Tiêu đề 10.MECHANICAL PROPERTIES OF FLUIDS - Explanations.pdf ✅

1 (b) Force on the base of the vessel = pressure ×area of the base = h ρ g × A = 0.4 × 900 × 10 × 2 × 10−3 = 7.2 N 2 (a) The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second. Volume of water flowing out per second = Av = A√2gh ...(i) Volume of water flowing in per second = 70 cm3 /sec ...(ii) From (i) and (ii) we get A√2gh = 70 ⇒ 1 × √2gh = 70 ⇒ 1 × √2 × 980 × h = 70 ∴ h = 4900 1960 = 2.5 cm 3 (a) The aerofils are so designed that Pupper side < Plower side So that the aerofils get a lifting force in upward direction. According to Bernoulli’s theorem, where the pressure is large, the velocity will be minimum or vice-versa Thus, Vupper side > vlower side 4 (c) Volume of liquid flowing at first point = A1v1. Similarly, volume of liquid flowing at second point = A2v2 From equation of continuity, A1v1 = A2v2 or v1 v2 = A2 A1 5 (c) Figure shows the flow speed profile for laminar flow of a viscous fluid in a long cylindrical pipe. The speed is greatest along the axis and zero at the pipe walls, 6 (d) Tension in spring T = upthrust – weight of sphere = Vσg − Vρg = Vηρg − Vρg [As σ = ηρ] = (η − 1)Vρg = (η − 1)mg 7 (a) In air, force of gravity acts on metals. Thus, these have their actual weight. Atomic weight of steel, ie, iron is 56 and that of aluminium is 27. Hence, it can be said that in air the weight of aluminium is half the weight of steel. 8 (d) The velocity gradient = ΔV Δr = 8 0.1 = 80s −1 9 (a) Volume of liquid flowing per second through each of the two tubes in series will be the same. So V = π p1 R 4 8 η L = πp2 (R/2) 4 8η(L/2) or p1 p2 = 1 4 10 (d) Let depth of lake is x cm. ∴ p1V1 = p2V2 (pdg + xdg) ( 4 3 πr 3) = pdg [ 4 3 π(2r) 3 ] (p + x)r 3 = p(8r 3) x = 8p − p x = 7p 11 (a) According to principle of continuity, Av = constant or A1v1 = A2v2 or πr1 2v1 = πr2 2v2 Given, r1 = 4 2 cm = 0.02 m, r2 = 2 2 cm = 0.01m, v1 = 3ms−1 ∴ π(0.02) 2 × 3 = π(0.01) 2v2 or v2 = ( 0.02 0.01) 2 × 3 = 12 ms−1 12 (a) If r is the radius of small droplet and R is the radius of big drop, then according to question, 4 3 πR 3 = 106 × 4 3 πr 3 or r = R 100 = 0.01 R = 0.01 × 10−2m = 10−4m Work done = surface tension × increase in area = 35 × 10−2 [106 × 4π × (10−4) 2 − 4π × (10−3) 2 ] = 4.35 × 10−2 J 13 (a) As shown in figure, in the two arms of a tube pressure remains same on surface PP’. Hence, 8 × ρy × g × 2 × ρHg × g = 10 × ρx × g ∴ 8ρy + 2 × 113.6 = 10 × 3.36 or ρy = 36.6−27.2 8 = 0.8 g cc −1 15 (c)
The onset of turbulence in a liquid is determined by a dimensionless parameter is called Reynold’s number 16 (a) Pressure is independent of area of cross-section 18 (d) Given that surface tension = 0.075N m−1 ; diameter = 30 cm = 0.30 m ∴ Force = 0.075 × 0.30 = 0.0225 N = 2.25 × 10−2 N 19 (b) Buoyant force = weight of the body in air – weight of the body in liquid = 4 − 3 = 1 N 20 (d) Pressure applied by liquid column p = hρg Ie, the pressure depends on the height of liquid column no on its size, so pressure at the bottom of A and B is same. 22 (a) The excess pressure inside a liquid drop is ∆p = 2T r or ∆p ∝ T r The excess pressure inside a liquid drop is directly proportional to surface tension (T ) and inversely proportional to radius (r). 23 (b) If the level in narrow tube goes down by h1 then in wider tube goes up to h2 Now, πr 2h1 = π(nr) 2h2 ⇒ h1 = n 2h2 Now, pressure at point A = pressure at point B hρg = (h1 + h2 )ρ′g ⇒ h = (n 2h2 + h2)s (As s = ρ ′ ρ ) ⇒ h2 = h (n2+1)s 24 (c) The excess pressure inside the soap bubble is inversely proportional to radius of soap bubble ie, p ∝ 1/r, r being radius of soap bubble. If follows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow from smaller bubble grows at the expense of the smaller one. 25 (d) 2 Sl = F Or S = F⁄2l = (2 × 10−2)/2 × 0.10 = 0.1 Nm−1 26 (a) Excess pressure is given by p = 4T r ⟹ r = 4T p ∴ r1 r2 = p2 p1 = 1.02 1.01 = 102 101 Ratio of volume′ s = 4 3 πr1 3 4 3 πr2 3 = (102) 3 (101) 3 ≈ 2 27 (c) Form principle of continuity A1v1 = A2v2 πR 2 ∙ v = π(2R) 2 ∙ v2 v2 = v 4 28 (b) In case of soap bubble W = T × 2 × ∆A = 0.03 × 2 × 40 × 10−4 = 2.4 × 10−4 J 29 (c) Depth of p below the free surface of water in the vessel= (1 + h). Since the liquid exerts equal pressure in all direction at one level, hence the pressure at p = (H − h) ρ g 30 (b) Retarding force acting on a ball falling in to a viscous fluid F = 6πɳRv where R = radius of ball, v = velocity of ball, and ɳ = coefficient of viscosity ∴ F ∝ R and F ∝ v Or in words, retarding force is directly proportional to both R and v. 32 (c) Weight of block = Weight of displaced oil + Weight of displaced water ⇒ mg = V1ρ0g + V2ρwg ⇒ m = (10 × 10 × 6) × 0.6 + (10 × 10 × 4) × 1 = 760 gm 33 (b) Let the volume of the ball be V . Force on the hall due to upthrust =Vdg Net upward force = Vdg =VDg If upward acceleration is a, then Vda = Vdg − VDg ∴ a = ( d−D D )g h Water Mercury h1 h2 A B
Velocity on reaching the surface, v = √2ah Further v = √2ah ∴ 2ah = 2gH or H = ah 8 = ( d−D D ) h = ( d D − 1) h 34 (d) Reading of the spring balance = Apparent weight of the block = Actual weight – upthrust = 12 − Vinσg = 12 − 500 × 10−6 × 103 × 10 = 12 − 5 = 7N 35 (d) If we have m gram of ice which is a floating in a liquid of density 1.2 and 9 L will displaces volume mcc 1.2 < mcc. After melting it occupics mcc. 36 (d) hρg = 2S r or h = 2S rρg = 2 × 75 0.005 × 1 × 1000 = 30 cm 38 (c) For the floatation V0d0g = Vin d g ⇒ Vin = V0 d0 d ∴ Vout = V0 − Vin = V0 − V0 d0 d = V0 [ d − d0 d ] ⇒ Vout V0 = d − d0 d 39 (d) Given, l1 = l2 = 1, and r1 r2 = 1 2 V = πP1r1 4 8ηl = πP2r2 4 8ηl ⇒ P1 P2 = ( r2 r1 ) 4 = 16 ⇒ P1 = 16P2 Since both tubes are connected in series, hence pressure difference across combination 40 (b) When a ball is given anticlockwise rotation along with linear motion towards RHS then it will have maximum flight 41 (a) When air is blown in the horizontal tube, the pressure of air decreases in the tube. Due to which the water will rise above the tube A 43 (c) For parallel combination 1 Reff = 1 R1 + 1 R2 ⇒ πr 4 8ηl = πr 4 8ηl1 + πr 4 8ηl2 ⇒ 1 l = 1 l1 + 1 l2 ∴ l = l1l2 l1 + l2 44 (d) Vσf = 0.6Vσ1g Vσg = 0.4Vσ2g ∴ 1 = 6 4 σ1 σ2 ∴ σ2 σ1 = 3 2 = 1.5 45 (b) Let specific gravities of concrete and saw dust are ρ1 and ρ2 respectively According to principle of floatation weight of whole sphere = upthrust on the sphere 4 3 π(R 3 − r 3)ρ1g + 4 3 πr 3 ρ2g = 4 3 πR 3 × 1 × g ⇒ R 3ρ1 − r 3ρ1 + r 3ρ2 = R 3 ⇒ R 3(ρ1 − 1) = r 3(ρ1 − ρ2 ) ⇒ R 3 r 3 = ρ1 − ρ2 ρ1 − 1 ⇒ R 3 − r 3 r 3 = ρ1 − ρ2 − ρ1 + 1 ρ1 − 1 ⇒ (R 3 − r 3 )ρ1 r 3ρ2 = ( 1 − ρ2 ρ1 − 1 ) ρ1 ρ2 ⇒ Mass of concrete Mass of saw dust = ( 1 − 0.3 2.4 − 1 ) × 2.4 0.3 = 4 47 (a) In level flight of aeroplane, mg = pA Or p = mg A = 3×104×10 120 Pa = 2.5 kPa 48 (d) Here, R = 2.8/2 = 1.4 mm = 0.14 cm = 4 3 πR 3 = 125 × 4 3 πr 3 Or r = R/5 = 0.14/5 = 0.028 cm Change in energy = surface tension × increase in area = 75 × (125 × 4πr 2 − 4πR 2 ) = 74 erg 49 (b) Apply Bernoullis, theorem p1 + 0ρgH = p2+ 1 2 ρv 2 + ρgH p1 − p2 = 1 2 ρv 2 3 × 105 − 1 × 105 = 1 2 ρv 2 2 × 105 = 1 2 ρv 2 2 × 105 = 1 2 × 103 × v 2 v 2 = 400 v = √400 50 (c) The air pressure inside a soap bubble is p = 4T R Which is greater than the atmospheric pressure. If a hole is made at A, air will flow outside through A. then the thread becomes convex looking from A and from B towards A it is concave. Hence,
becoming concave or convex, depends on size of A with respect to B. 51 (b) Let V0,Vt = Volume of the metal ball at 0°C and t°C respectively, ρ0, pt = density of alcohol at 0°C and t°C respecitvely. Then W1 = W0 − V0ρ0g W2 = Wt − Vtρtg Where Vt = V0(1 + γmt) and ρt = ρ0 (1+γat) g = V0ρ0 (1+γml) (1+γal) As γm < γa, hence upthrust at t°C is less than at 0°C. It means upthrust has been decreased with increase in temperature. Due to which the W2 > W1 52 (d) Let radii of two soap bubble are a and b respectively and radius of single larger bubble is c . As excess pressure for a soap bubble is 4T r and external pressure p pi = p + 4T r So, pa = p + 4T a , pb = p + 4T b and pc = p + 4T c ... . . (i) and Va = 4 3 πa 3 , Vb = 4 3 πb 3 and Vc = 4 3 πc 3 ... ... . (ii) Now as mass is conserved. μa + μb = μc ie, paVa R Ta + pbVb R Tb + pcVc R Tc (as p V = μR T) As temperature is constant, ie, Ta = Tb = Tc So, paVa + pbVb = pcVc Which in the light of Eqs. (i) and (ii) becomes, (p + 4T a ) ( 4 3 πa 3) + (p + 4T b ) ( 4 3 πb 3) = (p + 4T c ) ( 4 3 πc 3) ie, 4T(a 2 + b 2 − c 2) = p(c 3 − a 3 − b 3) ... ... . (iii) Now, V = 4 3 π(a 3 + b 3 − c 3) and A = 4π(a 2 + b 2 − c 2) ∴ TA π = − 3 4π = Vp or 4TA + 3pV = 0 53 (d) Let the volume of iceberg inside sea is x, then Volume of iceberg inside sea Total volume of iceberg = Density of ice Density of sea water or x V = 0.92 1.03 so x = 0.92 1.03 V Percentage of total volume of iceberg above the level of sea water is = ( V − x V ) × 100 = ( V − ( 0.92 1.03)V V ) × 100% = 0.11 1.03 × 100% = 11% (nearly) 55 (d) From the formula the viscous force is given by F = 6πɳrv = 6 × 22 7 × 2 × 10−4 × 0.35 × 10−3 × 1 = 13.2 × 10−7N 56 (c) Force exerted by the liquid on the base of the vessel is F = mg Here, mA = mB = MC ∴ FA = FB = FC 57 (b) ρoil < ρ < ρwater Oil is the least dense of them so it should settle at the top with water at the base. Now the ball is denser than oil but less denser than water. So, it will sink through oil but will not sink in water. So it will stay at the oil-water interface. 58 (c) Rate of flow of water through a capillary tube is V = πRr 4 8ηl As P, η remain the same ∴ V ′ V = (2r) 4 (r) 4 × (l) (2l) = 16 2 = 8 ⇒ V ′ = 8V 59 (a) In hydraulic life, the pressure of smaller piston = pressure of bigger piston = F/A = (3000 × 9.8)/(4.25 × 10−2 ) = 6.92 × 105 Nm−2 61 (c) Excess of pressure inside a soap bubble p = 4T R or p1 p2 = R2 R1 Given, p1 = 3p2