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FLUID MECHANICS:--
EXERCISES Q-31 TO Q-35
31. Water flows through a tube shown in figure (13-E8). The areas of the cross-section at A and B are 1 cm² and 0.5 cm² respectively. The height difference between A and B is 5 cm. If the speed of the water at A is 10 cm/s find (a) the speed at B and (b) the difference in pressure at A and B.
Figure for Q-31
ANSWER: (a) Let the reference for height be the level of B. Height of A = h = 5 cm =0.05 m. Vₐ = 10 cm/s =0.10 m/s.
Area of cross-section at A =1 cm² = 1/10000 m².
Hence the discharge Q = 0.10 *1/10000 m³ =10⁻⁵ m³
Area of cross-section at B =0.5 cm² =0.5/10000 m²
The velocity at B = Vᵦ =10⁻⁵/(0.5/10000) m/s =1/5 m/s
=100*1/5 cm/s
=20 cm/s
(b) Let the pressures at A and B be Pₐ and Pᵦ. From the Bernoulli's theorem,
Pₐ+ρgh+½ρVₐ² = Pᵦ+½ρVᵦ²
Pᵦ -Pₐ = ρgh-½ρ(Vᵦ²-Vₐ²)
=1000*10*0.05 - ½*1000*(0.2²-0.1²) N/m²
=500 - 500*(0.04-0.01) N/m²
=500 - 15 N/m²
=485 N/m²
32. Water flows through a horizontal tube as shown in figure (13-E9). If the difference of heights of the water column in the vertical tubes is 2 cm, and the areas of the cross-section at A and B are 4 cm² and 2 cm² respectively. Find the rate of flow of water across any section.
The figure for Q-32
ANSWER: The height difference =2 cm =0.02 m
The difference of pressure between A and B =Pₐ-Pᵦ =ρgh
=1000*10*0.02 =200 N/m²
Let the rate of flow of water be Q cc/s i.e. =Q*10⁻⁶ m³/s
Area of the cross-section at A =4 cm² =4/10000 m²
The speed at A = Vₐ =Q*10⁻⁶/(4/10000) m/s =Q/400 m/s
Area of the cross-section at B =2 cm² =2/10000 m²
The speed at B = Vᵦ =Q*10⁻⁶/(2/10000) m/s =Q/200 m/s
Since the height of both the points are the same, from Bernoulli's theorem,
Pₐ+½ρVₐ² = Pᵦ +½ρVᵦ²
→Pₐ-Pᵦ = ½ρ(Vᵦ²-Vₐ²) =½*1000*Q²{(1/200)²-(1/400)²}
→Pₐ-Pᵦ =(1/20)*Q²{1/4 - 1/16} =3Q²/320
→200 = 3Q²/320
→Q² = 200*320/3 =21333
→Q = 146 cc/s
33. Water flows through the tube shown in figure (13-E10). The areas of the cross-section of the wide and the narrow portions of the tube are 5 cm² and 2 cm² respectively. The rate of flow of water through the tube is 500 cm³/s. Find the difference of mercury levels in the U-tube.
Figure for Q-33
ANSWER: The Discharge Q = 500 cm³/s,
The speed at the wide section Vₐ =500/5 =100 cm/s =1 m/s
The speed at the narrow section Vᵦ =500/2 =250 cm/s =2.5 m/s
Let the difference in height levels of mercury =h
The pressure difference at the two points =ρ'gh
If the pressure at the wide section =Pₐ and at the narrow section =Pᵦ, then from the Bernoulli's theorem
Pₐ+½ρVₐ² = Pᵦ+½ρVᵦ²
→½ρ(Vᵦ²-Vₐ²) =Pₐ -Pᵦ
→½*1000*(2.5²-1²) =ρ'gh =13600*9.8*h
→h = 10*(6.25-1)/(2*136*9.8) m
→h =52.5*100/2665.6 cm
→h = 5250/2665.6 =1.97 cm
34. Water leaks out from an open tank through a hole of area 2 mm² in the bottom. Suppose water is filled up to a height 80 cm and the area of the cross-section of the tank is 0.4 m². The pressure at the open surface and at the hole are equal to the atmospheric pressure. Neglect the small velocity of the water near the open surface in the tank. (a) Find the initial speed of the water coming out of the hole. (b) Find the speed of water coming out when half of the water has leaked out. (c) Find the volume of water leaked out during a time interval dt after the height remained is h. Thus find the decrease in height dh in terms of h and dt. (d) From the result of part (c) find the time required for half of the water to leak out.
ANSWER: (a) From the Bernoulli's theorem,
P+ρgh+½ρv² = P'+ρgh'+½ρv'²
Here P = P', h = 80 cm =0.80 m, h'=0, v =0, hence
→ρgh =½ρv'²
→v'² = 2gh
→v' =√(2gh) =√(2*10*0.8) =√16 =4 m/s
{Taking g =10 m/s²}
(b) When half the water has leaked out, h = 0.40 m
Now v' = √(2gh) =√(2*10*0.40) =√8 m/s.
(c) If at any instant t, the height remaining in the tank is h, then the volume of the water leaked in a small interval dt is dQ.
dQ =Discharge rate*dt =Area*speed of water*dt
dQ =A*v'*dt = (2 mm²)√(2gh)dt
The figure for Q-34
And the decrease in height dh = The water leaked out in time dt divided by the open area of the tank
→dh = dQ/(0.4 m²)
→dh = (2 mm²)√(2gh)dt/(0.40 m²)
→dh = (2 m² *10⁻⁶)√(2gh)dt/(0.40 m²)
→dh = √(2gh)(2*10⁻⁶*10/4)dt
→dh = √(2gh)*5*10⁻⁶dt
(d) From above i.e. dh = √(2gh)*5*10⁻⁶dt
→dt = 2*10⁵/√(2gh)dh
Integrating between the limits for height from h to h/2 we get
T = ∫dt =∫{2*10⁵/√(2gh)}dh
={2*10⁵/√(2g)}∫h⁻1/2dh
=[{2*10⁵/√(2g)}2√h]
=[{2√2*10⁵/√g}√h], putting the limits H = h to H = h/2 we get
T = [{2√2*10⁵/√g}{√h-√(h/2)]
={(2√2*10⁵/√g)(√2-1)/√2}√h
=2(√2-1)*10⁵*√h/√g
Putting h = 0.80 m, g =10 m/s²
T = 2*0.414*10⁵*√(0.8/10)/3600 hours
=6.50 hours
35. The water level is maintained in a cylindrical vessel up to a fixed height H. The vessel is kept on a horizontal plane. At what height above the bottom should a hole be made in the vessel so that the water stream coming out of the hole strikes the horizontal plane at the greatest distance from the vessel (figure 13-E11).
Figure for Q-35
ANSWER: Let the hole be at a height h from the bottom. The height of the water above the hole =H-h. The speed of the water at the hole v =√{2g(H-h)}
Let the time taken by the water to strike the floor is t. Then for the vertical fall,
h = 0*t+½gt²
→t² = 2h/g
→t = √(2h/g)
Let the distance of the striking point from the hole be X. Then,
X = v*t = √{2g(H-h)}*√(2h/g) =√{2²*(H-h)h} =2√(H-h)h
→X = 2√(H-h) * √h
For maximum X, dX/dh = 0
→2*[√(H-h)*d√h/dh + √h * d√(H-h)/dh ] = 0
→√(H-h)*{1/(2√h)} + √h *1/{2√(H-h)*(-1)} = 0
→√(H-h)/(√h) = √h/{√(H-h)}
→H-h = h
→2h = H
→h = H/2
Hence the required height of the hole from the bottom for the greatest horizontal distance of water to strike is half the height of the water level.
31. Water flows through a tube shown in figure (13-E8). The areas of the cross-section at A and B are 1 cm² and 0.5 cm² respectively. The height difference between A and B is 5 cm. If the speed of the water at A is 10 cm/s find (a) the speed at B and (b) the difference in pressure at A and B.
 |
| Figure for Q-31 |
ANSWER: (a) Let the reference for height be the level of B. Height of A = h = 5 cm =0.05 m. Vₐ = 10 cm/s =0.10 m/s.
Area of cross-section at A =1 cm² = 1/10000 m².
Hence the discharge Q = 0.10 *1/10000 m³ =10⁻⁵ m³
Area of cross-section at B =0.5 cm² =0.5/10000 m²
The velocity at B = Vᵦ =10⁻⁵/(0.5/10000) m/s =1/5 m/s
=100*1/5 cm/s
=20 cm/s
(b) Let the pressures at A and B be Pₐ and Pᵦ. From the Bernoulli's theorem,
Pₐ+ρgh+½ρVₐ² = Pᵦ+½ρVᵦ²
Pᵦ -Pₐ = ρgh-½ρ(Vᵦ²-Vₐ²)
=1000*10*0.05 - ½*1000*(0.2²-0.1²) N/m²
=500 - 500*(0.04-0.01) N/m²
=500 - 15 N/m²
=485 N/m²
32. Water flows through a horizontal tube as shown in figure (13-E9). If the difference of heights of the water column in the vertical tubes is 2 cm, and the areas of the cross-section at A and B are 4 cm² and 2 cm² respectively. Find the rate of flow of water across any section.
 |
| The figure for Q-32 |
ANSWER: The height difference =2 cm =0.02 m
The difference of pressure between A and B =Pₐ-Pᵦ =ρgh
=1000*10*0.02 =200 N/m²
Let the rate of flow of water be Q cc/s i.e. =Q*10⁻⁶ m³/s
Area of the cross-section at A =4 cm² =4/10000 m²
The speed at A = Vₐ =Q*10⁻⁶/(4/10000) m/s =Q/400 m/s
Area of the cross-section at B =2 cm² =2/10000 m²
The speed at B = Vᵦ =Q*10⁻⁶/(2/10000) m/s =Q/200 m/s
Since the height of both the points are the same, from Bernoulli's theorem,
Pₐ+½ρVₐ² = Pᵦ +½ρVᵦ²
→Pₐ-Pᵦ = ½ρ(Vᵦ²-Vₐ²) =½*1000*Q²{(1/200)²-(1/400)²}
→Pₐ-Pᵦ =(1/20)*Q²{1/4 - 1/16} =3Q²/320
→200 = 3Q²/320
→Q² = 200*320/3 =21333
→Q = 146 cc/s
33. Water flows through the tube shown in figure (13-E10). The areas of the cross-section of the wide and the narrow portions of the tube are 5 cm² and 2 cm² respectively. The rate of flow of water through the tube is 500 cm³/s. Find the difference of mercury levels in the U-tube.
 |
| Figure for Q-33 |
ANSWER: The Discharge Q = 500 cm³/s,
The speed at the wide section Vₐ =500/5 =100 cm/s =1 m/s
The speed at the narrow section Vᵦ =500/2 =250 cm/s =2.5 m/s
Let the difference in height levels of mercury =h
The pressure difference at the two points =ρ'gh
If the pressure at the wide section =Pₐ and at the narrow section =Pᵦ, then from the Bernoulli's theorem
Pₐ+½ρVₐ² = Pᵦ+½ρVᵦ²
→½ρ(Vᵦ²-Vₐ²) =Pₐ -Pᵦ
→½*1000*(2.5²-1²) =ρ'gh =13600*9.8*h
→h = 10*(6.25-1)/(2*136*9.8) m
→h =52.5*100/2665.6 cm
→h = 5250/2665.6 =1.97 cm
34. Water leaks out from an open tank through a hole of area 2 mm² in the bottom. Suppose water is filled up to a height 80 cm and the area of the cross-section of the tank is 0.4 m². The pressure at the open surface and at the hole are equal to the atmospheric pressure. Neglect the small velocity of the water near the open surface in the tank. (a) Find the initial speed of the water coming out of the hole. (b) Find the speed of water coming out when half of the water has leaked out. (c) Find the volume of water leaked out during a time interval dt after the height remained is h. Thus find the decrease in height dh in terms of h and dt. (d) From the result of part (c) find the time required for half of the water to leak out.
ANSWER: (a) From the Bernoulli's theorem,
P+ρgh+½ρv² = P'+ρgh'+½ρv'²
Here P = P', h = 80 cm =0.80 m, h'=0, v =0, hence
→ρgh =½ρv'²
→v'² = 2gh
→v' =√(2gh) =√(2*10*0.8) =√16 =4 m/s
{Taking g =10 m/s²}
(b) When half the water has leaked out, h = 0.40 m
Now v' = √(2gh) =√(2*10*0.40) =√8 m/s.
(c) If at any instant t, the height remaining in the tank is h, then the volume of the water leaked in a small interval dt is dQ.
dQ =Discharge rate*dt =Area*speed of water*dt
dQ =A*v'*dt = (2 mm²)√(2gh)dt |
| The figure for Q-34 |
And the decrease in height dh = The water leaked out in time dt divided by the open area of the tank
→dh = dQ/(0.4 m²)
→dh = (2 mm²)√(2gh)dt/(0.40 m²)
→dh = (2 m² *10⁻⁶)√(2gh)dt/(0.40 m²)
→dh = √(2gh)(2*10⁻⁶*10/4)dt
→dh = √(2gh)*5*10⁻⁶dt
(d) From above i.e. dh = √(2gh)*5*10⁻⁶dt
→dt = 2*10⁵/√(2gh)dh
Integrating between the limits for height from h to h/2 we get
T = ∫dt =∫{2*10⁵/√(2gh)}dh
={2*10⁵/√(2g)}∫h⁻1/2dh
=[{2*10⁵/√(2g)}2√h]
=[{2√2*10⁵/√g}√h], putting the limits H = h to H = h/2 we get
T = [{2√2*10⁵/√g}{√h-√(h/2)]
={(2√2*10⁵/√g)(√2-1)/√2}√h
=2(√2-1)*10⁵*√h/√g
Putting h = 0.80 m, g =10 m/s²
T = 2*0.414*10⁵*√(0.8/10)/3600 hours
=6.50 hours
35. The water level is maintained in a cylindrical vessel up to a fixed height H. The vessel is kept on a horizontal plane. At what height above the bottom should a hole be made in the vessel so that the water stream coming out of the hole strikes the horizontal plane at the greatest distance from the vessel (figure 13-E11).
 |
| Figure for Q-35 |
ANSWER: Let the hole be at a height h from the bottom. The height of the water above the hole =H-h. The speed of the water at the hole v =√{2g(H-h)}
Let the time taken by the water to strike the floor is t. Then for the vertical fall,
h = 0*t+½gt²
→t² = 2h/g
→t = √(2h/g)
Let the distance of the striking point from the hole be X. Then,
X = v*t = √{2g(H-h)}*√(2h/g) =√{2²*(H-h)h} =2√(H-h)h
→X = 2√(H-h) * √h
For maximum X, dX/dh = 0
→2*[√(H-h)*d√h/dh + √h * d√(H-h)/dh ] = 0
→√(H-h)*{1/(2√h)} + √h *1/{2√(H-h)*(-1)} = 0
→√(H-h)/(√h) = √h/{√(H-h)}
→H-h = h
→2h = H
→h = H/2
Hence the required height of the hole from the bottom for the greatest horizontal distance of water to strike is half the height of the water level.
===<<<O>>>===
Links to the Chapters
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Links to the Chapters
CHAPTER- 13 - Fluid Mechanics
EXERCISES Q-11 TO Q-20
EXERCISES Q-21 TO Q30
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 13 - Fluid Mechanics
EXERCISES Q-11 TO Q-20
EXERCISES Q-21 TO Q30
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY
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Click here for → Exercise(43-54)
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HC Verma's Concepts of Physics, Chapter-7, Circular Motion
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HC Verma's Concepts of Physics, Chapter-6, Friction
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For more practice on problems on friction solve these "New Questions on Friction" .
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For more practice on problems on friction solve these "New Questions on Friction" .
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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion
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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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HC Verma's Concepts of Physics, Chapter-4, The Forces
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
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Click here for "The Forces" - Exercises
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Thank you.. It was very useful
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