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FLUID MECHANICS:--
EXERCISES Q1 TO Q10
1. The surface of the water in a water tank on the top of a house is 4 m above the tap level. Find the pressure of water at the tap when the tap is closed. Is it necessary to specify that the tap is closed? Take g = 10 m/s².
ANSWER: The pressure P = ρgh
ρ =1000 kg/m³
g =10 m/s²
h =4 m
∴ P = 1000*10*4 N/m² =40000 N/m².
Yes, it is necessary to specify that the tap is closed because for the open tap the pressure will be different.
2. The heights of mercury surfaces in the two arms of the manometer shown in figure (13-E1) are 2 cm and 8 cm. Atmospheric pressure = 1.01 x 10⁵ N/m². Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.
Figure for Q-2
ANSWER: (a) The pressure of the gas = Pressure of mercury column (difference) + atmospheric pressure.
=13.6*(10⁶/1000)*9.8*{(8-2)/100}+ 1.01 x 10⁵ N/m²
=8000+1.01 x 10⁵ N/m²
=0.08 x 10⁵+1.01 x 10⁵ N/m²
=1.09 x 10⁵ N/m²
(b) The pressure of the mercury at the bottom of the U-tube
=The pressure of mercury column on the open arm + Atmospheric pressure
=13.6(10⁶/1000)*9.8*0.08+1.01 x 10⁵ N/m²
=0.107 x10⁵ + 1.01 x 10⁵ N/m²
≈1.12 x 10⁵ N/m²
3. The area of the cross-section of the wider tube shown in figure (13-E2) is 900 cm². If the boy standing on the piston weighs 45 kg, find the difference in the levels of water in the two tubes.
Figure for Q-3
ANSWER: Cross-section area A = 900 cm² =900/10000 m²
=0.09 m²,
Weight of the boy =45*9.8 N =441 N
The pressure below the boy =441/0.09 N/m² =4900 N/m²
Hence the pressure due to difference of water column also =4900 N/m²
Let the difference of the water column = X
Hence 1000*9.8*X = 4900 N/m²
→X = 4900/(1000*9.8) = 0.5 m =50 cm
4. A glass full of water has a bottom of area 20 cm², top of area 20 cm², height 20 cm and volume half a liter. (a) Find the force exerted by the water on the bottom. (b) Considering the equilibrium of the water, find the resultant force exerted by the sides of the glass on the water. Atmospheric pressure = 1.0 x 10⁵ N/m². Density of water = 1000 kg/m³ and g = 10 m/s². Take all numbers to be exact.
Figure for Q-4
ANSWER: (a) The pressure of water at the bottom =ρgh+Atmospheric Pressure
=1000*10*(20/100) + 1.0 x 10⁵ N/m²
=2000 + 1.0 x 10⁵ N/m²
=0.02 x10⁵ +1.0 x 10⁵ N/m²
=1.02 x 10⁵ N/m²
Hence the force by the water on the bottom
=1.02 x 10⁵ N/m²*(20/10000) m²
=204 N
(Downwards)
(b) The resultant horizontal force by the sides of the glass = zero, because the horizontal pressure by the water is balanced by the equal reaction by the wall. The vertical force by the sides and the bottom of the glass F =Weight of the water + Force due to atmospheric pressure
=mg + A*Pₐ
=0.5*10 + (20/10000)*1.0 x 10⁵ N
=5 + 200 N
=205 N
Force by the bottom of the glass on the water is equal but opposite to that of the force by the water on the bottom P = 204 N
Hence the force by the sides of the glass on the water = F - P
=205-204 = 1 N upwards
5. Suppose the glass of the previous problem is covered by a jar and the air inside the jar is completely pumped out. (a) What will be the answers to the problem? (b) Show that the answers do not change if a glass of different shape is used provided the height, the bottom area and the volume are unchanged.
ANSWER: (a) In this case the atmospheric pressure on the surface of the water = zero.
Hence, the pressure of water at the bottom of the glass =ρgh
=1000*10*(20/100) N/m²
=2000 N/m²
Hence the force by the water on the bottom
=2000 N/m²*(20/10000) m²
=4 N Downwards.
(b) Whatever be the shape of the glass the resultant horizontal force by the glass on the water will be zero and the resultant verical force by the glass on the water = weight of the water = 0.50*10 N = 5 N.
Hence the force by the side of the glass on the water
= 5 N -1000*10* (20/100)*(20/10000) N
= 5 N - 4 N = 1 N upward.
As we see that the force by the bottom of the glass on the water depends on the height of water and the bottom area, also the total vertical force by the glass on the water depends only on the volume of the water. These do not depend on the shape of the glass. Thus the force by the glass side on the water will remain the same if the height, the bottom area and the volume are unchanged.
6. If water be used to construct a barometer, what would be the height of water-column at standard atmospheric pressure (76 cm of mercury)?
ANSWER: The pressure of the water column should be the same as that of 76 cm of the mercury. Let the required height of water be H cm.
→ 1000*g*H/100 = 13600*g*76/100
→H = 13600*76/1000 =13.6*76 = 1033.6 cm
7. Find the force exerted by the water on a 2 m² plane surface of a large stone placed at the bottom of a sea 500 m deep. Does the force depend on the orientation of the surface?
ANSWER: The pressure of the water at 500 m depth
=1000*10*500 N/m² {Taking g =10 m/s²},
=5 x 10⁶ N/m²
Hence the force exerted by the water on the 2 m² of the stone
=2* 5 x 10⁶ N
=10⁷ N
This force will remain the same irrespective of the orientation of the surface because the pressure at a point in the liquid is the same in all directions and acts perpendicular to the surface.
8. Water is filled in a rectangular tank of size 3 m x 2 m x 1 m. (a) Find the total force exerted by the water on the bottom surface of the tank. (b) Consider a vertical side of area 2 m x 1 m. Take a horizontal strip of width ẟx meter in this side, situated at a depth of x meter from the surface of the water. Find the force by the water on this strip. (c) Find the torque of the force calculated in part (b) about the bottom of edge of this side. (d) Find the total force by the water on this side. (e) Find the total torque by the water on the side about the bottom edge. Neglect the atmospheric pressure and take g = 10 m/s².
ANSWER: (a) The dimensions of the tank are Length L = 3 m,
Breadth B = 2 m, Height H = 1 m. Since the tank is filled, the height of water in the tank = H = 1 m.
The pressure of water at the bottom = ρgH
=1000*10*1 =10000 N/m²
Area of the bottom of the tank = L x B = 3 m x 2 m = 6 m²
So the force on the bottom = Pressure * Area
=10000*6 = 60000 N
(b) The pressure at x m below the surface of the water = ρgx
Area of a horizontal strip ẟx wide and 2 m length =2*ẟx
The force on this strip F =Area*pressure
=2* ẟx*ρgx
=2*ẟx*1000*10*x
=20000.x.ẟx N
Figure for Q-8
(c) The perpendicular distance of this force from the bottom of edge of this side = H-x =1-x m
Hence the torque of the force = Force*perpendicular distance
=20000.x.ẟx*(1-x) N-m
=20000.x(1-x)ẟx N-m
(d) The total force of the water on this side =∫Fda, {from x=0 to x=1 m, where F is the pressure at the depth x m and da=area of the strip}
=∫ρgx*(2*dx) {dx is the width of the strip}
=2ρg∫x.dx
=2ρg*[x²/2] {Limit from x=0 to x=H=1 m}
=ρgH²
=1000*10*1²
=10000 N
(e) The variation of the pressure will be triangular as shown in the figure.
We can also find out the force on the side of the tank by calculating the area of this triangle.
Total Force =½*ρgH*H*(2 m)
=ρgH² =1000*10*1² =10000 N
The height of this resultant force from the bottom of edge of the side will be one-third of H (height of the C.G. from the bottom) =H/3
Hence the torque =10000*H/3 =10000*1/3 N-m
=10000/3 N-m
9. An ornament weighing 36 g in air, weighs only 34 g in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. The specific gravity of gold is 19.3 and that of copper is 8.9.
ANSWER: Let the amount of copper in the ornament be x grams, then the volume of the ornament
V= Volume of copper + volume of gold
=(x/8.9) +(36-x)/19.3 cm³
When the ornament is fully dipped into the water it replaces V volume of water and the weight of the V volume of water is the buoyancy force on the ornament or the reduction in the weight. The weight of V volume of the water =V*1*g =Vg. From the given data,
Vg =(36-34)g {g is acceleration due to gravity}
→(x/8.9) +(36-x)/19.3 = 2
→19.3x + (36-x)8.9 =2*8.9*19.3
→19.3x-8.9x =2*8.9*19.3-36*8.9
→10.4x =8.9(38.6-36) =8.9*2.6
→x = 8.9*2.6/10.4 =8.9/4 ≈2.2 grams
10. Refer to the previous problem. Suppose, the goldsmith argues that he has not mixed copper or any other material with gold, rather some cavities might have been left inside the ornament. Calculate the volume of the cavities left that will allow the weights given in that problem.
ANSWER: Let the volume of the cavities be v,
The total volume of the ornament V =volume of gold+volume of cavities
=36/19.3 +v cm³
The volume of the water displaced is also V cm³
The mass of the water displaced =V*1 gram =V gram
The weight of the water displaced =Vg
Hence Vg=(36-34)g
→V=2
→36/19.3 +v =2
→v =2-36/19.3 =2-1.865
→v =0.135 cm³
→v =0.135 cm³
(Note: the result has some variation from the answer in the book which is 0.112 cm³)
1. The surface of the water in a water tank on the top of a house is 4 m above the tap level. Find the pressure of water at the tap when the tap is closed. Is it necessary to specify that the tap is closed? Take g = 10 m/s².
ANSWER: The pressure P = ρgh
ρ =1000 kg/m³
g =10 m/s²
h =4 m
∴ P = 1000*10*4 N/m² =40000 N/m².
Yes, it is necessary to specify that the tap is closed because for the open tap the pressure will be different.
2. The heights of mercury surfaces in the two arms of the manometer shown in figure (13-E1) are 2 cm and 8 cm. Atmospheric pressure = 1.01 x 10⁵ N/m². Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.
 |
| Figure for Q-2 |
ANSWER: (a) The pressure of the gas = Pressure of mercury column (difference) + atmospheric pressure.
=13.6*(10⁶/1000)*9.8*{(8-2)/100}+ 1.01 x 10⁵ N/m²
=8000+1.01 x 10⁵ N/m²
=0.08 x 10⁵+1.01 x 10⁵ N/m²
=1.09 x 10⁵ N/m²
(b) The pressure of the mercury at the bottom of the U-tube
=The pressure of mercury column on the open arm + Atmospheric pressure
=13.6(10⁶/1000)*9.8*0.08+1.01 x 10⁵ N/m²
=0.107 x10⁵ + 1.01 x 10⁵ N/m²
≈1.12 x 10⁵ N/m²
3. The area of the cross-section of the wider tube shown in figure (13-E2) is 900 cm². If the boy standing on the piston weighs 45 kg, find the difference in the levels of water in the two tubes.
 |
| Figure for Q-3 |
ANSWER: Cross-section area A = 900 cm² =900/10000 m²
=0.09 m²,
Weight of the boy =45*9.8 N =441 N
The pressure below the boy =441/0.09 N/m² =4900 N/m²
Hence the pressure due to difference of water column also =4900 N/m²
Let the difference of the water column = X
Hence 1000*9.8*X = 4900 N/m²
→X = 4900/(1000*9.8) = 0.5 m =50 cm
4. A glass full of water has a bottom of area 20 cm², top of area 20 cm², height 20 cm and volume half a liter. (a) Find the force exerted by the water on the bottom. (b) Considering the equilibrium of the water, find the resultant force exerted by the sides of the glass on the water. Atmospheric pressure = 1.0 x 10⁵ N/m². Density of water = 1000 kg/m³ and g = 10 m/s². Take all numbers to be exact.
 |
| Figure for Q-4 |
ANSWER: (a) The pressure of water at the bottom =ρgh+Atmospheric Pressure
=1000*10*(20/100) + 1.0 x 10⁵ N/m²
=2000 + 1.0 x 10⁵ N/m²
=0.02 x10⁵ +1.0 x 10⁵ N/m²
=1.02 x 10⁵ N/m²
Hence the force by the water on the bottom
=1.02 x 10⁵ N/m²*(20/10000) m²
=204 N
(Downwards)
(b) The resultant horizontal force by the sides of the glass = zero, because the horizontal pressure by the water is balanced by the equal reaction by the wall. The vertical force by the sides and the bottom of the glass F =Weight of the water + Force due to atmospheric pressure
=mg + A*Pₐ
=0.5*10 + (20/10000)*1.0 x 10⁵ N
=5 + 200 N
=205 N
Force by the bottom of the glass on the water is equal but opposite to that of the force by the water on the bottom P = 204 N
Hence the force by the sides of the glass on the water = F - P
=205-204 = 1 N upwards
5. Suppose the glass of the previous problem is covered by a jar and the air inside the jar is completely pumped out. (a) What will be the answers to the problem? (b) Show that the answers do not change if a glass of different shape is used provided the height, the bottom area and the volume are unchanged.
ANSWER: (a) In this case the atmospheric pressure on the surface of the water = zero.
Hence, the pressure of water at the bottom of the glass =ρgh
=1000*10*(20/100) N/m²
=2000 N/m²
Hence the force by the water on the bottom
=2000 N/m²*(20/10000) m²
=4 N Downwards.
(b) Whatever be the shape of the glass the resultant horizontal force by the glass on the water will be zero and the resultant verical force by the glass on the water = weight of the water = 0.50*10 N = 5 N.
Hence the force by the side of the glass on the water
= 5 N -1000*10* (20/100)*(20/10000) N
= 5 N - 4 N = 1 N upward.
As we see that the force by the bottom of the glass on the water depends on the height of water and the bottom area, also the total vertical force by the glass on the water depends only on the volume of the water. These do not depend on the shape of the glass. Thus the force by the glass side on the water will remain the same if the height, the bottom area and the volume are unchanged.
6. If water be used to construct a barometer, what would be the height of water-column at standard atmospheric pressure (76 cm of mercury)?
ANSWER: The pressure of the water column should be the same as that of 76 cm of the mercury. Let the required height of water be H cm.
→ 1000*g*H/100 = 13600*g*76/100
→H = 13600*76/1000 =13.6*76 = 1033.6 cm
7. Find the force exerted by the water on a 2 m² plane surface of a large stone placed at the bottom of a sea 500 m deep. Does the force depend on the orientation of the surface?
ANSWER: The pressure of the water at 500 m depth
=1000*10*500 N/m² {Taking g =10 m/s²},
=5 x 10⁶ N/m²
Hence the force exerted by the water on the 2 m² of the stone
=2* 5 x 10⁶ N
=10⁷ N
This force will remain the same irrespective of the orientation of the surface because the pressure at a point in the liquid is the same in all directions and acts perpendicular to the surface.
8. Water is filled in a rectangular tank of size 3 m x 2 m x 1 m. (a) Find the total force exerted by the water on the bottom surface of the tank. (b) Consider a vertical side of area 2 m x 1 m. Take a horizontal strip of width ẟx meter in this side, situated at a depth of x meter from the surface of the water. Find the force by the water on this strip. (c) Find the torque of the force calculated in part (b) about the bottom of edge of this side. (d) Find the total force by the water on this side. (e) Find the total torque by the water on the side about the bottom edge. Neglect the atmospheric pressure and take g = 10 m/s².
ANSWER: (a) The dimensions of the tank are Length L = 3 m,
Breadth B = 2 m, Height H = 1 m. Since the tank is filled, the height of water in the tank = H = 1 m.
The pressure of water at the bottom = ρgH
=1000*10*1 =10000 N/m²
Area of the bottom of the tank = L x B = 3 m x 2 m = 6 m²
So the force on the bottom = Pressure * Area
=10000*6 = 60000 N
(b) The pressure at x m below the surface of the water = ρgx
Area of a horizontal strip ẟx wide and 2 m length =2*ẟx
The force on this strip F =Area*pressure
=2* ẟx*ρgx
=2*ẟx*1000*10*x
=20000.x.ẟx N |
| Figure for Q-8 |
(c) The perpendicular distance of this force from the bottom of edge of this side = H-x =1-x m
Hence the torque of the force = Force*perpendicular distance
=20000.x.ẟx*(1-x) N-m
=20000.x(1-x)ẟx N-m
(d) The total force of the water on this side =∫Fda, {from x=0 to x=1 m, where F is the pressure at the depth x m and da=area of the strip}
=∫ρgx*(2*dx) {dx is the width of the strip}
=2ρg∫x.dx
=2ρg*[x²/2] {Limit from x=0 to x=H=1 m}
=ρgH²
=1000*10*1²
=10000 N
(e) The variation of the pressure will be triangular as shown in the figure.
We can also find out the force on the side of the tank by calculating the area of this triangle.
Total Force =½*ρgH*H*(2 m)
=ρgH² =1000*10*1² =10000 N
The height of this resultant force from the bottom of edge of the side will be one-third of H (height of the C.G. from the bottom) =H/3
Hence the torque =10000*H/3 =10000*1/3 N-m
=10000/3 N-m
9. An ornament weighing 36 g in air, weighs only 34 g in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. The specific gravity of gold is 19.3 and that of copper is 8.9.
ANSWER: Let the amount of copper in the ornament be x grams, then the volume of the ornament
V= Volume of copper + volume of gold
=(x/8.9) +(36-x)/19.3 cm³
When the ornament is fully dipped into the water it replaces V volume of water and the weight of the V volume of water is the buoyancy force on the ornament or the reduction in the weight. The weight of V volume of the water =V*1*g =Vg. From the given data,
Vg =(36-34)g {g is acceleration due to gravity}
→(x/8.9) +(36-x)/19.3 = 2
→19.3x + (36-x)8.9 =2*8.9*19.3
→19.3x-8.9x =2*8.9*19.3-36*8.9
→10.4x =8.9(38.6-36) =8.9*2.6
→x = 8.9*2.6/10.4 =8.9/4 ≈2.2 grams
10. Refer to the previous problem. Suppose, the goldsmith argues that he has not mixed copper or any other material with gold, rather some cavities might have been left inside the ornament. Calculate the volume of the cavities left that will allow the weights given in that problem.
ANSWER: Let the volume of the cavities be v,
The total volume of the ornament V =volume of gold+volume of cavities
=36/19.3 +v cm³
The volume of the water displaced is also V cm³
The mass of the water displaced =V*1 gram =V gram
The weight of the water displaced =Vg
Hence Vg=(36-34)g
→V=2
→36/19.3 +v =2
→v =2-36/19.3 =2-1.865
→v =0.135 cm³
→v =0.135 cm³
(Note: the result has some variation from the answer in the book which is 0.112 cm³)
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES- Q11 TO Q20
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- Q11 TO Q20
CHAPTER- 12 - Simple Harmonic Motion
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES- Q11 TO Q20
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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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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