Friday, July 17, 2015

HC Verma solutions, Concepts of Physics, Part 1,Chapter 3, REST AND MOTION : KINEMATICS",'Objective-I'

OBJECTIVE-I 

1. A motor car is going due north at a speed of 50 km/h. It makes a 90° left turn without changing the speed. The change in the velocity of the car is about

  (a) 50 km/h towards west

  (b) 70 km/h towards south-west

  (c) 70 km/h towards north-west

  (d) zero.   

Answer:   (b)

Explanation:   Initial velocity is Vn=50 km/h due north, Final velocity is Vw=50 km/h due west. 

So change in velocity= Final minus initial velocity 

                                  = Vw-Vn =Vw+(-Vn) 

-Vn is equal in magnitude but opposite in direction to Vn  

As is clear from the picture below, The direction of the above resultant is south-west. and its magnitude is = √(50²+50²) ≈ 70 km/h  

The figure for Question 1


2. The figure for (3-Q2) shows the displacement-time graph of a particle moving on the X-axis.

Pic for Q- 2

  (a) the particle is continuously going in the positive x direction

  (b) the particle is at rest

  (c) the velocity increases up to a time to and then becomes constant

  (d) the particle moves at a constant velocity up to a time to, and the stops. 


Answer:   (d)


Explanation:   Initially the displacement increases proportionately with time up to time to, so it is a constant velocity. After this instant the displacement is stagnant because the graph is parallel to the time axis, it shows that the particle has stopped.



3. A particle has a velocity u towards the east at t=0. Its acceleration is towards the west and is constant. Let xA and xB be the magnitudes of displacements in the first 10 seconds and the next 10 seconds 

  (a)  xA<  xB    

  (b) xA=  xB 

  (c)  xA>  xB

  (d) the information is insufficient to decide the relation of xAwith  xB .


Answer:   (d)


Explanation:   Magnitude of displacement depends upon initial velocity, time Duration, and acceleration. In this case, only time is known while the magnitude of initial velocity and acceleration is unknown. These two displacements could only be comparable if directions of initial velocity and acceleration had been the same. Since in this case, directions are opposite, their magnitudes need to be known, So there is not sufficient information.    



4. A person traveling on a straight line moves with a uniform velocity v1 for some time and with uniform velocity v2 for the next equal time. The average velocity v is given by  

  (a)  v= (v1+v2)/2

  (b)  v=√v1v2  

  (c)  2/v=(1/v1+1/v2) 

  (d) 1/v=(1/v1+1/v2)  


Answer:   (a)


Explanation:   Let t be the time related to each velocity. Total distance travelled =v1t+v2t. Total time taken=2t. Hence average velocity = v1t+v2t/2t  =(v1+v2)/2.          



5. A person traveling on a straight line moves with a uniform velocity v1 for a distance x and with a uniform velocity v2 for the next equal distance. The average velocity v is given by  

  (a)  v= (v1+v2)/2

  (b)  v=√v1v2  

  (c)  2/v=(1/v1+1/v2) 

  (d) 1/v=(1/v1+1/v2)  


Answer:   (c) 


Explanation:   Time taken during first phase =  x/v1 and during second phase =  x/v2.

Total distance =2x 
Total time taken = (x/v1)+(x/v2) = x(1/v1 +1/v2)   
Average velocity =Total distance/Total time taken   
                                         2x       
                     v = ---------------------
                               x(1/v1 +1/v2)   
                                         2       
       =>         v = ---------------------
                               (1/v1 +1/v2)   
       =>      2/v  = (1/v1 +1/v2)    

{Multiplying both sides by (1/v1 +1/v2)/v  }
  


6.  A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is 

  (a) a upward

  (b) (g-a) upward

  (c) (g-a) downward

  (d) g downward.


Answer:   (d) 


Explanation:    As soon as the stone is released from the lift the acceleration 'a' of the lift has no effect on the stone, it is now only under the influence of the gravitational pull. So only acceleration due to gravity 'g' acts.                           



7. A person standing near the edge of the top of a building throws two balls A and B. The ball A is thrown vertically upward and B is thrown vertically downward with the same speed. The ball A hits the ground with a speed vA and the ball B hits the ground with a speed vB. We have

  (a)  vA>vB

  (b)  vA<vB         

  (c)  vA=vB

 (d)  the relation between vA and vB depends on the height of the building above the ground. 


Answer:   (c) 


Explanation:    The ball 'A' thrown vertically upward will return back to the initial point with the same thrown speed but its direction will be downwards. This return speed is exactly the same as the speed of ball 'B'. So both of them will hit the ground with equal velocity.   



8. In a projectile motion the velocity

  (a) is always perpendicular to the acceleration

  (b) is never perpendicular to the acceleration   

  (c) is perpendicular to the acceleration for one instant only
  (d) is perpendicular to the acceleration for two instants.


Answer:   (c) 


Explanation:    The acceleration is downwards always while the velocity continuously changes the direction. The only instant when the velocity is horizontal is at the "Maximum height" and it is the only instant when both are perpendicular. 
9. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first?
  (a) the faster one
  (b) the slower one
  (c) both will reach simultaneously 
  (d) depends on the masses.


Answer:   (c) 


Explanation:    The component of vertical velocity will be the same for both bullets, so both will reach the ground simultaneously.
10. The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45° its range will be  
  (a) 25 m
  (b) 37 m
  (c) 50 m
  (d) 100 m


Answer:   (d) 


Explanation:  In the first case Range = 50 = (u2 /g) Sin(2x15°) = (u2 /2g) 

=> (u2 /g)=100 
In the second case range = (u2 /g) Sin(2x45°) = (u2 /g) Sin90° 
=100 m
*Otherwise, a projectile has maximum range when thrown at 45°, so in this case, its range will be more than 50 m, which is only in option (d). 

11. Two projectiles A and B are projected with angles of projection 15° for the projectile A and 45° for the projectile B. If  RA  and RB be the horizontal range for the two projectiles, then 
  (a) RA  < RB  
  (b) RA  = RB  
  (c) RA  > RB  
  (d) the information is insufficient to decide the relation of RA  with RB .  


Answer:   (d) 


Explanation: The range of a projectile depends upon the projected velocity and the angle of projection as is clear from the formula
R= (u2 /g) Sin2Θ
But in this question, only the angle of projection is given, not the projected velocity.
12. A river is flowing from west to east at a speed of 5 meters per minute. A man on the south bank of the river, capable of swimming at 10 meters per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction
  (a) due north
  (b) 30° east of north 
  (c) 30° north of west 
  (d) 60° east of north 

Answer: (a) 

 Explanation: Since the river flows from west to east the component of its velocity along the width (due north) is zero. So river velocity cannot help the man to cross the river. For the man to cross the river in the shortest time the component of his own velocity due north should be maximum. So he must swim due north with his full capacity.
(Note that in doing so he will cross in shortest time but will not reach the other side just opposite to his initial point but some point downstream because his resultant velocity will be at some angle east of north.)  



13. In the arrangement shown in figure (3-Q3), the ends P and Q of an in-extensible string move downwards with uniform speed u. Pulleys A and B are fixed. The mass M moves upwards with a speed 
  (a) 2uCosθ  (b) u/Cosθ  (c) 2u/Cosθ (d) uCosθ.
The figure for Question no. 13


 Answer: (b) 


 Explanation:  Speed of Q is u downwards so in unit time it goes distance u downwards. Since the strings are in-extensible the other end of the string near mass M will shorten by a length of u. Consider the portion of the above figure as shown below. 
Movement of point X to Y
When Q moves to Q', X should move to X' so that QQ'=XX'=u. But the mass moves upwards, so X will actually move upwards to Y so as BX'=BY. For the small angle X'BY the line X'Y 丄 XX'. So ΔXX'Y is a right-angled triangle. 
(XX'/XY)=Cosθ 
XY=(XX'/Cosθ) = u/Cosθ
It is the distance traveled by mass M in unit time i.e., the speed of M. Hence the answer.
===<<<O>>>===

My Channel on YouTube  →  SimplePhysics with KK

For links to 

other chapters - See bottom of the page

Or click here → kktutor.blogspot.com 





CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion



EXERCISES- Q1 TO Q10

EXERCISES- Q11 TO Q20

EXERCISES- Q21 TO Q30

EXERCISES- Q31 TO Q40

EXERCISES- Q41 TO Q50

EXERCISES- Q51 TO Q58 (2-Extra Questions)

CHAPTER- 11 - Gravitation



EXERCISES -Q 31 TO 39

CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 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)

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)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

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

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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

CHAPTER- 4 - The Forces

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


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

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


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"

Posted by at
Labels: , , , ,

5 comments:

  1. Miss esotericMarch 10, 2020 at 6:36 AM

    Sir in Q 13 why we have only considered one string I.e string Q .why we have not considered string P as well ?

    ReplyDelete
    Replies
    1. K K MISHRAMarch 10, 2020 at 7:27 AM

      Dear student, the system is symmetric about the vertical line through M. Hence the string P will have similar displacements.

      Delete
  2. My BlogMarch 22, 2020 at 11:00 PM

    This comment has been removed by the author.

    ReplyDelete
  3. My BlogMarch 22, 2020 at 11:02 PM

    Dear Sir / Madam,
    In Q13, why aren't we adding up the velocities of both strings (P and Q) to get u/Cosθ + u/Cosθ = 2u/Cosθ as the total upward velocity?
    Thanks in advance.

    ReplyDelete
    Replies
    1. K K MISHRAMarch 28, 2020 at 8:28 AM

      Tough the example is given for one string, it is same for the other. Due to attachment of both strings, the common end moves u/cos(theta).
      It is not the effort of one string but the effort of both strings , we can not add as you suggested.

      Delete

Subscribe to: Post Comments (Atom)