Physics 312 Final Notes

312 0000 CHAPTER: 3 LAWS OF MOTION Question 1: charge in momentum of a body is: [PYQ Oct 2022] (a) Force × displacement (b) mass × displacement (c) force × time (d) force × velo...

312 0000 CHAPTER: 3 LAWS OF MOTION Question 1: charge in momentum of a body is: [PYQ Oct 2022] (a) Force × displacement (b) mass × displacement (c) force × time (d) force × velocity Answer: (c) force × time Explanation: According to Newton’s second law ∆p Fex ∝  ∆p = Fex × ∆t ∆t Hence, change in momentum = ∆p = force × time Question 2: Define the term impulse. Give its S.I. unit. [PYQ Oct 2022] Answer: The impulse is defined as “the product of force and time, which is change in momentum of the body.” I = F. Δt S.I. unit of impulse is Newton × sec. Question 3: A body of mass m is thrown vertically up with velocity v. it returns back to the thrower with the same velocity. Calculate: (i) Change in momentum. (ii) Change in magnitude of momentum of the body. [PYQ April 2022] Answer: (i) P = mv Pinitial = mv and Pfinal = mv Change in momentum = Pinitial – Pfinal = mv – mv = 0 (ii) change in magnitude of momentum = 0 Question 4: State Newton’s three laws of motion. [PYQ April 2022] Answer: Newton's three laws of motion are: All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 Page 1 1. First law of motion or law of inertia: An object at rest will stay at rest, and an object in motion will stay in motion with the same speed and in the same direction, unless acted upon by an unbalanced force. 2. Second law of motion or law of acceleration: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. 3. Third law of motion or law of action to reaction: For every action, there is an equal and opposite reaction. Question 5: Starting from the third law of motion, derive the law of conservation of linear momentum. [PYQ April 2022] Answer: Newton’s third law states that for a force applied by an object A on object B, object B exerts back an equal force in magnitude, but opposite in direction. Consider two colliding particles A and B whose masses are m1 and m2 with initial and final velocities as u1 and v1 of A and u2 and v2 of B. The time of contact between two particles is given as t. m1u1 + m2u2 is the momentum of A and B before collision and m1v2 + m2v2 is the momentum of A and B after collision. So we can say that momentum is conserved. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 Page 2 Question 6: A ball of mass ‘m’ strikes a rigid wall with the speed ‘u’ and rebounds back with the same speed. The impulse imparted to the ball by the wall is: (a) 2mu (b) mu (c) 0 (d) -2mu [PYQ Oct 2021] Answer: (d) -2mu Explanation: Pinitial = mu and Pfinal = -mu We know that impulse = change in momentum  Impulse = Pfinal - Pinitial = – mu – mu = – 2mu Question 7: Define: (a) coefficient of friction and (b) coefficient of kinetic friction. [PYQ Oct 2021] Answer: Coefficient of friction –: The coefficient of friction (μ) between two surfaces is the ratio of their limiting frictional force to the normal force between them. 𝐥𝐢𝐦𝐢𝐭𝐢𝐧𝐠 𝐟𝐫𝐢𝐜𝐭𝐢𝐨𝐧𝐚𝐥 𝐟𝐨𝐫𝐜𝐞 𝐅 μ= 𝐧𝐨𝐫𝐦𝐚𝐥 𝐫𝐞𝐚𝐜𝐭𝐢𝐨𝐧 = 𝐑 Coefficient of kinetic friction –: the coefficient of kinetic friction (μs) between two surfaces is the ratio of force of kinetic friction to the normal force between them. 𝐤𝐢𝐧𝐞𝐭𝐢𝐜 𝐟𝐫𝐢𝐜𝐭𝐢𝐨𝐧𝐚𝐥 𝐟𝐨𝐫𝐜𝐞 𝐅𝐤 μs = 𝐧𝐨𝐫𝐦𝐚𝐥 𝐫𝐞𝐚𝐜𝐭𝐢𝐨𝐧 = 𝐑 Question 8: A block of mass ‘m’ is held on a rough inclined surface of inclination θ, show in diagram various forces acting on the block. [PYQ Oct 2021] Answer: All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 Page 3 Question 9: State Newton’s first law of motion. Define inertia. Which physical quantity is a measure of the inertia of a body? Is it correct to say that a body always moves in the direction of external force acting on it? Give reasons therefore. [PYQ Oct 2021] Answer: Newton’s first law of motion: An object at rest remains at rest, or if in motion, remains in motion until an external force does not applied on it. Inertia: Everybody has a property that it always opposes its change in position , this property is called Inertia. Inertia is the measure of mass of the object. It is not correct to say that a body always moves in the direction of external force acting on it. The statement is true only for a body which was at rest before the application of force. Question 10: State Newton’s third law of motion. What is the vectorial form of Newton’s third law of motion? [PYQ Oct 2021] Answer: Newton’s third law of motion –: 1. It states that “To every action, there is an equal and opposite reaction”. 2. When two objects interact with each other, the force exerted by the first object on the second is referred to as action. The reaction force is the force exerted by the second body on the first body. Therefore, the action and reaction are opposite and equal in magnitude. Vectorial form of third law –: 𝐅𝟏𝟐 = – 𝐅𝟐𝟏 Question 11: A body of mass m1 = 10 kg is placed on a smooth horizontal table. It is connected to a pulley string which passes over a frictionless pulley and carries at the other end a body m2 of mass 5kg. Calculate. (i) Acceleration of the bodies and (ii) Tension in the string when m2 is let free. Take g = 9.8 N/k. [PYQ JAN 2021] All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 Page 4 Answer: refer to fig.→ According to Newton’s second law of motion: F = ma Here, (m1 + m2) a = m2g m 2g  a= (m 1 +m 2 ) 5 × 9.8 49  a= = = 3.27 m/s2 (10+5) 15 Question 12: An object is moving at a constant speed in a circular path. Does the object have constant momentum? Give reason for your answer. Answer: No. Though the speed is constant, the velocity of the object changes due to change in direction. Hence its momentum will not be constant. Question 13: Two masses m1 and m2 (m1 > m2) are connected at the two ends of a light inextensible string that passes over a light frictionless fixed pulley. Find the acceleration of the masses and the tension in the string connecting them when the masses are released. Answer: When the masses are released, the acceleration of the system can be found using the equation: F = ma  (m1 - m2) × g = (m1 + m2) a 𝐦𝟏 −𝐦𝟐 𝐠  a= 𝐦𝟏 + 𝐦𝟐 where m1 and m2 are the masses, and g is the acceleration due to gravity. The tension in the string can be found using the equation: 𝐦𝟏 −𝐦𝟐 𝐠 tension = m1 × a = m1 × 𝐦𝟏 + 𝐦𝟐 All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 Page 5 Question 14: In which case will there be larger change in momentum of a 2 kg object: (a) When 10 N force acts on it for 1s? (b) When 10 N force acts on it for 1m? Calculate change in momentum in each case. Answer: The change in momentum of an object can be calculated using the formula: Change in momentum = force × time In option A, Change in momentum = 10 N × 1 s = 10 kg·m/s In option B, the force is 10 N and the distance is 1 meter. To calculate the change in momentum, we need to use the formula: Change in momentum = force × distance In option B, Change in momentum = 10 N × 1 m = 10 kg·m/s So, in both cases, the change in momentum of the 2 kg object would be the same, which is 10 kg·m/s. Question 15: A 5 kg block is resting on a horizontal surface for which μk = 0.1. What will be the acceleration of the block if it is pulled by a 10 N force acting on it in the horizontal direction? Answer: Acceleration = force / mass Substituting the values, we get: a = 10 N / 5 kg = 2 m/s² Therefore, the acceleration of the block will be 2 m/s² when a 10 N force is applied to it in the horizontal direction. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 Page 6  03 LAWS OF MOTION Short summary What do we learn Quick Revision  Force and motion  First law of motion To move any object or body we have Everybody continues to be in its state to exert force on it. It is exerted by of rest or of uniform motion in a external source. straight line unless compelled by some external force to act otherwise.  Newton’s laws of motion Simply, “If an object is at rest or in uniform motion, then it will remain at After failure of Galileo and Aristotle, rest or in uniform motion until an Isaac Newton gave three laws of external force does not applied to it.” motion: 1. First law of motion or law of inertia  Types of inertia 2. Second law of motion or law of Inertia are of three types : acceleration. a) INERTIA OF REST: If an object is at 3. Third law of motion or action to rest, then it will remain in at rest till an reaction law. external force doesn’t apply to it. Example -: when a bus is at rest and if suddenly it starts moving then passengers feel a force backward.  Inertia b) INERTIA OF MOTION: If an object is in Everybody has a property that it uniform motion, then it will remain in always opposes its change in position, motion till an external force doesn’t this property is called Inertia. apply on it. Example -: A moving bus suddenly stops and Two bodies of equal mass possess passengers feel a force forward. same inertia because it is a factor of c) INERTIA OF DIRECTION: if an object is mass only. moving in certain direction, then it will Inertia is the measure of mass of remain in this direction unless an external force does not apply to it. the object. Example -: When a vehicle tier moves in mud then, particles of mud move in tangential direction. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 1  Momentum  Applications of second law of Momentum of a body is the quantity motion of motion contained in the body. 1. Cricket player lowers his hand It is measured as the product of the while catching the ball. mass of the body and its velocity. 2. A karate player can break a pile of Momentum = mass × velocity tiles with a single blow of his hand. If a body of mass 𝑚 is moving with 3. In a high jump athletic event, the velocity v then its momentum v is given athletes are allowed to fall either by: 𝐩 = 𝒎𝐯 on a sand bed or cushioned bed. It is a vector quantity and it’s direction is the same as the direction of velocity of the body.  Impulse 𝒌𝒈 × 𝒎 Forces acting for a short time are Unit: 𝒔𝒆𝒄 called IMPULSIVE FORCE. Dimension: [MLT–1] it is measured as “the product of force and time, which is change in  Second law of motion momentum of the body.” The rate of change of momentum of a Impulse = force × time body is directly proportional to the duration applied force and takes place in the I = F. ∆t direction in which the force acts. Impulse is a vector quantity. Its Simply expressed as “the external direction is same as that of force. force apply to an object is directly Unit: N × sec proportional to the change in the Dimensional formula: (MLT-1). momentum.” Force-time graph- Impulse is equal ∆𝒑 ∆𝒑 Fex ∝ Fex = k to the area under F-t curve. ∆𝒕 ∆𝒕 Where k is the constant of proportionality. k = 1 for S.I. & C.G.S. unit F = ma All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 2  Third law of motion  Law of Conservation of Third law is states as “to every action, momentum there is always an equal and opposite If external force on an object is zero reaction.” then linear momentum of the object remains conserved. This is called the Let us note some important points law of conservation of momentum. about the third law –: Or, “the total momentum of an 1. Forces always occur in pairs. Force on isolated system of interchanging a body A by B is equal and opposite to particles is conserved.” the force on the body B by A. 2. There is no cause effect relation implied in the third law. The force on A by B and the force on B by A act at the same instant.  Equilibrium of particle 3. Action and reaction forces act on different bodies, not on the same body. Equilibrium of a particle in FAB = -FBA mechanics refers to the situation when the net external force on the (Force on A by B) = -(Force on B by A) particle is zero. So this is clear that the net force is zero.  Examples of third law 1. Firing of bullet. 2. Swimming and walking. 3. Kick the ball.  Friction 4. Rocket launch. It is an opposing force acting at the 5. Pushing a car and a truck. surface of contact between the two objects or surfaces, which always  Applications of third law opposes relative motion. 1. Recoiling of a gun Examples of friction 2. To walk, we press the ground in 1. Driving of a vehicle on a surface. backward direction with foot 2. Applying breaks to stop a moving vehicle. 3. Skating. 4. Flying of aero planes. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 3  Types of friction  Frame of reference 1. Kinetic: objects are in rest. A frame in which an observer is 2. Static: objects are in motion. situated and makes his observations is known as his ‘Frame of reference’. 3. Rolling: objects are in rolling.  Coefficient of friction There are two types of it: (i) Inertial frame of reference The coefficient of friction (µ) between two (ii) Non – inertial frame of surfaces is the ratio of their limiting frictional force to the normal force between them. reference 1. Inertial frame: If the observer’s 𝐥𝐢𝐦𝐢𝐭𝐢𝐧𝐠 𝐟𝐫𝐢𝐜𝐭𝐢𝐨𝐧𝐚𝐥 𝐟𝐨𝐫𝐜𝐞 𝐅 frame of reference is stationary or µ= = moving with a constant velocity 𝐧𝐨𝐫𝐦𝐚𝐥 𝐫𝐞𝐚𝐜𝐭𝐢𝐨𝐧 𝐑 with respect to the object (another frame of reference), such frame is called Inertial frame.  Simple pulley An inertial frame is one in which an The pulley is defined as a mechanical isolated object has zero acceleration. device that consists of a wheel and a rope. The rope is connected with 2. Non- inertial frame: if the frame is wheel at it is surrounding the wheel. accelerating, such frame is called Non- Inertial frame. Newton’s laws of motion are applicable only in an inertial frame of reference. Let a be the acceleration in this system, which is given by: (𝐦𝟏 −𝐦𝟐 )𝐠 a= 𝐦𝟏 − 𝐦𝟐  Free body diagram 𝟐 𝐦𝟏 𝐦𝟐 𝐠 By using the free body diagram, it Tension is given by: T = 𝐦𝟏 +𝐦𝟐 became easy to use the Newton’s laws to solve the problems in mechanics.  Limiting frictional force A diagram which shows all the forces acting on a body in a given situation is The maximum value of static friction called free body diagram(FBD). upto which body does not move is called limiting friction. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 4  MCQs 1. The maximum velocity (in m/s) with which a car driver must transverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is: (a) 60 (b) 30 (c) 15 (d) 25 Answer: (b) 30 Explanation: ν = μrg = 0.6 × 150 × 10 = 30 m/ s 2. The tension in string shown in fig m 2m F is: (a) F/3 (b) F/6 (c) F/2 (d) 2F Answer: (a) F/3 Explanation:2ma = F – T  T = F/3 3. A particle moves in x-y plane under the action of force F such that the value of its linear momentum p at any instant t is px = 2 cos t and py = 2 sin t. The angle θ between F and p at a given time t will be –: (a) 90° (b) 0° (c) 180° (d) 30° Answer: (a) 90° 4. The objects at rest suddenly explodes into three parts with the mass ratio 2:1:1. The parts of equal masses moves at right angles to each other with equal speeds. The speed of the third part after the explosion will be: v (a) 2v (b) 2 v (c) (d) 2v 2 𝐯 Answer: (b) 𝟐 5. Two Iron blocks of equal masses but with double surface area slide down an Inclined plane with coefficient of friction μ. If the first block with surface All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 5  area A experience a friction force f, then the second block with surface area 2A will experiences a frictional force. (a) f/2 (b) f (c) 2f (d) 4f Answer: (b) f Explanation: Frictional force is independent of area of contact. 6. If the tension in the cable supporting an elevator is equal to the weight of elevator, the elevator may be (a) going up with increasing speed (b) going down with increasing speed (c) going up with uniform speed (d) elevator falls freely under gravity Answer: (c) going up with uniform speed Explanation: Impulse = F dt 7. A graph is drawn with a force along y-axis and time along x-axis. The area under the graph represent– (a) Momentum (b) Couple (c) Moment of the force (d) Impulse of the force Answer: (d) impulse of the force 8. In a game of tug of wars, a condition of equilibrium exists. Both the team pull the rope with a force of 104 N. The tension in the rope is –: (a) 104 N (b) 108 N (c) 0 N (d) 2 × 104 N Answer: (a) 104 N 9. A bullet of mass m moving with a speed v strikes a wooden block of mass M & gets embedded into the block. The final speed is –: M m (a) V (b) V m+M M +m Mv V (c) (d) m+M 2 𝐌 Answer: (c) 𝐦+𝐌 V Explanation: By law of conservation of momentum All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 6  M v + M (0) = (m + M) V Mv  v= m+M 10. Which of the four arrangements in the figure correctly shows the vector addition of two forces F1 & F2 to yield the third force F3? Answer: (a) 11. If two forces are acting at a point such that the magnitude of each force is 2N and the magnitude of their resultant is also 2N, then the angle between the two forces is (a) 120° (b) 60° (c) 90° (d) 0° Answer: (a) 120° Explanation: P = Q = 2N, R = 2N R2 = P2 + Q2 + 2PQ Cos θ  4 = 4 + 4 + 2(4) Cos θ −4 −1  Cos θ = = 8 2  θ = 120° 12. The dimensions of action are –: (a) [MLT-2] (b) [M2LT-3] (c) [MLT-1] (d) [ML2T-1] Answer: (a) [MLT-2] Explanation: action = force  [MLT-2] 13. A car when passes through a bridge exerts a force on it which is equal to: mv 2 mv 2 (a) mg + (b) r r m v2 (c) mg – (d) none of these r 𝐦𝐯 𝟐 Answer: (c) mg – 𝐫 All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 7 14. A particle revolves round a circular path. The acceleration of the particle is inversely proportional to –: (a) radius (b) velocity (c) mass of particle (d) both (b) & (c) Answer: (a) radius V2 1 Explanation: a = a∝ r r 15. If Maximum and minimum values of the resultant of two forces acting at a point are 7N and 3N respectively, the smaller force is equal to –: (a) 4N (b) 5N (c) 3N (d) 2N Answer: (d) 2N Explanation: F max. = P + Q = 7N –––––– (1) F min. = P – Q = 3N –––––– (2) From (1) & (2) 2P = 10 N  P = 5 N On putting P = 5 N in in (1): 5 N + Q = 7 N  Q = 2N 16. The pulley & strings shown in figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be –: (a) 0° (b) 30° (c) 45° (d) 60° Answer: (c) 45° Explanation: For equilibrium of mass m, T = mg –––– (1) For equilibrium of mass 2 m 2T cos θ = 2 mg –––––– (2) Solve (1) & (2) 1  cos θ = , θ = 45°. 2 All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 8 17. Three block of masses m1, m2 & M3 are connected by mass less stings as shown on a frictionless table. They are pulled with a force T3 = 40N, If m1= 10 kg, m2= 6 kg, & m3= 4 kg, then tension T2 will be (a) 20 N (b) 40 N (c) 10 (d) 32 N Answer: (d) 32 N T3 40 Explanation: net acceleration = = = 2 m/s2 m 1 +m 2 +m 3 20 T2 = (m1 + m2)a = 16 × 2 = 32 N 18. Three forces A = i + j − k , B = 2i - j + 3k are acting on a body to keep it in equilibrium. Then C (a) – (3i + 4k) (b) – (4i + 3k) (c) 3i + 4k (d) 2i + 3k Answer: (a) – (3𝐢 + 4𝐤) 19. A body of mass M starts sliding down an Inclined plane where critical angle is < ACB = 30° as shown in figure, the coefficient of friction will be Mg (a) (b) 3 mg 3 (c) 3 (d) None of these Answer: (c) 𝟑 Explanation:  = tan 60° = 3 20. In the figure given, the position-time graph of a particle of mass 0.1 kg is shown the impulse at t = 2s is –: (a) 0.2 kg m/s (b) – 0.2 kg m/s (c) 0.1 kg m/s (d) – 0.4 kg m/s Answer: (b) – 0.2 kg m/s All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 9 Explanation: Impulse = change in momentum = m (v – u)= 0.1 (0 – 2) = – 0.2 kg m/s 21. If μs, μk and μr are coefficients of static friction, kinetic friction and rolling friction, then (a) μs < μk < μr (b) μk < μr < μs (c) μr < μk < μs (d) μr = μk = μs Answer: (c) μr < μk < μ 22. ‘Net force acting on an object is found to be zero.’ It can be inferred that the object (a) May be at rest (b) May be in uniform motion (c) May be in uniformly accelerated motion (d) Both a & b Answer: (d) Both a & b 23. The coefficient of static friction between two surfaces depends upon (a) the normal reaction (b) the shape of the surface in contact (c) the area of contact (d) None of the these Answer: (a) the normal reaction 24. Which of the following has the greatest inertia? (a) A single atom (b) a molecule (c) A one-rupee coin (d) a cricket ball Answer: (d) a cricket ball Explanation: The heavier an object is, the more difficult it is to move it or the more force is necessary to move it, resulting in a higher inertia. 25. If the force applied to a body is doubled and the mass is cut in half, What would be the accelerations’ ratio? (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1 All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 10 Answer: (c) 1:4 Explanation: 26. On the roof of a train travelling on horizontal rails, a simple pendulum swing. If the pendulum’s string is pointing towards the front, the train is- (a) Moving at a steady speed (b) Moving with acceleration (c) Moving with retardation (d) At rest Answer: (c) moving with retardation 27. The quicker the momentum changes –: (a) The force is less (b) The force is greater (c) The force is zero (d) There is no force. Answer: (b)the force is greater 28. A monkey is sitting at a spring balance that reads 60 kilograms. What effect will the monkey having jumped off the balance have on the reading? (a) Decrease (b) No change in reading (c) Increase (d) First increase and then becomes zero Answer: (d) First increase and then becomes zero 29. A shell of mass 200 g is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, calculate the recoil speed of the gun, (a) 16 cm/s (b) 18 cm/s (c) 4 cm/s (d) 16 cm/s Answer: (a) 16 cm/s 30. A bullet of mass 0.04 kg moving with a speed of 90 1 ms- enters a heavy wooden block and stopped after 3s. What is the average resistive force exerted by the block on the bullet? (a)1N (b)1.2N (c)2N (d)3N Answer: (b) 1.2 N 31. Every action has an equal and opposite reaction, which suggests that All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 11  (a) action and reaction always act on different bodies (b) the forces of action and reaction cancel to each other (c) the forces of action and reaction cannot cancel to each other (d) Both (a) and (c) Answer: (d) Both (a) and (c) 32. Inertia of an object is directly dependent on………. (a) impulse (b)momentum (c) mass (d)density Answer: (c) mass Explanation: The term inertia means resistance of any physical object. It is defined as the tendency of a body to remain in its position of rest or uniform motion. So, it is dependent on mass of the body. 33. An initially stationary device lying on a frictionless floor explodes into two pieces and slides across the floor. One piece is moving in positive x- direction then other piece is moving in –: (a) positive y – direction (b) negative y – direction (c) negative x – direction (d) at angle from x-direction Answer: (c) negative x – direction 34. A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is -: (a) frictional force along westward (b) muscle force along southward (c) frictional force along south – west (d) muscle force along south-west Answer: (c) frictional force along south – west 35. A constant retarding force of 50N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long time does the body take to stop? All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 12  (a) 6s (b) 8s (c) 9s (d) 10s Answer: (a) 6s Explanation: Given, F = 50N, m = 20kg, v = 15 ms1 mv Impulse, F = ∆t mv 20 ×15 Time, ∆t =  , ∆t = = 6 sec F 50 36. A body of mass 6kg is acted on by a force so that its velocity changes from 3ms-1 to 5ms-1, then change in momentum is –: (a) 48N-s (b) 24N-s (c) 30N-s (d) 12N-s Answer: (d) 12N-s 37. A batsman hits back at ball straight in the direction of the bowler without changing its initial speed of 12 ms-1. If the mass of the ball is 0.15 kg, find the impulse imparted to the ball. (Assume linear motion of the ball) (a) 1.8N-s (b) 3.6N-s (c) 3.6N-m (d) 1.8N-m Answer: (b) 3.6N-s 38. Three concurrent coplanar forces 1N, 2N and 3N are acting along different directions on a body can keep the body in equilibrium, if –: (a) 2N and 3N act at right angle (b) 1N and 2N act at acute angle (c)1N and 2N act at right angle (d) Cannot be possible Answer: (d) cannot be possible 39. If a box is lying in the compartment of an accelerating train and box is stationary relative to the train. What force cause the acceleration of the box? All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 13  (a) Frictional force in the direction of train (b) Frictional force in the opposite direction of train (c) Force applied by air (d) None of the above Answer: (a) Frictional force in the direction of train Explanation: Frictional force in the direction of train causes the acceleration of the box lying in the compartment of an accelerating train. 40. Apparent weight of a body = actual weight of the body, if the body (a) Moves with increasing velocity (b) Remains at rest (c) Remains in a state of uniform motion (d) Option (b) and (c) Answer: (d) Option (b) and (c) Explanation: Apparent weight of a body is equal to its actual weight if the body is either in a state of rest or in a state of uniform motion. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 14  06 Work, Energy And Power Short summary Quick Revision What do we learn  Work  Positive & Negative work Work is said to be done when a force Positive work: If the force is apply in (push or pull) applied to an object same direction of displacement then causes a displacement of the object. work done is called positive work. θ = 0° , W = Fd cos 0° = Fd (1) = Fd “The work done by a force is the product of the magnitude of force Negative work: If the force is apply in component in the direction of opposite direction of displacement displacement and the displacement then work done is called negative of this object.” work. θ = 180°  W = Fd cos180° = Fd (-1) = -Fd  Vector form of work (A) Work done shall be negative for θ If force F is acting at angle θ with lying between 90° and 270°. respect to the displacement d of the object, its component along d will be (B) Work done is zero when θ = 90°. F cos θ. 1  Work done by gravity Then work done by force F is given by: W = F cos θ. d In vector form: W = F.d In Fig. A, work is done against the force mg (downwards) and the Unit of work: N × m = Joule (SI) displacement is (upward)  θ = 180° Dimension: [M1L2T-2] W = Fd cos180° = mgh (-1) = -mgh 1 joule: “1 joule is defined as the In fig. B, mass is being lowered. mg amount of work done when a force of and d are in same direction.  θ = 0° 1 Newton displaces a body 1 meter.” W = Fd cos 0° = W = mgh (1) = +mgh All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 1  Work done by a variable force  Energy W= ∆𝐖  W = limΔx →0 𝐅 (𝐱) 𝚫𝐱 “Capacity to do work is called Energy.” If a system or object has Example –: spring experiences energy, it has ability to do work. variable force. It is also scalar quantity. spring will exert a force on the block S.I. unit of energy is also joule. given by F = -kx , where x is C.G.S. unit of energy is ‘erg’. compression or elongation in spring Practical units: electron volt (eV), and k is a constant called spring Kilowatt hour (KWh), Calories (Cal) constant. Work done by spring force: 𝟏 Relation between different units: W=- k𝒙𝟐𝒎 1 Joule = 107 erg 𝟐 1 eV = 1.6 × 10–19 Joule S.I unit of spring constant is Nm-1. 1 KWh = 3.6 × 106 Joule Value of k depends inversely on un 1 Calorie = 4.18 Joule stretched length and nature of material of spring.  Types of energy  Power There are two types of mechanical Energy: “The rate at which work is done is ① Kinetic energy called power.” ② Potential energy The average power is defined as:  Kinetic energy 𝐰𝐨𝐫𝐤 𝐝𝐨𝐧𝐞 Average power = 𝐭𝐢𝐦𝐞 𝐭𝐚𝐤𝐞𝐧 “The energy possessed by a body by virtue of its motion is called kinetic ∆𝐖 Mathematically, P = energy.” ∆𝐭 Instantaneous power is defined as: Let m = mass of the body, v = velocity of the body then 𝟏 ∆𝐖 𝐝𝐖 K.E. = mv2 P = limit Δt→0 = 𝟐 ∆𝐭 𝐝𝐭 S.I. unit of power is watt. Work-energy theorem: It states that 1 watt - : work done by “When 1 joule of work is done in a force acting on a body is equal to 1 second then we can say that power the change produced in the kinetic energy of the body. of doing work is 1 watt.” W = kf – ki 1 2 -3 Dimensional formula: [M L T ] All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 2  Potential energy  Law of Conservation of energy “Objects possess another kind of According to the law of conservation energy due to their position in space. of energy, the total energy of an This energy is known as Potential isolated system does not change. Energy.” Energy may be transformed from one P.E. Or U = mgh form to another but the total energy of isolated system remains constant.  Types of potential energy “Energy can neither be created, nor ① Gravitational potential energy destroyed.” ② Elastic potential energy U1 + K1 = U2 + K2 Besides mechanical energy, the  Gravitational potential energy energy may manifest itself in many other forms. Some of these forms are: Gravitational potential energy of a thermal energy, electrical energy, body is the energy possessed by the chemical energy, visual light energy, body by virtue of its position above nuclear energy etc. the surface of the earth. It is given by U = mgh  Equivalence of mass and where m = mass of a body energy g = acceleration due to gravity h = height through which the body is According to Einstein, mass and raised. energy are inter-convertible. That is, mass can be converted into energy and energy can be converted into  Elastic potential energy mass. When an elastic body is displaced from its equilibrium position, work is needed to be done against the m = mass transformed to photon energy restoring elastic force. The work done M = remained mass is stored up in the body in the form of c = speed of light its elastic potential energy. 𝟏 2 v = speed of M It is given by U = 𝟐 kx E = Energy generated by the transformation of Where, k = spring constant ‘m’ mass to photon All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 3  Collisions  Conservative force Collision is defined as an isolated If the work done by or against force is event in which two or more colliding dependent only on initial and final bodies exert relatively strong forces position of the body and not on the path followed by body, then this type on each other for a relatively short of force is called Conservative force. time. Collision between particles has been Examples of conservative force –: divided broadly into two types. – Gravitational force (i) Elastic collisions – Magnetic force (ii) Inelastic collisions – Electrostatic force etc. NOTE: Work done by the conservative  Elastic Collisions force in a closed path is zero. In a collision if both the linear momentum and the kinetic energy of the system remain conserved, that is  Non – Conservative force called elastic collision. If the work done by or against force is dependent on the path followed by Example: Collisions between atomic body ,then this type of force is called particles, atoms, marble balls and Non – conservative force. billiard balls. Examples of conservative force –: – Friction force – Viscous force  Inelastic Collisions – Induction force in cyclotron etc. A collision is said to be inelastic if the linear momentum of the system NOTE: Work done by the conservative remains conserved but its kinetic force in a closed path is not zero. energy is not conserved. Example: When we drop a ball of wet putty on to the floor then the collision between ball and floor is an inelastic collision. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 4  MCQs 1. A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 rev/min. it's kinetic energy would be: (a) 250 π2 (b) 100 π2 (c) 5 π2 (d) 0 2 Answer: (a) 250 π 2π Explanation: V = ωR, ω = T = 10π rad/s V = 10π (1) ms-1 = 10π ms-1 1 1 K.E. = mV2 = × 5 × V2 = 250 π2 J 2 2 2. A man is squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is (a) constant and equal to 'mg' in magnitude. (b) constant and greater than 'mg' in magnitude. (c) variable but always greater than 'mg' (d) at first greater than 'mg' and later becomes equal to 'mg' Answer: (d) at first greater than 'mg' and later becomes equal to 'mg' Explanation: In squatting, he is tilted somewhat, hence also has to balance frictional force besides his weight in this case R = friction + mg  R > mg When he set straight up in that case, R = mg. 3. A body of mass 0.5 kg travels in straight line with velocity V = ax 3/2 where a = 5 m –½ s –1. The work done by the net force during it's displacement from x = 0 to x = 2m is (a) 15 J (b) 50 J (c) 10 J (d) 100 j Answer: (b) 50 J 4. An Athlete in the Olympic games covers a distance of 100 m in 10s, this kinetic energy can be estimated to be in the range (assume m = 60 kg) (a) 200J – 500 J (b) 2 × 105J – 3 × 105J (c) 20000J – 50000J (d) 2000J – 5000J Answer: (d) 2000J – 5000J All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 5  S Explanation: Vav = t = 10 ms-1, m = 60 kg  Av. K.E. 3000 J 5. A block of mass 0.5 kg is moving with a speed of 2m/s on a smooth surface. It strikes another mass of 1 kg at rest and then they move together as a single body. The energy loss during the collision is (a) 0.16J (b) 1.00 J (c) 0.67 J (d) 0.34J Answer: (c) 0.67 J 6. A bullet fired in to a fixed target losses half of it's velocity after penetrating distance of 3 cm. How much further it will penetrate before coming to rest assuming that if faces constant resistance to it's motion? (a) 3 cm (b) 2.0 cm (c) 1.5 cm (d) 1.0 cm Answer: (d) 1.0 cm 7. A uniform chain of length 2m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table? (a) 7.2 J (b) 3.6 J (c) 120 J (d) 1200 J Answer: (b) 3.6 J 4 kg Explanation: Mass per unit length = 2 m = 2 kg m-1 Mass of 60 cm length = 1.2 kg Weight of hanging part = 1.2 × 10 = 12N W = F × S = 12 × 0.3 = 3.6 J. 8. If the linear momentum is increased by 50%, then kinetic energy will increase by (a) 50% (b) 100% (c) 125% (d) 25% Answer: (c) 125% 9. A block having mass 'm' collides with another stationary block having mass 2m. The lighter block comes to rest after collision. If the velocity of first block is V, than the value of co-efficient of restitution will be: All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 6  (a) 0.5 (b) 0.4 (c) 0.6 (d) 0.8 Answer: (a) 0.5 10. A body of mass 50 kg is at rest. The work done to accelerate it by 20 m/s in 10 s is: (a) 103 J (b) 104 J (c) 2 × 103 J (d) 4 × 104 J Answer: (b) 104 J v−u 1 Explanation: a = = 2 ms-2, S = ut + at2 = 100 m t 2 W = F × S = 104 J 11. A spring of force constant 800 Nm–1 has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is (a) 16 J (b) 8 J (c) 35 J (d) 24 J Answer: (b) 8 J 1 Explanation: W = 2 × 800 (0.152 – 0.052) = 8 J 12. A particle is projected at an angle of 60° to the horizontal with a kinetic energy E. The kinetic energy at the highest point is (a) E (b) E/4 (c) E/2 (d) Zero Answer: (b) E/4 1 Explanation: K.E. at highest point = 2 m (4cos 60°)2 = E/4 13. A child is sitting on a swing. Its minimum and maximum heights from the ground 0.75 m and 2 m respectively, it's maximum speed will be (a) 10 m s–1 (b) 5 m s–1 (c) 8 m s–1 (d) 15 m s–1 Answer: (b) 5 m s–1 Explanation: Maximum K.E. = Drop in P.E. 1 2 m Vmax = mg (h2 – h1)  Vmax = 5 ms-1 2 All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 7 14. 300 J of work is done in sliding a 2 kg block up on inclined plane of height 10 m. Work done against friction is (g = 10 ms–2) (a) 1000 J (b) 200 J (c) 100 J (d) Zero Answer: (c) 100 J Explanation: Total work done = Gain in P.E.+ Work done against friction 300 = 2 × 10 × 10 + W  W = 100 J. 15. During inelastic collision between two bodies, which of the following quantities always remain conserved (a) Total kinetic energy (b) Total mechanical energy (c) Total linear momentum (d) Speed of each body Answer: (c) Total linear momentum 16. Two bodies with kinetic energies in the ratio 4:1 are moving with equal linear momentum. The ratio of their masses is (a) 4 : 1 (b) 1 : 1 (c) 1 : 2 (d) 1 : 4 Answer: (d) 1:4 17. Which of the following is not a conservative force? (a) Gravitational force (b) Frictional force (c) Spring force (d) None of these Answer: (b) Frictional force Explanation: As work done by frictional force over a closed path is not zero, therefore, it is non-conservative force. 18. A position dependent force, F = 7 – 2x + 3x2 N acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m. The work done in joule is: (a) 135 (b) 270 (c) 35 (d) 70 Answer: (a) 135 19. A ball is dropped from height 'h' on the ground where co-efficient of restitution is 'e'. After one bounce the maximum height is All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 8  (a) e2h (b) e h (c) eh (d) eh 2 Answer: (a) e h 20. How much water a pump of 2 KW can raise in one minute to a height of 10 m? (g = 10 ms) (a) 1000 liters (b) 1200 liters (c) 10 liters (d) 2000 liters Answer: (b) 1200 liters W Explanation: P =  W = Pt = mgh  m = 1200 kg t 21. Four particles given, have same momentum. Which has maximum kinetic energy? (a) Proton (b) Electron (c) Deutron (d) Alpha-particles Answer: (b) Electron 22. A bomb of mass 30 kg at rest explodes in to two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 ms–1. The kinetic energy of the other mass is (a) 324 J (b) 486 J (c) 256 J (d) 5245 J Answer: (b) 486 J Explanation: By conservation of momentum 30 × 0 = 18 × 6 + 12 + V → V = - 9 ms-1 1 K.E. = mV2 = 486 J 2 23. If a light body and heavy body have same kinetic energy, then which one has greater linear momentum? (a) Lighter body (b) Heavier body (c) Both have same momentum (d) Can’t be predicted Answer: (b) Heavier body All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 9 24. What is the power utilized when work of 1000 J is done in 2 seconds? (a) 100 W (b) 200 W (c) 20 W (d) 500 W Answer: (d) 500 W Explanation: W = 1000J, t = 2 seconds Power = work/time = 1000/2 = 500 W 25. Work is always done on a body when (a) a force acts on it (b) it moves through a certain distance (c) it experiences an increase in energy through a mechanical influence (d) None of these Answer: (c) it experiences an increase in energy through a mechanical influence 26. In which of the following work is being not done? (a) Shopping in the supermarket (b) Standing with a basket of fruit on the head (c) Climbing a tree (d) Pushing a wheel barrow Answer: (b) Standing with a basket of fruit on the head 27. A wooden cube having mass 10 kg is dropped from the top of a building. After 1 s, a bullet of mass 20 g fired at it from the ground hits the block with a velocity of 1000 m / s at an angle of 30 to the horizontal moving upwards and gets imbedded in the block. The velocity of the block/bullet system immediately after the collision is (a) 17 m/ s (b) 27 m/s (c) 52 m / s (d) 10 m/s Answer: (a) 17 m/ s 28. An electric heater of rating 1000 W is used for 5 hrs per day for 20 days. What is the electrical energy utilized? All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 10  (a) 100 kWh (b) 200 kWh (c) 120 kWh (d) 500 kWh Answer: (a) 100 kWh Explanation: The power of the electric heater is 1000 W, and the time period is 20 × 5 = 100 hr. Electrical energy = Power × Time Electrical energy = 1000 × 100 = 100000 Wh Electrical energy = 100 kWh 29. The temperature at the bottom of a high water fall is higher than that at the top because (a) by itself heat flows from higher to lower temperature (b) the difference in height causes a difference in pressure (c) thermal energy is transformed into mechanical energy (d) mechanical energy is transformed into thermal energy. Answer: (b) the difference in height causes a difference in pressure 30. A ball moves in a frictionless inclined table without slipping. The work done by the table surface on the ball is: (a) Negative (b) Zero (c)Positive (d) None of the options Answer: (b) Zero Explanation: The work done by a ball when it moves on a frictionless inclined table without slipping is zero. 31. A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further will it penetrate before coming to rest assuming that it faces constant resistance to motion? (a) 2.0 cm (b) 3.0 cm (c) 1.0 cm (d) 1.5 cm Answer: (c) 1.0 cm All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 11 32. When the force retards the motion of body, the work done is (a) zero (b) negative (c) positive (d) none of these Answer: (b) negative 33. The coefficient of restitution e for a perfectly inelastic collision is (a) 1 (b) 0 (c) infinity (d) –1 Answer: (b) 0 Explanation: For a perfectly inelastic collision, e = 0. 34. If two particles are brought near one another, the potential energy of the system will (a) increase (b) decrease (c) remains the same (d) equal to the K.E Answer: (a) increase 35. If the speed of an object is doubled, how does its kinetic energy change? (a) It becomes half (b) It remains the same (c) It becomes double (d) It becomes quadruple Answer: (c) It becomes double 36. In an inelastic collision (a) momentum is not conserved (b) momentum is conserved but kinetic energy is not conserved (c) both momentum and kinetic energy are conserved (d) neither momentum nor kinetic energy is conserved Answer: (b) Momentum is conserved but kinetic energy is not conserved. 37. In an elastic collision, what is conserved? (a) Kinetic energy (b) Momentum (c) Both (a) and (b) (d) Neither (a) nor (b) All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 12 Answer: (c) Both (a) and (b) 38. No work is done if (a) displacement is zero (b) force is zero (c) force and displacement are mutually perpendicular (d) All of these Answer: (d) All of these 39. A force → F = - K ( y i + x j ) ( where K is a positive constant ) acts on a particle moving in the 𝑥𝑦 plane. Starting from the origin, the particle is taken along the positive x – axis to the point ( a, 0 ) and then parallel to the y – axis to the point ( a, a ). The total work done by the force F on the particle is: (a) - 2Ka2 (b) 2Ka2 (c) - Ka2 (d) Ka2 Answer: (c) - Ka2 40. The magnitude of work done by a force (a) depends on frame of reference (b) does not depend on frame of reference (c) cannot be calculated in non-inertial frames. (d) both (a) and (b) Answer: (d) both (a) and (b) 41. A force F = (5i + 3 j + 2k) N is applied over a particle which displaces it from its origin to the point r = (2i – j) m. The work done on the particle in joules is (a) - 7 (b) 7 (c) 10 (d) 13 Answer: (b) 7 42. The work energy theorem states that: (A) Work done on an object is equal to its mass times acceleration All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 13  (b) Work done on an object is equal to the change in its kinetic energy (c) Work done on an object is equal to its potential energy (d) Work done on an object is independent of its velocity Answer: (b). Work done on an object is equal to the change in its kinetic energy 43. What happens to the velocity of objects involved in a completely inelastic collision? (a) They move apart after the collision. (b) They stick together and move with a common velocity after the collision. (c) They bounce off each other after the collision. (d) Their velocities remain unchanged after the collision. Answer:: (b) They stick together and move with a common velocity after the collision. 44. If a moving object collides elastically with a stationary object of the same mass, what is the result? (a) The stationary object gains all the momentum of the moving object. (b) Both objects come to rest after the collision. (c) The moving object continues with its initial velocity, and the stationary object moves with double the initial velocity of the moving object. (d) Both objects move with half the initial velocity of the moving object after the collision. Answer: (c) The moving object continues with its initial velocity, and the stationary object moves with double the initial velocity of the moving object. 45. When is momentum conserved in a collision? (a) Only in perfectly elastic collisions. (b) Only in perfectly inelastic collisions. (c) In both perfectly elastic and perfectly inelastic collisions. (d) Momentum is never conserved in collisions. Answer: (c) In both perfectly elastic and perfectly inelastic collisions. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 14 46. Work done by a conservative force is equal to: (a) Force times distance (b) Change in kinetic energy of an object (c) Negative of the change in potential energy of an object (d) Force times velocity Answer: (c) Negative of the change in potential energy of an object 47. If an object is moved on a closed loop and the net work done by all the forces is zero, then: (a) The only force acting on the object is conservative. (b) The only force acting on the object is non-conservative. (c) The object is not moving. (d) The object is moving at a constant velocity. Answer: (a) The only force acting on the object is conservative. 48. Which of the following forces is conservative in nature? (a) Frictional force (b) Tension force in a rope (c) Magnetic force (d) Normal force Answer: (b) Tension force in a rope 49. When a ball is thrown vertically upwards, which of the following forces does work on it? (a) Gravity (b) Both gravity and air resistance (c) Air resistance (d) Neither gravity nor air resistance Answer: (c) Air resistance 50. If a car moves in a complete circle at a constant speed, the work done by gravitational force is: (a) Positive (b) Negative (c) Zero (d) Variable Answer: (c) Zero All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 15  09 Properties of fluids Short summary What do we learn Quick Revision  Fluids  Hydrostatic pressure Liquids and gases have a property to The pressure exerted by a fluid at rest flow. Hence the liquids and gases are is called hydrostatic pressure. collectively called FLUIDS.  Pressure is a scalar quantity.  Properties of fluids  Dimensional formula is [M1L-1T-2].  S.I. Unit of pressure is Nm-2 and is Hydrostatic pressure, buoyancy, called Pascal (Pa). surface tension, viscosity and some very important laws given by great physicist such as Pascal’s law,  Pressure at a point inside the Archimedes principle, Bernoulli’s principle, strokes law etc. cylinder  If the cylinder is in equilibrium the resultant force is zero.  Thrust of liquid  the pressure P at the bottom of a “The total force exerted by a liquid on column of liquid of height h is any surface in contact with it is called given by: thrust.” P=ρgh  Pressure “Effect of force on unit area is called  Atmospheric pressure pressure.” Or, “The pressure of atmosphere of the Normal force or thrust per unit area earth is termed as Atmospheric exerted by fluid is called pressure. pressure.”  Normal atmospheric pressure at 𝑻𝒉𝒓𝒖𝒔𝒕 𝑭𝒐𝒓𝒄𝒆 P= OR P= sea level (an average value) is 𝑨𝒓𝒆𝒂 𝑨𝒓𝒆𝒂 1.013 × 105 Pa. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 1  Mercury barometer  Archimedes principle A mercury barometer is an instrument “When an object is submerged used to measure atmospheric partially or fully in a fluid, the pressure. It operates on the principle magnitude of the buoyant force on it that atmospheric pressure can is always equal to the weight of the support a column of mercury in a fluid displaced by the object.” vacuum.  Components  A body is said to float in a liquid - Glass Tube when the average density of the body is less than that of the liquid. - Mercury Reservoir  Working Principle  The weight of the liquid displaced - Filling the Barometer by the immersed part of the body - Balancing Atmospheric Pressure should be the same as the weight - Reading the Pressure of the body.  Atmospheric pressure is measured in millimeters of mercury (mmHg)  The center of gravity of the body and center of buoyancy should be or inches of mercury (in Hg). along the same vertical line. Applications: Mercury barometers are used in weather  Pascal’s law forecasting, aviation, and scientific research. Pascal’s law can be explained simply as –: “If we apply pressure at any point on a container which is filled  Buoyancy with liquid , then pressure will be equally distributed at all points inside “Buoyancy is the tendency of an the liquid and as well as wall of object to float in a fluid.” container”. 𝐅𝟏 𝐅𝟐  All liquids and gases in the P = P1 = P2 = P3 = & 𝐀𝟏 𝐀𝟐 presence of gravity exert an upward force known as Buoyant Force.  Applications of Pascal’s law  Buoyant force is affected due to 1. Hydraulic Press/Balance/Jack/Lift density, volume & acceleration 2. Hydraulic Jack or Car Lifts due to gravity. 3. Hydraulic Brakes All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 2  Surface tension  Surface energy  Surface tension is the property by “The work done to increase the virtue of which the free surface of surface area of the liquid is termed as a liquid at the rest behaves like surface energy.” W = T × A elastic to occupy minimum surface area. This work done by the external force It is a property of the liquid is stored as the potential energy of surface due to which it has tendency the new surface and is called as to decrease its surface area. surface energy.  The surface tension of a liquid can be defined as “the force per unit Here, a new formula is derived for length in the plane of liquid surface tension (T). 𝐖 T= surface”. 𝐅 𝐀 T= 𝐋 - S.I. unit of surface tension is Nm–1. - Dimensional formula is [MT–2].  Applications of surface tension - It is a scalar quantity. 1. Mosquitoes sitting on water.  Intermolecular force 2. Excess pressure on concave side of a spherical surface. The force between the molecules of matter is termed as Intermolecular 3. Detergents and surface tension. force. There are two types of it as follows –: 4. Wax – duck floating on water. ①Cohesive force The force of attraction between the  Angle of contact molecules of same substance is called cohesive force. Angle of contact is defined as “the Ex. –: H2O – H2O and Hg – Hg angle between the tangent to the liquid surface and the wall of the ②Adhesive force container at the point of contact as measured from within the liquid.” The force of attraction between the Angle of contact depends on –: molecules of different substances is - Temperature called adhesive force. - Pair of solid and liquid Ex. -: H2O – SiO2 and H2O – Hg All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 3  v  Capillary action  Velocity gradient Capillary tube: A tube which is very The velocity changes by dv in going fine bore (hair like), or through a distance dx perpendicular its cross – sectional area is very small, to surface. The quantity dv/dx is is termed as ‘Capillary tube’. called the velocity gradient. 𝐝𝐯 Capillary action: Phenomenon of the Velocity gradient = 𝐝𝐱 elevation or depression of a liquid in a capillary The viscous force F between two tube is basically due to surface layers of the fluid is proportional to tension and is known as ‘Capillary area (A) and velocity gradient. action’ or ‘capillarity’. 𝐝𝐯 FαA  F = - A (dv/dx) 𝐝𝐱 If the radius of tube is less (in a very fine bore capillary), liquid rise will be high.  Coefficient of viscosity The liquid surface in a capillary tube The constant of proportionality  is is either concave or convex. This called coefficient of viscosity. curvature is due to surface tension. The rise in capillary is given by 𝟐 𝑻 𝐜𝐨𝐬 𝜽 h= 𝒓𝒅𝒈  Streamline flow When a fluid flows such that each particle of the liquid passing a given  Viscosity point moves along the same path and the same velocity as its perpendicular, “The property of a fluid by virtue of the flow is called ‘Streamline flow’. which it opposes the relative motion in its adjacent layers is known as viscosity.”  Laminar flow  Examples of Viscosity If a liquid is flowing a horizontal surface with a steady flow and moves 1. When we swim in a pool of water, in the form of layers of different we experience some resistance to velocities which do not mix with each our motion, this is due to viscous other, then the flow of liquid is called force of motion of water. ‘laminar flow’. 2. The cloud particles fall down very slowly account of viscosity of air and hence seeing floating in sky. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 4  Turbulent flow  Stokes’ law When a liquid moves with a velocity Stokes’ law states that tangential greater than its critical velocity, the backward viscous force acting on a motion of the particles of liquid spherical mass of radius r falling with becomes disordered or irregular. Such velocity ‘v’ in a liquid of coefficient of a flow is called a turbulent flow. viscosity η is given by F= 6π η r v  Critical velocity  Terminal velocity The critical velocity is that velocity of liquid flow up to which its flow is “Terminal velocity is defined as the streamlined and above which its flow highest velocity attained by an object becomes turbulent. falling through a fluid.” 𝟐 𝐫 𝟐 (𝛒−𝛔)𝐠 V0 = The value of critical velocity of any 𝟗 liquid depends on the: - coefficient of viscosity ( η ) of the liquid - diameter of the tube (d) through  Bernoulli’s principle which the liquid flows Bernoulli’s Principle states that - density of the liquid (ρ) “where the velocity of a fluid is high, 𝛈 the pressure is low and where the Therefore, vc = R. 𝛒 velocity of the fluid is low, pressure is high.” Where R is constant of proportionality and is called Reynolds’s Number. It has no dimensions.  Bernoulli’s theorem According to this theorem the total energy (pressure energy, potential  Equation of continuity energy and kinetic energy) per unit For an incompressible, streamlined volume or mass of an incompressible and non-viscous liquid product of area and non-viscous fluid in steady flow of cross section of tube and velocity through a pipe remains constant of liquid remains constant. throughout the flow. It is expressed mathematically as: A1V1 = A2V2 𝟏 P+ 𝛒𝐯 𝟐 + ρgh = constant This expression is called “Equation of 𝟐 continuity”. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 5  Question – Answer  Very short answer questions 1. What is pressure? Answer: Pressure is defined as force per unit area and is measured in Pascal. 2. What causes pressure in a fluid? Answer: Pressure in a fluid is caused by the collisions of its molecules with the walls of the container. 3. Define hydrostatic pressure. Answer: Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity acting on the fluid. 4. What is the equation for hydrostatic pressure? Answer: P = P0+ ρgh where P is the hydrostatic pressure, P0 is the atmospheric pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the fluid. 5. How does hydrostatic pressure change with depth in a fluid? Answer: Hydrostatic pressure increases with depth in a fluid due to the increasing weight of the fluid above the given depth. 6. What is Pascal's law? Answer: Pascal's law states that in a confined fluid at rest, any change in pressure applied at any point in the fluid is transmitted undiminished throughout the fluid in all directions. 7. What is atmospheric pressure at sea level? Answer: Atmospheric pressure at sea level is approximately 101.3 k Pa or 1 atmosphere (atm). All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 6 8. What is a mercury barometer? Answer: A mercury barometer is a device used to measure atmospheric pressure by balancing the weight of a column of mercury against the atmospheric pressure. 9. Why is mercury used in barometers instead of water? Answer: Mercury is used in barometers because it has a high density, allowing for a reasonably sized column in the barometer, making it sensitive to atmospheric pressure changes. 10. Define buoyancy. Answer: Buoyancy is the upward force exerted by a fluid on an object placed in it, counteracting the weight of the object and allowing it to float or experience an apparent loss of weight. 11. What determines the buoyant force acting on an object submerged in a fluid? Answer: The buoyant force is equal to the weight of the fluid displaced by the submerged object and depends on the volume of the object submerged. 12. State Archimedes' Principle. Answer: Archimedes' Principle states that a body submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. 13. When does an object float in a fluid? Answer: An object floats in a fluid when its density is less than the density of the fluid, causing the buoyant force to be greater than the object's weight. 14. What happens to the apparent weight of an object when it is submerged in a fluid? All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 7 Answer: The apparent weight of an object submerged in a fluid decreases due to the buoyant force acting in the upward direction, reducing the net weight experienced by the object. 15. Define surface energy. Answer: Surface energy is the energy required to increase the surface area of a liquid by a unit amount. It is also the work done to increase the surface area. 16. What causes surface tension in a liquid? Answer: Surface tension is caused by the cohesive forces between molecules at the surface of a liquid, pulling them inward and creating a 'skin' on the surface. 17. Which of the following has more excess of pressure - An air bubble in water of radius 1cm, surface tension of water being 727 × 10–3 Nm–1 or a soap bubble in air of radius 4cm, surface tension of soap solution being 25 × 10–3 Nm–1. [PYQ NIOS , Jan 2021] Answer: For air bubble in water Hence, an air bubble has more excess of pressure. 18. Differentiate between cohesive and adhesive forces. Answer: Cohesive forces are the intermolecular forces between similar molecules within a substance, while adhesive forces are the forces between different molecules of different substances. 19. Give example of cohesive force and adhesive force separately. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 8 Answer: Water droplets forming on a leaf due to the cohesive forces between water molecules. Water sticking to the surface of a glass due to adhesive forces between water molecules and glass molecules. 20. How does soap reduce surface tension? Answer: Soap molecules have a hydrophilic (water-attracting) and a hydrophobic (water-repelling) end. When added to water, they disrupt the cohesive forces between water molecules, reducing surface tension. 21. How does surface tension affect capillary action? Answer: Surface tension pulls the liquid upward along the walls of a narrow tube, aiding capillary action. 22. Give an example of capillary action in plants. Answer: Capillary action helps water move upward from the roots to the leaves in plants, providing nutrients to the entire plant. 23. What is the application of surface tension in the formation of droplets? Answer: Surface tension allows liquids to form spherical droplets, which is essential in raindrop formation and various industrial processes like inkjet printing. 24. How is surface tension used in the creation of soap bubbles? Answer: Surface tension in soap films forms bubbles by minimizing the surface area, allowing soap bubbles to hold their shape. 25. What is viscosity? Give an example of a viscous liquid. Answer: Viscosity is a measure of a fluid's resistance to flow. It quantifies how easily a fluid can deform or flow. Honey is an example of a viscous liquid due to its high resistance to flow. 26. What is critical velocity? All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 9 Answer: Critical velocity is the minimum velocity required for a liquid to flow steadily without turbulence through a pipeline. 27. State the equation of continuity. Answer: The equation of continuity states that the product of the cross – sectional area and velocity of a fluid remains constant in a continuous flow. 28. What is stokes’ law ? Answer: Stokes' law states that the viscous drag force experienced by a small spherical object moving through a viscous fluid is proportional to its velocity and radius. 29. Give an application of Stokes' law. Answer: Stokes' law is used to determine the viscosity of fluids and is also applied in the study of sedimentation rates. 30. What is Bernoulli's principle? Answer: Bernoulli's principle states that as the speed of a fluid increases, its pressure decreases, and vice versa, when the flow is continuous and the fluid is incompressible. 31. Give an application of Bernoulli's principle. Answer: An airplane's lift is generated based on Bernoulli's principle. Faster air over the curved wing creates lower pressure, allowing the airplane to lift off the ground. Short Answer Questions 32. Explain why (a) The blood pressure in humans is greater at the feet than at the brain. (b) Atmospheric pressure at the height of about 6 km decreases to nearly half of its value at sea level, though the height of the atmosphere is more than 100 km. (c) Hydrostatic pressure is a scalar quantity even though the pressure is force divided by area. All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 10 Answer: (a).The blood column to the feet is at a greater height than the head. Thus, the blood pressure in the feet is greater than that in the brain. (b) The density of the atmosphere does not decrease linearly with the increase in altitude, in fact, most of the air molecules are close to the surface. Thus, there is this nonlinear variation of atmospheric pressure. (c) In hydrostatic pressure, the force is transmitted equally in all directions in the liquid; thus, there is no fixed direction of pressure, making it a scalar quantity. 33. How does hydrostatic pressure enable the functioning of a hydraulic lift? Answer: In a hydraulic lift, applying a small force to a small piston generates pressure in the fluid. This pressure, transmitted equally in all directions, moves a larger piston, exerting a much greater force. The principle behind this amplification of force is Pascal's law, based on hydrostatic pressure transmission. 34. Why does hydrostatic pressure increase with depth in a fluid? Answer: Hydrostatic pressure increases with depth because the weight of the fluid above a certain point creates pressure. Deeper levels have more fluid above, resulting in a higher weight and, consequently, higher hydrostatic pressure at greater depths. 35. Explain Pascal's Law and its significance in hydraulic systems. Answer: Pascal's Law states that any change in pressure applied to an enclosed fluid is transmitted undiminished in all directions. This principle is fundamental in hydraulic systems, allowing a small force applied at one point to be transmitted and amplified as a larger force at another point in the system. 36. Two pistons of a hydraulic press have diameter of 30.0 cm and 2.5 cm. Find the force exerted on the longer piston when 50.0 kg. wt. is placed on smaller piston. [PYQ NIOS Oct 2022] All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 11 Answer: As in hydraulic press, pressure will be equally transferred ⇒ P1= P2 ⇒ 50 × g × A1= m × g× A2 ⇒ 50 (1.25)2× π = m (15)2× π ⇒m = 7200kg So, 7200kg weight should be put on a larger piston. 37. On what factors does the critical velocity of the liquid depend? Answer: critical velocity (vc) of liquid is: 1. Diretly proportional to the coefficient of viscosity of liquid, i.e. vc ∝ . 1 2. Inversely proportional to the density of liquid, i.e. vc ∝. ρ 3. Inversely proportional to the diameter of the tube through which it 1 flows, i.e. vc ∝. D 38. How does buoyancy work, and what factors affect the buoyant force experienced by an object submerged in a fluid? Answer: Buoyancy is the upward force exerted on an object submerged in a fluid, counteracting the force of gravity. The buoyant force depends on the density of the fluid, the volume of the displaced fluid, and the acceleration due to gravity. An object will float if its density is less than the density of the fluid it is placed in. 39. A barometer reads 760 mm Hg at sea level. If the barometer is taken to the top of a mountain at an altitude where the atmospheric pressure is 650 mm Hg, calculate the height of the mountain. (Density of mercury = 13.6 g/cm3) Answer: Using the barometric formula, the height of the mountain can be P 0 −P calculated as h = , where P0 and P are atmospheric pressures at sea level ρ ×g All Rights Reserved © Manish Verma, For More Visit – http://manishvermaofficial.com/ 8368259468 12 and mountain respectively, ρ is the density of mercury, and g is acceleration due to gravity. (760−650) Substituting the values, h = = 7.89 km 13.6 ×9.8 40. A hydraulic lift has a small piston with an area of 0.02 m2 connected to a larger piston with an area of 0.1 m2. If a force of 2000 N is applied to the small piston, calculate the pressure exerted on the larger piston. (According to Pascal's law) Answer: Using Pascal's law F1 /A1= F2/A2, where F1 and A1 are the force and area of the small piston, and F2 and A2 are the force and area of the large piston. Given → A1 = 0.02 m2 , A2 = 0.1 m2 , F1 = 2000 N F1 Solving for P2, the pressure on the larger piston, P2 = × A2