HC Verma Solutions: The Forces (Chapter 4) - Notes | Study Physics Class 11 - NEET

 Page 1


4.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 4
1. m = 1 gm = 1/1000 kg
F = 6.67 × 10
–17
N ? F = 
2
2 1
r
m Gm
? 6.67 × 20
–17 
= 
2
11
r
) 1000 / 1 ( ) 1000 / 1 ( 10 67 . 6 ? ? ?
?
? r
2
= 
17
17
17
6 11
10
10
10 64 . 6
10 10 67 . 6
?
?
?
? ?
?
?
? ?
= 1
? 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
And g = acceleration due to gravity on the surface of earth = 10 m/s
2
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 
= 
2
9
2
2 1
o
r
1
10 9
r
q q
4
1
? ?
??
The force of attraction is equal to the weight 
Mg = 
2
9
r
10 9 ?
? r
2
= 
m
10 9
10 m
10 9
8 9
?
?
?
?
    [Taking g=10 m/s
2
]
? r = 
m
10 3
m
10 9
4 8
?
?
?
mt
For example, Assuming m= 64 kg, 
r = 
4
4
10
8
3
64
10 3
?
?
= 3750 m
4. mass = 50 kg
r = 20 cm = 0.2 m
04 . 0
2500 10 67 . 6
r
m m
G F
11
2
2 1
G
? ?
? ?
?
Coulomb’s force         F
C
= 
2
2 1
o
r
q q
4
1
??
= 9 × 10
9 
04 . 0
q
2
Since, F
G
= F
c
= 
04 . 0
q 10 9
04 . 0
2500 10 7 . 6
2 9 11
? ?
?
? ?
?
? q
2
=  25
10 9
10 7 . 6
04 . 0
2500 10 7 . 6
9
9 11
?
?
?
?
? ?
? ?
= 18.07 × 10
–18
q = 
-18
10 18.07 ? = 4.3 × 10
-9 
C.
m 1 = 1 gm m 2 = 1 gm
r
Page 2


4.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 4
1. m = 1 gm = 1/1000 kg
F = 6.67 × 10
–17
N ? F = 
2
2 1
r
m Gm
? 6.67 × 20
–17 
= 
2
11
r
) 1000 / 1 ( ) 1000 / 1 ( 10 67 . 6 ? ? ?
?
? r
2
= 
17
17
17
6 11
10
10
10 64 . 6
10 10 67 . 6
?
?
?
? ?
?
?
? ?
= 1
? 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
And g = acceleration due to gravity on the surface of earth = 10 m/s
2
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 
= 
2
9
2
2 1
o
r
1
10 9
r
q q
4
1
? ?
??
The force of attraction is equal to the weight 
Mg = 
2
9
r
10 9 ?
? r
2
= 
m
10 9
10 m
10 9
8 9
?
?
?
?
    [Taking g=10 m/s
2
]
? r = 
m
10 3
m
10 9
4 8
?
?
?
mt
For example, Assuming m= 64 kg, 
r = 
4
4
10
8
3
64
10 3
?
?
= 3750 m
4. mass = 50 kg
r = 20 cm = 0.2 m
04 . 0
2500 10 67 . 6
r
m m
G F
11
2
2 1
G
? ?
? ?
?
Coulomb’s force         F
C
= 
2
2 1
o
r
q q
4
1
??
= 9 × 10
9 
04 . 0
q
2
Since, F
G
= F
c
= 
04 . 0
q 10 9
04 . 0
2500 10 7 . 6
2 9 11
? ?
?
? ?
?
? q
2
=  25
10 9
10 7 . 6
04 . 0
2500 10 7 . 6
9
9 11
?
?
?
?
? ?
? ?
= 18.07 × 10
–18
q = 
-18
10 18.07 ? = 4.3 × 10
-9 
C.
m 1 = 1 gm m 2 = 1 gm
r
Chapter-4
4.2
5. The limb exerts a normal force 48 N and frictional force of 20 N. Resultant magnitude of 
the force, 
R = 
2 2
) 20 ( ) 48 ( ?
= 400 2304 ?
= 2704
= 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 
? kx = 150
? k = 
2 . 0
150
=
2
1500
= 750 N/m
7. Suppose the height is h.
At earth station F = GMm/R
2
M = mass of earth
m = mass of satellite
R = Radius of earth
F= 
2
) h R (
GMm
?
= 
2
R 2
GMm
? 2R
2
= (R + h)
2
? R
2
– h
2
– 2Rh = 0
? h
2
+ 2Rh – R
2
= 0
H = 
2
R 4 R 4 R 2
2 2
?
?
?
?
?
?
? ? ?
= 
2
R 2 2 R 2 ? ?
    = –R ± R 2 = R ? ? 1 2 ?
    = 6400 × (0.414)
    = 2649.6 = 2650 km
8. Two charged particle placed at a sehortion 2m. exert a force of 20m.
F
1
= 20 N. r
1
= 20 cm
F
2
= ? r
2
= 25 cm
Since, F = 
2
2 1
o
r
q q
4
1
??
, F ?
2
r
1
2
1
2
2
2
1
r
r
F
F
? ? F
2
= F
1 
× 
2
2
1
r
r
?
?
?
?
?
?
?
?
= 20 × 
2
25
20
?
?
?
?
?
?
= 20 × 
25
16
= 
5
64
= 12.8 N = 13 N.
9. The force between the earth and the moon, F= G 
2
c m
r
m m
F = 
? ?
2
8
24 22 11
10 8 . 3
10 6 10 36 . 7 10 67 . 6
?
? ? ? ? ?
?
= 
? ?
16 2
35
10 8 . 3
10 36 . 7 67 . 6
?
? ?
    = 20.3 × 10
19
=2.03 × 10
20
N = 2 ×10
20
N
10. Charge on proton = 1.6 × 10
–19
? F
electrical
= 
2
2 1
o
r
q q
4
1
?
??
= 
? ?
2
38 2 9
r
10 6 . 1 10 9
?
? ? ?
mass of proton = 1.732 × 10
–27
kg
48N
x
F
F
Page 3


4.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 4
1. m = 1 gm = 1/1000 kg
F = 6.67 × 10
–17
N ? F = 
2
2 1
r
m Gm
? 6.67 × 20
–17 
= 
2
11
r
) 1000 / 1 ( ) 1000 / 1 ( 10 67 . 6 ? ? ?
?
? r
2
= 
17
17
17
6 11
10
10
10 64 . 6
10 10 67 . 6
?
?
?
? ?
?
?
? ?
= 1
? 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
And g = acceleration due to gravity on the surface of earth = 10 m/s
2
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 
= 
2
9
2
2 1
o
r
1
10 9
r
q q
4
1
? ?
??
The force of attraction is equal to the weight 
Mg = 
2
9
r
10 9 ?
? r
2
= 
m
10 9
10 m
10 9
8 9
?
?
?
?
    [Taking g=10 m/s
2
]
? r = 
m
10 3
m
10 9
4 8
?
?
?
mt
For example, Assuming m= 64 kg, 
r = 
4
4
10
8
3
64
10 3
?
?
= 3750 m
4. mass = 50 kg
r = 20 cm = 0.2 m
04 . 0
2500 10 67 . 6
r
m m
G F
11
2
2 1
G
? ?
? ?
?
Coulomb’s force         F
C
= 
2
2 1
o
r
q q
4
1
??
= 9 × 10
9 
04 . 0
q
2
Since, F
G
= F
c
= 
04 . 0
q 10 9
04 . 0
2500 10 7 . 6
2 9 11
? ?
?
? ?
?
? q
2
=  25
10 9
10 7 . 6
04 . 0
2500 10 7 . 6
9
9 11
?
?
?
?
? ?
? ?
= 18.07 × 10
–18
q = 
-18
10 18.07 ? = 4.3 × 10
-9 
C.
m 1 = 1 gm m 2 = 1 gm
r
Chapter-4
4.2
5. The limb exerts a normal force 48 N and frictional force of 20 N. Resultant magnitude of 
the force, 
R = 
2 2
) 20 ( ) 48 ( ?
= 400 2304 ?
= 2704
= 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 
? kx = 150
? k = 
2 . 0
150
=
2
1500
= 750 N/m
7. Suppose the height is h.
At earth station F = GMm/R
2
M = mass of earth
m = mass of satellite
R = Radius of earth
F= 
2
) h R (
GMm
?
= 
2
R 2
GMm
? 2R
2
= (R + h)
2
? R
2
– h
2
– 2Rh = 0
? h
2
+ 2Rh – R
2
= 0
H = 
2
R 4 R 4 R 2
2 2
?
?
?
?
?
?
? ? ?
= 
2
R 2 2 R 2 ? ?
    = –R ± R 2 = R ? ? 1 2 ?
    = 6400 × (0.414)
    = 2649.6 = 2650 km
8. Two charged particle placed at a sehortion 2m. exert a force of 20m.
F
1
= 20 N. r
1
= 20 cm
F
2
= ? r
2
= 25 cm
Since, F = 
2
2 1
o
r
q q
4
1
??
, F ?
2
r
1
2
1
2
2
2
1
r
r
F
F
? ? F
2
= F
1 
× 
2
2
1
r
r
?
?
?
?
?
?
?
?
= 20 × 
2
25
20
?
?
?
?
?
?
= 20 × 
25
16
= 
5
64
= 12.8 N = 13 N.
9. The force between the earth and the moon, F= G 
2
c m
r
m m
F = 
? ?
2
8
24 22 11
10 8 . 3
10 6 10 36 . 7 10 67 . 6
?
? ? ? ? ?
?
= 
? ?
16 2
35
10 8 . 3
10 36 . 7 67 . 6
?
? ?
    = 20.3 × 10
19
=2.03 × 10
20
N = 2 ×10
20
N
10. Charge on proton = 1.6 × 10
–19
? F
electrical
= 
2
2 1
o
r
q q
4
1
?
??
= 
? ?
2
38 2 9
r
10 6 . 1 10 9
?
? ? ?
mass of proton = 1.732 × 10
–27
kg
48N
x
F
F
Chapter-4
4.3
F
gravity 
= 
2
2 1
r
m m
G = 
? ?
2
54 11
r
10 732 . 1 10 67 . 6
? ?
? ? ?
? ?
? ?
2
54 11
2
38 2 9
g
e
r
10 732 . 1 10 67 . 6
r
10 6 . 1 10 9
F
F
? ?
?
? ? ?
? ? ?
? = 
? ?
? ?
65 2
29 2
10 732 . 1 67 . 6
10 6 . 1 9
?
?
? ?
= 1.24 × 10
36
11. The average separation between proton and electron of Hydrogen atom is r= 5.3 10
–11
m.
a) Coulomb’s force = F = 9 × 10
9
× 
2
2 1
r
q q
= 
? ?
? ?
2
11
2
19 9
10 3 . 5
10 0 . 1 10 9
?
?
?
? ? ?
= 8.2 × 10
–8
N.
b) When the average distance between proton and electron becomes 4 times that of its ground state
Coulomb’s force F = 
? ?
2
2 1
o r 4
q q
4
1
?
??
= 
? ?
? ?
22 2
2
19 9
10 3 . 5 16
10 6 . 1 10 9
?
?
? ?
? ? ?
= 
? ?
? ?
7
2
2
10
3 . 5 16
6 . 1 9
?
?
?
?
= 0.0512 × 10
–7
= 5.1 × 10
–9
N.
12. The geostationary orbit of earth is at a distance of about 36000km.
We know that, g ? = GM / (R+h)
2
At h = 36000 km. g ? = GM / (36000+6400)
2
? 0227 . 0
106 106
256
42400 42400
6400 6400
g
` g
?
?
?
?
?
?
? g ? = 0.0227 × 9.8 = 0.223
[ taking g = 9.8 m/s
2
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
* * * *