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.
NF=
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
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4.3
10 7