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Fluid Dynamics Problem Solutions | Physics Help

Solve physics problems related to fluid dynamics involving water flow, pressure calculations, buoyant force, and equilibrium conditions. Detailed solutions provided for better understanding.

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Fluid Dynamics Problem Solutions | Physics Help

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  1. Ch-14 Help-Session

  2. Q23.: Water is pumped out of a swimming pool at a speed of 5.0 m/s through a uniform hose of radius 1.0 cm. Find the mass of water pumped out of the pool in one minute. (Density of water = 1000 kg/m3).(Ans: 94 kg) Q24. A large tank open to atmosphere is filled with water. Fig 6 shows this tank with a stream of water flowing through a hole (open to atmosphere) at a depth of 4.00 m. The speed of water, v2, leaving the hole is: ( Ans: 8.85 m/s) CH-14-072 Mass flow rate= Rm=Av Rm= x r2 x v =1000 x  (0.01)2 x 5.0 =1.57 kg/s Water mass pumped in one min M= Rm x t=1.57 x 60=94 kg P0+v12/2+gh=p0+v22/2+gy2 V1=0; y2=0 then gh=v22/2; v2=2gh= (2x9.8x4) v2=(2x9.8x4)=8.85 m/s

  3. Q25.A 10 kg spherical object with a volume of 0.10 m3 is held in static equilibrium under water by a cable fixed to the bottom of a water tank. What is the tension T in the cable? (See Fig. 7) (Ans: 880 N) CH-14-072-071 T071 Q8.The pressure of a gas in a tank is measured with a mercury manometer (Fig. 5). The mercury is 36.2 cm higher in the outside arm than in the arm connected to the gas cell. Find the gauge pressure of the gas cell. (Density of mercury is 13.6 g/cm3) (Ans: 4.82 × 104 Pa) FB-T-mg=0; T=FB-mg; Buoyant Force FB=Vxgx FB = 0.1x9.8x1000=980 N T=FB-mg =980-10x9.8=882 N p= gh = 13.6 x 103 x 9.8 x 0.362 =48247.4 Pa

  4. CH-14-071 Q9.: A cube of wood of side = 10 cm has a density of 700 kg/m3. As shown in Fig. 6, the cube is held in equilibrium under water by a string tied to the BOTTOM of a container. Find the tension in the string. (Ans: 2.94 N) Q10. A garden hose has an inner diameter of 16 mm. The hose can fill a 10 liter bucket in 20 s. Find the speed of the water at the end of the hose (1 Liter = 10-3 m3).(Ans: 2.5 m/s) Rv=Av=Volume/time=10x10-3/20 Rv =5x10-4 m3/s; A=r2= (8x10-3)2= 2x10-4 m2 V=RV/A= 5/2=2.5 m/s FB-T-mg=0; T=FB-mg; Buoyant Force FB=Vxgx FB = ( 0.1)3x9.8x1000 =9.8 N mg= ( 0.1)3x9.8x700=6.86 N T=FB-mg =9.80-6.86=2.94 N

  5. CH-14-071 Q11.Water flows through a pipe as shown in Fig. 7. At the lower elevation, the water’s speed (vA) is 5.0 m/s and the gauge pressure (PA) is 7.5 × 104 Pa. Find the gauge pressure at the higher elevation (PB). (Diameter at A = 6.0 cm, diameter at B = 4.0 cm and the elevation of B relative to A is 2.0 m) (Ans: 4.60 kPa) PA+vA2/2+gyA= PB+ vB2/2+gyB; PB=PA-([-vA2/2+ vB2/2]-g[yA-yB)]; yA-yB=-2.0 m; vB=vA (AA/AB)=5x(32/22)= 11.25 m/s PB=7.5 x 104 -1000([-52/2+11.252/2]-9.8[-2)]= 4.6 kPa

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