SOLUTIONS TO CONCEPTS
CHAPTER – 4
1. m = 1 gm = 1/1000 kg F = 6.67 × 10–17 N F = 6.67 × 20 r =
2 –17
Gm1m2 r2
m1 = 1 gm r
m2 = 1 gm
6.67 10 11 (1/ 1000 ) (1/ 1000 ) = r2
6.67 10 11 10 6 10 17 17 = 1 6.64 10 17 10
2.
=
1 4 o
q1 q 2 1 9 10 9 2 2 r r
The force of attraction is equal to the weight Mg =
2
For example, Assuming m= 64 kg,
4.
mass = 50 kg r = 20 cm = 0.2 m
FG G
M
r=
3 10 4
3 10 4 = 3750 m 8 64
m1m2 6.67 10 11 2500 0.04 r2
Coulomb’s force Since, FG = Fc = q =
2
6.7 10 11 2500 6.7 10 9 25 0.04 9 10 9
= 18.07 × 10
–18 -9
q=
18.07 10 -18 = 4.3 × 10 C.
4.1
yC
r=
9 10 8 3 10 4 mt m m
q2 1 q1 q2 9 FC = = 9 × 10 0.04 4 o r 2
6.7 10 11 2500 9 10 9 q2 0.04 0.04
ol
r =
9 10 9 9 10 m 10 m
8
le
9 109 r2
ge
[Taking g=10 m/s ]
2
Ba
3.
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
g.
in
Chapter-4 5. The limb exerts a normal force 48 N and frictional force of 20 N. Resultant magnitude of the force, R = = = 6.
( 48 )2 (20)2
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 k=
F
x
F
7.
2R = (R + h) R – h – 2Rh = 0 2 2 h + 2Rh – R = 0
2
2
2
2
= –R ±
2R = R
8.
M
= 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 F2 = ? r2 = 25 cm 1 q1q2 1 Since, F = , F 2 4 o r 2 r
yC
F1 r2 F 2 = F1 × F2 r12
9.
2
r1 16 64 20 = 20 × = 20 × = = 12.8 N = 13 N. r 25 5 25 ...