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04.SOLUTIONS TO CONCEPTS

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					                                            SOLUTIONS TO CONCEPTS
                                                          CHAPTER – 4

1.   m = 1 gm = 1/1000 kg
                                                                                   m1 = 1 gm       m2 = 1 gm
                       –17               Gm1m2
     F = 6.67 × 10           NF=
                                          r2                                                   r
                      –17     6.67 10 11  (1/ 1000 )  (1/ 1000 )
      6.67 × 20            =
                                               r2
          2    6.67 10 11 10 6 10 17
     r =                          17 = 1
                  6.64 10 17     10
      r = 1 = 1 metre.
     So, the separation between the particles is 1 m.
2.   A man is standing on the surface of earth
     The force acting on the man = mg ………(i)
     Assuming that, m = mass of the man = 50 kg
                                                                          2
     And g = acceleration due to gravity on the surface of earth = 10 m/s
     W = mg = 50× 10= 500 N = force acting on the man
     So, the man is also attracting the earth with a force of 500 N
3.   The force of attraction between the two charges
            1         q1 q 2             1
     =                        9  10 9 2
          4  o       r 2
                                        r
     The force of attraction is equal to the weight
              9 109
     Mg =
                r2
                9  10 9   9  10             8
     r =
          2                                                                  2
                                                              [Taking g=10 m/s ]
                m  10        m
                   9 10 8 3 10 4
     r=                          mt
                      m       m
     For example, Assuming m= 64 kg,
           3  10 43
     r=            10 4 = 3750 m
               64  8
4.   mass = 50 kg
     r = 20 cm = 0.2 m
                    m1m2 6.67  10 11  2500
          FG  G        
                     r2          0.04
                                               1 q1 q2           q2
                                                               9
          Coulomb’s force               FC =           = 9 × 10 0.04
                                             4 o r 2
                                  6.7 10 11  2500 9 10 9  q2
          Since, FG = Fc =                          
                                        0.04            0.04
               2      6.7 10 11  2500 6.7 10 9
          q =                                      25
                            0.04          9 10 9
                                      –18
                      = 18.07 × 10
                                                  -9
          q=       18.07  10 -18 = 4.3 × 10 C.

                                                                  4.1
                                                                                                   Chapter-4
5.   The limb exerts a normal force 48 N and frictional force of 20 N. Resultant magnitude of
     the force,
     R =          ( 48 )2  (20)2

          =       2304  400

          =       2704                                                                                 48N
        = 52 N
6.   The body builder exerts a force = 150 N.
     Compression x = 20 cm = 0.2 m
     Total force exerted by the man = f = kx                                                  F          F
                                                                                                   x
      kx = 150
            150 1500
     k=          =       = 750 N/m
             0 .2     2
7.   Suppose the height is h.
                                 2
     At earth station F = GMm/R
     M = mass of earth
     m = mass of satellite
     R = Radius of earth
           GMm        GMm
     F=           2
                    =
          (R  h)     2 R2
              2                2      2         2
      2R = (R + h)  R – h – 2Rh = 0
        2          2
      h + 2Rh – R = 0
          2R  4R 2  4R 2 
                            
     H=                      =  2R  2 2R
                 2                     2
        = –R ±         2R = R        2  1
       = 6400 × (0.414)
       = 2649.6 = 2650 km
8.   Two charged particle placed at a sehortion 2m. exert a force of 20m.
     F1 = 20 N.         r1 = 20 cm
     F2 = ?             r2 = 25 cm
                         1 q1q2                              1
     Since, F =                  ,                  F
                       4 o r 2                            r2
                  2                             2                    2
     F1 r2                            r1       20      16   64
            F 2 = F1 ×               = 20 ×   = 20 ×
                                     r                      =    = 12.8 N = 13 N.
     F2 r12                           2        25      25    5
                                                                                    mmmc
9.   The force between the earth and the moon, F= G
                                                                                        r2
          6.67 10 11  7.36 10 22  6 10 24                          6.67  7.36  10 35
     F=                                                          =
                            3.8 10     8 2                                3.8 2 1016
                          19                        20           20
      = 20.3 × 10 =2.03 × 10 N = 2 ×10                                   N
                                –19
10. Charge on proton = 1.6 × 10
                                        9 109  1.6  10 38
                                                       2
                           1    qq
      Felectrical =            12 2 =
                         4 o   r                 r 2

                                                     –27
     mass of proton = 1.732 × 10                           kg



                                                                                  4.2
                                                                                                                  Chapter-4

                   m1m2   6 .67  10    11
                                               1 .732  10  54
    Fgravity = G        =
                    r 2
                                               r2
           9 10 9  1.6  10 38
                           2

                                           9  1.6 10 29
                                                    2
    Fe                 r2                                                 36
                                      =                      = 1.24 × 10
    Fg 6.67 10 11  1.732 10  54   6.67 1.732 10  65
                                                      2

                       r2
                                                                                    –11
11. The average separation between proton and electron of Hydrogen atom is r= 5.3 10 m.

    a) Coulomb’s force = F = 9 × 10 ×
                                       9      q1q2
                                                   =
                                                                     
                                                     9 109  1.0 10 19          2
                                                                                        = 8.2 × 10
                                                                                                        –8
                                                                                                             N.
                                               r2           
                                                         5.3  10 11
                                                                      2
                                                                             
    b) When the average distance between proton and electron becomes 4 times that of its ground state

    Coulomb’s force F =
                             1   qq
                                 1 2 =
                                                        
                                        9 10 9  1.6 10 19            
                                                                         2
                                                                             =
                                                                                 9  1.6 
                                                                                           2
                                                                                                   10  7
                           4 o 4r     16  5.3 10  22                    16  5.3 
                                      2             2                                         2

                                                                –7                –9
                                          = 0.0512 × 10 = 5.1 × 10 N.
12. The geostationary orbit of earth is at a distance of about 36000km.
                                   2
    We know that, g = GM / (R+h)
                                                2
    At h = 36000 km. g = GM / (36000+6400)
        g`   6400  6400   256
                               0.0227
        g 42400  42400 106 106
     g = 0.0227 × 9.8 = 0.223
                        2
    [ taking g = 9.8 m/s at the surface of the earth]
    A 120 kg equipment placed in a geostationary satellite will have weight
    Mg` = 0.233 × 120 = 26.79 = 27 N


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                                                            4.3

				
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Description: HC VERMA SOLUTIONS PHYSICS