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 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
* * * *

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FAQs on HC Verma Solutions: Chapter 4 - The Forces - Physics Class 11 - NEET

1. What are the different types of forces?

Ans. The different types of forces include gravitational force, electromagnetic force, nuclear force, frictional force, and applied force.

2. How is gravitational force calculated?

Ans. Gravitational force can be calculated using Newton's law of universal gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

3. What is the difference between mass and weight?

Ans. Mass is a measure of the amount of matter in an object and remains constant regardless of the location, while weight is the force exerted on an object due to gravity and can vary depending on the location.

4. How does frictional force affect motion?

Ans. Frictional force opposes the motion of an object and can either slow it down or bring it to a stop. It can also cause objects to heat up due to the energy dissipated during friction.

5. What is the principle of superposition of forces?

Ans. The principle of superposition of forces states that the net force acting on an object is the vector sum of all the individual forces acting on it. This principle is based on the fact that forces are vector quantities and can be added or subtracted based on their direction and magnitude.

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