Sunday, October 7, 2018

Solutions to Problems on "FLUID MECHANICS" - H C Verma's Concepts of Physics, Part-I, Chapter-13, OBJECTIVE-II

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FLUID MECHANICS:--
OBJECTIVE-II

1. A solid floats in a liquid in a partially dipped position.
(a) The solid exerts a force equal to its weight on the liquid.
(b) The liquid exerts a force of buoyancy on the solid which is equal to the weight of the solid.
(c) The weight of the displaced liquid equals the weight of the solid.
(d) The weight of the dipped part of the solid is equal to the weight of the displaced liquid. 

ANSWER: (a), (b), (c)

EXPLANATION:  (a) The weight acts downward so the floating solid will exert a force equal to its weight on the liquid.
(b) In the case of floating solid the force of buoyancy is equal to the weight of the solid, that is why it floats.
(c) The weight of the displaced liquid is equal to the force of buoyancy which is equal to the weight of the solid. Archimedes principle.
(d) is not true because it is not according to the explanation (c) above. 

2. The weight of an empty balloon on a spring balance is W₁. The weight becomes W₂ when the balloon is filled with air. Let the weight of the air itself be w. Neglect the thickness of the balloon when it is filled with air. Also, neglect the difference in the densities of air inside and outside the balloon.

(a) W₂ = W₁
(b) W₂ = W₁ + w
(c) W₂ < W₁ + w
(d) W₂ > W₁


ANSWER: (a), (c)
Figure for Q-2

EXPLANATION:  The weight of the empty balloon is W₁. The weight of the balloon filled with air should be W₁+w but the balloon is immersed fully in the air, so a force of buoyancy must act on it so the apparent weight W₂ < W₁+w, hence (c) is correct. Since the densities of air inside and outside are same so the weight of the displaced air by the balloon will also be w. Thus the apparent weight of air-filled balloon W₂ = (W₁+w)-w = W₁. Hence (a) is correct.
Other options are not true. 

3. A solid is completely immersed in a liquid. The force exerted by the liquid on the solid will

(a) increase if it is pushed deeper inside the liquid 
(b) change if its orientation is changed
(c) decrease if it is taken partially out of the liquid
(d) be in the vertically upward direction. 

ANSWER: (c), (d)

EXPLANATION:  The force exerted by the liquid on the solid is equal to the volume of the liquid displaced, and the volume of the liquid displaced will not change if it is pushed deeper inside of the liquid or its orientation inside the liquid is changed. Hence (a) and (b) is not true.
  
(c) is correct because if it is taken partially out the volume of the displaced liquid will decrease.
(d) is correct because of the pressure in a horizontal plane in the liquid is same, so no net horizontal force on the solid. But the pressure near the top of the solid is less than near the bottom of the solid. Hence the force exerted by the liquid on the solid will be in the vertically upward direction.

4. A closed vessel is half filled with water. There is a hole near the top of the vessel and air is pumped out from this hole.

(a) The water level will rise up in the vessel.
(b) The pressure at the surface of the water will decrease.
(c) The force by the water on the bottom of the vessel will decrease.
(d) The density of the liquid will decrease. 

ANSWER: (b), (c)

EXPLANATION:  The option (a) is not true because due to gravity the water level will remain the same in the closed vessel. 
            Initially, the air over the water surface is at atmospheric pressure so the pressure at the surface of the water is equal to the atmospheric pressure. When air is pumped out the pressure at the surface of the water decreases to zero because there is a vacuum now. Option (b) is true.  
          Initially, the force by the water on the bottom of the vessel is equal to the force by the air plus the force by the water. When the air is pumped out the force by the air is absent. Therefore the total force by the water on the bottom of the vessel will decrease. Option (c) is correct. 
         Since the water is liquid and we assume it to be incompressible there will be hardly any change in the volume. Mass will also remain the same so there will be no change in the density when the air above the water is pumped out. Option (d) is not true.

5. In a streamline flow,

(a) the speed of a particle always remains the same.
(b) the velocity of a particle always remains the same
(c) the kinetic energy of all the particles arriving at a given point are the same.
(d) the momenta of all the particles arriving at a given point are the same. 

ANSWER: (c), (d)

EXPLANATION:  The speed and the velocity of a particle in a streamline flow may change as it moves ahead so (a) and (b) are not true.
         All the particles arriving at a point in a streamline flow have the same velocity. Hence all the particles arriving at that point will have the same kinetic energy and the same momentum. Options (c) and (d) are true.

6. Water flows through two identical tubes A and B. A volume V₀ of water passes through the tube A and 2V₀ through B in a given time. Which of the following may be correct?

(a) Flow in both the tubes are steady.
(b) Flow in both the tubes are turbulent.
(c) Flow is steady in A but turbulent in B.
(d) Flow is steady in B but turbulent in A. 

ANSWER: (a), (b), (c)

EXPLANATION:  Since double the volume passes in B at the same time, the velocity of flow in B will be twice that of A. There may be a limit of velocity after which the steady flow may turn into turbulent flow. If the velocities in both the tubes are below this limit option (a) may be correct.
     If the velocities in both the tubes are above this limit velocity option (b) may be true.
       If the velocity in A is less than the limit velocity but in B more than the limit then option (c) may be correct.
       If the velocity in B is less than the limit velocity then only the flow in it will be steady, the velocity in A will be half of B so it will also be steady, not turbulent. Hence the option (d) will never be correct.       

7. Water is flowing in streamline motion through a tube with its axis horizontal. Consider two points A and B in the tube at the same horizontal level.

(a) The pressure in A and B are equal for any shape of the tube.
(b) The pressures are near equal.
(c) The pressures are equal if the tube has a uniform cross-section.
(d) The pressures may be equal if the tube has a non-uniform cross-section.

ANSWER: (c), (d)

EXPLANATION:  According to the Bernoulli's theorem if the height of the points is the same the pressure will be more where the speed is less. 
         If the tube is of any shape the velocities at A and B may vary. Hence pressure at A and B will not be equal. Option (a) is not true.
        If we do not know the shape of the tube and the cross-section at A and B, the pressures at A and B may vary. Option (b) is not true.
         If the tube has uniform cross-section the velocity, in this case, will be equal everywhere so the pressures are also equal. Option (c) is true.
         If the cross-section of the tube is non-uniform there may be a chance that the areas of the cross-sections at A and B are the same. In that case, the velocities at A and B will be the same and hence the pressure. Option (d) is true.

8. There is a small hole near the bottom of an open tank filled with a liquid. The speed of the water ejected does not depend on

(a) area of the hole
(b) the density of the liquid
(c) the height of the liquid from the hole
(d) acceleration due to gravity. 

ANSWER: (a), (b)


EXPLANATION:  The velocity of the liquid coming out of a hole in a tank is given as V =√(2gh) where h is the height of the liquid from the hole and g is the acceleration due to gravity. Clearly, V depends on g and h only. It does not depend on the area of the hole and the density of the liquid.
       Hence the options (a) and (b) are true but the options (c) and (d) are not true.

 ===<<<O>>>=== 

Links to the Chapters


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 → Exercises (11-20)

Click here for → Exercises (21-30)
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)

HC Verma's Concepts of Physics, Chapter-6, Friction

Click here for → Friction OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II

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" .

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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion

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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HC Verma's Concepts of Physics, Chapter-3, Kinematics-Rest and Motion:---

Click here for "OBJECTIVE-I"

Click here for EXERCISES (Question number 1 to 10)

Click here for EXERCISES (Question number 11 to 20)

Click here for EXERCISES (Question number 21 to 30)

Click here for EXERCISES (Question number 31 to 40)

Click here for EXERCISES (Question number 41 to 52)

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Chapter -2, "Vector related Problems"

Click here for "Questions for Short Answers"

Click here for "OBJECTIVE-II"


Click here for "Exercises"   


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