Friday, September 18, 2015

HC Verma solutions, Concepts of Physics, Part 1,Chapter 4, THE FORCES, EXERCISES

"EXERCISES"

1. The gravitational force acting on a particle of 1g due to a similar particle is equal to 6.67x10-17 N. Calculate the separation between the particles.  

  

Answer: Let the separation between the particles be r. Mass of particles m=1 g =1x10-3 kg, Force between them F= 6.67x10-17 N, 
From the gravitational equation we have  
F=Gm²/r²   →  6.67x10-17 = 6.67x10-11(1x10-3)²/r²   

→ r²= 1       

→ r=1 m           

                 


2. Calculate the force with which you attract the earth.    


Answer: The magnitude of force with which we attract the earth is exactly the same as the earth attracts us, that is our weight. So if my mass is m kg then this force ie my weight is mg N.                     



3. At what distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight?

   

Answer:  Let the distance be r. 

The force between two charges  

F = (1/4 π ε0)q1 q2 /r²  

The term 1/4 π ε0 =9x109 N-m²/C²,  q1 = q2 = 1 C, If my mass be m kg then here F=mg N, Now the equation becomes  

mg= 9x109 x1² /r²  → r²= 9x109 /mg = 9x108/m    

(if g=10 m/s²)

→ r = 3x104/√m 

If m=64 kg ; r=30000/8 m =3.75 km.




4. Two spherical bodies, each of mass 50 kg are placed at a separation of 20 cm. Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude. Find the magnitude of the charge placed on either body.  


Answer: r=20 cm =0.20 m, mass m = 50 kg each

Let q C be the equal charge on each of the spheres.

Coulomb force of repulsion = 9x109 q²/r²  N 

Gravitational force of attraction = 6.67x10-11(50)²/r²   

Equating these two we get 

q²= 6.67x10-11(50)² /9x109 =1853x 10-20     

→ q=√(1853x 10-20 ) =43.0 x 10-10 C = 4.3 x 10-9 C 

Note, the separation r=20 cm is a redundant data which cancels out in the equation.                               

   

5. A monkey is sitting on a tree limb. The limb exerts a normal force of 48 N and a frictional force of 20 N. Find the magnitude of the total force exerted by the limb on the monkey.
The figure for Question no. - 5
  

Answer: A diagram has been drawn for this problem on the right. The Normal force of 48 N will act perpendicular to the tree limb surface on the monkey. The force of friction 20 N acts parallel to the surface of the limb on the monkey opposing the tendency to slip. So these two forces are perpendicular to each other. The magnitude of the total force can be calculated as,

F²=48²+20² 

   = 2304+400 

   = 2704  

F= √2704 = 52 N      



6. A bodybuilder exerts a force of 150 N against a bullworker and compresses it by 20 cm. Calculate the spring constant of the spring of the bullworker.     


Answer: Spring constant of a spring is the force required to compress/elongate the spring by unit length. Here 150 N force compresses the spring by 20 cm = 0.20 m.  

So Spring constant of this spring = 150 N/ 0.20 m = 750 N/m.   



7. A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station? (Radius of the earth is 6400 km.) 


Answer:  Let the mass of the satellite be m kg. Force on the satellite due to earth at the earth station =mg, where g is the acceleration due to gravity. 

Given Radius of the earth R = 6400 km = 64x105  m   

From the gravity equation, we have 

mg=GMm/R² .......     (i)

(G= Universal constant of gravity, M= mass of the earth)

Let at a distance of r from the earth's surface this force reduces to half that is mg/2, Now distance of the satellite from the center of the earth = R+r, Now the equation for this situation is  

mg/2 =GMm/(R+r)² 

→ mg = 2GMm/(R+r)²   ..........................(ii)

Equating the RHS expressions of equations (i) and (ii) we get,

GMm/R² = 2GMm/(R+r)²  

→ (R+r)² = 2R²  

→ R+r = R√2  

→ r = (√2-1)R = 0.414 x 64x105  m 

      = 2649600 m ≈ 2650 km.                                        


8. Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other. What will be the force if the separation is increased to 25 cm?

                    

Answer: Coulomb force  F= 9x109 q1q2  /r²     N

→   9x109 q1q2  = Fr² = 20x(0.20)²  = 0.80  N-m²   {r= 20cm = 0.20 m}                      

When separation is changed to r = 25 cm = 0.25 m  

Force = 9x109 q1q2  /(0.25)²    

          = 0.80/(0.25)² = 0.80x 16 = 12.8  N ≈ 13 N


Alternately:   Since the force is inversely proportional to the square of the distance between the charges, the two forces F1, F2 and the separations r1, r2 can be expressed as       

   F1/F2 =  (r2 / r1)² 

→ F2  = ( r1 / r2 )²  F1 =(0.20/0.25)²x 20 N = (4/5)²x20 N = 12.8 N ≈13N                                          


9. The force with which the earth attracts an object is called the weight of the object. Calculate the weight of the moon from the following data:       

The universal constant of gravitation G=6.67x10-11  N-m²/kg², the mass of the moon = 7.36x 1022  kg, the mass of the earth 6x 1024 kg and the distance between the earth and the moon = 3.8x  105 km.   


Answer: We have  M1  = 6x 1024 kg ,  M2  = 7.36x 1022 kg ,   

r = 3.8x  105 km =  3.8x  108 m, G=6.67x10-11  N-m²/kg²,   

Now the weight of the moon = G M1 M2 /r²    

 =  6.67x10-11 x 6x 1024  x 7.36x 1022 /(3.8x  108)² 

=(6.67x6x7.36/3.8²) x10-11+24+22-16    

=20.4x 1019     

=2.04 x 1020 N  

≈ 2 x 1020 N             


                       

10. Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons. 


Answer:  Let the separation between protons be 'r'. 

Electric force F = 9x109 q2 /r²  N {where q is the charge on each proton}  

Gravitational force P = Gm²/r²  N {Where m is the mass of each proton and G is the universal constant of gravity}  

We have q=1.6 x10-19 C , m= 1.672 × 10-27 kg , G=6.67x10-11  N-m²/kg²

So the required ratio F/P = 9x109 q2 /Gm² 

 = 9x109 (1.6 x10-19)²/6.67x10-11 (1.672 × 10-27)² 
= (9x1.6²/6.67x1.672²)x 109-38+11+54

= 1.235x 1036 ≈ 1.24x 1036              

(So it is clear that gravitational force between two protons is nothing in comparison to electric force)   


                   

11. The average separation between the proton and the electron in a hydrogen atom in the ground state is 5.3x10-11 m. (a) Calculate the Coulomb force between them at this separation. (b) When the atom goes into its first excited state the average separation between the proton and the electron increases to four times its value in the ground state. What is the Coulomb force in this state?   

Answer: (a) Charge on each of them = q = 1.6 x10-19 C  

Separation between them = r1 = 5.3x10-11 m 

So coulomb force between them 

F = 9x109 q2 /r 1 ²  N  =    9x109 (1.6 x10-19 /5.3x10-11)²   N 

   = (9x2.56/28.09) x 10-7 N = 0.82 x  10-7 N  = 8.2 x10-8 N

(b) Since the Coulomb force is inversely proportional to the square of the separation between the charges, Hence when the separation increases four times the force decreases sixteen times. So in the excited state, the force will be  

=(8.2/16) x  10-8 N  = 5.1 x  10-9 N.     


                  

12. The geostationary orbit of the earth is at a distance of 36000 km from the earth's surface. Find the weight of a 120-kg equipment placed in a geostationary satellite. The radius of the earth is 6400 km.  


Answer: The weight of the equipment is the gravitational force between it and the earth which is balanced by the centrifugal force on the equipment due to its revolution around the earth. In case of a geostationary orbit, the time of revolution is exactly equal to 24 hrs. So if we calculate the centrifugal force on the equipment, it will give its weight. 

Hence weight = Centrifugal force = mω²r 

We have m=120 kg, r=6400+36000 km =42400 km =4.24x 107 m, 

Time period T=24 h = 24x3600 s=86400 s

ω= 2π/T= 2π/86400  rad/s     

Now weight = 120x(2π/86400)²x 4.24x 107 N 

= 26.88 N ≈ 27 N 


We can solve it in another way too, as follows:--    

Weight of equipment on the surface of the earth = mg = 120x9.8 N 

=1176 N    {Taking g=9.8 m/s² at the surface of the earth}  

We know that the gravitational force is inversely proportional to the square of the distance. So if the distance is increased n times, the force will be reduced by n² times. Here this ratio n=42400/6400 =6.625. So the weight of the equipment in the satellite   

=1176/n²   N =1176/6.625² N = 26.80 N ≈27 N

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Links to the Chapters



CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




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".

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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)

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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


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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"

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3 comments:

  1. UnknownJune 13, 2019 at 8:33 AM

    Sir why in question no. 6 total force is calculated just like a Pythagoras method

    ReplyDelete
    Replies
    1. UnknownJune 13, 2019 at 8:38 AM

      In question 5 not 6

      Delete
    2. K K MISHRAJune 16, 2019 at 5:22 AM

      Dear Student,
      Force is a vector, hence its addition or finding the resultant vector is as per vector rules. Please revise the chapter on vectors.

      Delete

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