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FLUID MECHANICS:--
QUESTIONS FOR SHORT ANSWER
1. Is it always true that the molecules of a dense liquid are heavier than the molecules of a lighter liquid?
ANSWER: No, the density of a liquid depends upon how tightly the molecules are packed, not on the weight of the individual molecules.
2. If someone presses a pointed needle against your skin, you are hurt. But if someone presses a rod against your skin with the same force, you easily tolerate. Explain.
ANSWER: It is because of the pressure in the case of the needle is more than the pressure in the case of the rod. Suppose the same force F is applied in both cases. If the area of needlepoint =a and the area of the rod =A so that A>a. The pressure of needlepoint F/a >F/A. So the needlepoint hurts.
3. In the derivation of P₁ - P₂ =ρgz, it was assumed that the liquid is incompressible. Why will this equation not be strictly valid for a compressible liquid?
ANSWER: With the assumption that the liquid is incompressible, the density ρ becomes constant and the result is valid. For a compressible liquid, ρ will change as the pressure increase (due to change in volume). So the equation is not completely valid.
4. Suppose the density of air at Madras is ρ₀ and atmospheric pressure is P₀. If we go up, the density and pressure both decrease. Suppose we wish to calculate the pressure at a height 10 km above Madras. If we use the equation P₀ -P =ρ₀gz, will we get the pressure more than the actual or less than the actual? Neglect the variation in g. Does your answer change if you also consider the variation in g?
ANSWER: P₀ -P =ρ₀gz
→P = P₀ -ρ₀gz
Suppose the average density of air up to the height z =10,000 m is
ρ <ρ₀, since g is taken constant,
ρ₀gz >ρgz
Therefore we are getting P = P₀ - ρ₀gz less than the actual.
If we consider the variation in g, then the average value of g will also be less than g at ground level. The value of ρgz will be even smaller. So the answer will not change.
5. The free surface of the liquid resting in an inertial frame is horizontal. Does the normal to the free surface pass through the center of the earth? Think separately if the liquid is (a) at the equator (b) at the pole (c) somewhere else.
ANSWER: The line of the gravitational force pass through the center of the earth but there is also a centrifugal force on the body at the surface due to rotation of the earth. If the direction of the centrifugal force is different from the gravitational force, the line of action of the net resultant force will not pass through the center of the earth. And this is the case at places other than the poles or equator.
(a) At the equator:-
The centrifugal force due to the rotation of the earth is vertically upward. This results in a reduction in the magnitude of the net force but the line of action passes through the center of the earth. So the normal to the free surface of the liquid will pass through the center of the earth.
(b) At the poles:-
The magnitude of the centrifugal force due to rotation of the earth is zero because the radius of the rotating circle is zero. So there is no change in the magnitude and the direction of the gravitational force here. Therefore the normal to the free surface will pass through the center of the earth.
(c) Somewhere else:-
As stated in the beginning, the net force on a body will not pass through the center of the earth anywhere else, the plumb-line will also not pass through the center of the earth. Since the horizontal line or the free surface of a still liquid is perpendicular to the plumb line So the normal to the free surface of the liquid will not pass through the center of the earth.
6. A barometer tube reads 76 cm of mercury. If the tube is gradually inclined keeping the open end immersed in the mercury reservoir, will the length of mercury column be 76 cm, more than 76 cm or less than 76 cm?
ANSWER: The vertical height of the mercury column in the barometer tube remains 76 cm even if it is tilted. It is because the pressure at the reservoir level depends on the vertical height of the mercury (ρgh). The length of mercury in the tube will vary accordingly.
Tilted tube of the barometer
7. A one-meter long glass tube is open at both ends. One end of the tube is dipped into a mercury cup, the tube is kept vertical and the air is pumped out of the tube by connecting the upper end to a suction pump. Can mercury be pulled up into the pump by this process?
ANSWER: The pump sucks the air in the upper part of the tube, thus the pressure in the upper part gets down from the atmospheric pressure at the surface of the mercury in the cup. So the mercury in the tube goes up towards the lower pressure. The height of the mercury column in the tube balances the pressure in the tube at the level of the free surface in the cup. The maximum suction in the upper part of the tube will be zero pressure i.e. vacuum. Thus the maximum pressure difference in the tube will be equal to one atmospheric pressure which will be balanced by 76 cm of the mercury column in the tube. Above the height of 76 cm the mercury will not rise and hence it will not reach the pump because the tube is 1 m = 100 cm long.
8. A satellite revolves around the earth. Air pressure inside the satellite is maintained at 76 cm of mercury. What will be the height of mercury column in a barometer tube 1 m long being placed in the satellite?
ANSWER: The barometer tube is initially fully filled and the open end is closed temporarily, dipped into the mercury reservoir and then opened in the liquid. On the earth's surface, the atmospheric pressure at the level of the free surface of the mercury is balanced by the weight of the mercury column in the barometer tube.
In the satellite, the value of g is zero, that is the things there are weightless. So the weight of the mercury column (initially 1 m) is zero. It means the column has no pressure at the reservoir level to balance the outer pressure of 76 cm of mercury. Therefore the level of the mercury in the barometer tube will not fall and it will remain 1 m.
9. Consider the barometer shown in figure (13-Q1). If a small hole is made at a point P in the barometer tube, will the mercury come out from this hole?
Figure for Q-9
ANSWER: No. The pressure outside the hole is equal to the atmospheric pressure say P₀. The pressure inside the hole P is equal to P₀ - ρgh (where h is the height of the hole above the reservoir level). So P < P₀. Since the outer pressure is more the mercury will not come out of the hole.
10. Is Archimedes' principle valid in an elevator accelerating up? In a car accelerating on a level road?
ANSWER: The Archimedes' principle states that the loss in the apparent weight of a fully or partially submerged body is equal to the weight of the liquid displaced. In an elevator accelerating up neither we get the true weight of the body nor the true weight of the displaced liquid. The weight of a body with mass m in the elevator going up with acceleration a = m(a+g) and the buoyancy force B = m'(g+a) where m' is the mass of the liquid displaced. So the Archimedes' principle is not valid.
Let us see what happens in an accelerating car. The liquid and the body both will have a pseudo force in the opposite direction in this non-inertial frame. If the acceleration of the car is 'a' the weight of the body will be m√(g²+a²) recorded in a spring balance. The direction will be along the resultant of g and a. Similarly, the force of buoyancy will be m'√(g²+a²) in a direction opposite to the weight. So the Archimedes' principle is not valid even in this case.
11. Why is it easier to swim in sea water than in fresh water?
ANSWER: Since the force of buoyancy is equal to the water displaced, the force of the buoyancy will be more in seawater than in the freshwater because the density of the sea-water is more than the fresh water. Therefore the swimmer in the seawater will feel lighter and easy to swim.
12. A glass of water has an ice cube floating in the water. The water level just touches the rim of the glass. Will the water overflow when the ice melts?
ANSWER: Let the mass of the floating ice = m. The mass of the displaced water is also equal to m. The volume of the displaced water = m/d, where d is the density of the water. Now when the whole ice cube melts its mass remains the same =m and now its density is also equal to d. The volume of the melted ice cube =m/d which is the same as the initially displaced water. So the volume of the water will not change and the water will not overflow when the ice cube melts.
13. A ferry boat loaded with rocks has to pass under a bridge. The maximum height of the rocks is slightly more than the height of the bridge so that the boat just fails to pass under the bridge. Should some of the rocks be removed or some more rocks be added?
ANSWER: Some more rocks should be added to the ferry boat. It will increase the weight of the boat and to balance it more water in the river would be displaced. To displace more water the boat will sink a little and it will pass under the bridge. Of course, the rocks should not be placed over the topmost rock but at a level below it.
14. Water is slightly coming out from a vertical pipe. As the water descends after coming out, its area of cross section reduces. Explain this on the basis of the equation of continuity.
ANSWER: In the continuity of flow, the rate of flow of the volume of the liquid remains constant. That is, A*v = constant. If the cross-sectional area of the pipe is A and the velocity of the liquid V, also the cross-sectional area of the flow after descending some distance A' and velocity V' then from the continuity equation
A'V' =AV
→A' =A*(V/V')
Since the coming out liquid is falling under gravity, V'>V
→V/V' <1
Thus A' < A.
So as the water descends the area of the cross section reduces.
15. While watering a distant plant, a gardener partially closes the exit hole of the pipe by putting his finger on it. Explain why this results in the water streams going to a larger distance.
ANSWER: As the equation of continuity states that at any section
Area*velocity =constant.
So the velocity at any cross-section is inversely proportional to the Area. When the gardener partially closes the exit hole of the pipe, the cross-sectional area reduces, thus the velocity increases and the water streams go to a larger distance.
16. A Gipsy car has a canvas top. When the car runs at high speed, the top bulges out. Explain.
ANSWER: The Bernoulli's equation is
P+ρgh+½ρv² =contant.
Taking ρ almost constant, in this case, h is also constant. Thus this equation becomes P +½ρv² =constant = k (say)
→P =k-½ρv²
This shows that if at a cross-section speed increases, the pressure decreases. When the car runs at high speed, the air just above the canvas is at a high speed but just below is static (with respect to the canvas). Thus the pressure above the canvas top is less in comparison to the inside. This results in bulging out of the canvas.
1. Is it always true that the molecules of a dense liquid are heavier than the molecules of a lighter liquid?
ANSWER: No, the density of a liquid depends upon how tightly the molecules are packed, not on the weight of the individual molecules.
2. If someone presses a pointed needle against your skin, you are hurt. But if someone presses a rod against your skin with the same force, you easily tolerate. Explain.
ANSWER: It is because of the pressure in the case of the needle is more than the pressure in the case of the rod. Suppose the same force F is applied in both cases. If the area of needlepoint =a and the area of the rod =A so that A>a. The pressure of needlepoint F/a >F/A. So the needlepoint hurts.
3. In the derivation of P₁ - P₂ =ρgz, it was assumed that the liquid is incompressible. Why will this equation not be strictly valid for a compressible liquid?
ANSWER: With the assumption that the liquid is incompressible, the density ρ becomes constant and the result is valid. For a compressible liquid, ρ will change as the pressure increase (due to change in volume). So the equation is not completely valid.
4. Suppose the density of air at Madras is ρ₀ and atmospheric pressure is P₀. If we go up, the density and pressure both decrease. Suppose we wish to calculate the pressure at a height 10 km above Madras. If we use the equation P₀ -P =ρ₀gz, will we get the pressure more than the actual or less than the actual? Neglect the variation in g. Does your answer change if you also consider the variation in g?
ANSWER: P₀ -P =ρ₀gz
→P = P₀ -ρ₀gz
Suppose the average density of air up to the height z =10,000 m is
ρ <ρ₀, since g is taken constant,
ρ₀gz >ρgz
Therefore we are getting P = P₀ - ρ₀gz less than the actual.
If we consider the variation in g, then the average value of g will also be less than g at ground level. The value of ρgz will be even smaller. So the answer will not change.
5. The free surface of the liquid resting in an inertial frame is horizontal. Does the normal to the free surface pass through the center of the earth? Think separately if the liquid is (a) at the equator (b) at the pole (c) somewhere else.
ANSWER: The line of the gravitational force pass through the center of the earth but there is also a centrifugal force on the body at the surface due to rotation of the earth. If the direction of the centrifugal force is different from the gravitational force, the line of action of the net resultant force will not pass through the center of the earth. And this is the case at places other than the poles or equator.
The centrifugal force due to the rotation of the earth is vertically upward. This results in a reduction in the magnitude of the net force but the line of action passes through the center of the earth. So the normal to the free surface of the liquid will pass through the center of the earth.
The magnitude of the centrifugal force due to rotation of the earth is zero because the radius of the rotating circle is zero. So there is no change in the magnitude and the direction of the gravitational force here. Therefore the normal to the free surface will pass through the center of the earth.
As stated in the beginning, the net force on a body will not pass through the center of the earth anywhere else, the plumb-line will also not pass through the center of the earth. Since the horizontal line or the free surface of a still liquid is perpendicular to the plumb line So the normal to the free surface of the liquid will not pass through the center of the earth.
6. A barometer tube reads 76 cm of mercury. If the tube is gradually inclined keeping the open end immersed in the mercury reservoir, will the length of mercury column be 76 cm, more than 76 cm or less than 76 cm?
ANSWER: The vertical height of the mercury column in the barometer tube remains 76 cm even if it is tilted. It is because the pressure at the reservoir level depends on the vertical height of the mercury (ρgh). The length of mercury in the tube will vary accordingly.
 |
| Tilted tube of the barometer |
7. A one-meter long glass tube is open at both ends. One end of the tube is dipped into a mercury cup, the tube is kept vertical and the air is pumped out of the tube by connecting the upper end to a suction pump. Can mercury be pulled up into the pump by this process?
ANSWER: The pump sucks the air in the upper part of the tube, thus the pressure in the upper part gets down from the atmospheric pressure at the surface of the mercury in the cup. So the mercury in the tube goes up towards the lower pressure. The height of the mercury column in the tube balances the pressure in the tube at the level of the free surface in the cup. The maximum suction in the upper part of the tube will be zero pressure i.e. vacuum. Thus the maximum pressure difference in the tube will be equal to one atmospheric pressure which will be balanced by 76 cm of the mercury column in the tube. Above the height of 76 cm the mercury will not rise and hence it will not reach the pump because the tube is 1 m = 100 cm long.
8. A satellite revolves around the earth. Air pressure inside the satellite is maintained at 76 cm of mercury. What will be the height of mercury column in a barometer tube 1 m long being placed in the satellite?
ANSWER: The barometer tube is initially fully filled and the open end is closed temporarily, dipped into the mercury reservoir and then opened in the liquid. On the earth's surface, the atmospheric pressure at the level of the free surface of the mercury is balanced by the weight of the mercury column in the barometer tube.
In the satellite, the value of g is zero, that is the things there are weightless. So the weight of the mercury column (initially 1 m) is zero. It means the column has no pressure at the reservoir level to balance the outer pressure of 76 cm of mercury. Therefore the level of the mercury in the barometer tube will not fall and it will remain 1 m.
9. Consider the barometer shown in figure (13-Q1). If a small hole is made at a point P in the barometer tube, will the mercury come out from this hole?
 |
| Figure for Q-9 |
ANSWER: No. The pressure outside the hole is equal to the atmospheric pressure say P₀. The pressure inside the hole P is equal to P₀ - ρgh (where h is the height of the hole above the reservoir level). So P < P₀. Since the outer pressure is more the mercury will not come out of the hole.
10. Is Archimedes' principle valid in an elevator accelerating up? In a car accelerating on a level road?
ANSWER: The Archimedes' principle states that the loss in the apparent weight of a fully or partially submerged body is equal to the weight of the liquid displaced. In an elevator accelerating up neither we get the true weight of the body nor the true weight of the displaced liquid. The weight of a body with mass m in the elevator going up with acceleration a = m(a+g) and the buoyancy force B = m'(g+a) where m' is the mass of the liquid displaced. So the Archimedes' principle is not valid.
Let us see what happens in an accelerating car. The liquid and the body both will have a pseudo force in the opposite direction in this non-inertial frame. If the acceleration of the car is 'a' the weight of the body will be m√(g²+a²) recorded in a spring balance. The direction will be along the resultant of g and a. Similarly, the force of buoyancy will be m'√(g²+a²) in a direction opposite to the weight. So the Archimedes' principle is not valid even in this case.
11. Why is it easier to swim in sea water than in fresh water?
ANSWER: Since the force of buoyancy is equal to the water displaced, the force of the buoyancy will be more in seawater than in the freshwater because the density of the sea-water is more than the fresh water. Therefore the swimmer in the seawater will feel lighter and easy to swim.
12. A glass of water has an ice cube floating in the water. The water level just touches the rim of the glass. Will the water overflow when the ice melts?
ANSWER: Let the mass of the floating ice = m. The mass of the displaced water is also equal to m. The volume of the displaced water = m/d, where d is the density of the water. Now when the whole ice cube melts its mass remains the same =m and now its density is also equal to d. The volume of the melted ice cube =m/d which is the same as the initially displaced water. So the volume of the water will not change and the water will not overflow when the ice cube melts.
13. A ferry boat loaded with rocks has to pass under a bridge. The maximum height of the rocks is slightly more than the height of the bridge so that the boat just fails to pass under the bridge. Should some of the rocks be removed or some more rocks be added?
ANSWER: Some more rocks should be added to the ferry boat. It will increase the weight of the boat and to balance it more water in the river would be displaced. To displace more water the boat will sink a little and it will pass under the bridge. Of course, the rocks should not be placed over the topmost rock but at a level below it.
14. Water is slightly coming out from a vertical pipe. As the water descends after coming out, its area of cross section reduces. Explain this on the basis of the equation of continuity.
ANSWER: In the continuity of flow, the rate of flow of the volume of the liquid remains constant. That is, A*v = constant. If the cross-sectional area of the pipe is A and the velocity of the liquid V, also the cross-sectional area of the flow after descending some distance A' and velocity V' then from the continuity equation
A'V' =AV
→A' =A*(V/V')
Since the coming out liquid is falling under gravity, V'>V
→V/V' <1
Thus A' < A.
So as the water descends the area of the cross section reduces.
15. While watering a distant plant, a gardener partially closes the exit hole of the pipe by putting his finger on it. Explain why this results in the water streams going to a larger distance.
ANSWER: As the equation of continuity states that at any section
Area*velocity =constant.
So the velocity at any cross-section is inversely proportional to the Area. When the gardener partially closes the exit hole of the pipe, the cross-sectional area reduces, thus the velocity increases and the water streams go to a larger distance.
16. A Gipsy car has a canvas top. When the car runs at high speed, the top bulges out. Explain.
ANSWER: The Bernoulli's equation is
P+ρgh+½ρv² =contant.
Taking ρ almost constant, in this case, h is also constant. Thus this equation becomes P +½ρv² =constant = k (say)
→P =k-½ρv²
This shows that if at a cross-section speed increases, the pressure decreases. When the car runs at high speed, the air just above the canvas is at a high speed but just below is static (with respect to the canvas). Thus the pressure above the canvas top is less in comparison to the inside. This results in bulging out of the canvas.
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Links to the Chapters
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ALL LINKS
CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q-10
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
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CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q-10
EXERCISES- Q11 TO Q20
CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q-10
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
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
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Click here for → Exercise(43-54)
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Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
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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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