NEET Exam > NEET Notes > Physics Class 11 > HC Verma Solutions: Chapter 4 - The Forces
HC Verma Solutions: Chapter 4 - The Forces | Physics Class 11 - NEET PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
* * * *
Read More
|
97 videos|378 docs|103 tests
|
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.
About this Document
|
7.8K Views |
|
4.80/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: HC Verma Solutions: Chapter 4 - The Forces for NEET 2024 is part of Physics Class 11 preparation.
The notes and questions for HC Verma Solutions: Chapter 4 - The Forces have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 4 - The Forces covers topics
like and HC Verma Solutions: Chapter 4 - The Forces Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 4 - The Forces.
Introduction of HC Verma Solutions: Chapter 4 - The Forces in English is available as part of our Physics Class 11
for NEET & HC Verma Solutions: Chapter 4 - The Forces in Hindi for Physics Class 11 course.
Download more important topics related with notes, lectures and mock test series for NEET
Exam by signing up for free. NEET: HC Verma Solutions: Chapter 4 - The Forces | Physics Class 11 - NEET
Description
Full syllabus notes, lecture & questions for HC Verma Solutions: Chapter 4 - The Forces | Physics Class 11 - NEET - NEET | Plus excerises question with solution to help you revise complete syllabus for Physics Class 11 | Best notes, free PDF download
Information about HC Verma Solutions: Chapter 4 - The Forces
In this doc you can find the meaning of HC Verma Solutions: Chapter 4 - The Forces defined & explained in the simplest way possible. Besides explaining types of
HC Verma Solutions: Chapter 4 - The Forces theory, EduRev gives you an ample number of questions to practice HC Verma Solutions: Chapter 4 - The Forces tests, examples and also practice NEET
tests
|
Explore Courses for NEET exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Semester Notes
,Viva Questions
,ppt
,video lectures
,Previous Year Questions with Solutions
,Extra Questions
,HC Verma Solutions: Chapter 4 - The Forces | Physics Class 11 - NEET
,Objective type Questions
,HC Verma Solutions: Chapter 4 - The Forces | Physics Class 11 - NEET
,MCQs
,study material
,Sample Paper
,HC Verma Solutions: Chapter 4 - The Forces | Physics Class 11 - NEET
,practice quizzes
,mock tests for examination
,Exam
,Summary
,shortcuts and tricks
,Free
,Important questions
,past year papers
;
Additional Information about HC Verma Solutions: Chapter 4 - The Forces for NEET Preparation
HC Verma Solutions: Chapter 4 - The Forces Free PDF Download
The HC Verma Solutions: Chapter 4 - The Forces is an invaluable resource that delves deep into the core of the NEET exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the HC Verma Solutions: Chapter 4 - The Forces now and kickstart your journey towards success in the NEET exam.
Importance of HC Verma Solutions: Chapter 4 - The Forces
The importance of HC Verma Solutions: Chapter 4 - The Forces cannot be overstated, especially for NEET aspirants.
This document holds the key to success in the NEET exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
HC Verma Solutions: Chapter 4 - The Forces Notes
HC Verma Solutions: Chapter 4 - The Forces Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to HC Verma Solutions: Chapter 4 - The Forces.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, HC Verma Solutions: Chapter 4 - The Forces Notes on EduRev are your ultimate resource for success.
HC Verma Solutions: Chapter 4 - The Forces NEET Questions
The "HC Verma Solutions: Chapter 4 - The Forces NEET Questions" guide is a valuable resource for all aspiring students preparing for the
NEET exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study HC Verma Solutions: Chapter 4 - The Forces on the App
Students of NEET can study HC Verma Solutions: Chapter 4 - The Forces alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the HC Verma Solutions: Chapter 4 - The Forces,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of HC Verma Solutions: Chapter 4 - The Forces is prepared as per the latest NEET syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup