11.SOLUTIONS TO CONCEPTS

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11.SOLUTIONS TO CONCEPTS Powered By Docstoc
					                                                    SOLUTIONS TO CONCEPTS
                                                                     CHAPTER 11

1.   Gravitational force of attraction,
         GMm
     F=
           r2
         6.67  10 11  10  10            –7
     =                           = 6.67 × 10 N
                 (0.1)2
2.   To calculate the gravitational force on ‘m’ at unline due to other mouse.
              G  m  4m     8Gm2                                                                            A        E
     FOD =                 =                                                                            m
                                                                                                                              B
                                                                                                                                  2m
               (a / r 2 )2    a2
             G  m  2m     6Gm2
     FOI =                =
              (a / r 2 )2    a2                                                                                       m
                                                                                                                                  3m
              G  m  2m                     4Gm2                                                       4m
                                                                                                                          F   C
     FOB =                 2 2
                                         =                                                                   D
               (a / r )                       a2
              Gmm                          2Gm2
     FOA =             2 2
                                     =
               (a / r )                       a2
                                                    2                    2
                                        Gm2     Gm2                             Gm2
     Resultant FOF =                 64 2   36 2                        = 10
                                        a       a                                a2
                                                    

                                                    2                2
                                        Gm2    Gm 2                              Gm 2
     Resultant FOE =                 64 2   4 2 
                                        a      a                         = 2 5
                                                                                  a2
     The net resultant force will be,
                                 2                      2
                Gm2     Gm 2   Gm2 
     F=     100 2   20 2   2 2   20 5
                a       a      a 
                                   
                   2                                             2

     =
           Gm2 
          
           a2 
                       
                 120  40 5 =                          Gm2 
                                                        
                                                         a2 
                                                               (120  89.6)
                                                           
        Gm2                   Gm 2
     =      2
                 40.4 = 4 2 2
          a                    a
3.   a) if ‘m’ is placed at mid point of a side
                       4Gm2
     then FOA =             in OA direction                                                                 A m
                        a2
             4Gm 2                                                                                      m
     FOB =          in OB direction
              a2                                                                                            O
     Since equal & opposite cancel each other                                                       B                              C
                       2                       2                                                        m                         m
               Gm                        4Gm
     Foc =                       =             in OC direction
             r / 2a
               3            2
                                          3a 2
                                                                                                    A   m
                                                        4Gm 2
     Net gravitational force on m =
                                                            a2
     b) If placed at O (centroid)
                                                                                                                  m
                    Gm 2       3Gm2                                                                              O
     the FOA =               =                                                              B                                      C
                   (a / r3 )    a2                                                              m                                 m
                                                                                11.1
                                                                                                 Chapter 11

                   3Gm2
     FOB =
                    a2
                                         2           2
                                 3Gm 2         2            2
     Resultant F =              2         2 3Gm   1 = 3Gm
                                  a2         a2  2       a 2
                                                  
                    3Gm2
     Since FOC =           , equal & opposite to F, cancel
                     a2
     Net gravitational force = 0
               Gm2             Gm2
4.   FCB =          cos 60 ˆ 
                           i        sin 60ˆ
                                          j
               4a 2            4a 2                                                          M
                                                                                                 A       B
                        2                2
               Gm                Gm
     FCA =            cos 60 ˆ 
                             i        sin 60ˆ
                                            j
                4a 2            4a 2
                                                                                                    C
     F = FCB + FCA
        2Gm 2            2Gm2 r3      r Gm 2
     =         sin 60ˆ =
                     j                = 3 2
         4a 2              4a 2 2         4a
5.   Force on M at C due to gravitational attraction.
               Gm2 ˆ
     FCB =          j
               2R 2                                                                              A           B
                            2
                GM ˆ
     FCD =           i                                                                               R
                4R 2
                                                                                                 D           C
             GM2              GM2
     FCA =          cos 45 ˆ 
                           j        sin 45ˆ
                                          j
              4R 2             4R 2
     So, resultant force on C,
      FC      = FCA + FCB + FCD
           GM2      1 ˆ GM2                1 ˆ
     =         2    i               2    j
           4R 2 
                    2
                           4R 2         
                                             2
                                                


     FC =
               GM2
               4R 2
                    2 2 1         
                                     mv 2
     For moving along the circle, F =
                                       R

     or
          GM2
          4R 2
                   
               2 2 1 =
                        MV 2
                         R
                                
                             or V =
                                                 GM  2 2  1 
                                                    
                                                  R    4 
                                                              
                                                             
          GM            6.67  10 11  7.4  10 22     49.358  1011
6.                  =                               =
     R  h   2                         2
                         (1740  1000 )  10    6
                                                      2740  2740  10 6
        49.358  1011            –2         2
     =                = 65.8 × 10 = 0.65 m/s
         0.75  1013
7.   The linear momentum of 2 bodies is 0 initially. Since gravitational force is internal, final momentum is
     also zero.
     So (10 kg)v1 = (20 kg) v2
     Or v1 = v2                  …(1)
     Since P.E. is conserved
                 6.67  10 11  10  20            –9
     Initial P.E. =                       = –13.34×10 J
                            1
     When separation is 0.5 m,

                                                              11.2
                                                                                                                               Chapter 11

                             –9              13.34  10 9                2                2
     –13.34 × 10                  +0=                       + (1/2) × 10 v1 + (1/2) × 20 v2     …(2)
                                                (1/ 2)
                                  –9                  –9       2        2
      – 13.34 × 10 = -26.68 ×10 + 5 v1 + 10 v2
                   –9           –9      2
      – 13.34 × 10 = -26.68 ×10 + 30 v2
           2  13.34  10 9             –10
      v2 =                 = 4.44 × 10
                   30
                     –5
      v2 = 2.1 × 10 m/s.
                       –5
     So, v1 = 4.2 × 10 m/s.
8.   In the semicircle, we can consider, a small element of d then R d = (M/L) R d = dM.
          GMRdm
     F=                                                                                     M
             LR 2                                                                                                                         
                                                                                 
                          2GMm
     dF3 = 2 dF since =          sin  d
                            LR                                                        d                                          d
                /2
                        2GMm             2GMm                                                                                        R
     F =                   sin d        cos 0 / 2 
                                                       
                                                                                                   L
                         LR               LR                                                                          m
                    0

            GMm             2GMm      2GMm        2GMm
      = –2         ( 1) =        =           =
              LR              LR     L L / A         L2
9.   A small section of rod is considered at ‘x’ distance mass of the element = (M/L). dx = dm
           G(dm)  1
     dE1 =
                    
             d2  x 2
                        = dE2
                                  
     Resultant dE = 2 dE1 sin                                                                                            
            G(dm)            d          2  GM  d dx                                                      dE2                 dE1
     =2× 2
                
            d  x2
                     
                           2
                          d x 2
                                 =
                                                    
                                   L d  x 2  d2  x 2 
                                      2
                                                                                                            d
                                                       
     Total gravitational field                                                                                                            a
                                                                                                   M                                 
           L/2
                        2Gmd dx
     E=      Ld
            0
                         2
                              x2     
                                      3/2
                                                                                                       x
                                                                                                                      O
                                                                                                                               x
                                                                                                                                          dx

     Integrating the above equation it can be found that,
             2GM
     E=               
          d L2  4d2
10. The gravitational force on ‘m’ due to the shell of M2 is 0.
                        R  R2
    M is at a distance 1                                                                                                             R2   m2
                            2                                                                                        M1
                                                                                                                          R1
    Then the gravitational force due to M is given by
         GM1m          4GM1m                                                                                              m
    =               =
      (R1  R 2 / 2   (R1  R 2 )2
                                                 3
11. Man of earth M = (4/3) R 
    Man of the imaginary sphere, having
                            3
    Radius = x, M = (4/3)x 
          M   x3
     or      = 3
          M   R                                                                                                                m

                                                     GMm
      Gravitational force on F =                                                                                                    x
                                                      m2
                    GMx 3m   GMmx
     or F =           3 2
                           =      
                     R x      R3


                                                                         11.3
                                                                                                           Chapter 11
12. Let d be the distance from centre of earth to man ‘m’ then
                   R2 
     D=      x2      
                   4  = (1/2)        4x 2  R2                                                                                  m
                                                                                                                          x
                      
                                                                                                                    R/2
                                                                                                                                  d
     M be the mass of the earth, M the mass of the sphere of radius d/2.
                       3                                                                                                 O
     Then M = (4/3) R 
                 3
     M = (4/3)d 
       M     d3
     or   = 3
       M     R
      Gravitational force is m,
                                                                                                           n
          Gmm        Gd3Mm       GMmd                                                                              
     F=       2
                   =            =                                                                                  d F
            d          R 3 d2       R3                                                                 x            
                                                                                                                   R/2
    So, Normal force exerted by the wall = F cos.
        GMmd R            GMm
    =                 =            (therefore I think normal force does not depend on x)
          R3       2d      2R 2
13. a) m is placed at a distance x from ‘O’.
    If r < x , 2r, Let’s consider a thin shell of man
                                                                                                                          R M
                   m      4      mx 3
     dm =                 x 3 = 3
             ( 4 / 3)r 2 3       r
                        mx 3
     Thus       dm =
                         r3
                                                                                                                         m
                                                                                                                              r
                                                                                                                              O
                                   G md m      Gmx 3 / r 3     Gmx
     Then gravitational force F =       2
                                            =       2
                                                           = 
                                      x           x             r3
     b) 2r < x < 2R, then F is due to only the sphere.
             Gmm
     F=
             x  r 2
     c) if x > 2R, then Gravitational force is due to both sphere & shell, then due to shell,
            GMm
     F=
           x  R 2
                               Gmm
     due to the sphere =
                               x  r 2
                                 Gm m           GMm
     So, Resultant force =                   +
                                 x  r 2       x  R 2
                                                               GM             GM
14. At P1, Gravitational field due to sphere M =                         =
                                                             3a  a 
                                                                     2
                                                                             16a 2                             a
     At P2, Gravitational field is due to sphere & shell,                                         49
                                    GM  1     1                                                   P1            a
          GM             GM                           61  GM
     =            2
                    +              = 2          =       2
     (a  4a  a)     ( 4a  a ) 2   a  36 25       900  a                                        P2          a

15. We know in the thin spherical shell of uniform density has gravitational field at its internal point is zero.
     At A and B point, field is equal and opposite and cancel each other so Net field is
                                                                                                                          A
     zero.                                                                                        A

     Hence, EA = EB                                                                                B

16. Let 0.1 kg man is x m from 2kg mass and (2 – x) m from 4 kg mass.                                                     B

      2  0 .1     4  0 .1
              =–
        x 2
                  ( 2  x )2


                                                                 11.4
                                                                                                     Chapter 11

          0.2       0 .4
     or       =–
          x 2
                 ( 2  x )2
          1         2               2      2
     or      =            or (2 – x) = 2 x
          x2   ( 2  x )2
    or 2 – x = 2 x or x(r2 + 1) = 2
                2
    or x =            = 0.83 m from 2kg mass.
             2.414
                                                                            m
17. Initially, the ride of  is a
    To increase it to 2a,                                               a       a
                            2        2           2
                Gm     Gm        3Gm
     work done =              =                                                m
                 2a      a         2a                            m       a                              100g
18. Work done against gravitational force to take away the particle from sphere,
                                                                                                             10cm
       G  10  0.1       6.67  10 11  1             –10
     =                =                      = 6.67 × 10 J
         0.1 0.1            1 10 1
    
19. E = (5 N/kg) ˆ + (12 N/kg) ˆ
                    i                  j
                                                                                                    10kg
    a) F = E m
    = 2kg [(5 N/kg) ˆ + (12 N/kg) ˆ ] = (10 N) ˆ + (12 N) ˆ
                        i                  j          i     j
     
    F = 100  576 = 26 N
            
    b) V = E r
                                            
    At (12 m, 0), V = – (60 J/kg) ˆ V = 60 J
                                         i
                                   
     At (0, 5 m), V = – (60 J/kg) ˆ V = – 60 J
                                  j
                (1,2,5 )

                      E mdr =  (10N)ˆi  (24N)ˆj r 
                                                     (12,5 )
     c)  V =                                          ( 0,0 )
                 ( 0,0 )

      = – (120 J ˆ + 120 J ˆ ) = 240 J
                 i          i
                                   
                                0,5m 
     d)  v = – r(10N ˆ  24Nˆ) 12m,0 
                      i       j

     = –120 ˆ + 120 ˆ = 0
            j       i
20. a) V = (20 N/kg) (x + y)
     GM         MLT 2      M1L3 T 2M1      ML2 T 2
           =           L or              =
      R           M               L             M
           0 2 –2       0 2 –2
    Or M L T = M L T
     L.H.S = R.H.S
        
    b) E( x, y ) = – 20(N/kg) ˆ – 20(N/kg) ˆ
                              i             j
               
    c) F = E m
    = 0.5kg [– (20 N/kg) ˆ – (20 N/kg) ˆ = – 10N ˆ - 10 N ˆ
                            i             j            i  j
         
     | F | = 100  100 = 10 2 N
     
21. E = 2 ˆ + 3 ˆ
             i      j
     The field is represented as
     tan 1 = 3/2
                                                                       5/3                                    3j
     Again the line 3y + 2x = 5 can be represented as                                 2
                                                                                                   
     tan 2 = – 2/3                                                                                     2j
                                                                               5/2
     m1 m2 = –1
     Since, the direction of field and the displacement are perpendicular, is done by the particle on the line.
                                                                 11.5
                                                                                           Chapter 11
22. Let the height be h
            GM        GM
    (1/2) 2 =
             R      (R  h)2
             2         2
    Or 2R = (R + h)
    Or 2 R = R + h
    Or h = (r2 – 1)R
23. Let g be the acceleration due to gravity on mount everest.
           2h 
    g = g1   
              R
              17696                                   2
    =9.8 1            = 9.8 (1 – 0.00276) = 9.773 m/s
          6400000 
24. Let g be the acceleration due to gravity in mine.
                  d
    Then g= g 1  
                R
                  640                                   2
    = 9.8 1             3
                              = 9.8 × 0.9999 = 9.799 m/s
            6400  10 
25. Let g be the acceleration due to gravity at equation & that of pole = g
               2
    g= g –  R
                        –5 2              3
    = 9.81 – (7.3 × 10 ) × 6400 × 10
    = 9.81 – 0.034
                 2
    = 9.776 m/s
                               2
    mg = 1 kg × 9.776 m/s
    = 9.776 N or 0.997 kg
    The body will weigh 0.997 kg at equator.
                             2
26. At equator, g = g –  R        …(1)
    Let at ‘h’ height above the south pole, the acceleration due to gravity is same.
                       2h 
    Then, here g = g 1                 …(2)
                          R
           2       2h 
     g -  R = g 1   
                      R
             2R      2h
    or 1        = 1
              g       R

    or h =
             2R 2
                   =
                                 
                                 2
                      7.3  10  5  6400  10 3  
                                                  2
                                                 = 11125 N = 10Km (approximately) 
               2g                 2  9.81
27. The apparent ‘g’ at equator becomes zero.
                   2
    i.e. g = g –  R = 0
              2
    or g =  R
                 g        9 .8
                                       1.5  10 6 = 1.2 × 10
                                                                –3
    or  =         =               =                                 rad/s.
                 R     6400  10 3
          2      2  3.14              –6
    T=        =               = 1.5 × 10 sec. = 1.41 hour
                1.2  10  3
28. a) Speed of the ship due to rotation of earth v = R
                               2
    b) T0 = mgr = mg – m R
                        2                                                                      To
     T0 – mg = m R
    c) If the ship shifts at speed ‘v’                                                 A            A
                    2
    T = mg – m R


                                                       11.6
                                                                                               Chapter 11

            v  R 2 
    = T0 -             R
                R2     
                       
            v 2  2R 2  2Rv 
    = T0 –                      m
                     R         
                               
     T = T0 + 2v m
29. According to Kepler’s laws of planetary motion,
     2    3
    T R
          2                  3
     Tm             R ms
                              
     Te 2           R es 3
                3                    2
      Rms          1.88 
     
     R          
                           
      es           1 
        R ms          2/3
            = (1.88) = 1.52
        R es

                     r3
30. T = 2
                     GM

    27.3 = 2 × 3.14
                                          3.84  10   5 3

                                     6.67  10 11  M

    or 2.73 × 2.73 =
                                                    
                                     2  3.14  3.84  10 5          
                                                                     3


                                         6.67  10 11  M
                    2  (3.14 )2  (3.84)3  1015            24
    or M =                          11      2
                                                  = 6.02 × 10 kg
                       3.335  10 (27.3 )
                                                                             24
     mass of earth is found to be 6.02 × 10                                      kg.
                         3
                     r
31. T = 2
                     GM

     27540 = 2 × 3.14
                                               9.4  10   3
                                                           103         3

                                                          11
                                                 6.67  10  M

    or (27540) = (6.28)
                         2                2      9.4  10    6 2

                                              6.67  10 11  M
                    (6.28)2  (9.4 )3  1018            23
    or M =                   11            2
                                              = 6.5 × 10 kg.
                    6.67  10  (27540 )
                     GM                   gr 2
32. a) V =                =
                     r h                r h
            9.8  ( 6400  10 3 )2                               3
    =                6
                                                 = 6.9 × 10 m/s = 6.9 km/s
               10  (6.4  2)
                                     2
    b) K.E. = (1/2) mv
                             6             10
    = (1/2) 1000 × (47.6 × 10 ) = 2.38 × 10 J
                GMm
    c) P.E. =
                (R  h)
              6.67  10 11  6  10 24  10 3    40  1013              10
     =–                                        =–           = – 4.76 × 10 J
                 (6400  2000 )  10 3             8400
                2(r  h)   2  3.14  8400  10 3            2
    d) T =                =                 3
                                                   = 76.6 × 10 sec = 2.1 hour
                   V               6.9  10

                                                                                        11.7
                                                                                                             Chapter 11
33. Angular speed f earth & the satellite will be same
    2    2
       =
    Te    Ts

             1                         1                                               (R  h)3
    or             =                                       or 12 I 3600 = 3.14
         24  3600                    (R  h)3                                           gR 2
                                 2
                                        gR 2

         (R  h)2   (12  3600 )2                                (6400  h)3  109      (12  3600 )2
    or            =                                        or                         =
           gR 2       (3.14 )2                                  9.8  (6400 )2  10 6     (3.14 )2
        (6400  h)3  10 9            4
    or               9
                           = 432 × 10
           6272  10
                  3                     4
    or (6400 + h) = 6272 × 432 × 10
                                    4 1/3
    or 6400 + h = (6272 × 432 × 10 )
                              4 1/3
    or h = (6272 × 432 × 10 ) – 6400
    = 42300 cm.
    b) Time taken from north pole to equator = (1/2) t
                                 ( 43200  6400)3                    ( 497 )3  10 6
    = (1/2) × 6.28                             2       6
                                                           = 3.14
                               10  (6400 )  10                      (64)2  1011
              497  497  497
    = 3.14                      = 6 hour.
               64  64  10 5
34. For geo stationary satellite,
                4
    r = 4.2 × 10 km
    h = 3.6 × 104 km
    Given mg = 10 N
              R2 
    mgh = mg           
              R  h2 
                       
         
    = 10 
                       
                 6400  103
                             2
                                                    
                                                      =
                                                          4096
                                                               = 0.23 N
              
          6400  10 3  3600  10 3
                                                  
                                                   2
                                                     
                                                     
                                                         17980

                   R 23
35. T = 2
                   gR12
                             3
                        R2
    Or T = 4
          2
                               2
                       gR1
                           3
                  42 R 2
    Or g =
                  T 2 R12
                                                                                  3
                                                                         42 R 2
     Acceleration due to gravity of the planet is =
                                                                         T 2 R12                        A
                                                                                                                      Colatitude
36. The colattitude is given by .                                                                          
    OAB = 90° – ABO
    Again OBC =  = OAB
                6400       8                                                                                             O
     sin  =          =
               42000      53                                                                                     
                                                                                                        B
                  –1    8     –1                                                                                        C
     = sin             = sin 0.15.                                                            
                        53 


                                                                         11.8
                                                                               Chapter 11
37. The particle attain maximum height = 6400 km.
    On earth’s surface, its P.E. & K.E.
                   2     GMm 
     Ee = (1/2) mv +                             …(1)
                        R 
     In space, its P.E. & K.E.
           GMm 
     Es =     +0
           Rh 
           GMm 
     Es =                 …(2)                  ( h = R)
              2R 
     Equating (1) & (2)
       GMm 1             GMm
              mv 2 = 
         R      2         2R
                2       1    1
     Or (1/2) mv = GMm      
                          2R R 
            2   GM
     Or v =
                 R
         6.67  10 11  6  10 24
     =
              6400  10 3
        40.02  1013
     =
         6.4  10 6
                7          8
     = 6.2 × 10 = 0.62 × 10
                                             4
    Or v = 0.62  10 8 = 0.79 × 10 m/s = 7.9 km/s.
38. Initial velocity of the particle = 15km/s
    Let its speed be ‘v’ at interstellar space.
                                       GMm
                                         
                          3 2   2
    (1/2) m[(15 × 10 ) – v ] =               dx
                                      R  x2
                                                       
                       3 2   2         1
      (1/2) m[(15 × 10 ) – v ] = GMm  
                                       x R
                               6     2       GMm
      (1/2) m[(225 × 10 ) – v ] =
                                              R
                    6   2  6.67  10 11  6  10 24
                         2
      225 × 10 – v =
                               6400  10 3
        2           6   40.02        8
     v = 225 × 10 –            × 10
                          32
        2           6           8       8
     v = 225 × 10 – 1.2 × 10 = 10 (1.05)
                    4
    Or v = 1.01 × 10 m/s or
       = 10 km/s
                                     24
39. The man of the sphere = 6 × 10 kg.
                             8
    Escape velocity = 3 × 10 m/s
                2GM
     Vc =
                 R
                2GM
     Or R =
                 Vc 2
         2  6.67  10 11  6  10 24       80.02     –3         –3
     =                                   =         × 10 = 8.89× 10 m  9 mm.
                  3  10 
                         8 2                   9

                                                           

                                                               11.9

				
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Description: HC VERMA SOLUTIONS PHYSICS