 = 1512.87 N
\n\u2234 The lift on the wings is about 1.512 \u00d7 103<\/sup>N<\/p>\n

Question 15.
\nFigures 10.23(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?
\n\"1st
\nAnswer:
\nFigure (a) is incorrect. From the equation of continuity AV = constant. Thus, when the area of constriction is narrow, velocity of streamline is high. If the velocity is high, we can say that the pressure head is low at that point (from Bernoulli\u2019s equation). Correspondingly the height of the liquid column should be lesser at the narrow construction than that of the wide construction. Hence figure (a) is incorrect.<\/p>\n

Question 16.
\nThe cylindrical tube of a spray pump has a cross-section of 8.0 cm2<\/sup> one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the. tube is 1.5 m min-1<\/sup>, what is the speed of ejection of the liquid through the holes?
\nAnswer:
\nNumber of holes = 40
\nDiameter of each hole, D = 10-3<\/sup>m
\nArea of cross section of each hole
\n\\(=\\frac{\\pi D^{2}}{4}\\)
\n\\(=\\frac{\\pi \\times 10^{-6}}{4}\\)
\nTotal area of cross section of 40 holes,
\na2<\/sub> = \\(\\frac{40 \\times \\pi \\times 10^{-6}}{4}\\) = \u03c0 \u00d7 10-5<\/sup>m2<\/sup>
\nspeed of liquid inside the tube,
\nV1<\/sub> = 1.5 m\/min = \\(\\frac{1.5}{60}\\) m\/s
\nArea of cross section of tube,
\na1<\/sub> = 8.0cm2<\/sup> = 8 \u00d7 10-4<\/sup>m2<\/sup>
\nIf V2<\/sub> is the velocity of ejection of liquid through the holes then from equation of
\ncontinuity,
\na1<\/sub> v1<\/sub> = a2<\/sub> v2<\/sub>
\n\u21d2 v2<\/sub> = \\(\\frac{a_{1} v_{1}}{a_{2}}=\\frac{8 \\times 10^{-4} \\times 1.5}{60 \\times \\pi \\times 10.5}\\)
\n\u21d2 v2<\/sub> = 0.637 m\/s
\n\u2234 the speed of ejection is 0.637 m\/s.<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nA U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 x 10-2<\/sup>N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
\nAnswer:
\nThe soap film has two free surfaces.
\n\u2234 Total length of the film to he supported,
\nl = 2 \u00d7 30 = 60 cm = 0.6
\nTotal force on the slider due to surface tension will be, F = S \u00d7 2l = S \u00d7 0.6N
\nAt equilibrium , the force F on slider due to surface tension should balance the weight mg
\n\u2234 F = mg = 1.5 \u00d7 10-2<\/sup> N
\n\u21d2 S \u00d7 0.6 = 1.5 \u00d7 10-2<\/sup>
\n\u21d2 S = 2.5 \u00d7 10-2<\/sup> N\/m.<\/p>\n

Question 18.
\nFigure 10.24 (a) shows a thin liquid film supporting a small weight = 4.5 \u00d7 10-2<\/sup> N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.
\n\"1st
\nAnswer:
\nIn all three figures, the liquid is at the same. Temperature is also the same for each case. Hence, the surface tension in figure (b) and (c) is the same as in figure.
\n1. Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 \u00d7 10-2<\/sup> N.<\/p>\n

Question 19.
\nWhat is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? The surface tension of mercury at that temperature (20 \u00b0C) is 4.65 \u00d7 10-1<\/sup> N m-1<\/sup>. The atmospheric pressure is 1.01 \u00d7 105<\/sup> Pa. Also, give the excess pressure inside the drop.
\nAnswer:
\nRadius of the mercury drop,
\nr = 3mm = 3 \u00d7 10-3<\/sup>m Surface tension of mercury, S = 4.65 \u00d7 10-1<\/sup> Nm-1<\/sup>. Atmospheric pressure, Po<\/sub> = 1.01 \u00d7 105<\/sup> Pa
\nTotal pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure
\n= \\(\\frac{2 \\mathrm{S}}{\\mathrm{r}}\\) + Po<\/sub>
\n= \\(\\frac{2 \\times 4.65 \\times 10^{-1}}{3 \\times 10^{-3}}\\) + 1.01 \u00d7 105<\/sup>
\nTotal Pressure = 1.013 \u00d7 105<\/sup> Pa
\nExcess pressure = \\(\\frac{2 \\mathrm{S}}{\\mathrm{r}}\\)
\n\\(=\\frac{2 \\times 4.65 \\times 10^{-1}}{3 \\times 10^{-3}}\\)
\n= 310 Pa.<\/p>\n

Question 20.
\nWhat is the excess pressure inside a bubble of soap solution of radius 5 mm, given that the surface tension of soap solution at the temperature (20 \u00b0C) is 2.50 \u00d7 10-2<\/sup> N\/m? If an air bubble of the same dimension were formed at a depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 \u00d7 105<\/sup> Pa).
\nAnswer:
\nRadius of the soap bubble, r = 5mm
\n= 5 \u00d7 103<\/sup>m
\nSurface tension of the soap solution,
\nS = 2.5 \u00d7 10-2<\/sup> N\/m
\nSince the soap Bubble has two surfaces,
\nExcess pressure, P = \\(\\frac{4 \\mathrm{S}}{\\mathrm{r}}\\)
\n\\(=\\frac{4 \\times 2.5 \\times 10^{-2}}{5 \\times 10^{-3}}\\)
\nP = 20 Pa
\nExcess pressure inside the soap bubble is 20 Pa
\nRadius of air bubble, r = 5 mm = 5 \u00d7 10-3<\/sup>
\nDepth, h = 40 c = 0.4 m
\nDensity of soap solution, \u03c1 = 1.2 \u00d7 10-3<\/sup> kg\/m2<\/sup>
\nAtmospheric pressure, Po<\/sub> = 1.01 \u00d7 105 Pa
\ng = 9.8 m\/s2<\/sup>
\nExcess pressure inside air bubble,
\nP1<\/sub> = \\(\\frac{25}{r}\\)
\n\u21d2 P1<\/sub> = \\(\\frac{2 \\times 2.5 \\times 10^{-2}}{5 \\times 10^{-3}}\\)
\nP1<\/sub> = 10Pa
\nAt a depth of 0.4m, the total pressure inside the air bubble,
\nPTotal<\/sub> = Po<\/sub> + \u03c1gh + P1<\/sub>
\nPTotal<\/sub> = 1.01 \u00d7 105<\/sup> + 1.2 \u00d7 103<\/sup> \u00d7 9.8 \u00d7 0.4 + 10
\n\u21d2 PTotal<\/sub> = 1.057 \u00d7 105<\/sup>Pa
\n\u2234 The pressure inside the air bubble is 1.06 \u00d7 105<\/sup>Pa.<\/p>\n

Question 21.
\nA tank with a square base of area 1.0 m2<\/sup> is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2<\/sup>. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4 m. compute the force necessary to keep the door closed.
\nAnswer:
\nAssume g = 9.8 m\/s2<\/sup>
\nDensity of water, \\(\\rho_{1}\\) = 103<\/sup> kg\/m3<\/sup>
\nheight of water column, h1<\/sub> = 4m
\n\u2234 Pressure due to water, P1<\/sub> = \\(\\rho_{1}\\) gh1<\/sub>
\n= 103<\/sup> \u00d7 9.8 \u00d7 4
\n= 3.92 \u00d7 104<\/sup> Pa
\nDensity of acid
\n= Relative, densityx density of water = 1.7 \u00d7 103<\/sup> kg\/m3<\/sup>
\nheight of acid column, h2<\/sub> = 4m
\n\u2234 Pressure due to acid, P2<\/sub> = \\(\\rho_{2}\\) gh2<\/sub>
\n= 1.7 \u00d7 103<\/sup> \u00d7 9.8 \u00d7 4
\n\u21d2 P2<\/sub> = 6.664 \u00d7 104<\/sup>Pa
\nPressure difference between the water and acid columns :
\n\u2206 P = P2<\/sub> – P1<\/sub>
\n\u2206 P = 6.664 \u00d7 104<\/sup> – 3.92 \u00d7 104<\/sup>
\n\u21d2 \u2206 P = 2.744 \u00d7 104<\/sup> Pa
\nArea of the door, a = 20 cm2<\/sup>
\n= 20 \u00d7 10-4<\/sup>m2<\/sup>
\n\u2234 Force exerted on the door, F = \u2206 P \u00d7 a
\n\u21d2 F = 2.744 \u00d7 104<\/sup> \u00d7 20 \u00d7 10-4<\/sup>
\nF = 54.88 N
\nTherefore, the force necessary to keep the door closed is 54.88 N.<\/p>\n

Question 22.
\nA manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury,<\/p>\n

    \n
  1. Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), In units of cm of mercury.<\/li>\n
  2. How would the levels change in case of (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).<\/li>\n<\/ol>\n

    \"1st
    \nAnswer:
    \n1. Difference between the mercury levels in the two limbs gives gauge pressure
    \n\u2234 Gauge pressure, Pa<\/sub> = 20 cm of mercury
    \nAtmospheric pressure, Po<\/sub> = 76 cm of mercury
    \n\u2234 Absolute pressure = Atmospheric pressure + Gauge pressure
    \n= 76 cm + 20 cm
    \n= 96 cm of mercury<\/p>\n

    2. Difference between the levels of mercury in the two limbs = – 18 cm
    \n\u2234 Gauge pressure = – 18cm of mercury
    \n\u2234 Absolute pressure = Atmospheric pressure + Gauge pressure
    \n= 76cm – 18 cm
    \n= 58 cm of mercury.<\/p>\n

    Question 23.
    \nTwo vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
    \nAnswer:
    \nTwo vessels having water-filled up to the same height, exert equal pressure on their bases. Also since the base area is the same, they exert an equal amount of force on the base. (F = PA). Force exerted on the sides of the vessels have non zero vertical components. When there vertical components are added, the total force experienced by one vessel will be different than the other vessel depending on the shape of the vessels. That is why the two vessels filled with water to the same vertical height show different readings on a weighing machine.<\/p>\n

    \"KSEEB<\/p>\n

    Question 24.
    \nDuring a blood transfusion, the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein?
    \nAnswer:
    \nGauge pressure, P = 2000 Pa
    \nDensity of blood, \u03c1 = 1.06 \u00d7 103<\/sup> kg\/m3<\/sup>
    \ng = 9.8 m\/s2<\/sup>
    \nLet the height of the blood container be h
    \n\u2234 pressure of the blood container, P = \u03c1gh
    \n\u21d2\u00a0 \u03c1gh = 2000
    \nh = \\(\\frac{2000}{1.06 \\times 10^{3} \\times 9.8}\\)
    \n\u21d2 h = 0.1925 m
    \nThe blood may just enter the vein if the the blood container is kept at a height slightly greater than 0.1925 m.<\/p>\n

    Question 25.
    \nIn deriving Bernoulli\u2019s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.<\/p>\n

      \n
    1. What is the largest average velocity of blood flow in an artery of diameter 2 \u00d7 10-3<\/sup> m if the flow must remain laminar?<\/li>\n
    2. How does the pressure change as the fluid moves along the tube if dissipative forces are present?<\/li>\n
    3. Do the dissipative forces become more important as the fluid velocity Increases? Discuss qualitatively.<\/li>\n<\/ol>\n

      Answer:
      \n1. Diameter of artery = 2 \u00d7 10-3<\/sup>m
      \nViscosity of blood, = 2.084 \u00d7 10-3<\/sup>Pa s
      \nDensity of blood, \u03c1 = 1.06 \u00d7 103<\/sup>kg\/m3<\/sup>
      \nReynolds number for laminar flow,
      \nNR<\/sub> = 2000
      \nThe largest average velocity of blood is
      \n\\(\\mathrm{V}_{\\mathrm{avg}}=\\frac{\\mathrm{N}_{\\mathrm{R}} \\eta}{\\rho \\mathrm{d}}\\)
      \n\\(=\\frac{2000 \\times 2.084 \\times 10^{-3}}{1.06 \\times 10^{3} \\times 2 \\times 10^{-3}}\\)
      \n= 1.966 m\/S
      \nThe largest average velocity of blood is 1.966 m\/s.<\/p>\n

      2. If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces due to which the pressure drop becomes large.<\/p>\n

      3. The dissipative forces become more important with increasing flow velocity. This is because of the rise of turbulence. Turbulent flow causes dissipative losses in fluid.<\/p>\n

      Question 26.<\/p>\n

        \n
      1. What is the largest average velocity of blood flow in an artery of radius 2 \u00d7 10-3<\/sup>m if the flow must remain laminar?<\/li>\n
      2. What is the corresponding flow rate? (Take viscosity of blood to be 2.084 \u00d7 10-3<\/sup> Pas).<\/li>\n<\/ol>\n

        Answer:
        \n1. Diameter of artery, d = 2 \u00d7 2 \u00d7 10-3<\/sup> m
        \n= 4 \u00d7 10-3<\/sup> m
        \nViscosity of blood, h = 2.084 \u00d7 10-3<\/sup> m Pas
        \nDensity of blood, \u03c1 = 1.06x 10-3<\/sup> kg\/m3<\/sup>
        \nReynold\u2019s number for laminar flow,
        \nNR<\/sub> = 2000
        \nThe largest average velocity of blood is,
        \n\\(\\mathrm{V}_{\\mathrm{avg}}=\\frac{\\mathrm{N}_{\\mathrm{R}} \\eta}{\\rho \\mathrm{d}}\\)
        \n\\(V_{\\text {avg }}=\\frac{2000 \\times 2.084 \\times 10^{-3}}{1.06 \\times 10^{3} \\times 2 \\times 10^{-3}}\\)
        \nV = 0.983 m\/s
        \nHence, the largest average velocity of blood is 0.983 m\/s.<\/p>\n

        2. Flow rate is given by
        \nR= \u03c0r2<\/sup>Vavg<\/sub>
        \nR = 3.14 \u00d7 (2 \u00d7 10-3<\/sup>)2<\/sup> \u00d7 0.983
        \n\u21d2 R = 1.235 \u00d7 10-5<\/sup>m3<\/sup>\/s
        \nTherefore, the corresponding flow rate is 1.235 \u00d7 10-5<\/sup>m3<\/sup>\/s<\/p>\n

        Question 27.
        \nA plane is in level flight at constant speed and each of Its two wings has an area of 25 m2<\/sup>. If the speed of the air is 180 km\/h over the lower wing and 234 km\/h over the upper wing surface, determine the plane\u2019s mass. (Take air density to be 1 kg m-3<\/sup>).
        \nAnswer:
        \nTotal area of wings, A = 2 \u00d7 25
        \n= 50m2<\/sup>
        \nspeed of air lover the longer wing,
        \nV1<\/sub> = 180 km\/h = 50 m\/s
        \nspeed of air over the upper wings,
        \nV2<\/sub> = 243 km\/h = 65 m\/s
        \nDensity of air, \u03c1 =1kg\/m3<\/sup>
        \nLet P1<\/sub> and P2<\/sub>be the pressure of air over the lower wing and upper wing respectively.
        \nFrom Bimoull\u2019s equation,
        \nP1<\/sub> + \\(\\frac{1}{2} \\rho V_{1}^{2}\\) = P2<\/sub> + \\(\\frac{1}{2} \\rho V_{2}^{2}\\)
        \nP1<\/sub> – P2<\/sub> = \\(\\frac{1}{2} \\rho\\left(V_{2}^{2}-V_{1}^{2}\\right)\\)
        \n\u2234 pressure difference, \u2206 P
        \n= \\(\\frac{1}{2} \\rho\\left(V_{2}^{2}-V_{1}^{2}\\right)\\)
        \n= \\(\\frac{1}{2}\\) \u00d7 1 \u00d7 (652<\/sup> – 502<\/sup>)
        \n= 862.5 Pa
        \nThe upward force on the plane, F = ( \u2206 P) A
        \n= 862.5 \u00d7 50
        \n=43125 N
        \nWe know that the upward force balances the weight of the plane
        \n\u2234 F = mg
        \n43125 = m \u00d7 9.8
        \n\u21d2 m = 4400.51 kg.<\/p>\n

        Question 28.
        \nIn Millikan\u2019s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 \u00d7 10-5<\/sup> m and density 1.2 \u00d7 103<\/sup> kg m-3<\/sup>? Take the viscosity of air at the temperature of the experiment to be 1.8 \u00d7 10-5<\/sup> Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.(Take g = 9.8 ms2)
        \nAnswer:
        \nRadius of the given unchanged drop, r = 2 \u00d7 10-5<\/sup> m
        \nDensity of the uncharged drop,
        \n\u03c1 = 1.2 \u00d7 103<\/sup> kg\/m-3<\/sup>
        \nViscosity of air, \\(\\eta\\) = 1.8 \u00d7 10-5<\/sup> Pa S
        \nThe density of air (\\(\\rho_{0}\\)) can be taken as zero in order to neglect the buoyancy of air.
        \nTerminal velocity (v) is given by the relation:
        \n\"1st
        \nV = 5.807 \u00d7 10-2<\/sup> m\/s
        \nV = 5.8 cm\/s
        \nThe terminal speed of the drop is 5.8 cm\/s
        \nViscous force, F = \u03c0 6 hrv
        \n\u2234 F = 6 \u00d7 3.14 \u00d7 1.8 \u00d7 10-5<\/sup> \u00d7 2 \u00d7 105<\/sup> \u00d7 5.8 \u00d7 102<\/sup>
        \n\u21d2 F = 3.9 \u00d7 10-10<\/sup>N
        \nThe viscous force on the drop is
        \n= 3.9 \u00d7 10-10<\/sup> N<\/p>\n

        Question 29.
        \nMercury has an angle of contact equal to 140\u00b0 with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? The surface tension of mercury at the temperature of the experiment is 0.465 N m. Density of mercury = 13.6 \u00d7 103<\/sup> kg m-3<\/sup>.
        \nAnswer:
        \nAngle of contact 1 \u03b8 = 140\u00b0
        \nRadius of the tube, r = 1 mm = 10-3<\/sup>m
        \nSurface tepsion of mercury S = 0.465 N\/m
        \nDensity of mercury, \u03c1 = 13.6 \u00d7 103<\/sup> kg\/m3<\/sup>
        \ng = 9.8 m\/s2<\/sup>
        \nLet the Dip in height of mercury be h Surface tension is given by,
        \nS= \\(\\rho_{o} e^{-y \/ y_{o}}\\)
        \n\u21d2 h = \\(\\frac{2 \\mathrm{S} \\cos \\theta}{\\rho \\mathrm{gr}}\\)
        \n\u21d2 h = \\(\\frac{2 \\times 0.465 \\times \\cos 140^{\\circ}}{13.6 \\times 10^{3} \\times 9.8 \\times 10^{-3}}\\)
        \nh = – 0.00534 m = -5.34 mm
        \nNegative sign indicates that mercury level, dips by 5.34 mm.<\/p>\n

        \"KSEEB<\/p>\n

        Question 30.
        \nTwo narrow bores of diameters 3.0 mm and 6.0 mm are Joined together to form a U-tube open at both ends. If the U-tube contains water, what Is the difference In its levels in the two limbs . of the tube? Surface tension of water at the temperature of the experiment is 7.3 \u00d7 10-2<\/sup> N\/m. Take the angle of contact to be zero and density of water to be 1.0 \u00d7 103<\/sup> kg m-3<\/sup> (g= 9.8 m s-2<\/sup>).
        \nAnswer:
        \nRadius of the first bore, r1<\/sub> = \\(\\frac{3 \\mathrm{mm}}{2}\\)
        \n= 1.5 \u00d7 10-3<\/sup> m
        \nRadius of the second bore, r2<\/sub> = \\(\\frac{6 \\mathrm{mm}}{2}\\)
        \n= 3 \u00d7 10-3<\/sup> m
        \nSurface tension of water, S
        \n= 7.3 \u00d7 10-2<\/sup> N\/m
        \nAngle of contact, \u03b8 = 0\u00b0
        \n(give in the question) Density of water, \u03c1 = 1 \u00d7 103<\/sup> kg\/m3<\/sup>
        \nLet h1<\/sub> and h2<\/sub> he the heights to which the water rises in the 3 mm and 6 mm diameter bores respectively.We have,
        \nh1<\/sub> = \\(\\frac{2 \\mathrm{S} \\cos \\theta}{\\mathrm{r}_{1} \\rho \\mathrm{g}}\\)
        \nh2<\/sub> = \\(\\frac{2 \\mathrm{S} \\cos \\dot{\\theta}}{\\mathrm{r}_{2} \\rho \\mathrm{g}}\\)
        \nDifference in water level = h1<\/sub> – h2<\/sub>
        \n\"1st
        \n= 4.966 \u00d7 10-3<\/sup>m
        \n= 4.97 mm
        \nThe difference in the water levels in the cores is 4.97 mm.<\/p>\n

        Question 31.
        \n1. It is known that density P of air decreases with height y as\\(\\rho_{\\mathrm{o}} \\mathrm{e}^{-y \/ \\mathrm{y}_{0}}\\) where \\(\\rho_{0}\\) = 1.25 kg\/m-3<\/sup> is the density at sea level, and yo<\/sub> is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of the atmosphere remains a constant (isothermal conditions). Also, assume that the value of g remains constant.
        \n2. A large He balloon of volume 1425 m3<\/sup> is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as It rises. How high does it rise?
        \n[Take y0<\/sub> = 8000m and \\(\\rho_{\\mathrm{He}}\\) = 0.18kg\/m3<\/sup>].
        \nAnswer:
        \n1. Volume of the balloon, V = 1425m3<\/sup>
        \nPayload mass, m = 400 kg
        \ng = 9.8 m\/s2<\/sup>
        \nGive that, yo<\/sub> = 8000m
        \n\\(\\rho_{\\mathrm{He}}\\) =0.18 kg\/m3<\/sup>
        \n\\(\\rho_{\\mathrm{o}}\\) = 1.25 kg\/m3<\/sup>
        \nDensity of the balloon = \u03c1
        \nLet the height to which the balloon rises be y Density (\u03c1) of air decreases with height (y) as,
        \n\"1st
        \nwe can infer from (1), that rate of decrease of density with height is directly proportional to \u03c1,i.e.,
        \n\"1st
        \n\\(\\frac{\\mathrm{d} \\rho}{\\rho}=-\\mathrm{kdy}\\)
        \nwhere, K is the constant of proportionality Height changes from o to y, while density changes from \\(\\rho_{\\mathrm{o}}\\)
        \nto \u03c1 by integrating.
        \nwe get:
        \n\"1st
        \nfrom (1) and (2),
        \nyo<\/sub> = 1\/K
        \n\u21d2 K = \\(\\frac{1}{y_{0}}\\)
        \nFrom (2) and (3) we get,
        \n\"1st<\/p>\n

        2. Density,
        \n\"1st
        \n\u21d2 \u03c1 = 0.46 kg\/m3<\/sup>
        \nwe have proved that,
        \n\"1st
        \n\u21d2 y = – 8000 \u00d7 loge<\/sub> \\(\\frac{0.46}{1.25}\\)
        \n\u21d2 y = 8000m = 8km .
        \nHence balloon will rise to a height of 8 km.<\/p>\n

        1st PUC Physics Mechanical Properties Of Fluids One Mark Questions and Answers<\/h3>\n

        Question 1.
        \nPascal\u2019s law is valid for dynamic fluids. True\/False.
        \nAnswer:
        \nFalse. Valid only for static fluid.<\/p>\n

        Question 2.
        \nWhat is the relation between absolute pressure and gauge pressure?
        \nAnswer:
        \nAbsolute pressure = atmospheric pressure \u00b1 gauge pressure.<\/p>\n

        Question 3.
        \nAngle of contact for a glass tube dipped in mercury is obtuse. True \/ False.
        \nAnswer:
        \nTrue.<\/p>\n

        \"KSEEB<\/p>\n

        Question 4.
        \nWater is an ideal fluid. True\/False.
        \nAnswer:
        \nFalse. It is real fluid.<\/p>\n

        Question 5.
        \nDefine thrust.
        \nAnswer:
        \nThe total normal force exerted by the fluid at rest on a surface is called thrust.<\/p>\n

        Question 6.
        \nWhat is the dimensional formula for pressure?
        \nAnswer:
        \nML-1<\/sup>T-2<\/sup><\/p>\n

        Question 7.
        \nWhat happens to the viscosity of a gas when the temperature is increased?
        \nAnswer:
        \nViscosity of gases increases as the temperature is increased.<\/p>\n

        \"KSEEB<\/p>\n

        Question 8.
        \nHow does the coefficient of viscosity of liquid vary with temperature?
        \nAnswer:
        \nThe coefficient of viscosity of liquids decreases as temperature is increased.<\/p>\n

        1st PUC Physics Mechanical Properties Of Fluids Two Marks Questions and Answers<\/h3>\n

        Question 1.
        \nName two applications of Pascal\u2019s law.
        \nAnswer:
        \nHydraulic lifts, hydraulic breaks.<\/p>\n

        Question 2.
        \nWhat is streamline? What are its properties?
        \nAnswer:
        \nThe path taken by a fluid particle under a steady flow is a streamline.
        \nProperties:<\/p>\n

          \n
        1. The tangent at any point in streamline gives direction of fluid velocity at that point.<\/li>\n
        2. No two streamlines can cross.<\/li>\n<\/ol>\n

          Question 3.
          \nDefine the specific gravity of a fluid. What is its unit?
          \nAnswer:
          \nSpecific gravity is defined as a ratio of a density of a fluid to the density of water at 4\u00b0C. It has no unit.<\/p>\n

          Question 4.
          \nWhich are the factors on which viscous force depends?
          \nAnswer:
          \nViscous force acting between two layers of liquids depends on,<\/p>\n

            \n
          1. area of the layers<\/li>\n
          2. relative velocity of the two layers and<\/li>\n
          3. distance between two layers.<\/li>\n<\/ol>\n

            \"KSEEB<\/p>\n

            Question 5.
            \nA force 6f 1 kN is applied on a road surface by the tyre of a car. If the contact area of the tyre is 10 cm2<\/sup>, find the pressure on the contact area.
            \nAnswer:
            \nPressure = \\(\\frac{\\text { Force }}{\\text { area }}\\)
            \n\\(\\frac{1000 N}{10 \\times 10^{-4} M^{2}}\\) = 106<\/sup> Pa<\/p>\n

            Question 6.
            \nState whether true or false :<\/p>\n

              \n
            1. An ideal liquid has non zero viscosity but zero compressibility.<\/li>\n
            2. With increase in temperature, the viscosity of liquid decreases but the viscosity of gases increases.<\/li>\n<\/ol>\n

              Answer:<\/p>\n

                \n
              1. False<\/li>\n
              2. True.<\/li>\n<\/ol>\n

                Question 7.
                \nA bigger raindrop falls faster than a smaller one (True\/False) Explain.
                \nAnswer:
                \nTrue.
                \nThe raindrop moves with terminal velocity due to viscous drag of the air. The terminal velocity of drop varies as the square of its radius. Hence, a bigger drop has a higher terminal velocity than a smaller one.<\/p>\n

                Question 8.
                \nMachine parts get jammed during winter. Why?
                \nAnswer:
                \nIn winter, due to low temperature, the viscosity of oil between the machine parts increases considerably resulting in jamming of machine parts.<\/p>\n

                Question 9.
                \nDiscuss the effect of temperature on the viscosity of gases and liquids.
                \nAnswer:
                \nViscosity of liquid decreases with increase in temperature and viscosity of gases increases with increase in temperature.<\/p>\n

                Question 10.
                \nWhat is an ideal liquid?
                \nAnswer:<\/p>\n

                  \n
                1. An ideal liquid is incompressible meaning density of liquid remains irrespective of the pressure.<\/li>\n
                2. An ideal liquid is non-viscous. No tangential forces between layers of liquid in relative motion.<\/li>\n
                3. Ideal liquid cannot withstand any shearing stress.<\/li>\n<\/ol>\n

                  \"KSEEB<\/p>\n

                  Question 11.
                  \nExplain the effect of density and pressure on the viscosity of liquids\/gases.
                  \nAnswer:
                  \nLiquids:
                  \n1. Viscosity increases with increase in density of the liquid. Viscosity increases with increase in pressure (except water).<\/p>\n

                  2. Gases:
                  \nViscosity decreases with increase in density. Viscosity decreases with increase in pressure.<\/p>\n

                  Question 12.
                  \nThe excess pressure inside a soap bubble is four times the excess pressure inside a moving soap bubble. What is the ratio of the first and second bubble?
                  \nAnswer:
                  \nLet r1<\/sub> and r2<\/sub> be radius of soap bubbles. Excess pressure in them are
                  \n\\(\\frac{4 \\sigma}{r_{1}}\\) and \\(\\frac{4 \\sigma}{r_{2}}\\) where \u03c3 is surface tension.
                  \nGiven P1<\/sub> = 4P2<\/sub> \\(\\frac{4 \\sigma}{r_{1}}\\) = 4 \\(\\frac{4 \\sigma}{r_{2}}\\)
                  \nratio of volume of I to II
                  \n\\(=\\frac{r_{1}^{3}}{r_{2}^{3}}=\\frac{r_{1}^{3}}{64 r_{1}^{3}}=\\frac{1}{64}\\)<\/p>\n

                  Question 13.
                  \nExplain why water with detergent dissolved in it should have small angles of contact.
                  \nAnswer:
                  \nCloth has narrow spaces inform of capillaries. The rise of liquid in a capillary tube is directly proportional to cos e.
                  \n\u03b8 \u221d \\(\\frac{1}{\\cos \\theta}\\). Hence e should be less in order that detergent penetrates more in clothes.<\/p>\n

                  Question 14.
                  \nWhat is a venturi meter? Name two applications.
                  \nAnswer:
                  \nVenturi meter is a device to measure the flow speed of incompressible fluid. The principle is used in filter pumps, Bunsen burner, atomiser.<\/p>\n

                  Question 15.
                  \nWater is filled as shown In following containers. In which case highest force is exerted on the base of the container? Why?
                  \n\"1st
                  \nAnswer:
                  \nWater exerts same pressure at the bottom of the container, regardless of its shape. Since height, his constant, equal pressure is exerted.
                  \nP = \u03c1gh (h = constant)<\/p>\n

                  Question 16.
                  \nIs it better to wash clothes in hot soap solution? Why?
                  \nAnswer:
                  \nYes. Surface tension decreases with temperature. Hence, in hot soap solution, spreading of solution over the surface of clothes happens easily. Hot soap solution can penetrate and clean clothes better.<\/p>\n

                  Question 17.
                  \nExplain why a drop of liquid under no external force is always spherical in shape.
                  \nAnswer:
                  \nDue to surface tension, a drop of liquid tries to acquire a shape with minimum surface area. For a given volume, surface area of sphere is least. Hence, liquid drop takes a spherical shape.<\/p>\n

                  Question 18.
                  \nPoint out any two properties required for a fluid to be used in a barometer.
                  \nAnswer:<\/p>\n

                    \n
                  1. Density of fluid should be high so as to avoid use of long barometer,<\/li>\n
                  2. Temperature variation of fluid should be minirial.<\/li>\n<\/ol>\n

                    \"KSEEB<\/p>\n

                    Question 19.
                    \nWhy does water flow faster that honey?
                    \nAnswer:
                    \nThe coefficient of viscosity of water is less compared to that of honey. This makes water to move faster.<\/p>\n

                    Question 20.
                    \nWhat are the factors affecting viscosity?
                    \nAnswer:<\/p>\n

                      \n
                    1. Increase in temperature decreases viscosity.<\/li>\n
                    2. Increases in pressure increases viscosity in liquids. In water, it decreases whereas, in gases, it remains the same.<\/li>\n<\/ol>\n

                      Question 21.
                      \nWhat do you mean by angle of contact of a liquid with a solid surface. What are the factors that effect it?
                      \nAnswer:
                      \nThe angle made by the tangent drawn to the meniscus from the point of contact with the walls of the container measured from within the liquid is called angle of contact. It depends on the atmosphere pressure, adhesive and cohesive forces of the liquid in the tube.<\/p>\n

                      Question 22.
                      \nBernoulli\u2019s equation can be used to describe the flow of a river rapidly. True\/False. Explain why?
                      \nAnswer:
                      \nFalse. Bernoulli\u2019s equation can be applied only to streamline flow.<\/p>\n

                      Question 23.
                      \nArchimedes\u2019s Principle holds good in a vessel at free fall. True\/False. Give reason.
                      \nAnswer:
                      \nFalse
                      \nThe principle does not hold good in this particular case as the vessel in free fall is in a condition of weightlessness, where the buoyant force accounting for the Archimedes\u2019s principle does not exist.<\/p>\n

                      Question 24.
                      \nWhy are cars and aeroplanes streamlined?
                      \nAnswer:
                      \nThe shape of the aeroplane is streamlined because when it flies in air. The velocity on the top surface is more than the bottom surface. Hence the pressure on the top surface decreases. This causes an upward thrust on the wings of the place which gives uplift to the aeroplane.<\/p>\n

                      Question 25
                      \nWhy do some liquids rise and some liquids dip in a capillary tube? Explain
                      \nAnswer:
                      \nThe rise or fall of liquid in a capillary tube is due to surface tension. If the liquid wets the glass then there is rise in the liquid inside the capillary tube.<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 26.
                      \nDefine cohesive force and adhesive, force of molecules.
                      \nAnswer:
                      \nForce of attraction between molecules of the same substance is called cohesive force. Force of attraction between molecules of the different substances is called adhesive force.<\/p>\n

                      Question 27.
                      \nDiameter of a ball A is thrice that of B. What will be the ratio of their terminal velocities in water?
                      \nAnswer:
                      \nTerminal velocity \u00d7 (rad. of ball)2<\/sup> Required ratio is \\(\\frac{3^{2}}{1}\\) = 9<\/p>\n

                      Question 28.
                      \nA hole of area 2 mm opens near bottom of a large water storage tank and stream of water shoots from it. If top of the water in the tank is to be kept 30 m above leaking point, how much water in litres Is should be added to reservoir tank to keep this level? Take g = 10 m\/s2<\/sup>
                      \nAnswer:
                      \nVelocity of the outflowing water
                      \nv = \\(\\sqrt{2 g h}=\\sqrt{2 \\times 10 \\times 30}\\) = 24.49 m\/s
                      \nQuantity of water flowing out per sec.
                      \n= av = 10-5<\/sup> \u00d7 4 \u00d7 24.49
                      \n= 97.96 \u00d7 10-6<\/sup>
                      \n= 9.796 \u00d7 10-5<\/sup> \u00d7 103<\/sup> litres\/s
                      \n= 97.96 ml\/s.<\/p>\n

                      Question 29.
                      \nA small ball of mass m and density \u03c1 is dropped in a viscous liquid having density \\(\\rho_{0}\\). After some time, ball falls with a constant velocity. Calculate viscous force acting on the ball.
                      \nAnswer:
                      \nVolume of the ball, v = m\/\u03c1; mass of liquid displaced by the ball.
                      \n\\(\\mathrm{m}^{\\prime}=\\left(\\frac{\\mathrm{m}}{\\rho}\\right) \\rho_{0}\\)
                      \n\u03c1 = density of ball
                      \nWhen the ball falls with a constant velocity
                      \n\u21d2 viscous force = effective weight of the ball or F = mg = m’g = (m – m’)g
                      \n\\(=\\left(\\mathrm{m}-\\frac{\\mathrm{m} \\rho_{0}}{\\rho}\\right) \\mathrm{g}=\\mathrm{mg}\\left(1-\\frac{\\rho_{0}}{\\rho}\\right)\\)<\/p>\n

                      Question 30.
                      \nWhat is Reynold\u2019s number? What is its significance?
                      \nAnswer:
                      \nTurbulent flow is less likely for viscous fluid flowing at low rates. A dimensionless number, whose value gives one an approximate idea whether the flow would be turbulent. This number is called the Reynold\u2019s Re<\/sub>.
                      \n\\(R_{e}=\\rho v d \/ \\eta\\)
                      \n\u03c1 \u2192 density of fluid
                      \nspeed v,d stands for the dimension of the pipe.
                      \n\\(\\eta\\) \u2192 viscosity of the fluid. Significance is that,
                      \nflow is lamina\u2019s\/streamline for Re<\/sub> <1000 flow is turbulent for Re<\/sub> > 2000.<\/p>\n

                      Question 31.
                      \nA wooden log 100 mm \u00d7 100 mm \u00d7 5 m hangs vertically from a vertical string so that 3 m length of log Is submerged in water. Find the tension in string. Take specific gravity of wood = 0.65.
                      \nAnswer:
                      \n\"1st
                      \nFB<\/sub> + T = W
                      \n(0.1 \u00d7 0.1 \u00d7 3) \u00d7 9810 + T
                      \n= (0.1 \u00d7 0.1 \u00d7 5) \u00d7 0.65 \u00d7 9810
                      \nT = 24.52 N
                      \n(0.1 \u00d7 0.1 \u00d7 3) \u00d7 9810 + T
                      \n= (0.1 \u00d7 0.1 \u00d7 5) \u00d7 0.65 \u00d7 9810
                      \n\u2234 T = 24.52 N<\/p>\n

                      Question 32.
                      \nA piston of small cross-section A1<\/sub> is used to exert a force F1<\/sub> on liquid to transfer it from the existing cylinder to a larger cylinder attached with a larger piston of area A2<\/sub>. What is the weight, say if a truck that can be placed on a platform supported by the larger piston due to the force exerted on the liquid in existing cylinder?
                      \nAnswer:
                      \nF2<\/sub> = \\(\\frac{\\mathrm{F}_{1} \\mathrm{A}_{2}}{\\mathrm{A}_{1}}=\\frac{100 \\mathrm{kN} \\times 1 \\mathrm{m}^{2}}{10 \\times 10^{-4} \\mathrm{m}^{2}}\\)
                      \n= 105<\/sup> kN
                      \n\u2234 Weight of truck = 105<\/sup> kN.<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 33.
                      \nThe pressure inside a droplet of water is 10.3 N\/m2<\/sup> in excess of atmospheric pressure. Its diameter is 50 mm. Find surface tension of the water film.
                      \nAnswer:
                      \nP = \\(\\frac{2 \\lambda}{r}\\)
                      \n\u03bb = \\(\\frac{P r}{2}=\\frac{10.3 \\times 25 \\times 10^{3}}{2}\\)
                      \n= 128.75 N\/m<\/p>\n

                      1st PUC Physics Mechanical Properties Of Fluids Four\/Five Marks Questions and Answers<\/h3>\n

                      Question 1.
                      \nDefine viscosity. Discuss the cause the viscosity.
                      \nAnswer:
                      \nViscosity is the property of a fluid, (liquid or gas) by virtue of which an internal frictional force comes into play when the fluid is in motion and opposes the relative motion of its different layers. Viscosity is due to the intermolecular forces which are effective when the different layers of the liquid are moving with different velocities.<\/p>\n

                      These forces are of Vander waal type. Due to these forces, every fast-moving liquid layers tends to accelerate the adjoining slow-moving layers and every slow-moving layer tends to retard the adjoining fast-moving layer of liquid. As a result a backward dragging force called viscous drag comes into play which accounts for viscosity of liquid.<\/p>\n

                      Question 2.
                      \nDerive the dimensional formula for co-efficient of viscosity and hence its unit.
                      \nAnswer:
                      \n\"1st
                      \nUnits:
                      \nWe have \\(\\eta=\\frac{F}{A d v \/ d n}\\)
                      \nin cgs system, the unit of \\(\\eta\\) is called poise.
                      \n1 poise \\(=\\frac{1 \\mathrm{dyne}}{1 \\mathrm{cm}^{2} \\times(1(\\mathrm{cm} \/ \\mathrm{s}) \/ \\mathrm{cm})}\\)
                      \n= dyne cm-2<\/sup> sec.
                      \nThe S.I.unit of \\(\\eta\\) is Pa.s (Pascal second) or deca poise.
                      \n1 dec poise = \\(\\frac{1 \\mathrm{N}}{1 \\mathrm{m}^{2}\\left(1 \\mathrm{ms}^{-1} \/ \\mathrm{m}\\right)}\\) = 1Nsm-2<\/sup><\/p>\n

                      Question 3.
                      \nState Stoke\u2019s law? What are the factors on which viscous drag depends?
                      \nAnswer:
                      \nStoke\u2019s law states the backward dragging force (F) acting on a small spherical body of radius r, moving through a medium having coefficient of viscosity \\(\\eta\\) and velocity v is given by
                      \nF = 6\u03c0\\(\\eta\\)rv
                      \nviscous drag (F) depends on<\/p>\n

                        \n
                      1. coefficient of viscosity of the medium \\(\\eta\\)<\/li>\n
                      2. velocity of the body (v)<\/li>\n
                      3. radius of the spherical body (r)<\/li>\n<\/ol>\n

                        Question 4.
                        \nHow is the rise of liquid affected if top of the capillary is closed?
                        \nAnswer:
                        \nThere will be a small rise in the capillary tube if top of capillary tube is closed. Because the rise of liquid in capillary tube due to surface tension will be opposed by the downward force exerted by the compressed air above the liquid in tube. This downward force increases with increase in height of liquid column. Therefore only a small rise of liquid column is possible in a capillary tube with a closed top.<\/p>\n

                        Question 5.
                        \nWhat is terminal velocity? What are the factors on which terminal velocity depends?
                        \nAnswer:
                        \nThe terminal velocity of an object is the maximum constant velocity acquired by the object while falling freely in a viscous medium.
                        \nTerminal velocity (v) of an object of radius (r) density \\((\\rho)\\) moving through a viscous medius of viscosity \\(\\eta\\) and density \\(\\rho_{0}\\) is given by
                        \n\\(\\mathrm{v}=\\frac{2}{9} \\frac{\\mathrm{r}^{2}}{\\eta}\\left(\\rho-\\rho_{0}\\right) \\mathrm{g}\\)
                        \nHence, terminal velocity depends on<\/p>\n

                          \n
                        1. radius of the object<\/li>\n
                        2. coefficient of viscosity of the medium<\/li>\n
                        3. density of the object.<\/li>\n
                        4. density of the medium<\/li>\n<\/ol>\n

                          Question 6.
                          \nDerive expression for capillary rise in a tube in terms of surface tension, radius of tube angle of contact and density of fluid.
                          \nAnswer:
                          \nConsider a tube of radius \u2018a\u2019 dipped in a container filled with fluid of density of \u03c1 as shown.
                          \n\"1st
                          \nh = capillary rise
                          \ns = surface tension
                          \n\u03b8 = angle of contact
                          \na = radius of tube
                          \nThe weight of column of liquid of height h in the tube is balanced by vertical component of the surface tensile force of fluid. Vertical component of the surface tensile force
                          \n= (S cos \u03b8) \u00d7 (length over which it acts)
                          \n= (S cos \u03b8) \u00d7 circumference of tube
                          \n= S cos \u03b8 \u00d7 2\u03c0a ……..(1)
                          \nWeight of column of fluid balanced by surface tension = volume of fluid \u00d7 density \u00d7 g
                          \n= (\u03c0a2<\/sup>h) \u00d7 \u03c1g ……(2)
                          \nEquating (1)and(2),
                          \nS cos\u03b8 \u00d7 2\u03c0a = \u03c0a2<\/sup>h \u03c1g
                          \nh \\(=\\frac{S \\cos \\theta \\times 2}{a \\rho g}\\)
                          \n\\(=\\frac{2 \\mathrm{S} \\cos \\theta}{\\mathrm{a} \\rho \\mathrm{g}}\\)<\/p>\n

                          Question 7.
                          \nState and prove the Archimedes principle?
                          \nAnswer:
                          \nArchimedes principle states that when a body is wholly partially immersed in a liquid at rest. The loss of weight of the body in the liquid is equal to the weight of the liquid displaced by the immersed part of the body.
                          \nProof:
                          \n\"1st
                          \nLet A be the cross-sectional area of the top or bottom face of the object.
                          \nVolume of liquid displaced, v = volume of the object = Ah
                          \n\u2234 mass of liquid displaced, m = v\u03c1 = Ah\u03c1
                          \n\u03c1 is the density of the liquid.
                          \nLiquid pressure on the top of the body
                          \nP1<\/sub> = x\u03c1g
                          \nVertical downward thrust (Force f1<\/sub>) on top face of the object.
                          \nF1<\/sub> = P1<\/sub>A = xgA\u03c1
                          \nLiquid pressure on the bottom face of body P2<\/sub> = (x + h) \u03c1g
                          \nVertical upward thrust (Force F2<\/sub>) on bottom force
                          \nF2<\/sub> = P2<\/sub> A(X + h) \u03c1gA.
                          \nSince F2<\/sub> >F1<\/sub>, the net upward force acting is F = F2<\/sub> – F = (x + h) \u03c1 gA – x \u03c1gA
                          \n= h\u03c1gA = hA(\u03c1g)
                          \n= mg
                          \nweight of liquid displaced.
                          \nTrue weight of the object = Mg
                          \nUpward thrust (Force F) on the body F = mg
                          \nApparent weight of the body in liquid.
                          \n= W-F = Mg – mg
                          \nThis means that observed weight at the body (object) immersed in a liquid is less than its true weight by an amount equal to weight of the liquid displaced by body.<\/p>\n

                          Question 8.
                          \nState and prove Bernoulli’s theorem.
                          \nAnswer:
                          \nBernoulli theorem states that for streamlined flow of an ideal liquid, the total energy (Pressure energy + potential energy + kinetic energy) per unit mass remains constant at every cross-section throughout the flow.
                          \nProof:
                          \nConsider a tube AB of varying cross-section through which an ideal liquid is in streamlined flow.
                          \n\"1st
                          \nLet P1<\/sub> be the pressure applied on the liquid at A and P2<\/sub> be the pressure against which liquid has to move out at B
                          \na1<\/sub>, a2<\/sub> be the cross-sectional area at tube at A and B
                          \nh1<\/sub> and h2<\/sub> are the mean height at section A and B from the reference level.
                          \nv1<\/sub> and v2<\/sub> are the velocity of liquid flow at A and B
                          \n\u03c1 is the density of the ideal liquid flowing through the tube.
                          \nThe liquid flows from A to B hence,
                          \nP1<\/sub> > P2<\/sub>
                          \nThe mass m of the liquid crossing per second through any section at the tube with respect to equation of continuity is,
                          \na1<\/sub>v1<\/sub> \u03c1 = a2<\/sub>v2<\/sub>\u00a0= \u03c1m
                          \na1<\/sub>v1<\/sub> = a2<\/sub>v2<\/sub> = V (say)
                          \nForce on liquid at A = P1<\/sub>a1<\/sub>
                          \nForce on liquid at B = P2<\/sub>a2<\/sub>
                          \nWork done per second by liquid at A = P1<\/sub>a1<\/sub> \u00d7 v1<\/sub> = P1<\/sub>V
                          \nWork done per second by liquid at B = P2<\/sub>a2<\/sub> \u00d7 v2<\/sub>\u00a0= P2<\/sub>V
                          \nNet work done per second on liquid by pressure energy moving from A to B is
                          \n= P1<\/sub>V- P2<\/sub>V = v (P1<\/sub>-P2<\/sub>)
                          \nIncrease in potential energy per second as the liquid flows from A to B
                          \n= mgh2<\/sub> – mgh,
                          \n(since v2<\/sub> is greater than v1<\/sub>)
                          \nIncrease in kinetic energy per second of liquid from A to B = 1\/2 m\\(v_{2}^{2}\\) – 1\/2 m\\(v_{1}^{2}\\)
                          \nAccording to work – energy principle, work done per second by pressure energy
                          \n= increase in P.E per second + increase in K.E per second
                          \nor P1<\/sub>v – P2<\/sub>v = (mgh2<\/sub> – mgh1<\/sub>) + \\(\\left(\\frac{1}{2} m v_{2}^{2}-\\frac{1}{2} m v_{1}^{2}\\right)\\)
                          \ni.e., P1<\/sub>v + mgh1<\/sub> + \\(\\frac{1}{2} m v_{1}^{2}\\)
                          \n= P2<\/sub>v + mgh2<\/sub> + \\(\\frac{1}{2} m v_{2}^{2}\\)
                          \nDividing by m1<\/sub>
                          \n\"1st
                          \nHence, \\(\\frac{\\mathrm{P}}{\\rho}+\\mathrm{gh}+\\frac{1}{2} \\mathrm{v}^{2}\\) = constant.
                          \nBut, \\(\\frac{P}{\\rho}\\) is the pressure energy per unit mass, gh is the potential energy per unit mass and \\(\\frac{1}{2} v^{2}\\) is the kinetic energy
                          \nper unit mass Hence,
                          \nPressure Energy + Potential Energy + Kinetic Energy remains constant.
                          \nHence proved.<\/p>\n

                          Question 9.
                          \nDerive the equation at continuity.
                          \nAnswer:
                          \nConsider a non-viscous liquid in streamlined flow through a tube AB at varying cross-section.
                          \n\"1st
                          \nLet a1<\/sub>, a2<\/sub> be the area of cross-section of the tube at A and B.
                          \nv1<\/sub>, v2<\/sub> = velocity of flow of liquid at A and B.
                          \nVolume of the liuid entering per second at A = a1<\/sub>v1<\/sub>
                          \nMass of liquid entering per second at A = a1<\/sub>v1<\/sub>\\(\\rho_{1}\\)
                          \nSimilarly mass of liquid leaving per second at B = a2<\/sub>v2<\/sub> \\(\\rho_{2}\\)
                          \nAssuming, no loss of liquid in the tube and
                          \nsteady flow, then mass of liquid entering at A\/sec = mass of liquid leaving at B\/sec.
                          \n\u21d2 a1<\/sub>v1<\/sub>\\(\\rho_{1}\\) = a2<\/sub>v2<\/sub>\\(\\rho_{2}\\)
                          \nAssuming liquid is incompressible,
                          \n\\(\\rho_{1}\\) = \\(\\rho_{2}\\)
                          \n\u21d2 a1<\/sub>v1<\/sub> = a2<\/sub>v2<\/sub>
                          \nav = constant
                          \nThis is the equation of continuity.<\/p>\n

                          \"KSEEB<\/p>\n

                          Question 10.
                          \nExplain the limitations of Bernoulli theorem.
                          \nAnswer:<\/p>\n

                            \n
                          1. While deriving the Bernoulli theorem, it is assumed that the velocity of every particle of liquid across any crosssection is uniform. Practically this is not possible.<\/li>\n
                          2. Viscous drag of the liquid which comes into play when the liquid is in motion has not been considered.<\/li>\n
                          3. While deriving the equation, it is assumed that there is no loss of energy when liquid is in motion. In fact, some KE is converted to heat energy.<\/li>\n
                          4. If a liquid is flowing along a curved path, energy due to centrifugal force should also be considered.<\/li>\n<\/ol>\n

                            Question 11.
                            \nExplain why the angle of contact of water with glass is acute while that mercury with glass is obtuse?
                            \nAnswer:
                            \n\"1st
                            \nSSL<\/sub> + SLA<\/sub> cos \u03b8 = SSA<\/sub>
                            \nor cos \u03b8 = \\(\\frac{S_{S A}-S_{S L}}{S_{L A}}\\)
                            \nIn case of mercury – glass, SSA<\/sub> < SSL<\/sub>
                            \n\u2234 cos \u03b8 is negative or \u03b8>90\u00b0 i.e., obtuse.
                            \nIn case of water – glass, SSA<\/sub> > SSL<\/sub>
                            \ncos \u03b8 is positive or \u03b8<90\u00b0 i.e., acute.
                            \nNote:
                            \nSSL<\/sub>:
                            \nSurface tension corresponding to solid-liquid interface.
                            \nSLA<\/sub>:
                            \nSurface tension corresponding to liquid – air interface.
                            \nSSA<\/sub>:
                            \nSurface tension corresponding to solid-air interface.<\/p>\n

                            Question 12.
                            \nDerive an expression for velocity of fluid at wide neck in a venturi meter.
                            \nAnswer:
                            \nSpeed of fluid flowing through tube at broad neck = V1<\/sub>
                            \nSpeed of fluid flowing through tube at narrow neck = V2<\/sub>
                            \nBy equation of continuity,
                            \nA1<\/sub>V1<\/sub> = A2<\/sub>V2<\/sub>
                            \nV2<\/sub> = \\(\\frac{A_{1} V_{1}}{A_{2}}=\\frac{A V_{1}}{a}\\)
                            \nFrom Bernoulli\u2019s equation,
                            \n\"1st<\/p>\n

                            Question 13.
                            \nWhat is the Magnus effect? Explain.
                            \nAnswer:
                            \nA ball Which is spinning drags air along with it. If the surface is rough, more air will be dragged.
                            \n\"1st
                            \nBall is moving forward and relative to it air is moving backwards. This results in a larger velocity of air relative to the ball above it and a smaller velocity below it. This difference in velocities of air results in pressure difference between lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called the Magnus effect.<\/p>\n

                            Question 14.
                            \nDefine terminal velocity. Give reasons why a sphere attains this velocity. Derive a relation for the terminal velocity.
                            \nAnswer:
                            \nThe constant velocity with which a body drops down after initial acceleration in a dense liquid or fluid is called terminal velocity. This is attained when the apparent weight is compensated by the viscous force. It is given by
                            \nv = \\(\\frac{2}{9} \\frac{r^{2} g}{\\eta}(\\rho-\\sigma)\\), where \u03c1 and \u03c3 are densities of the body and liquid respectively, \\(\\eta\\) is the coefficient of viscosity of liquid and r is the radius of the spherical body.
                            \nThe net force on the sphere becomes zero as the viscous force equals the apparent weight.
                            \nConsider a long column of dense liquid-like glycerine. As the ball is dropped in it, the forces experienced are;<\/p>\n

                              \n
                            1. weight = mg = \\(\\frac{4}{3} \\pi r^{3} \\rho g\\), where \u03c1 is the density of the ball.<\/li>\n
                            2. upthrust = U = \\(\\frac{4}{3} \\pi r^{3} \\rho_{1} g\\), where\\(\\rho_{1}\\) is the density of the liquid.<\/li>\n
                            3. viscous force Fv<\/sub> = \\(6 \\pi \\eta \\rho v\\), where v is the terminal velocity.<\/li>\n<\/ol>\n

                              Net force and the acceleration should be 0.
                              \n\u2234 mg – U – Fv<\/sub> = 0
                              \n\"1st<\/p>\n

                              Question 15.
                              \nWrite the differences between laminas\/streamline flow and turbulent flow.
                              \nAnswer:
                              \n1. Laminar Flow<\/p>\n

                                \n
                              • Layered flow with one layer of fluid sliding smoothly over the other.<\/li>\n
                              • Reynold\u2019s No. < 1000<\/li>\n
                              • Observed usually in flow through porous materials.<\/li>\n<\/ul>\n

                                2. Turbulent<\/p>\n

                                  \n
                                • Haphazard flow with rapid and continuous mixing of fluid resulting in momentum transfer.<\/li>\n
                                • Reynold\u2019s No. > 2000<\/li>\n
                                • Pipeline flow, river flow, etc.<\/li>\n<\/ul>\n

                                  \"KSEEB<\/p>\n

                                  Question 16.
                                  \nA razor blade can be made to float on water. What forces act on this blade? Is Archimedes\u2019s principle applicable?
                                  \nAnswer:
                                  \nThree forces are acting on the blade when it is made to float on water:<\/p>\n

                                    \n
                                  1. Weight of blade acting vertically downwards.<\/li>\n
                                  2. Reaction in blade exerted by the liquid surface acting vertically upwards.<\/li>\n
                                  3. Force of surface tension on circumference, of the blade acting tangentially to the liquid surface.
                                    \nIn this case, as no portion of razor blade is immersed in water, hence Archimedes\u2019s principle is not applicable.<\/li>\n<\/ol>\n

                                    1st PUC Physics Mechanical Properties Of Fluids Numerical Problems Questions and Answers<\/strong><\/p>\n

                                    Question 1.
                                    \nA capillary tube of diameter 1.5 mm is dipped in<\/p>\n

                                      \n
                                    1. mercury.<\/li>\n
                                    2. water.<\/li>\n<\/ol>\n

                                      Find capillary rise for each case. Surface tension for water and mercury can be taken as 0.07 N\/m and 0.52 N\/m respectively. The contact angle for water and mercury can be taken as 0\u00b0 and 130\u00b0.
                                      \nAnswer:
                                      \n1. h = \\(\\frac{4 \\lambda \\cos \\theta}{\\mathrm{dr}}\\)
                                      \n\\(=\\frac{4 \\times 0.52 \\times \\cos (130)}{1.5 \\times 10^{-3} \\times 13.65 \\times 9810}\\)
                                      \n= 6.66 mm (fall)<\/p>\n

                                      2. h = \\(\\frac{4 \\lambda \\cos \\theta}{\\mathrm{dr}}\\)
                                      \n\\(=\\frac{4 \\times 0.07 \\times 1}{1.5 \\times 10^{-3} \\times 9810}\\)
                                      \n= 19 mm (rise).<\/p>\n

                                      Question 2.
                                      \nCalculate the energy evolved when 8 droplets of water (surface tension 0.72 N\/m) of radius 0.5 mm combine into one.
                                      \nAnswer:
                                      \nHere, s = 0.072 N\/m
                                      \nr = 0.5 mm = 0.5 \u00d7 10-3<\/sup> m.
                                      \nLet R be the radius of big drop formed.
                                      \nVolume of the big drop = volume of 8 small drops.
                                      \n\\(\\frac{4}{3}\\) \u03c0 R3<\/sup> = 8 \u00d7 \\(\\frac{4}{3}\\) \u03c0r3<\/sup>
                                      \nor R = 2 r = 2 \u00d7 0.5 \u00d7 10-3<\/sup>= 10-3<\/sup> m
                                      \nSurface area of big drop :
                                      \n= 4\u03c0 R2<\/sup> = 4\u03c0 \u00d7 (10-3<\/sup>)2<\/sup>
                                      \n= 4\u03c0 \u00d7 10-6<\/sup>m2<\/sup>
                                      \nSurface area of 8 small drops:
                                      \n= 8 \u00d7 4\u03c0 \u00d7 r2<\/sup>
                                      \n= 8 \u00d7 4\u03c0 \u00d7 (0.5 \u00d7 10-3<\/sup>)2<\/sup>
                                      \n= 8\u03c0 \u00d7 10-6<\/sup>m2<\/sup>
                                      \nEnergy evolved = S.T \u00d7 decrease in area.
                                      \n= 0.072 \u00d7 4\u03c0 \u00d7 10-6<\/sup> = 9.05 \u00d7 107<\/sup>J.<\/p>\n

                                      \"KSEEB<\/p>\n

                                      Question 3.
                                      \nA plate 0.025 mm distant from a fixed plate moves at 60 cm\/s and requires a force of 2 N per meter square to maintain its speed. Determine fluid viscosity between the plates.
                                      \nAnswer:
                                      \ndy = 0.025 mm
                                      \n= 0.025 \u00d7 10-3<\/sup>m
                                      \nVelocity of upper plate .
                                      \n= 60 cm\/s = 0.6 m\/s
                                      \nForce on upper plate
                                      \n= 2\\(\\frac{\\mathrm{N}}{\\mathrm{m}^{2}}\\) = Shear stress , \u03c4
                                      \ndu = change of velocity = u – 0 = 0.6 m\/s
                                      \ndy = change of distance = 0.025 \u00d7 10-3<\/sup>m
                                      \nViscosity,
                                      \n\u00b5 = \\(\\frac{\\tau}{\\mathrm{du} \/ \\mathrm{dy}}=\\frac{2}{\\left(0.6 \/ 0.25 \\times 10^{-3}\\right)}\\)
                                      \n8.33 \u00d7 10-5<\/sup> \\(\\frac{\\mathrm{Ns}}{\\mathrm{m}^{2}}\\) = 8.33 \u00d7 10-5<\/sup> Pas<\/p>\n

                                      Question 4.
                                      \n\"1st
                                      \nIn the above figure, if area of plate is A, viscosity of fluid is \u00b5, distance of plate from top and bottom plane surfaces are y1<\/sub> and y2<\/sub> respectively, find the expression for force F required to drag the plate at a velocity v.
                                      \nAnswer:
                                      \n\"1st
                                      \nRequired force F = F1<\/sub> + F2<\/sub>
                                      \n\\(=\\frac{\\mu_{1} A_{1} v_{1}}{y_{1}}+\\frac{\\mu_{2} A_{2} v_{2}}{y_{2}}\\)
                                      \n\\(=\\mu \\mathrm{Av}\\left(\\frac{1}{\\mathrm{y}_{1}}+\\frac{1}{\\mathrm{y}_{2}}\\right)\\)<\/p>\n

                                      Question 5.
                                      \nAn open tank contains water up to a depth of 3m and above it an oil of sp.gr 0.9 for a depth of 1m. Find pressure intensity<\/p>\n

                                        \n
                                      1. at the interface of two liquids<\/li>\n
                                      2. at the bottom of the tank.<\/li>\n<\/ol>\n

                                        Answer:
                                        \n\"1st
                                        \nHeight of water, z1<\/sub> = 3 m
                                        \nHeight of oil, z2<\/sub> = 1 m
                                        \nDensity of oil
                                        \n= Sp.gr of oil \u00d7 Density of water
                                        \n= 0.9 \u00d7 1000 kg\/m3<\/sup> = 9000 kg\/m3<\/sup>
                                        \n= \\(\\rho_{2}\\)
                                        \nDensity of water = \\(\\rho_{1}\\) = 1000 kg\/m3<\/sup><\/p>\n

                                        1. At the interface, i.e., at A,
                                        \nP= \\(\\rho_{2}\\) g \u00d7 z2<\/sub> = 900 \u00d7 9.81 \u00d7 1
                                        \n= 8829 N\/m2<\/sup><\/p>\n

                                        2. At the bottom i.e., at B,
                                        \np =\\(\\rho_{2}\\)gz2<\/sub> + \\(\\rho_{1}\\)gz1<\/sub>
                                        \n= 900 \u00d7 9.81 \u00d7 1 + 1000 \u00d7 9.81 \u00d7 3
                                        \n= 8829 + 29430 = 38259 N\/m2<\/sup><\/p>\n

                                        Question 6.
                                        \nDetermine pressure difference between two points A and B for the setup as shown in figure.
                                        \n\"1st
                                        \nAnswer:
                                        \nG = specific gravity,
                                        \nPA = pressure at A,
                                        \nPB = Pressure at B
                                        \nPressure at any point is the same horizontal line x – x should be same.
                                        \nHence, pressure at C in left column = pressure at D in right column.
                                        \nPressure at C
                                        \n= (2 \u00d7 1000 \u00d7 g \u00d7 2) + (1 \u00d7 1000 \u00d7 g \u00d7 4) + PA …..(1)
                                        \npressure at D
                                        \n= (4 \u00d7 1000 \u00d7 g \u00d7 2) + PB ……(2)
                                        \nEquating (1) and (2),
                                        \n8000 g + PA = 8000 g + PB
                                        \nPA – PB = 0<\/p>\n

                                        Question 7.
                                        \nA closed tank contains 1m of mercury, 2m of water and 3 m of oil of sp.gravity 0.6. There is an unknown fluid in space above oil. If gauge pressure at the bottom of the tank is 200 kPa. What is the pressure on the top surface of oil?
                                        \nAnswer:
                                        \n\"1st
                                        \nLet pressure on top of oil be PA
                                        \nTotal pressure = PA<\/sub> + Poil<\/sub> +Pwater<\/sub> + Pmercury<\/sub>
                                        \nP= \u03c1gh
                                        \n200 \u00d7 103<\/sup> =PA<\/sub> + 0.6 \u00d7 1000 \u00d7 g \u00d7 3 + 1000 \u00d7 g \u00d7 2 + 13.6 \u00d7 1000 \u00d7 g \u00d7 1
                                        \nPA<\/sub> = 200 \u00d7 103 – 170694
                                        \n= 29.306 kPa<\/p>\n

                                        Question 8.
                                        \nThe flow rate of water from a tap of diameter 1.2 cm is 0.48 L\/min. The coefficient of viscosity of H2<\/sub>0 is 10-3<\/sup> Pa s. After some time the flow rate is increased to 4 L\/min. Characterise the flow for both the flow rates.
                                        \nAnswer:
                                        \nLet the speed of the flow be v and the area of tap be d = 1.2 cm. The volume of the water flowing out per second is
                                        \nQ = v \u00d7 \u03c0d2<\/sup> \/4
                                        \nv = 4Q\/d2<\/sup>
                                        \nReynold\u2019s number is
                                        \nRe<\/sub> = 4P2<\/sup> \/ \u03c0d\\(\\eta\\)
                                        \n= 4 \u00d7 103<\/sup>kg\/m3<\/sup>\u00d7 Q\/(3.14 \u00d7 1.2 \u00d7 10-2<\/sup>m \u00d7 10-1<\/sup>Pa s)
                                        \n= 1.061 \u00d7 108<\/sup> m-3<\/sup>s\u00a0 \u00a0Q = 1.061 \u00d7 108<\/sup>Q
                                        \nInitially,
                                        \nQ = 0.48 L\/min = 8.0 cm3<\/sup>\/s
                                        \n= 8.00 \u00d7 10-6<\/sup>m3<\/sup>\/s
                                        \nwe obtain
                                        \nR =484.8
                                        \nSince this is below 1000, the flow is steady.
                                        \nAfter sometime when Q = 4L\/min,
                                        \n= 66.67 cm3<\/sup>\/s
                                        \n= 6.67 \u00d7 10-5<\/sup> m3<\/sup>s-1<\/sup>
                                        \nwe obtain
                                        \nR = 1.061 \u00d7 108<\/sup> \u00d7 6.67 \u00d7 10-5<\/sup>
                                        \n= 7076.9
                                        \nThe flow now is turbulent.<\/p>\n

                                        \"KSEEB<\/p>\n

                                        Question 9.
                                        \nCalculate the total energy possessed by one kg of water at a point where pressure is 30 gm wt\/sq.mm. Velocity of 0.1 ms-1<\/sup> and height is 60 cm above the ground.
                                        \nAnswer:
                                        \nGiven, P = 30 g wt\/sq.mm
                                        \n= \\(\\frac{30}{1000}\\) \u00d7 9.8 \u00d7 106<\/sup>N\/m2<\/sup>
                                        \nv = 0.1ms-1<\/sup> h = 0.6 m
                                        \nTotal energy per unit mass
                                        \n\\(\\frac{P}{\\rho}\\) + gh + 1\/2 v2<\/sup>
                                        \n= \\(\\frac{30 \\times 9.8 \\times 10^{3}}{10^{3}}\\) + 9.8 \u00d7 0.6 + 1\/2 \u00d7 (0.1)2<\/sup>
                                        \n= 294 + 5.88 + 0.005
                                        \n= 299.885 J.<\/p>\n

                                        Question 10.
                                        \nWater flows at the rate of 4 litres per second through an orifice at the bottom of the tank which contains water 720 cm deep. Find the rate of escape of water If additional pressure of 20 kg\/cm2<\/sup> is applied to the surface of water.
                                        \nAnswer:
                                        \nGiven, h = 720 cm v = \\(\\sqrt{2 g h}\\)
                                        \nv = \\(\\sqrt{2 \\times 980 \\times 720}\\) cm\/s ;
                                        \n= 1187.93 cm\/s v
                                        \nv = 4 litres
                                        \nAdditional pressure = 20 kg\/cm2<\/sup>
                                        \n= 20000 g\/cm2<\/sup>
                                        \n\\(\\frac{20000 \\times 980}{980}\\) Cm 0f Water
                                        \n= 20,000 cm of water column.
                                        \nNow, pressure head (h1<\/sub>) = 20,000 + 720
                                        \n= 20,720 cm
                                        \nNew velocity v1<\/sub> = \\(\\sqrt{2 \\mathrm{gh}_{1}}\\)
                                        \n\\(\\sqrt{2 \\times 980 \\times 20,720}\\) cm\/s
                                        \n= 6372.69 cm\/s
                                        \nAs v = av and V1<\/sub> = av1<\/sub>
                                        \nv1<\/sub> = V\\(\\frac{v_{1}}{v}\\) = 4 \u00d7 \\(\\sqrt{\\frac{20720}{720}}\\)
                                        \n= 21.45 litres\/s<\/p>\n

                                        Question 11.
                                        \nA venturi meter is connected to two points in the mains where its radii are 20 cm and 10 cm. and the levels of water column in the tubes differ by 10 cms. How much water flows through the pipe per minute?
                                        \nAnswer:
                                        \nVolume of water flowing per second
                                        \nv = a1<\/sub>a2<\/sub>\\(\\sqrt{\\frac{2 g h}{a_{1}^{2}-a_{2}^{2}}}\\)
                                        \nwhere a1<\/sub> = \\(\\pi r_{1}^{2}\\) = 22\/2 \u00d7 2012<\/sup>cm2<\/sup>
                                        \na2<\/sub> = \\(\\pi r_{2}^{2}\\) 22\/7 \u00d7 1012<\/sup>cm2<\/sup>
                                        \ng = 980 cm\/s2<\/sup>;
                                        \nh = 10 cm .
                                        \nVolume of water flowing per minute
                                        \n\"1st
                                        \n= 2726.58 litres\/minute.<\/p>\n

                                        Question 12.
                                        \nWhen a drop of mercury of radius R is split into n similar drops, What is the change in surface energy? Assume \u03c3 as surface tension of mercury.
                                        \nAnswer:
                                        \nVolume of initial mercury drop = \\(\\frac{4}{3}\\) \u03c0R3<\/sup>
                                        \nr = radius of smaller drops,
                                        \nvolume conservation
                                        \n\u21d2 \\(\\frac{4}{3}\\) \u03c0R3<\/sup> = n \\(\\frac{4}{3}\\) \u03c0r3<\/sup>
                                        \nr = Rn-1\/3<\/sup>
                                        \nInitial Surface Energy = \u03c3 \u00d7 surface area
                                        \n= \u03c3 \u00d7 4\u03c0R2<\/sup>
                                        \nFinal Surface Energy = n \u00d7 \u03c3 \u00d7 4\u03c0r2<\/sup>
                                        \n= n\u03c3 4\u03c0[Rn-1\/3<\/sup>]2<\/sup>
                                        \n= n\u03c3 4\u03c0R2<\/sup>n-2\/3<\/sup>
                                        \n= n1\/3<\/sup>4\u03c3 \u03c0R2<\/sup>
                                        \nChange in surface energy
                                        \n= 4\u03c3 \u03c0R2<\/sup>[1 – n1\/3<\/sup>]<\/p>\n

                                        Question 13.<\/p>\n

                                          \n
                                        1. The diameter of pipe at section (1) and (2) are 10 and 15 cm respectively. Find discharge through pipe if the velocity of water flowing through pipe at section (1) is 5 m\/s. Find also velocity at section (2)<\/li>\n
                                        2. Find the work done in blowing a soup bubble of surface tension 0.06 N\/m from a 2 cm radius to a 5 cm radius.<\/li>\n<\/ol>\n

                                          Answer:
                                          \n1.
                                          \n\"1st
                                          \nQ = A1<\/sub>V1<\/sub> = \u03c0 \u00d7 \\(\\frac{0.1^{2}}{4}\\) \u00d7 5
                                          \n= 0.0393 m3<\/sup>\/s
                                          \nV2<\/sub> = \\(\\frac{A_{1} V_{1}}{A_{2}}\\) (continuity equation)
                                          \n= \\(\\frac{0.0393}{\\frac{\\pi}{4} \\times 0.15^{2}}\\) = 2.22 m\/s<\/p>\n

                                          2. Here, s = 0.06 N\/m
                                          \nr1<\/sub> = 2 cm = 0.02 m ;
                                          \nr2<\/sub> = 5 cm = 0.05 m
                                          \nSince bubble has 2 surfaces, initial surface area of the bubble.
                                          \n= 2 \u00d7 4\u03c0r1<\/sub>2<\/sup>= 2 \u00d7 4\u03c0(0.02)2<\/sup>
                                          \n= 32\u03c0 \u00d7 10-4<\/sup>m2<\/sup>
                                          \nFinal surface area of the bubble
                                          \n= 2 \u00d7 4\u03c0r2<\/sub>2<\/sup> = 2 \u00d7 4\u03c0(0.05)2<\/sup>
                                          \n= 200\u03c0 \u00d7 10-4<\/sup>m2<\/sup>
                                          \nIncrease in surface area
                                          \n= 200\u03c0 \u00d7 10-4<\/sup> – 32\u03c0 \u00d7 10-4<\/sup>
                                          \n= 168\u03c0 \u00d7 10-4<\/sup>m2<\/sup>
                                          \n\u2234 work done = S \u00d7 increase in surface area
                                          \n= 0.06 \u00d7 168\u03c0 \u00d7 10-4<\/sup> = 0.003168 J<\/p>\n

                                          Question 14.
                                          \nA tank is filled with water to a height H. At a depth \u2018h\u2019 from the free surface a hole is made so that the water comes out of it. What is the velocity of efflux? Also, what is the maximum range and the position of hole for the same? Find the time taken by a water molecule to reach the ground. Determine the horizontal length covered by the molecule.
                                          \nAnswer:
                                          \n\"1st
                                          \nTake two points at the same height (H- h) from ground one inside and one outside the hole. Applying Bernoulli\u2019s theorem for the points, we have
                                          \n\\(\\frac{P_{0}+h p g}{\\rho g}\\) + 0 + (H – h) = (H – h) + \\(\\frac{v^{2}}{2 g}+\\frac{P_{0}}{\\rho g}\\)
                                          \nOn Solving, we get, v = \\(\\sqrt{2 \\mathrm{gh}}\\)
                                          \nThe velocity of efflux thus depends on the depth of which the hole is made from the surface of the liquid. Time taken to reach
                                          \nground = t = \\(\\sqrt{\\frac{2(\\mathrm{H}-\\mathrm{h})}{\\mathrm{g}}}\\),
                                          \nSince v = \\(\\sqrt{2 \\mathrm{gh}}\\) is horizontal and ax<\/sub> = 0;
                                          \nR = vt = \\(\\sqrt{2 \\mathrm{gh}}\\) \\(\\sqrt{\\frac{2(\\mathrm{H}-\\mathrm{h})}{\\mathrm{g}}}\\)
                                          \nR = \\(2 \\sqrt{\\mathrm{h}(\\mathrm{H}-\\mathrm{h})}\\)
                                          \nTo maximise Range, \\(\\frac{\\mathrm{d} \\mathrm{R}}{\\mathrm{dh}}\\) = 0
                                          \ni.e., h = H\/2
                                          \nMaximum Range,
                                          \n\\(R_{\\max }=2 \\sqrt{\\frac{H}{2}(H-H \/ 2)}=H\\).<\/p>\n

                                          \"KSEEB<\/p>\n

                                          Question 15.
                                          \nWater flows through a horizontal pipe of varying cross-section. If the pressure is 1 cm of mercury when the velocity is 0.35 m\/s. Find the pressure at a point where velocity is 0.65 m\/s.
                                          \nAnswer:
                                          \nAt 1st<\/sup> point, P1<\/sub> = 1 cm of mercury
                                          \n= 0.01m of Hg.
                                          \n= 0.01 \u00d7 (13.6 \u00d7 103<\/sup>) \u00d7 9.8 P a;
                                          \nv1<\/sub> = 0.35 m\/s
                                          \nAt the second nd point,
                                          \np2<\/sub> = ?
                                          \nv2 <\/sub>= 0.65 m\/s and
                                          \n\u03c1 = 103<\/sup> hg\/m3<\/sup>.
                                          \nAccording to Bernoulli\u2019s theorem,
                                          \np1<\/sub> + 1\/2 \\(\\rho \\mathrm{V}_{1}^{2}\\) = p2<\/sub> + 1\/2 \\(\\rho \\mathrm{V}_{2}^{2}\\)
                                          \nor P2<\/sub> = P1<\/sub> + 1\/2 \\(\\rho\\left(v_{2}^{2}-v_{1}^{2}\\right)\\)
                                          \n= 0.01 \u00d7 13.6 \u00d7 103<\/sup> \u00d7 9.8 – 1\/2 \u00d7 103<\/sup> \u00d7 [(0.65)2<\/sup> – (0.35)2<\/sup>]
                                          \n= 1182.8 Pa \u21d2 \\(\\frac{1182.8}{9.8\\left(13.6 \\times 10^{3}\\right)}\\)
                                          \n= 0.00887 m of Hg.<\/p>\n

                                          Question 16.
                                          \nIf excess pressure inside a soap bubble of diameter 2 cm is balanced by that due to a column of oil of sp.gravity 0.8, 2mm high, find the surface tension of the soap bubble.
                                          \nAnswer:
                                          \nR = 1 cm = 10-2<\/sup> m
                                          \n\u03c1 of oil = 0.8 \u00d7 103<\/sup>kg\/m3<\/sup>
                                          \nh = 2 mm = 2 \u00d7 10-3<\/sup>m
                                          \nPressure due to 2 mm column of oil,
                                          \nP = h\u03c1g – (2 \u00d7 10-3<\/sup>)(0.8 \u00d7 10-3<\/sup>) \u00d7 9.8
                                          \n= 2 \u00d7 0.8 \u00d7 9.8 Pa
                                          \nIn case of soap bubble, P = \\(\\frac{4 \\mathrm{s}}{r}\\)
                                          \nor s = \\(\\frac{P r}{4}=\\frac{2 \\times 0.8 \\times 9.8 \\times 10^{-2}}{4}\\)
                                          \n= 3.92 \u00d7 10-2<\/sup>N\/m<\/p>\n

                                          Question 17.
                                          \nA spray pump having a cylindrical tube of cross-section 8 cm2<\/sup> has 50 fine holes of radius 0.5 mm. Consider the flow of liquid inside the tube to be 1.5 m\/min. What is the speed of ejection of the liquid through the whole?
                                          \nAnswer:
                                          \nArea of cross-section of the tube:
                                          \na1<\/sub> = 8 cm2<\/sup> = 8 \u00d7 10-4<\/sup>m2<\/sup>
                                          \nNumber of holes = 50
                                          \nDiameter of each hole = 0.5 \u00d7 2 = 1 mm
                                          \n= 10-3<\/sup>
                                          \nor radius of the hole
                                          \n= 0.5 mm = 5 \u00d7 10-4<\/sup>m
                                          \nArea of cross-section of each hole
                                          \n= \u03c0 r2<\/sup> = \u03c0 (5 \u00d7 10-4<\/sup>)2<\/sup> m2<\/sup>
                                          \nTotal area of cross-section of 50 holes,
                                          \na2<\/sub>= 50 \u00d7 \u03c0(5 \u00d7 10-4<\/sup>) m2<\/sup>
                                          \nSpeed of the liquid inside the tube,
                                          \nv1<\/sub> = 1.5 m\/min = \\(\\frac{1.5}{60}\\) m\/s
                                          \nIf v2<\/sub> is the velocity of ejection of the liquid through the holes,
                                          \nthen, a1<\/sub>v1<\/sub> = a2<\/sub>v2<\/sub> or v2<\/sub> = a1<\/sub>v1<\/sub>\/a2<\/sub>
                                          \n\u2234 v2<\/sub> = \\(\\frac{\\left(8 \\times 10^{-4}\\right) \\times 1.5}{60 \\times 50 \\times \\pi \\times\\left(5 \\times 10^{-4}\\right)^{2}}\\)
                                          \n= 0.5096 m\/s.<\/p>\n

                                          \"KSEEB<\/p>\n

                                          Question 18.
                                          \nCalculate the total energy\/unit mass possessed by water at a point, where the pressures is 10gm f\/sq mm, velocity is 0.1 m\/s and height of water level from the ground is 0.2 m (g = 9.8 m\/s2<\/sup>).
                                          \nAnswer:
                                          \nHere, P = 10 gm f\/sq mm
                                          \n= \\(\\frac{10}{1000}\\) \u00d7 9.8 \u00d7 (103<\/sup>)2<\/sup>
                                          \n= 98 \u00d7 103<\/sup>Nm-2<\/sup>
                                          \nv = 0.1 m\/s ;
                                          \n\u03c1 = 103<\/sup> kg\/m3;
                                          \nh = 0.20 m
                                          \n1. Pressure energy \/ unit mass
                                          \n= \\(\\frac{P}{\\rho}=\\frac{9.8 \\times 10^{3}}{10^{3}}\\) = 98 J\/kg<\/p>\n

                                          2. Gravitational potential energy \/ unit mass = gh = 9.8 \u00d7 0.2 = 1.96 J\/kg<\/p>\n

                                          3. Kinetic energy \/ unit mass
                                          \n= 1\/2 v2<\/sup>= \\(\\frac{1}{2}\\)(0.1)2<\/sup> = 0.0053\/kg
                                          \n\u2234 Total energy \/ unit mass
                                          \n= \\(\\frac{P}{\\rho}\\) + gh + 1\/2 v2<\/sup> = 98 + 1.96 + 0.005
                                          \n= 99.965 J\/kg<\/p>\n

                                          Question 19.
                                          \nA mass of 15 kg is placed on the wider tube of a U tube as shown in the figure. Given that the area of the wider tube is 5 m2<\/sup>, find the difference in the water levels in the two tubes, (density of water = 1 kg\/m3<\/sup>).
                                          \nAnswer:
                                          \n\"1st<\/p>\n

                                          \"KSEEB
                                          \nArea of the wider tube, A = 500 cm2<\/sup>
                                          \n= 5 \u00d7 10-2<\/sup> m2<\/sup>
                                          \nThe pressure applied by the mass at A is transmitted through the liquid and leads to a difference in the water levels.
                                          \nSince the water level is same a A and B,
                                          \nPA<\/sub> = PB<\/sub> ……(1)
                                          \nBut PA<\/sub> = \\(\\frac{F}{A}=\\frac{m g}{A}=\\frac{15 \\times 9.8}{5}\\) = 29.4 Pa
                                          \nAlso we know that PB<\/sub> = \u03c1gh = 1 \u00d7 9.8 \u00d7 h
                                          \nFrom (1) PB<\/sub> = PA<\/sub>
                                          \n9.8 \u00d7 h = 29.4
                                          \nh = 3m
                                          \n\u2234 The difference in water level obtained is 3 m.<\/p>\n

                                          Question 20.
                                          \nIf two liquids of mass m1<\/sub> and m2<\/sub> and density \\(\\rho_{1}\\), and \\(\\rho_{2}\\) are mixed together, then what is the density of the resulting mixture?
                                          \nAnswer:
                                          \nVolume occupied by first liquid,
                                          \nV1<\/sub> = \\(\\frac{m_{1}}{\\rho_{1}}\\)
                                          \nVolume occupied by second liquid,
                                          \nV2<\/sub> = \\(\\frac{m_{2}}{\\rho_{2}}\\)
                                          \n\u2234 density of the mixture,
                                          \n\"1st<\/p>\n

                                          Question 21.
                                          \nThe container shown in the figure below is filled with a liquid of density \u03c1. Note that the container has a height h1<\/sub> and area of cross-section A1<\/sub> for the upper half height h2<\/sub> and area of cross-section A2<\/sub> for the lower half. Find:<\/p>\n

                                            \n
                                          1. The pressure at the base of the container.<\/li>\n
                                          2. Force exerted by the liquid on the base of the container.<\/li>\n<\/ol>\n

                                            \"1st
                                            \nAnswer:
                                            \n1. The pressure at the base of the container is due to the liquids both at the upper and lower half.
                                            \n\u2234 PTotal<\/sub> = P1<\/sub> + P2<\/sub>
                                            \nPTotal<\/sub> = \u03c1gh1<\/sub> + \u03c1gh2<\/sub>
                                            \nPTotal<\/sub> = \u03c1g(h1<\/sub> + h2<\/sub>)<\/p>\n

                                            2. Force exerted on the bottom of the container,
                                            \nP = PTotal<\/sub> \u00d7 A2<\/sub>
                                            \nF = \u03c1g(h1<\/sub> + h2<\/sub>) \u00d7 A2<\/sub>.<\/p>\n","protected":false},"excerpt":{"rendered":"

                                            You can Download Chapter 10 Mechanical Properties Of Fluids Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Physics Question Bank Chapter 10 Mechanical Properties Of Fluids 1st PUC Physics Mechanical Properties Of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10344"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=10344"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10344\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10344"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10344"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10344"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}