Tuesday, October 27, 2015

Solutions to Problems on "Newton's Laws of Motion"-'H C Verma's Concepts of Physics, Part-I, Chapter-5,OBJECTIVE-II

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OBJECTIVE-II

1. A reference frame attached to earth  

(a) is an inertial frame by definition

(b) cannot be an inertial frame because the earth is revolving around the sun

(c) is an inertial frame because Newton's laws are applicable in this frame

(d) cannot be an inertial frame because the earth is rotating about its axis.  

Answer: (b), (d).     

Reason: (b) and (d) are correct because theoretically earth is not an inertial frame. Due to its revolution around the sun and rotation about its own axis, its velocity changes continuously, means has acceleration. So the earth and any frame attached to it is non-inertial. 

2. A particle stays at rest as seen in a frame. We can conclude that

(a) The frame is inertial    

(b) resultant force on the particle is zero

(c) the frame may be inertial but the resultant force on the particle may be zero

(d) the frame may be non inertial but there is a nonzero resultant force.  

Answer: (c), (d). 

Reason:  (a) is incorrect because if the particle and the non-inertial frame both are moving with same acceleration, the particle will be seen at rest from this frame. So we cannot conclude that the frame is inertial. 

(b) is  also not correct. Suppose as viewed from an inertial frame a force F produces an acceleration a in a particle. If  this particle be seen from a non-inertial frame that too is moving with same acceleration in the same direction, it will seem to be at rest. But the force on the particle is not zero. 

(c) and (d) are possible conditions.     

3. A particle is found to be at rest when seen from a frame S1 and moving with a constant velocity when seen from another frame S2. Mark out the possible options.

(a) Both the frames are inertial

(b) Both the frames are non-inertial

(c) S1 is inertial and S2 is non-inertial

(d) S1 is non-inertial and S2 is inertial 

Answer: (a), (b). 

Reason: (a) is possible. Suppose in one inertial frame the particle is at rest. If viewed from another frame which is moving with a constant velocity the same particle will be seen as moving with a constant velocity. but this frame is also inertial. 

(b) is also possible. Suppose a non-inertial frame and the particle both are moving with same acceleration in the same direction. From this frame the particle will be seen at rest. If this particle is seen from another non-inertial frame moving with the same acceleration in the same direction but with different velocity at that instant, it will seem to be moving with a constant velocity. 

(c) and (d) are not true because this condition is not possible if one frame is inertial and another non-inertial.  

4.  Figure (5-Q3) shows the displacement of a particle going along the X-axis as a function of time. The force acting on the particle is zero in the region

(a) AB          (b) BC            (c) CD             (d) DE
Figure for problem number 4
     

Answer: (a), (c).

Reason: Only in the reason AB and CD the displacement is directly proportional to time, it means that the particle is moving with a uniform velocity. So in this reason force on the particle is zero.      

5. Figure (5-Q4) shows a heavy block kept on a friction-less surface and being pulled by two ropes of equal mass m. At t=0 the force on the left rope is withdrawn but the force on the right rope continues to act. Let F1 and F2 be the magnitude of the forces by the right rope and the left rope on the block respectively.

Figure for problem number 5

(a) F1 = F2 =F  for t<0          

(b) F1 = F2 =F + mg for t<0                  

(c) F1 = F, F2 =F  for t>0                       

(d) F1 < F, F2 =F  for t>0    

Answer: (a).    ,

Reason: (a) is correct because as seen in the figure for t<0, F1 = F2 =F. For the same reason (b) is incorrect. 

Since at t=0 the force on the left rope is withdrawn so at t>0, F1  F, F2 ≠ F. So (c) is in correct. 

Since force on the left rope is withdrawn at t=0, so at t>0 force by the right rope on the block will not change ie. F1 will not be less than F. So (d) is incorrect.     

6. The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator is 

(a) going up and slowing down          

(b) going up and speeding up                  

(c) going down and slowing down                       

(d) going down and speeding up    

Answer: (b), (c). 

Reason: If we are going to use Newton's laws of motion in a non-inertial frame like accelerating elevator, we have to apply a pseudo force on the object that is opposite in the direction of the acceleration and its magnitude is mass times the magnitude of this acceleration. Only in the cases of option (b) and (c) the acceleration of the elevator has direction upwards. In these cases the direction of the pseudo force will be downwards  which will add to the weight of the person.   

7. If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be  

(a) going up with increasing speed          

(b) going down with increasing speed                  

(c) going up with uniform speed                       

(d) going down with uniform speed     

Answer: (c), (d).

Reason: In the cases of  (a) and (b) the elevator is accelerating, so pseudo forces will have to be applied on it to apply the Newton's laws of motion. It will result in either increase or decrease in the apparent weight of the elevator. So the tension in the cable will be either more or less than the weight of the lift.

There are no such pseudo forces in the cases of (c) and (d). So the tension in the cable will be equal to the weight of the elevator.   

8. A particle is observed from two frames S1 and S2 . The frame S2 moves with respect to S1 with an acceleration a. Let F1 and F2  be the pseudo forces on the particle when seen from S1 and S2 respectively. Which of the following is not possible ?   

(a) F1 = 0, F2 ≠ 0  

(b) F1  0, F2 = 0 

(c) F1  0, F2 ≠ 0  

(d) F1 = 0, F2 = 0    

Answer: (d). 

Reason: (a), (b) and (c) are possible under different values of accelerations of the frames S1 and S2 with respect to another inertial frame. But (d) is not possible because due to difference in accelerations of frames the pseudo forces in both frames cannot be equal.      

9. A person says that he measured the acceleration of a particle to be nonzero while no force was acting on the particle. 

(a) He is a liar. 

(b) His clock might have run slow   

(c) His meter scale might have been longer than the standard.    

(d) He might have used non-inertial frame.           

Answer: (d). 

Reason: (d) is true because if a particle with no force acting on it is viewed from a non-inertial frame that is moving with a certain acceleration, then the particle will be seen moving with nonzero acceleration.   

===<<<O>>>===

 Links for the chapter
CHAPTER-7 - Circular Motion

CHAPTER-8 - Work and Energy


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

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Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)

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Click here for → EXERCISES (21-31)

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

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Click here for → Newton's Law's of Motion - Exercises(1 to12)

Click here for → Newton's Laws of Motion - Exercises(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" - 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)

===============================================

Chapter -2, "Vector related Problems"

Click here for "Questions for Short Answers"

Click here for "OBJECTIVE-II"

Click here for "Exercises"

                     

Friday, October 9, 2015

Solutions to Problems on "Newton's Laws of Motion"-'H C Verma's Concepts of Physics, Part-I, Chapter-5',OBJECTIVE-I

My Channel on You Tube  --->  SimplePhysics with KK

For links to 

other chapters - See bottom of the page

Or click here → kktutor.blogspot.com

OBJECTIVE-I

1. A body of weight wis suspended from the ceiling of a room through a chain of weight w2 . The ceiling pulls the chain by a force

(a) w1                (b) w2            (c) w1+w2                        (d) (w1+w2)/2

Answer: (c)

Reason: The ceiling is pulled by a force = w1+w2 and as per third law of motion the ceiling too pulls the chain by a force = w1+w2.  

2. When a horse pulls a cart, the force that helps the horse to move forward is the force exerted by 

(a) the cart on the horse      (b) the ground on the horse 

(c) the ground on the cart   (d) the horse on the ground   

Answer: (b)

Reason: The horse pushes the ground backwards and as per Newton's third law of motion the ground too pushes the horse forward equally.  

3. A car accelerates on a horizontal road due to the force exerted by 

(a) the engine of the car        (b) the driver of the car  

(c) the earth                           (d) the road  

Answer: (d)

Reason: The wheels of the car push the ground backwards and as per Newton's third law of motion the road pushes the wheels with equal and opposite force forwards that accelerates the car.  

4. A block of mass 10 kg is suspended through two light spring balances as shown in figure (5-Q2).
Figure for question number 4

(a) both the scales will read 10 kg (b) both the scales will read 5 kg  (c) the upper scale will read 10 kg and the lower zero.              

(d) the readings may be anything but their sum will be 10 kg  

Answer: (a)

Reason: At any point between the load and the ceiling the tension force will be equal to the load as per the third law of motion.  

5. A block of mass m is placed on a smooth inclined plane of inclination θ with the horizontal. The force exerted by the plane on the block has a magnitude  

(a) mg      (b) mg/cosθ      (c) mg cosθ        (d) mg tanθ    

Answer: (c)
Force exerted by plane

Reason:  See the figure on the right. Weight of the block is mg downwards. We resolve it in two components-along the plane and perpendicular to the plane. Former component does not exert force on the plane while the later component mg.cosθ is the force which the block exerts on the plane. If we take it as action then equal and opposite force mg.cosθ is exerted by the plane on the block as reaction satisfying the "Newton's Third law of Motion".      

6. A block of mass m is placed on a smooth wedge of inclination Î¸. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude 

(a) mg      (b) mg/cosθ      (c) mg cosθ        (d) mg tanθ    

Answer: (b)
Accelerated Plane

Reason: When the plane is accelerated to balance the block, a horizontal force F acts on the block such that its component along the plane Fcosθ  is equal and opposite to component of the weight along the plane mg.sinθ, see the figure on the right. So, Fcosθ = mg.sinθ → F= mg.tanθ.

The components of F and weight mg perpendicular to plane (Fsinθ and mgcosθ) together push the plane. Equal and opposite force is exerted by the plane on the block as per "Newton's Third Law of Motion" and its value is = Fsinθ + mgcosθ

mg.tanθ.sinθ + mgcosθ  =mg(sin2θ +cos2θ)/cosθ = mg/cosθ

7. Neglecting the effect of rotation of the earth. Suppose the earth suddenly stops attracting objects placed near its surface. A person standing on the surface of the earth will 

(a) fly up                                                    (b) slip along the surface 

(c) fly along a tangent to the earth's surface      (d) remain standing   

Answer: (d)

Reason:  The person will loose weight but remain standing. As no force acts on him, so according to "Newton's First Law of Motion" he will remain in his state of rest as before.        

8. Three rigid rods are joined to form an equilateral triangle ABC of side 1 m. Three particle carrying charges 20 µC each are attached to the vertices of the triangle. The whole system is at rest in an inertial frame. The resultant force on the charged particle at A has the magnitude 

(a) zero              (b) 3.6 N                  (c) 3.6/3 N                (d) 7.2 N 

Answer: (a)

Reason: Since the charged particles are attached to rigid triangle ABC which is in rest in an inertial frame, so the particle at A is also at rest. It means the magnitude of resultant of forces acting on it is zero.       

9. A force F1 acts on a particle so as to accelerate it from rest to a velocity v. The force F1 is then replaced by F2 which decelerates it to rest.              

(a) F1 must be equal to F2                     (b) F1 may be equal to F2  

(c) F1 must be unequal to F2                 (d) None of these 

Answer: (b)

Reason:  Neither the time nor the distance for accelerating and decelerating journey is mentioned. So both forces may or may not be equal.        

10. Two objects A and B are thrown upward simultaneously with the same speed. The mass of  A is greater than the mass of B. Suppose the air exerts a constant and equal force of resistance on the two bodies.   

(a) The two bodies will reach the same height                   

(b) A will go higher than B 

(c) B will go higher than A               (d) Any of the above three may happen depending on the speed with which the objects are thrown. 

Answer: (b)

Reason: The force of resistance of air will add to retardation of both the body but its magnitude will depend upon mass of the objects as per "Newton's Second Law of Motion" and it will be equal to Force/mass. Since the force is equal on both the objects, the object having greater mass (A) will be retarded less, so it will go higher than B.           

11. A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth's surface. A block B placed at the top of the wedge takes a time T to slide down the length of the wedge. If the block is placed at the top of the wedge and the cable supporting the chamber is broken at the same instant, the block will 

(a) take a time longer than T to slide down the wedge 

(b) take a time shorter than T to slide down the wedge 

(c) remain at the top of the wedge

(d) jump off the wedge.   

Answer: (c)

Reason: Both the wedge and block will be in free fall motion and have equal velocities downwards at any instant of time so there will be no relative motion.           

12. In an imaginary atmosphere, the air exerts a small force F on any particle in the direction of the particle's motion. A particle of mass m projected upward takes a time t1 in reaching the maximum height and t2 in the return journey to the original point. Then  

(a) t1 < t2    (b) t1 > t2   (c)  t1 = t2   (d) The relation between t1 and t2 depends on the mass of the particle.        

Answer: (b)

Reason: Since F is in the direction of particle's motion it will add an acceleration in the direction of motion. Let it be a. Now magnitude of upward acceleration =a-g and downward acceleration =a+g. Clearly a+g>a-g and the particle travels same distance/height. So the upward journey will take a longer time.              

13. A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in a time t1 if the elevator is stationary and in time t2 if it is moving uniformly. Then             

(a) t1 = t2    (b) t1 < t2   (c) t1 > t2    (d) t1 < t2  or t1 > t2 depending on whether the lift is going up or down.       

Answer: (a)

Reason: The stationary and uniformly moving elevator are both inertial frame so value of acceleration due to gravity will remain the same in both conditions. The distance traveled by the coin with respect to the frame of elevator is equal in both cases. So the time taken will also be equal.  

14. A free 238U nucleus kept in a train emits an α - particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x at time t after the decay. If the decay takes place while the train is moving at a uniform velocity v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay as measured by the passenger is 

(a) x+vt     (b)   x-vt   (c) x  (d) depends on the direction of the train.

Answer: (c)

Reason: In both of the cases the frame of reference is inertial so with reference to the frame of train the separation too will be equal.      

===<<<O>>>===

Links for the chapter - 
CHAPTER-7 - Circular Motion

CHAPTER-8 - Work and Energy


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)

--------------------------------------------------

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 Law's of Motion - Exercises(1 to 12)

Click here for → Newton's Laws of Motion - Exercises(13 to 27)

Click here for→ Newton's Laws of Motion-Exercises(Q.No.28 to 42)
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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" - Exercises


--------------------------------------------------------------------------------------------------------------

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)

===============================================

Chapter -2, "Vector related Problems"

Click here for "Questions for Short Answers"

Click here for "OBJECTIVE-II"

Click here for "Exercises"