Page 1 : Properties of Fluids, , Mechanical, , ————"”, , , , , , , , 10.1 Explain why, , @), ), , (), , Sol. (a), , The blood pressure in humans is greater, at the feet than at the brain?, Atmospheric pressure at a height of about, 6 km decreases to nearly half its value at, the sea level, though the 'height' of the, atmosphere is more than 100 km, Hydrostatic pressure is a scalar quantity, even though pressure is force divided by, area,, The height of the blood column in the, human body is more at feet than at the brain., That is why, the blood exerts more pressure, at the feet, than at the brain., , (. pressure =hp g), We know that the density ofair is maximum, near the surface of the carth and decreases, rapidly with height and at a height of about, 6 km it decreases to nearly half its value at, the sea level. Beyond 6 km height, the, density of air decreases very slowly with, height. Due to this reason, the atmospheric, pressure at a height of about 6 km, decreases to nearly half of its value at sea, level., Since due to applied force on liquid, the, pressure is transmitted equally in all, directions inside the liquid. That is why,, there is no fixed direction for the pressure, due to liquid. Hence hydrostatic pressure, isa scalar quantity., , 10.2. Explain why, , @), , ), , The angle of contact of mercury with glass, is obtuse, while that of water with glass is, acute?, , Water on a clean glass surface tends to, spread out while mercury on the same, surface tends to form drops?, , ©, @, ), , Sol. (a), , (b), , , , NCERT EXERCISES <3, , Surface tension of liquid is independent of, the area of the liquid surface?, Detergents should have small angles of, contact?, , A drop of liquid under no external forces, is always spherical in shape., , When a small quantity of liquid is poured, on solid, three interfaces namely liquid —, air, solid — air and solid—liquid are formed., The surface tensions corresponding to, these three layers are S, 4, Sg and Sg, respectively. Let 0 be the angle of contact, between the liquid and solid. The, molecules in the region, where the three, interfaces meet are in equilibrium, It means, net force acting on them is zero. For the, molecule at O to be in equilibrium, we have, , Sst Spa CosO = Sgq or cos0 =, , , , In case of mercury — glass, Soa «q Sg, therefore cos0 is negative or 0 > 90° ic, obtuse. In case of water — glass, S., >S,, therefore, cos 0 is positive 8 < 90° i.e., , , , , SL?, acute., , , , , , Sia, , , , , , Solid, , , , , , , , For mercury — glass, angle of contact is, obtuse. In order to achieve this obtuse, valuc of angle of contact, the mercury tends, to form a drop. In case of water — glass, the, angle of contact is acute. To achieve this, acute value of angle of contact, the water, tends to spread.

Page 2 : 10.3, , Sol., , (c), , (@), , Surface tension of liquid is the force acting, per unit length ona line drawn tangentially, to the liquid surface at rest. Since, this force, is independent of the area of liquid surface,, therefore, surface tension is also, independent of the area of the liquid, surface., , We know that the cloth has narrow spaces, in the form of capillaries. The rise of liquid, in a capillary tube is directly proportional, to cosO. If 8 is small cos@ will be large. Due, to which capillary rise will be more so the, detergent will penetrate more in cloth., , In the absence of external forces, the, surface of the liquid drop tends to acquire, the minimum surface area due to surface, tension. Since, for a given volume, the, surface area of sphere is least, hence the, liquid drop takes the spherical shape., , Fillin the blanks using the words from the list, appended with each statements:, , (@), , (b), , ©), , (@), , (©, , (a), 1), ©), @, ©, , Surface tension of liquids generally, with temperature., (increases/decreases), Viscosity of gases _____ with, temperature, whereas viscosity of liquids, with temperature., (increases/decreases), , For solids with elastic modulus of rigidity,, the shearing force is proportional to, __ __ while for fluids it is, porportional to :, , (shear strain/rate of shear strain), For a fluid in steady flow, the increases, in flow speed at a constriction follows, from while the decrease of, pressure there follows from, , (conservation of mass/Bernoulli’s, , principle), , For the model of a plane in a wind tunnel,, turbulence occurs ata speed, than the critical speed for turbulence for, an actual plane. (greater/smaller), decreases, increases; decreases, shear strain; rate of shear strain, conservation of mass: Bernoulli’s principle, greater., , , , 10.4 Explain why, , Sol., , (@, i), , (, , @, , (e), (@), , (6), , (e), , (a, , (e), , To keep a piece of paper horizontal, you, Should blow over, not under, it., , When we try to close a water tap with our, Singers, fast jets of water gush through, the openings between our fingers., , The size of a needle of a syringe controls, Slow rate better than the thumb pressure, exerted by a doctor while administering, an injection., , A fluid flowing out of a small hole ina, vessel results in a backward thurst on the, vessel., , a spinning cricket ball in air does not, follow a parabolic trajectory., , When we blow over the piece of paper, the, velocity of air increases. As a result, the, pressure on it decreases in accordance with, the Bernoulli’s theorem whereas the, pressure below remains the same, (atmospheric pressure). Thus, the paper, remains horizontal., , By doing so the area of outlet of water jet, is reduced, so velocity of water increases, according to equation of continuity av =, a constant., , For a constant height, the Bernoulli’s, theorem is expressed as, , , , Io, Ps abe = constant, , Tn this equation, the pressure P occur with, a single power whereas the velocity occurs, with a square power. Therefore, the velocity, has more effect compared to the pressure., It is for this reason that needle of the, springe controls flow rate better than the, thumb pressure exerted by the doctor., This is because of principle of conservation, of momentum. While the flowing fluid, carries forward momentum, the vessel gets, a backward momentum., , A spinning cricket ball would have, followed a parabolic trajectory has there, been no air. But because of air the Magnus, effect take place. Due to the Magnus effect, the spinning cricket ball deviates from its, parabolic trajectory.

Page 3 : 10.5, , Sol., , 10.6, , Sol., , 10.7, , Sol., , 10.8, , Sol., , A50 kg girl wearing high heel shoes balances, on a single heel. The heel is circular with a, diameter 1 cm. What is the pressure exerted by, the heel on the horizontal floor?, , Dil, Here, m=50kg; r= 77 yom= xoo™, , force _m, Pressure = = 78, area qr?, , 50x 9.8, , (22/7) «(1/200)°, Torricelli’s barometer used mercury. Pascal, duplicated it using French wine of density 984, kgm. Determine the height of the wine, column for normal atmospheric pressure., P=0.76 x (13.6 x 103) x 9.8 =h x 984 x 9.8, , n-2 76 x13.6x10° x9.8, 9849.8, A vertical off shore structure is built to, withstand a maximum stress of 10° Pa. Is the, structure suitable for putting upon top of an oil, well in Bombay High? Take the depth of the sea, to be roughly 3 km, and ignore the ocean, currents., Here, maximum stress= 10?Pa, h=3km=3 x 103m;, P (water) = 10° kg/m? and g = 9.8 m/s?. The, structure will be suitable for putting upon top of, an oil well provided the pressure exerted by sea, water is less than maximum stress it can bear., Pressure due to sea water, P=hpg, 3x 103 x 103 x 9.8=2.94 x 107 Pa, Since, the pressure of sea water is less than the, maximum stress of 10°Pa, the structure will be, suitable for putting upon tap of the oil well., A hydraulic automobile lift is designed to lift, cars with a maximum mass of 3000 kg. The, area of cross - section of the piston carrying, the load is 425 em?. What maximum pressure, would smaller piston have to bear?, The maximum force, which the bigger piston can, bear, F = 3000 kg, £=3000 x 9.8N, Area of piston, A= 425 em? = 425 x 10-4 m?, Maximum pressure on the bigger piston,, F _ 3000x9.8, , A 425x104, Since, the liquid transmits pressure equally,, , <, the maximum pressure the smaller piston, can bear is 6.92 x 105 Pa., , , , = 6.24 x 10°Nm?, , or =10.5m., , , , = 6.92 x 10° Pa, , 10.9 AU tube contains water and methylated spirit, , Sol., , 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 relative density of spirit?, For water column in onearm of U tube,, hy = 10.0 cm; P} (density) = lgem?, For the spirit column in the other arm of U tube,, h, = 12.5 cm; p,=?, As the mercury columns in the two arms of U, tube are in level,, , pressure exerted by each is equal., Hence, h,p)g = hp., , WO%I og a ome, =U; m, 2s °°, , hypy, or P,= =, 2 hy, , 3, , , , , , 0.8, Relative density of spirit =f2 = , PI, 10.10 In Q.10.9, if 15.0 em of water and spirit each, , Sol., , 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), , On pouring 15.0 cm of water and spirit each into, the respective arms of the U tube, the mercury, level will rise in the arm containing spirit. Let h, be the difference in the levels of mercury in the, two arms of U tube and let p be the density of, mercury., , The pressure exerted by h cm of mercury, column = difference in pressure exerted by water, and spirit., , hp g=h/P,g-h, Pyg, Here, h=?; p =13.6gcm 3; hy 15+ 10=25em;, p ,=Igem; h,=15+ 12.5=27.5 om;, P,=0.8gem*, , Putting values in (i), we get, hx 13.6 g=25* 1x g-27.5x08g=3¢, , or ates 0.22 cm., 13.6, , , , 10.11 Can Bernoulli's equation be used to describe, , Sol., , the flow of water throught a rapid motion ina, river? Explain., , Bernoulli’s theorem is applicable only for the, ideal fluids in streamlined motion. Since the flow, of water in ariver in arapid way cannot be treated, as streamlined motion, the theorem cannot be, used.

Page 4 : 10.12 Does it matter if one uses gauge instead of, , Sol., , absolute pressure in applying Bernoulli’s, equation? Explain., , No, it does not matter if one uses gauge instead, of absolute pressures in applying. Bernoulli’s, equation, provided the atmospheric pressure at, the two points where Bernoulli’s equation is, , applied are significantly different., , 10.13 Glycerine flows steadily through a horizontal, , Sol., , tube of length 1.5 m and radius 1.0 cm. If the, amount of glycerine collected per second at one, end is 4.0 x 10-3 kg s~!, what is the pressure, difference between the two ends of the tube?, (density of glycerine = 1.3 x 103 kg m3 and, viscosity of glycerine = 0.83 Pa., , Here, ¢ =1.5m;r=1.0em=10? m:, P=1.3x10 kg/m’;, , n =0.83 Pa, , Mass of glycerine flowing per sec,, , M=4x 10% kg/s, , Volume of glycerine flowing per second,, , 3, ye ML?, , P 13x10, If P is the difference of pressure between two, ends of the tube, then using Poiscuille's formula, we have, , , , prt Vx8yb, v=— or p=, 8yL ort, (4x10 3x15, p= 4x10 8x 0.83x1.5 =97537Pa, , , , , , , , = 5 saenn ee, 13x10) 3.142x(10-7, , 10.14 In a test experiment on a model aeroplane in a, , Sol., , wind tunnel, the flow speeds on the upper and, lower surfaces of the wing are 70 ms and 63, ms“ respectively. What is the lift on the wing if, its area is 2.5 m2? Take the density of air 1.3, kgm., , Let v,, v3 be the speeds on the upper and lower, of the wing of acroplane, and P, and P, are the, on upper and lower surfaces of the wing, respectively. Then v, = 70 ms!; v,=63 ms"! p, =13kem?., , From Bernoulli's theorem, , P, 1 P, Ll,, , i Feb i +eht ov, , , , P, P. 1, t-2=- (ve -v)?), pp 2, , 1 5, or PP PSH), , =2 «13 707° — (637, = 5 % 13 [(70)°- (63)], , = 605.15 Pa, The difference of pressure provides the lift to, the aeroplane., , So, lift on the aeroplane = pressure difference x, arca of wings = 605.15 x 2.5=1512.875N, = 1.51% 108N., , 10.15 Figures (a) and (b) refer to the steady flow of, , a(non-viscous) liquid. Which of the two figures, in incorrect? Why?, , — — — —, , Sol., , (a) {by, , Figure (a) is incorrect. It is because of the fact, that at the kink, the velocity of flow of liquid is, larrge and hence using the Bernoulli’s theorem, the pressure is less. Asa result, the water should, not rise higher in the tube where there is a kink, (i.e., where the area of cross-section is small)., , 10.16 The cylindrical tube of a spray pump has a, , Sol., , cross - section of 8.0 cm? one of which has 40, fine holes each of diameter 1.0 mm. If the liquid, flow inside the tube is 1.5 m per minute, what is, the speed of ejection of the liquid through the, holes?, , Area of cross-section of tube, a, = 8.0 cm? =8, 104 m2,, , No. of holes = 40; Diameter of each hole,, D=Imm=104m, , - Dot, «. Radius of hole, mS = 103m =5*10-+m, , Area of cross-section of cach hole =m 1? = 2 (5, 10+)? m?, Total area of cross section of 40 holes,, , a, =40* 0 (5 x 104) m?

Page 5 : Speed of liquid inside the tube, y= 1.5 m/min, , 3, =i? ms"!, , If v, is the velocity of the ejection of the liquid, through the holes, then a, v, =a, vy, (8x10~4) x 1.5, OO Ny ee, 60x 40x7x (5x10 ")~, , => v,=0.637 mst, , 10.17 A U-shaped wire is dipped in a soap solution an, , Sol., , removed thin soap film formed between the wire, and a light slider supports a weight of 1.5 x 10—, 2 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?, , Since, soap film has two free surfaces, so total, length of the film=2 ¢ =2 x 30=60cm=0.6m., Total force on the slider due to surface tension,, F=$x2¢=5%x0.6N, , In equilibrium, F = mg, , Sx 0.6= 1.5 x 107, , _15x10-7, 0.6, , =2.5*107Nm!, , 10.18. Figure (a) below shows a thin film supporting, , , , Sol., , a small weight = 4.5 x 107 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., , , , U, a ee, , , , , , , , (a) Here, length of the film supporting the weight, =40cm=0.4m., , Total weight supported (or force) = 4.5 x 10-2N, Film has two free surfaces,, , 45x10, , Surface tension, S =, 2x 0.4, , = 5.625 x10 Nm!, , Since the liquid is same for all the cases (a), (6), and (c), and temperature is also same, therefore, surface surface tension for cases (4) and (c) will, also be the same = 5.625 x 10°? In Fig. (a), (b), and (c), the length of the film supporting the, weight is also (a), hence the total weight, supported in cach case is 4.5 x 10-2N., , 10.19 What is the pressure inside a drop of mercury, , Sol., , of radius 3 mmat room temperature? Surface, tension of mercury at that temperature (20° C), is 4.65 x 10! Nov !. The atmospheric pressure, is 1.01 x 105 Pa. Also give the excess pressure, inside the drop., , Given, r=3mm=0.3 m, S=4.65 x 10-7! Nm! P, =1.01 ¥ 105 Pa, , Excess of pressure inside the drop of mercury is, , 28 2x 4.65x107), , given by, p= ~~ Sein =310Pa, , 10.20 What is excess pressure inside a bubble of, , Sol., , soap solution of radius 5 mm, given that the, surface tension of soap solution at the, temperature (20°C) is 2.5 x 10°? Nm!? If an, air bubble of the same dimension were formed, ata depth of 40 cm inside a container containing, the soap solution (of relative density 1.2), what, would be the pressure inside the bubble? (1 atm, = 1.01 x 105 Pa), , Given, S=2.5 x 10? Nov! r=5 mm=5* 107 m, Density of soap solution, p = 1.2 x 10° kg m3, Excess pressure inside the soap bubble,, , , , AS 4x25, , p=—, , ——— = 20 Pa., r 5x10”, , Excess pressure inside the air bubble,, , 28 _2x25x107, r 5x07, , p= =10Pa., , Total pressure inside the air bubble at depth hin, soap solution,, , =p'+atmospheric pressure +h p g, =10+ 1.01 x 10°+0.4« 1.2 «10? «9.8, = 1.06 x 10° Pa.