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FLUID MECHANICS:--
EXERCISES Q-11 TO Q-20
11. A metal piece of mass 160 g lies in equilibrium inside a glass of water (figure 13-E4). The piece touches the bottom of the glass at a small number of points. If the density of the metal is 8000 kg/m³, find the normal force exerted by the bottom of the glass on the metal piece.
Figure for Q-11
ANSWER: The mass of metal piece =160 g = 0.160 kg
The density of the metal piece = 8000 kg/m³
The volume of the metal piece = 0.160/8000 m³ =0.00002 m³
It is also the volume of water displaced. Hence the force of buoyancy =0.00002*1000*10 N =0.20 N
(Taking g = 10 m/s²)
So the net submerged weight of the metal piece =0.160*10-0.20 N
=1.60 -0.20 N =1.4 N
Hence the normal force exerted by the bottom of the glass on the metal piece =1.4 N
12. A ferry boat has internal volume 1 m³ and weight 50 kg. (a) Neglecting the thickness of the wood, find the fraction of the volume of the boat immersed in water. (b) If a leak develops in the bottom and water starts coming in, what fraction of the boat's volume will be filled with water before water starts coming in from the sides?
ANSWER: (a) The volume of the boat immersed in water will be that volume of water which weight =50 kg. Since the density of water =1000 kg/m³, this volume will be =50/1000 m³ =1/20 m³.
Hence the fraction of the volume of the boat immersed in water =(1/20 m³)/(1 m³) =1/20.
(b) In the leak condition, when the outer water level comes to the top of the side of the boat, the total volume of water displaced is equal to the internal volume of the boat = 1m³.
The force of buoyancy = weight of this water = 1000 kg. This force balances the weight of the boat plus the weight of the water inside. If the weight of the water inside be X kg, then
X+50 = 1000
→X =950 kg
Hence the volume of inside water =950/1000 m³ =0.950 m³
The inside water volume fraction =(0.950 m³)/(1 m³) =19/20.
13. A cubical block of ice floating in water has to support a metal piece weighing 0.50 kg. What can be the minimum edge of the block so that it does not sink in the water? The specific gravity of ice = 0.9.
ANSWER: Let the minimum edge of the block = X cm
The volume of the ice block =X³ cm³ =X³/10⁶ m³
Density of the ice = 0.9*1000 kg/m³ =900 kg/m³
The weight of the ice block =(X³/10⁶)*900 kg
Total weight of the ice block and the metal piece = (X³/10⁶)*900+0.50 kg
The force of the buoyancy = weight of the water displaced =(X³/10⁶)*1000 kg. (The volume of water displaced just before it sinks is equal to the volume of the ice cube = X³ cm³)
In the given condition the weight of ice cube and the metal = force of buoyancy,
→(X³/10⁶)*900+0.50 = (X³/10⁶)*1000
→(X³/10⁶)*(1000-900) =0.50
→X³ = 0.50*10⁶/100 =5000
→X ≈ 17 cm
14. A cube of ice floats partly in water and partly in K.oil (figure 13-E5). Find the ratio of the volume of ice immersed in water to that in K.oil. The specific gravity of K.oil is 0.8 and that of ice is 0.9.
Figure for Q-14
ANSWER: Let the volume of ice in water = X m³ and the volume of ice in K.oil = Y m³
So total weight of the ice cube =(X+Y)*900 kg
{Because density of the ice = 0.9*1000 =900 kg/m³}
In the floating condition, the total force of buoyancy by the water and the K.oil will be equal to the weight of the ice cube.
The total force of buoyancy = X*1000 + Y*800 kg
Equating the two we get,
(X+Y)*900 = X*1000+Y*800
→1000X-900X = 900Y-800Y
→100X = 100Y
→X = Y
So the volume of ice cube immersed in water is equal to the volume immersed in K.oil. Hence their ratio = 1:1.
15. A cubical box is to be constructed with iron sheets 1 mm in thickness. What can be the minimum value of the external edge so that the cube does not sink in water? Density of iron = 8000 kg/m³ and density of water = 1000 kg/m³.
ANSWER: Let the required external edge = X cm.
The surface area of the box = 6X² cm²
The volume of the iron in the box =6X²*0.1 cm³ =0.60X² cm³
The mass of iron in the box = (0.60X²/10⁶ m³)*8000 kg/m³
=4.8X²/1000 kg
The volume of the water displaced =X³ cm³
The force of buoyancy = (X³/10⁶ m³)*1000 kg/m³
=X³/1000 kg
In the given condition weight of the box = the force of buoyancy
→4.8X²/1000 = X³/1000
→X³ = 4.8 X²
→X = 4.8 cm.
16. A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just allow the block to float in water. The specific gravity of wood is 0.8 and that of lead is 11.3.
ANSWER: Let the mass of the lead piece =X g
Total weight of the wood and the lead = 200+X g
The volume of the lead piece = X/11.3 cm³
{Since the density of lead = 11.3 g/cm³}
When the wooden block is just allowed to float in water, it displaces a volume of water equal to its volume = 200/0.8 cm³ =250 cm³
{Because the density of wood =0.8*1 g/cm³ =0.8 cm³}
In the given condition total volume of the water displaced =(250 + X/11.3) cm³
Hence the force of buoyancy =(250 + X/11.3) cm³ * 1 g/cm³
=(250 + X/11.3) g
{NOTE: In these problems, the unit of force is taken as gram-weight or kg-weight, not dyne or Newton because while equating, the conversion factor g, the acceleration due to gravity, cancel out. So do not be confused}
Now while just allowed to float on the water, the weight of wood plus the lead piece = the force of buoyancy.
→ 200+X = (250 + X/11.3)
→ X-X/11.3 =250-200
→10.3X =50*11.3
→X = 50*11.3/10.3 = 54.8 g
17. Solve the previous problem if the lead piece is fastened on the top surface of the block and the block is to float with its upper surface just dipping in water.
ANSWER: In this case, the lead piece does not displace the water, only the wooden block displaces. So the force of buoyancy = (250 cm³)*(1 g/cm³) =250 g, this balances the total weight. Equating we get,
200 + X = 250
→X = 250-200 = 50 g
18. A cubical metal block of edge 12 cm floats in mercury with one-fifth of the height in mercury. Water is poured till the surface of the block is just immersed in it. Find the height of the water column to be poured. The specific gravity of mercury = 13.6.
ANSWER: The volume of the metal block =12³ cm³
The volume of the metal block in the mercury =the volume of the mercury displaced by the block =12³/5 cm³
The weight of the metal block = the weight of the volume of the mercury displaced =(12³/5)*13.6 g
=13.6*12³/5 g
Figure for Q-18
Let the required height of water column = H cm
Now the volume of the water displaced =12²*H cm³
The volume of the mercury displaced =12²(12-H) cm³
The force of the buoyancy = The weight of the water displaced + the weight of the mercury displaced
=12²*H +{12²(12-H)}*13.6 g
This should be equal to the weight of the metal cube. Equating,
12²*H +{12²(12-H)}*13.6 = 13.6*12³/5
→12²H(1-13.6) =13.6*12³/5 - 12³*13.6
→-12.6*12²H =13.6*12³(-4/5)
→12.6H =4*13.6*12/5
→H = 652.8/63 ≈ 10.4 cm
19. A hollow spherical body of inner and outer radii 6 cm and 8 cm respectively floats half submerged in water. Find the density of the material of the sphere.
ANSWER: The outer volume of the sphere =4π8³/3 cm³
The inner volume of the sphere =4π6³/3 cm³
The volume of the material =4π(8³-6³)/3 cm³
Let the density of the material of the sphere be D kg/m³.
The weight of the material ={4π(8³-6³)/(3*10⁶)}*D kg-weight
The volume of water displaced =half of the outer volume of the sphere
=2π8³/3 cm³
The weight of the water displaced ={2π8³/(3*10⁶)}*1000 kg-weight
At the floating condition, the weight of the sphere material = weight of the water displaced
→{4π(8³-6³)/(3*10⁶)}*D = {2π8³/(3*10⁶)}*1000
→D = ½*1000*8³/(8³-6³) =1000/2(1-0.75³)
=1000/1.156 =865 kg/m³
20. A solid sphere of radius 5 cm floats in water. If a maximum load of 0.1 kg can be put on it without wetting the load, find the specific gravity of the material of the sphere.
ANSWER: Let the specific gravity of the material of the sphere be S. Then the density of the material =1000S kg/m³
The volume of the sphere = 4π*5³/3 cm³ =500π/3 cm³
The weight of the sphere ={500π/(3*10⁶)}*1000S kg
Total weight of the sphere and the load =0.10 + {500π/(3*10⁶)}*1000S kg
The volume of the water displaced without wetting the load = the volume of the sphere =500π/3 cm³ = the force of buoyancy
The weight of the water displaced = {500π/(3*10⁶)}*1000 kg
Equating the weights we get,
0.10 + {500π/(3*10⁶)}*1000S ={500π/(3*10⁶)}*1000
→{500π/(3*10⁶)}*1000S ={500π/(3*10⁶)}*1000-0.10
→(500π/3000)*S =500π/3000-0.1
→S*π/6 = π/6-0.1
→S = 1-0.6/π =1-0.2 = 0.8
11. A metal piece of mass 160 g lies in equilibrium inside a glass of water (figure 13-E4). The piece touches the bottom of the glass at a small number of points. If the density of the metal is 8000 kg/m³, find the normal force exerted by the bottom of the glass on the metal piece.
|
| Figure for Q-11 |
ANSWER: The mass of metal piece =160 g = 0.160 kg
The density of the metal piece = 8000 kg/m³
The volume of the metal piece = 0.160/8000 m³ =0.00002 m³
It is also the volume of water displaced. Hence the force of buoyancy =0.00002*1000*10 N =0.20 N
(Taking g = 10 m/s²)
So the net submerged weight of the metal piece =0.160*10-0.20 N
=1.60 -0.20 N =1.4 N
Hence the normal force exerted by the bottom of the glass on the metal piece =1.4 N
12. A ferry boat has internal volume 1 m³ and weight 50 kg. (a) Neglecting the thickness of the wood, find the fraction of the volume of the boat immersed in water. (b) If a leak develops in the bottom and water starts coming in, what fraction of the boat's volume will be filled with water before water starts coming in from the sides?
ANSWER: (a) The volume of the boat immersed in water will be that volume of water which weight =50 kg. Since the density of water =1000 kg/m³, this volume will be =50/1000 m³ =1/20 m³.
Hence the fraction of the volume of the boat immersed in water =(1/20 m³)/(1 m³) =1/20.
(b) In the leak condition, when the outer water level comes to the top of the side of the boat, the total volume of water displaced is equal to the internal volume of the boat = 1m³.
The force of buoyancy = weight of this water = 1000 kg. This force balances the weight of the boat plus the weight of the water inside. If the weight of the water inside be X kg, then
X+50 = 1000
→X =950 kg
Hence the volume of inside water =950/1000 m³ =0.950 m³
The inside water volume fraction =(0.950 m³)/(1 m³) =19/20.
13. A cubical block of ice floating in water has to support a metal piece weighing 0.50 kg. What can be the minimum edge of the block so that it does not sink in the water? The specific gravity of ice = 0.9.
ANSWER: Let the minimum edge of the block = X cm
The volume of the ice block =X³ cm³ =X³/10⁶ m³
Density of the ice = 0.9*1000 kg/m³ =900 kg/m³
The weight of the ice block =(X³/10⁶)*900 kg
Total weight of the ice block and the metal piece = (X³/10⁶)*900+0.50 kg
The force of the buoyancy = weight of the water displaced =(X³/10⁶)*1000 kg. (The volume of water displaced just before it sinks is equal to the volume of the ice cube = X³ cm³)
In the given condition the weight of ice cube and the metal = force of buoyancy,
→(X³/10⁶)*900+0.50 = (X³/10⁶)*1000
→(X³/10⁶)*(1000-900) =0.50
→X³ = 0.50*10⁶/100 =5000
→X ≈ 17 cm14. A cube of ice floats partly in water and partly in K.oil (figure 13-E5). Find the ratio of the volume of ice immersed in water to that in K.oil. The specific gravity of K.oil is 0.8 and that of ice is 0.9.
 |
| Figure for Q-14 |
ANSWER: Let the volume of ice in water = X m³ and the volume of ice in K.oil = Y m³
So total weight of the ice cube =(X+Y)*900 kg
{Because density of the ice = 0.9*1000 =900 kg/m³}
In the floating condition, the total force of buoyancy by the water and the K.oil will be equal to the weight of the ice cube.
The total force of buoyancy = X*1000 + Y*800 kg
Equating the two we get,
(X+Y)*900 = X*1000+Y*800
→1000X-900X = 900Y-800Y
→100X = 100Y
→X = Y
So the volume of ice cube immersed in water is equal to the volume immersed in K.oil. Hence their ratio = 1:1.
15. A cubical box is to be constructed with iron sheets 1 mm in thickness. What can be the minimum value of the external edge so that the cube does not sink in water? Density of iron = 8000 kg/m³ and density of water = 1000 kg/m³.
ANSWER: Let the required external edge = X cm.
The surface area of the box = 6X² cm²
The volume of the iron in the box =6X²*0.1 cm³ =0.60X² cm³
The mass of iron in the box = (0.60X²/10⁶ m³)*8000 kg/m³
=4.8X²/1000 kg
The volume of the water displaced =X³ cm³
The force of buoyancy = (X³/10⁶ m³)*1000 kg/m³
=X³/1000 kg
In the given condition weight of the box = the force of buoyancy
→4.8X²/1000 = X³/1000
→X³ = 4.8 X²
→X = 4.8 cm.16. A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just allow the block to float in water. The specific gravity of wood is 0.8 and that of lead is 11.3.
ANSWER: Let the mass of the lead piece =X g
Total weight of the wood and the lead = 200+X g
The volume of the lead piece = X/11.3 cm³
{Since the density of lead = 11.3 g/cm³}
When the wooden block is just allowed to float in water, it displaces a volume of water equal to its volume = 200/0.8 cm³ =250 cm³
{Because the density of wood =0.8*1 g/cm³ =0.8 cm³}
In the given condition total volume of the water displaced =(250 + X/11.3) cm³
Hence the force of buoyancy =(250 + X/11.3) cm³ * 1 g/cm³
=(250 + X/11.3) g
{NOTE: In these problems, the unit of force is taken as gram-weight or kg-weight, not dyne or Newton because while equating, the conversion factor g, the acceleration due to gravity, cancel out. So do not be confused}
Now while just allowed to float on the water, the weight of wood plus the lead piece = the force of buoyancy.
→ 200+X = (250 + X/11.3)
→ X-X/11.3 =250-200
→10.3X =50*11.3
→X = 50*11.3/10.3 = 54.8 g17. Solve the previous problem if the lead piece is fastened on the top surface of the block and the block is to float with its upper surface just dipping in water.
ANSWER: In this case, the lead piece does not displace the water, only the wooden block displaces. So the force of buoyancy = (250 cm³)*(1 g/cm³) =250 g, this balances the total weight. Equating we get,
200 + X = 250
→X = 250-200 = 50 g
18. A cubical metal block of edge 12 cm floats in mercury with one-fifth of the height in mercury. Water is poured till the surface of the block is just immersed in it. Find the height of the water column to be poured. The specific gravity of mercury = 13.6.
ANSWER: The volume of the metal block =12³ cm³
The volume of the metal block in the mercury =the volume of the mercury displaced by the block =12³/5 cm³
The weight of the metal block = the weight of the volume of the mercury displaced =(12³/5)*13.6 g
=13.6*12³/5 g
|
| Figure for Q-18 |
Let the required height of water column = H cm
Now the volume of the water displaced =12²*H cm³
The volume of the mercury displaced =12²(12-H) cm³
The force of the buoyancy = The weight of the water displaced + the weight of the mercury displaced
=12²*H +{12²(12-H)}*13.6 g
This should be equal to the weight of the metal cube. Equating,
12²*H +{12²(12-H)}*13.6 = 13.6*12³/5
→12²H(1-13.6) =13.6*12³/5 - 12³*13.6
→-12.6*12²H =13.6*12³(-4/5)
→12.6H =4*13.6*12/5
→H = 652.8/63 ≈ 10.4 cm
19. A hollow spherical body of inner and outer radii 6 cm and 8 cm respectively floats half submerged in water. Find the density of the material of the sphere.
ANSWER: The outer volume of the sphere =4π8³/3 cm³
The inner volume of the sphere =4π6³/3 cm³
The volume of the material =4π(8³-6³)/3 cm³
Let the density of the material of the sphere be D kg/m³.
The weight of the material ={4π(8³-6³)/(3*10⁶)}*D kg-weight
The volume of water displaced =half of the outer volume of the sphere
=2π8³/3 cm³
The weight of the water displaced ={2π8³/(3*10⁶)}*1000 kg-weight
At the floating condition, the weight of the sphere material = weight of the water displaced
→{4π(8³-6³)/(3*10⁶)}*D = {2π8³/(3*10⁶)}*1000
→D = ½*1000*8³/(8³-6³) =1000/2(1-0.75³)
=1000/1.156 =865 kg/m³
20. A solid sphere of radius 5 cm floats in water. If a maximum load of 0.1 kg can be put on it without wetting the load, find the specific gravity of the material of the sphere.
ANSWER: Let the specific gravity of the material of the sphere be S. Then the density of the material =1000S kg/m³
The volume of the sphere = 4π*5³/3 cm³ =500π/3 cm³
The weight of the sphere ={500π/(3*10⁶)}*1000S kg
Total weight of the sphere and the load =0.10 + {500π/(3*10⁶)}*1000S kg
The volume of the water displaced without wetting the load = the volume of the sphere =500π/3 cm³ = the force of buoyancy
The weight of the water displaced = {500π/(3*10⁶)}*1000 kg
Equating the weights we get,
0.10 + {500π/(3*10⁶)}*1000S ={500π/(3*10⁶)}*1000
→{500π/(3*10⁶)}*1000S ={500π/(3*10⁶)}*1000-0.10
→(500π/3000)*S =500π/3000-0.1
→S*π/6 = π/6-0.1
→S = 1-0.6/π =1-0.2 = 0.8
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Links to the Chapters
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Links to the Chapters
CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q10
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
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q10
CHAPTER- 12 - Simple Harmonic Motion
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q10
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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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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HC Verma's Concepts of Physics, Chapter-4, The Forces
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