# 11.SOLUTIONS TO CONCEPTS

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```					                                                    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