Equation of Continuity
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The human heart forces 4500 cc of blood per min through the arteries, under a pressure of 15 cm of Hg. Calculate the hp of the heart.
Take 1 hp = 746 W and g = 980 cm/s2.
If the heart pushes of blood in under pressure the power of the heart is
- 5.8 m2
- 4.6 m2
- 16.5 m2
- 2.9 m2
- 5 : 3 : 7
- 7 : 3 : 5
- 21 : 35 : 15
- 1 : 1 : 1
- 6 m/s
- 1.5 m/s
- 3 m/s
- 2 m/s
In the figure shown, a liquid is flowing through a tube at the rate of 0.1 m3/sec. The tube is randed into two semicircular tubes of cross sectional area A3 and 2A3. The velocity of liquid at Q is VQ and Velocity at P is VP (The cross-sectional area of the main is A)
If VP = 10 m/s, then VQ = 5 m/s
If VP = 20 m/s, then VQ = 5 m/s
If VP = 20 m/s, then VQ = 10 m/s
If VP = 30 m/s, then VQ = 20 m/s
- 5×10−4 m2
- 1×10−4 m2
- 5×10−5 m2
- 2×10−5 m2
- 3.0 m/s
- 1.5 m/s
- 1.0 m/s
- 2.25 m/s
In the figure shown, a liquid is flowing through a tube at the rate of 0.1 m3/sec. The tube is randed into two semicircular tubes of cross sectional area A3 and 2A3. The velocity of liquid at Q is VQ and Velocity at P is VP (The cross-sectional area of the main is A)
If VP = 10 m/s, then VQ = 5 m/s
If VP = 20 m/s, then VQ = 5 m/s
If VP = 20 m/s, then VQ = 10 m/s
If VP = 30 m/s, then VQ = 20 m/s
- 1.50
- 1.70
- 2.35
- 3.0
- 1 m/s
- 0.25 m/s
- 20 m/s
- 0.5 m/s
Water flows in a horizontal tube. The pressure of water changes by 600 N m-2between A and B where the areas of cross section are 30 cm2 and 15 cm2 respectively. Find the rate of flow of water through the tube. Given √0.4= 0.63 approx.
1890 m3/s
2260 cm3/s
1890 cm3/s
1664 cm3/s
Water enters through end A with speed v1 and leaves through end B with speed v2 of a cylindrical tube AB. The tube is always completely filled with water. In case I tube is horizontal and in case II it is vertical with end A upwards and in case III it is vertical with end B upwards. We have v1 = v2 for
Case I
Case II
Case III
Each case
- 5 m/s
- 20 m/s
- 10 m/s
- 2.5 m/s
- 16 m/s
- 1 m/s
- 4 m/s
- 2 m/s
- 125 m/s
- 250 m/s
- 450 m/s
- 500 m/s
- 12.5 m
- 25 m
- 50 m
- 67.5 m
At Deoprayag (Garhwal, UP) river Alaknanda mixes with the river Bhagirathi and becomes river Ganga. Suppose Alaknanda has a width of 12 m, Bhagirathi has a width of 8 m and Ganga has a width of 16 m. Assume that the depth of water is same in the three rivers. Let the average speed of water in Alaknanda be 20 km h-1 and in Bhagirathi be 16 km h-1. Find the average speed of water in the river Ganga.
46 km/hr
23 m/s
23 km/hr
46 m/s
- 10×10−6
- 11×10−6
- 13×10−6
- 18×10−6
- 2 m/s
- 5 m/s
- 10 m/s
- 8 m/s
- 3.0 m/s
- 1.5 m/s
- 1.0 m/s
- 2.25 m/s
Water enters through end A with a speed v1 and leaves through end B with a speed v2 of a cylindrical tube AB. The tube is always completely filled with water. In case I the tube is horizontal, in case II it is vertical with the end A upward and in case III it is vertical with the end B upward. We have v1 = v2 for
Case I
Case II
Case III
each case
- 81256
- 916
- 34
- 316
- 2×10−3 m
- 4×10−3 m
- 6×10−3 m
- 8×10−3 m
- V2
- 2V
- V4
- V
- Only if the flow is compressible
- Only if the flow is laminar
- Only if the flow is rotational
- Never possible
- 1.50
- 1.70
- 2.35
- 3.0
Water flows through a horizontal tube of variable cross section. The area of cross section at A and B are 4 mm2 and 2 mm2 respectively. If 1 cc of water enters per second through A, find the speed of water at B?
25 cm/s
50 cm/s
25 mm/s
50 mm/s
- V2Rnr
- VR2n2r2
- VR2nr2
- VR2n3r2
- 125 m/s
- 250 m/s
- 450 m/s
- 500 m/s