04.SOLUTIONS TO CONCEPTS

Page 1

SOLUTIONS TO CONCEPTS CHAPTER – 4 1.

m = 1 gm = 1/1000 kg F = 6.67 × 10

–17

–17

 6.67 × 20 2

r =

2.

3.

NF=

m1 = 1 gm

Gm1m2 r2

r

6.67 10 11  (1/ 1000 )  (1/ 1000 ) = r2

6.67 10 11 10 6 10 17  17 = 1 6.64 10 17 10

 r = 1 = 1 metre. So, the separation between the particles is 1 m. 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 The force of attraction between the two charges =

1 4  o

q1 q 2 1  9  10 9 2 2 r r

The force of attraction is equal to the weight Mg =

9 109 r2

r =

9  10 9 9  10  m  10 m

r=

9 10 8 3 10 4 mt  m m

2

8 2

[Taking g=10 m/s ]

For example, Assuming m= 64 kg, r= 4.

3  10 4

3  10 4 = 3750 m 8 64

mass = 50 kg r = 20 cm = 0.2 m

FG  G

m1m2 6.67  10 11  2500  0.04 r2

q2 1 q1 q2 9 FC = = 9 × 10 0.04 4 o r 2

Coulomb’s force Since, FG = Fc = 2

q =

6.7 10 11  2500 9 10 9  q2  0.04 0.04

6.7 10 11  2500 6.7 10 9   25 0.04 9 10 9 = 18.07 × 10

q=

m2 = 1 gm

–18 -9

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,

( 48 )2  (20)2

R =

6.

=

2304  400

=

2704

48N

= 52 N The body builder exerts a force = 150 N. Compression x = 20 cm = 0.2 m Total force exerted by the man = f = kx  kx = 150

F

150 1500 = = 750 N/m 0 .2 2 Suppose the height is h. 2 At earth station F = GMm/R M = mass of earth m = mass of satellite R = Radius of earth k=

7.

F=

GMm GMm = 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     =  2R  2 2R H=  2 2 = –R ±

8.

2R = R

 2  1

= 6400 × (0.414) = 2649.6 = 2650 km Two charged particle placed at a sehortion 2m. exert a force of 20m. F1 = 20 N. r1 = 20 cm r2 = 25 cm F2 = ? Since, F =

1 q1q2 , 4 o r 2

2

F1 r2  F 2 = F1 ×  F2 r12 9.

F

1 r2

2

2

 r1  16 64  20    = 20 ×   = 20 × = = 12.8 N = 13 N. 25 5  25   r2 

The force between the earth and the moon, F= G F=

6.67 10 11  7.36 10 22  6 10 24

3.8 10 

8 2

19

20

=

= 20.3 × 10 =2.03 × 10 N = 2 ×10 –19 10. Charge on proton = 1.6 × 10  Felectrical =

20

mmmc

6.67  7.36  10 35

3.8 2 1016

N

2 1 qq 9 109  1.6  10 38  12 2 = 2 4 o r r

mass of proton = 1.732 × 10

–27

r2

kg

4.2

x

F


Chapter-4 Fgravity = G

6 .67  10 m1m2 = 2 r

 11

 1 .732  10  54 r2

9 10 9  1.6  10 38 2 Fe 9  1.6 10 29 36 r2 = = 1.24 × 10  2 Fg 6.67 10 11  1.732 10  54 6.67 1.732 10  65 r2 –11 11. The average separation between proton and electron of Hydrogen atom is r= 5.3 10 m. 2

9

a) Coulomb’s force = F = 9 × 10 ×

9 109  1.0 10 19 q1q2 = 2 r2 5.3  10 11

2 –8

= 8.2 × 10

N.

b) When the average distance between proton and electron becomes 4 times that of its ground state Coulomb’s force F =

1 qq 9 10 9  1.6 10 19  1 22 = 2 4 o 4r  16  5.3 10  22 –7

2

9  1.6 

2

=

16  5.3 

2

–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

 10  7


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