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FLUID MECHANICS:--
OBJECTIVE-I
1. A liquid can easily change its shape but a solid can not because
(a) the density of a liquid is smaller than that of a solid
(b) the forces between the molecules are stronger in solid than in liquids
(c) the atoms combine to form bigger molecules in a solid
(d) the average separation between the molecules is larger in solids.
ANSWER: (b)
EXPLANATION: The shape will change only if the molecules have some freedom of movement. In a solid the forces between the molecules are so strong that they cannot move easily, thus the shape is not changed.
2. Consider the equations
F
P = Lim---
Δs→0 Δs
and P₁ - P₂ =ρgz.
In an elevator accelerating upward
(a) both the equations are valid
(b) the first is valid but not the second
(c) the second is valid but not the first
(d) both are invalid.
ANSWER: (b)
EXPLANATION: The first equation gives the pressure at a point. In the upward accelerating elevator, it still gives the pressure at a point.
The second equation gives the pressure difference between two points in a liquid with a height difference z. In an upward accelerating elevator, this equation is not valid because the effective g becomes g+a where a is the acceleration of the elevator.
3. The three vessels shown in figure (13-Q2) have same base area. Equal volumes of liquid are poured in the three vessels. The forces on the base will be
(a) maximum in vessel A
(b) maximum in vessel B
(c) maximum in vessel C
(d) equal in all the vessels.
Figure for Q-3
ANSWER: (c)
EXPLANATION: Due to the shape of the vessels, the same volume of liquid will have the greatest height in C and the pressure at a point in the liquid depends upon the height above the point.
4. Equal mass of three liquids are kept are kept in three identical cylindrical vessels A, B and C. The densities are ρₐ, ρᵦ, ρₑ with ρₐ<ρᵦ< ρₑ. The force on the base will be
(a) maximum in vessel A
(b) maximum in vessel B
(c) maximum in vessel C
(d) equal in all the vessels.
ANSWER: (d)
EXPLANATION: Let the mass of the liquids be m. The volumes of the liquids are m/ρₐ, m/ρᵦ, m/ρₑ. If the base areas of the identical vessels are a then the heights of the liquids in the vessels are m/aρₐ, m/aρᵦ, m/aρₑ. Hence the pressures at the bottom are ρₐgm/aρₐ, ρᵦgm/aρᵦ, ρₑgm/aρₑ
ie, mg/a, mg/a, mg/a
So equal hence (d).
5. Figure (13-Q3) shows a siphon. The liquid shown is water. The pressure difference PB-PA between the points A and B is
(a) 400 N/m²
(b) 3000 N/m²
(c) 1000 N/m²
(d) zero.
Figure for Q-5
ANSWER: (d)
EXPLANATION: The pressures at these two points are equal to atmospheric pressure hence their difference is zero.
6. A beaker containing a liquid is kept inside a big closed jar. If the air inside the jar is continuously pumped out. The pressure in the liquid near the bottom of the liquid will
(a) increase
(b) decrease
(c) remain constant
(d) First decrease and then increase.
ANSWER: (b)
EXPLANATION: The pressure at the bottom of the liquid is the sum of atmospheric pressure and the pressure due to the height of the liquid. When the air inside the jar is continuously pumped out the pressure of the air part decreases. Hence the total pressure at the bottom of the liquid will decrease.
7. The pressure in a liquid at two points in the same horizontal plane are equal. Consider an elevator accelerating upward and a car accelerating on a horizontal road. The above statement is correct in
(a) the car only
(b) the elevator only
(c) both of them
(d) neither of them.
ANSWER: (b)
EXPLANATION: In the accelerating car on the horizontal road the pressure along the horizontal plane in the liquid is not the same.
8. Suppose the pressure at the surface of the mercury in a barometer tube is P₁ and the pressure at the surface of the mercury in the cup is P₂.
(a) P₁ =0, P₂ = atmospheric pressure
(b) P₁ = atmospheric pressure, P₂ =0
(c) P₁ = P₂ atmospheric pressure
(d) P₁ = P₂ =0.
ANSWER: (a)
EXPLANATION: Since there is nothing above the surface of the mercury in the tube the density above this surface =0, hence the pressure P₁ =0, and the pressure at the surface of the mercury in the cup is P₂ = atmospheric pressure.
9. A barometer kept in an elevator reads 76 cm when it is at rest. If the elevator goes up with increasing speed, the reading will be
(a) zero
(b) 76 cm
(c) <76 cm
(d) >76 cm.
ANSWER: (c)
EXPLANATION: The pressure of the atmosphere remains the same in both the conditions =ρg*(0.76). But in the upward accelerating elevator the pressure of the mercury column =ρ(g+a)h, since these two are equal so h = (0.76){g/(g+a)}; clearly h<0.76 m i.e. h<76 cm.
10. A barometer kept in an elevator accelerating upward reads 76 cm. The air pressure in the elevator is
(a) 76 cm
(b) <76 cm
(c) >76 cm
(d) zero.
ANSWER: (c)
EXPLANATION: Since the pressure of the mercury column 76 cm high in the upward accelerating elevator is more than the stationary elevator condition due to the effective g being g+a. So the air pressure in the given elevator is >76 cm.
11. To construct a barometer a tube of length 1 m is filled completely with mercury and is inverted in a mercury cup. The barometer reading on a particular day is 76 cm. Suppose a 1 m tube is filled with mercury up to 76 cm and then closed by a cork. It is inverted in a mercury cup and the cork is removed. The height of the mercury column in the tube over the surface in the cup will be
(a) zero
(b) 76 cm
(c) >76 cm
(d) <76 cm.
ANSWER: (d)
EXPLANATION: In this case, the space above the mercury is filled with air at atmospheric pressure initially. When the cork is removed, the atmospheric pressure at the surface of the mercury in the cup is balanced by the sum of the pressures of mercury column and the air in the tube. Since the air pressure outside is 76 cm of mercury a part of which is contributed by the air inside the tube, hence the height of the mercury in the tube < 76 cm.
12. A 20 N metal block is suspended by a spring balance. A beaker containing some water is placed on a weighing machine which reads 40 N. The spring balance is now lowered so that the block gets immersed in the water. The spring balance now reads 16 N. The reading of the weighing machine will be
(a) 36 N
(b) 60 N
(c) 44 N
(d) 56 N.
ANSWER: (c)
EXPLANATION: The water in the beaker applies upward buoyancy force on the metal block = 20 -16 = 4 N. According to Newton's third law the block applies a downward force = 4 N on the water. So the weight of the water =40+4 = 44 N shown in the weighing machine.
13. A piece of wood is floating in water kept in a bottle. The bottle is connected to an air pump. Neglect the compressibility of water. When more air is pushed into the bottle from the pump, the piece of wood will float with
(a) larger part in the water
(b) lesser part in the water
(c) same part in the water
(d) it will sink.
ANSWER: (c)
EXPLANATION: Pushing air inside the bottle will increase the air pressure. When the pressure of air inside the bottle is increased it does not increase the weight of the wood nor the density of the water. So there will be no change in the submerged part.
14. A metal cube is placed in an empty vessel. When water is filled in the vessel so that the cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube
(a) will increase
(b) will decrease
(c) will remain the same
(d) will become zero.
ANSWER: (c)
EXPLANATION: In this problem, we should be clear that the water is filled in the vessel just to completely submerge the cube and not that water is filled over the cube up to some height.
The most common and easy explanation is given assuming that water does nor enter beneath the cube (which is most unlikely) thus there is no buoyancy and the force will remain the same. But practically the water will creep beneath the cube because the surfaces of contact are not fully plane. The cube will be in contact with the bottom of the vessel through some molecules and the gap will be filled by the water. See the figure below.
Figure for Q-14
The pressure of water under the cube = ρga, and it will push the cube upward with a force B =ρga*a²=ρga³ =(ρa³)*g
Now, a³ is the volume of the cube and also the volume of the water displaced, ρa³g is the weight of the water displaced and the B is the force of buoyancy. So the apparent weight of the cube becomes
W = mg-B =mg-ρga³, this weight is transferred to the vessel's surface through contact molecules. The pressure of the water beneath the cube is in all directions and it also pushes down the area beneath the cube by a force B = ρga*a²=ρga³.
So the total force downward on the area a² beneath the cube
= W+B = mg-B+B = mg which is the weight of the cube before filling the vessel with water. So the force on the bottom of the vessel in contact with the cube remains the same.
15. A wooden object floats in water kept in a beaker. the object is near a side of the beaker (figure 13-Q4). Let P₁, P₂, P₃ be the pressures at the three points A, B and C of the bottom as shown in the figure.
Figure for Q-15
(a) P₁ = P₂ = P₃
(b) P₁ < P₂ < P₃
(c) P₁ > P₂ > P₃
(d) P₂ = P₃ ≠ P₁.
ANSWER: (a)
EXPLANATION: The pressure along a horizontal plane in a liquid at rest is the same.
16. A closed cubical box is completely filled with water and accelerated horizontally towards the right with an acceleration a. The resultant normal force by the water on the top of the box
(a) passes through the center of the top
(b) passes through a point to the right of the center
(c) passes through a point to the left of the center
(d) becomes zero.
ANSWER: (c)
EXPLANATION: Due to the acceleration a towards the right, a pseudo force ma acts on the water towards left. The pressure due to this force increases from right to left and the normal pressure on the top of the box by the water will vary as shown colored in the figure below. The pressure due to the gravitation on the top is zero. The total force (N) of this triangular pressure distribution will act at 1/3rd distance from the left side of the triangle. So, N will pass through a point left of the center at the top of the box.
Figure for Q-16
17. Consider the situation of the previous problem. Let the water push the left wall by a force F₁ and the right wall by a force F₂.
(a) F₁ = F₂
(b) F₁ > F₂
(c) F₁ < F₂
(d) The information is insufficient to know the relation between F₁ and F₂.
ANSWER: (b)
EXPLANATION: As the pressure depends on the depth of the liquid at rest, similarly in an accelerating liquid the pressure due to acceleration depends on the horizontal length from the leading surface.
18. 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 end B upward. We have v₁ = v₂ for
(a) Case I
(b) Case II
(c) Case III
(d) Each case.
ANSWER: (d)
EXPLANATION: Since the water is incompressible, the volume of water entering end A and leaving through end B per unit time will remain constant in each case. Since the area of the cross-section of the cylindrical tube will be constant, v₁ = v₂ for each case.
19. Bernoulli's theorem is based on the conservation of
(a) momentum
(b) mass
(c) energy
(c) angular momentum.
ANSWER: (c)
EXPLANATION: Bernoulli's theorem is based on the work-energy theorem which states that the total work done on a system is equal to the change in its kinetic energy. It is also the conservation of energy if no work is done.
20. Water is flowing through a long horizontal tube. Let Pₐ and Pᵦ be the pressures at two different two points A and B of the tube.
(a) Pₐ must be equal to Pᵦ
(b) Pₐ must be greater than Pᵦ
(c) Pₐ must be smaller than Pᵦ
(d) Pₐ = Pᵦ only if the cross-sectional area at A and B are equal.
ANSWER: (d)
EXPLANATION: Pₐ = Pᵦ only if the velocities are equal at A and B (From the Bernoulli's theorem it can be deduced that in a horizontal tube flow pressure is less where velocity is more). The velocities at A and B will be equal only if the cross-sectional areas at A and B are the same.
21. Water and mercury are filled in two cylindrical vessels up to the same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are v₁ and v₂ respectively.
(a) v₁ = v₂
(b) v₁ = 13.6v₂
(c) v₁ = v₂/13.6
(d) v₁ = √(13.6) v₂.
ANSWER: (a)
EXPLANATION: The velocity of the water v =√(2gh) which is independent of the density of the liquid.
22. A large cylindrical tank has a hole of area A at its bottom. Water is poured in the tank by a tube of equal cross-sectional area A ejecting water at the speed v.
(a) The water level in the tank will keep rising.
(b) No water can be stored in the tank.
(c) The water level will rise to a height v²/2g and then stop.
(d) The water level will oscillate.
ANSWER: (c)
EXPLANATION: Initially the velocity of the water coming out of the tank will be negligible due to the negligible height of the water. So the quantity of the water coming out of the tank per second is less than the coming in, so the level of the water in the tank will begin to rise and the velocity of the water coming out will begin to increase. The level of the water will rise up to a height h for which the velocity of the water coming out is equal to the velocity of the water coming in (= v). So, v =√(2gh)
→h =v²/2g.
1. A liquid can easily change its shape but a solid can not because
(a) the density of a liquid is smaller than that of a solid
(b) the forces between the molecules are stronger in solid than in liquids
(c) the atoms combine to form bigger molecules in a solid
(d) the average separation between the molecules is larger in solids.
ANSWER: (b)
2. Consider the equations
F
P = Lim---
Δs→0 Δs
and P₁ - P₂ =ρgz.
In an elevator accelerating upward
(a) both the equations are valid
(b) the first is valid but not the second
(c) the second is valid but not the first
(d) both are invalid.
ANSWER: (b)
EXPLANATION: The first equation gives the pressure at a point. In the upward accelerating elevator, it still gives the pressure at a point.
The second equation gives the pressure difference between two points in a liquid with a height difference z. In an upward accelerating elevator, this equation is not valid because the effective g becomes g+a where a is the acceleration of the elevator.
EXPLANATION: The first equation gives the pressure at a point. In the upward accelerating elevator, it still gives the pressure at a point.
The second equation gives the pressure difference between two points in a liquid with a height difference z. In an upward accelerating elevator, this equation is not valid because the effective g becomes g+a where a is the acceleration of the elevator.
3. The three vessels shown in figure (13-Q2) have same base area. Equal volumes of liquid are poured in the three vessels. The forces on the base will be
(a) maximum in vessel A
(b) maximum in vessel B
(c) maximum in vessel C
(d) equal in all the vessels.
 |
| Figure for Q-3 |
ANSWER: (c)
EXPLANATION: Due to the shape of the vessels, the same volume of liquid will have the greatest height in C and the pressure at a point in the liquid depends upon the height above the point.
4. Equal mass of three liquids are kept are kept in three identical cylindrical vessels A, B and C. The densities are ρₐ, ρᵦ, ρₑ with ρₐ<ρᵦ< ρₑ. The force on the base will be
(a) maximum in vessel A
(b) maximum in vessel B
(c) maximum in vessel C
(d) equal in all the vessels.
ANSWER: (d)
ie, mg/a, mg/a, mg/a
So equal hence (d).
5. Figure (13-Q3) shows a siphon. The liquid shown is water. The pressure difference PB-PA between the points A and B is
(a) 400 N/m²
(b) 3000 N/m²
(c) 1000 N/m²
(d) zero.
 |
| Figure for Q-5 |
ANSWER: (d)
EXPLANATION: The pressures at these two points are equal to atmospheric pressure hence their difference is zero.
EXPLANATION: The pressures at these two points are equal to atmospheric pressure hence their difference is zero.
6. A beaker containing a liquid is kept inside a big closed jar. If the air inside the jar is continuously pumped out. The pressure in the liquid near the bottom of the liquid will
(a) increase
(b) decrease
(c) remain constant
(d) First decrease and then increase.
ANSWER: (b)
EXPLANATION: The pressure at the bottom of the liquid is the sum of atmospheric pressure and the pressure due to the height of the liquid. When the air inside the jar is continuously pumped out the pressure of the air part decreases. Hence the total pressure at the bottom of the liquid will decrease.
EXPLANATION: The pressure at the bottom of the liquid is the sum of atmospheric pressure and the pressure due to the height of the liquid. When the air inside the jar is continuously pumped out the pressure of the air part decreases. Hence the total pressure at the bottom of the liquid will decrease.
7. The pressure in a liquid at two points in the same horizontal plane are equal. Consider an elevator accelerating upward and a car accelerating on a horizontal road. The above statement is correct in
(a) the car only
(b) the elevator only
(c) both of them
(d) neither of them.
ANSWER: (b)
EXPLANATION: In the accelerating car on the horizontal road the pressure along the horizontal plane in the liquid is not the same.
EXPLANATION: In the accelerating car on the horizontal road the pressure along the horizontal plane in the liquid is not the same.
8. Suppose the pressure at the surface of the mercury in a barometer tube is P₁ and the pressure at the surface of the mercury in the cup is P₂.
(a) P₁ =0, P₂ = atmospheric pressure
(b) P₁ = atmospheric pressure, P₂ =0
(c) P₁ = P₂ atmospheric pressure
(d) P₁ = P₂ =0.
ANSWER: (a)
9. A barometer kept in an elevator reads 76 cm when it is at rest. If the elevator goes up with increasing speed, the reading will be
(a) zero
(b) 76 cm
(c) <76 cm
(d) >76 cm.
ANSWER: (c)
EXPLANATION: The pressure of the atmosphere remains the same in both the conditions =ρg*(0.76). But in the upward accelerating elevator the pressure of the mercury column =ρ(g+a)h, since these two are equal so h = (0.76){g/(g+a)}; clearly h<0.76 m i.e. h<76 cm.
EXPLANATION: The pressure of the atmosphere remains the same in both the conditions =ρg*(0.76). But in the upward accelerating elevator the pressure of the mercury column =ρ(g+a)h, since these two are equal so h = (0.76){g/(g+a)}; clearly h<0.76 m i.e. h<76 cm.
10. A barometer kept in an elevator accelerating upward reads 76 cm. The air pressure in the elevator is
(a) 76 cm
(b) <76 cm
(c) >76 cm
(d) zero.
ANSWER: (c)
EXPLANATION: Since the pressure of the mercury column 76 cm high in the upward accelerating elevator is more than the stationary elevator condition due to the effective g being g+a. So the air pressure in the given elevator is >76 cm.
EXPLANATION: Since the pressure of the mercury column 76 cm high in the upward accelerating elevator is more than the stationary elevator condition due to the effective g being g+a. So the air pressure in the given elevator is >76 cm.
11. To construct a barometer a tube of length 1 m is filled completely with mercury and is inverted in a mercury cup. The barometer reading on a particular day is 76 cm. Suppose a 1 m tube is filled with mercury up to 76 cm and then closed by a cork. It is inverted in a mercury cup and the cork is removed. The height of the mercury column in the tube over the surface in the cup will be
(a) zero
(b) 76 cm
(c) >76 cm
(d) <76 cm.
ANSWER: (d)
EXPLANATION: In this case, the space above the mercury is filled with air at atmospheric pressure initially. When the cork is removed, the atmospheric pressure at the surface of the mercury in the cup is balanced by the sum of the pressures of mercury column and the air in the tube. Since the air pressure outside is 76 cm of mercury a part of which is contributed by the air inside the tube, hence the height of the mercury in the tube < 76 cm.
EXPLANATION: In this case, the space above the mercury is filled with air at atmospheric pressure initially. When the cork is removed, the atmospheric pressure at the surface of the mercury in the cup is balanced by the sum of the pressures of mercury column and the air in the tube. Since the air pressure outside is 76 cm of mercury a part of which is contributed by the air inside the tube, hence the height of the mercury in the tube < 76 cm.
12. A 20 N metal block is suspended by a spring balance. A beaker containing some water is placed on a weighing machine which reads 40 N. The spring balance is now lowered so that the block gets immersed in the water. The spring balance now reads 16 N. The reading of the weighing machine will be
(a) 36 N
(b) 60 N
(c) 44 N
(d) 56 N.
(a) 36 N
(b) 60 N
(c) 44 N
(d) 56 N.
ANSWER: (c)
EXPLANATION: The water in the beaker applies upward buoyancy force on the metal block = 20 -16 = 4 N. According to Newton's third law the block applies a downward force = 4 N on the water. So the weight of the water =40+4 = 44 N shown in the weighing machine.
EXPLANATION: The water in the beaker applies upward buoyancy force on the metal block = 20 -16 = 4 N. According to Newton's third law the block applies a downward force = 4 N on the water. So the weight of the water =40+4 = 44 N shown in the weighing machine.
13. A piece of wood is floating in water kept in a bottle. The bottle is connected to an air pump. Neglect the compressibility of water. When more air is pushed into the bottle from the pump, the piece of wood will float with
(a) larger part in the water
(b) lesser part in the water
(c) same part in the water
(d) it will sink.
(a) larger part in the water
(b) lesser part in the water
(c) same part in the water
(d) it will sink.
ANSWER: (c)
EXPLANATION: Pushing air inside the bottle will increase the air pressure. When the pressure of air inside the bottle is increased it does not increase the weight of the wood nor the density of the water. So there will be no change in the submerged part.
EXPLANATION: Pushing air inside the bottle will increase the air pressure. When the pressure of air inside the bottle is increased it does not increase the weight of the wood nor the density of the water. So there will be no change in the submerged part.
14. A metal cube is placed in an empty vessel. When water is filled in the vessel so that the cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube
(a) will increase
(b) will decrease
(c) will remain the same
(d) will become zero.
(a) will increase
(b) will decrease
(c) will remain the same
(d) will become zero.
ANSWER: (c)
EXPLANATION: In this problem, we should be clear that the water is filled in the vessel just to completely submerge the cube and not that water is filled over the cube up to some height.
The most common and easy explanation is given assuming that water does nor enter beneath the cube (which is most unlikely) thus there is no buoyancy and the force will remain the same. But practically the water will creep beneath the cube because the surfaces of contact are not fully plane. The cube will be in contact with the bottom of the vessel through some molecules and the gap will be filled by the water. See the figure below.
 |
| Figure for Q-14 |
The pressure of water under the cube = ρga, and it will push the cube upward with a force B =ρga*a²=ρga³ =(ρa³)*g
Now, a³ is the volume of the cube and also the volume of the water displaced, ρa³g is the weight of the water displaced and the B is the force of buoyancy. So the apparent weight of the cube becomes
W = mg-B =mg-ρga³, this weight is transferred to the vessel's surface through contact molecules. The pressure of the water beneath the cube is in all directions and it also pushes down the area beneath the cube by a force B = ρga*a²=ρga³.
So the total force downward on the area a² beneath the cube= W+B = mg-B+B = mg which is the weight of the cube before filling the vessel with water. So the force on the bottom of the vessel in contact with the cube remains the same.
15. A wooden object floats in water kept in a beaker. the object is near a side of the beaker (figure 13-Q4). Let P₁, P₂, P₃ be the pressures at the three points A, B and C of the bottom as shown in the figure.
 |
| Figure for Q-15 |
(a) P₁ = P₂ = P₃
(b) P₁ < P₂ < P₃
(c) P₁ > P₂ > P₃
(d) P₂ = P₃ ≠ P₁.
ANSWER: (a)
EXPLANATION: The pressure along a horizontal plane in a liquid at rest is the same.
EXPLANATION: The pressure along a horizontal plane in a liquid at rest is the same.
16. A closed cubical box is completely filled with water and accelerated horizontally towards the right with an acceleration a. The resultant normal force by the water on the top of the box
(a) passes through the center of the top
(b) passes through a point to the right of the center
(c) passes through a point to the left of the center
(d) becomes zero.
ANSWER: (c)
EXPLANATION: Due to the acceleration a towards the right, a pseudo force ma acts on the water towards left. The pressure due to this force increases from right to left and the normal pressure on the top of the box by the water will vary as shown colored in the figure below. The pressure due to the gravitation on the top is zero. The total force (N) of this triangular pressure distribution will act at 1/3rd distance from the left side of the triangle. So, N will pass through a point left of the center at the top of the box.
EXPLANATION: Due to the acceleration a towards the right, a pseudo force ma acts on the water towards left. The pressure due to this force increases from right to left and the normal pressure on the top of the box by the water will vary as shown colored in the figure below. The pressure due to the gravitation on the top is zero. The total force (N) of this triangular pressure distribution will act at 1/3rd distance from the left side of the triangle. So, N will pass through a point left of the center at the top of the box.
 |
| Figure for Q-16 |
17. Consider the situation of the previous problem. Let the water push the left wall by a force F₁ and the right wall by a force F₂.
(a) F₁ = F₂
(b) F₁ > F₂
(c) F₁ < F₂
(d) The information is insufficient to know the relation between F₁ and F₂.
(a) F₁ = F₂
(b) F₁ > F₂
(c) F₁ < F₂
(d) The information is insufficient to know the relation between F₁ and F₂.
ANSWER: (b)
EXPLANATION: As the pressure depends on the depth of the liquid at rest, similarly in an accelerating liquid the pressure due to acceleration depends on the horizontal length from the leading surface.
EXPLANATION: As the pressure depends on the depth of the liquid at rest, similarly in an accelerating liquid the pressure due to acceleration depends on the horizontal length from the leading surface.
18. 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 end B upward. We have v₁ = v₂ for
(a) Case I
(b) Case II
(c) Case III
(d) Each case.
(a) Case I
(b) Case II
(c) Case III
(d) Each case.
ANSWER: (d)
19. Bernoulli's theorem is based on the conservation of
(a) momentum
(b) mass
(c) energy
(c) angular momentum.
(a) momentum
(b) mass
(c) energy
(c) angular momentum.
ANSWER: (c)
EXPLANATION: Bernoulli's theorem is based on the work-energy theorem which states that the total work done on a system is equal to the change in its kinetic energy. It is also the conservation of energy if no work is done.
20. Water is flowing through a long horizontal tube. Let Pₐ and Pᵦ be the pressures at two different two points A and B of the tube.
(a) Pₐ must be equal to Pᵦ
(b) Pₐ must be greater than Pᵦ
(c) Pₐ must be smaller than Pᵦ
(d) Pₐ = Pᵦ only if the cross-sectional area at A and B are equal.
(a) Pₐ must be equal to Pᵦ
(b) Pₐ must be greater than Pᵦ
(c) Pₐ must be smaller than Pᵦ
(d) Pₐ = Pᵦ only if the cross-sectional area at A and B are equal.
ANSWER: (d)
EXPLANATION: Pₐ = Pᵦ only if the velocities are equal at A and B (From the Bernoulli's theorem it can be deduced that in a horizontal tube flow pressure is less where velocity is more). The velocities at A and B will be equal only if the cross-sectional areas at A and B are the same.
21. Water and mercury are filled in two cylindrical vessels up to the same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are v₁ and v₂ respectively.
(a) v₁ = v₂
(b) v₁ = 13.6v₂
(c) v₁ = v₂/13.6
(d) v₁ = √(13.6) v₂.
(b) v₁ = 13.6v₂
(c) v₁ = v₂/13.6
(d) v₁ = √(13.6) v₂.
ANSWER: (a)
EXPLANATION: The velocity of the water v =√(2gh) which is independent of the density of the liquid.
22. A large cylindrical tank has a hole of area A at its bottom. Water is poured in the tank by a tube of equal cross-sectional area A ejecting water at the speed v.
(a) The water level in the tank will keep rising.
(b) No water can be stored in the tank.
(c) The water level will rise to a height v²/2g and then stop.
(d) The water level will oscillate.
(a) The water level in the tank will keep rising.
(b) No water can be stored in the tank.
(c) The water level will rise to a height v²/2g and then stop.
(d) The water level will oscillate.
ANSWER: (c)
EXPLANATION: Initially the velocity of the water coming out of the tank will be negligible due to the negligible height of the water. So the quantity of the water coming out of the tank per second is less than the coming in, so the level of the water in the tank will begin to rise and the velocity of the water coming out will begin to increase. The level of the water will rise up to a height h for which the velocity of the water coming out is equal to the velocity of the water coming in (= v). So, v =√(2gh)
→h =v²/2g.
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Links to the Chapters
===<<<O>>>===
Links to the Chapters
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
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
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
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
HC Verma's Concepts of Physics, Chapter-7, Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
HC Verma's Concepts of Physics, Chapter-6, Friction
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
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
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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