# 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
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

****

4.3

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Description: HC VERMA SOLUTIONS PHYSICS