NCERT Solutions Class 9 Maths Chapter 10 Gravitation – Here are all the NCERT solutions for Class 9 Maths Chapter 10. This solution contains questions, answers, images, explanations of the complete Chapter 10 titled The Fun They Had of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 10 Gravitation. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 10 Gravitation in one place.

## NCERT Solutions Class 9 Science Chapter 10 Gravitation

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Science for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 10 Gravitation , Science, Class 9.

Class | 9 |

Subject | Science |

Book | Science |

Chapter Number | 10 |

Chapter Name |
Gravitation |

### NCERT Solutions Class 9 Science chapter 10 Gravitation

Class 9, Science chapter 10, Gravitation solutions are given below in PDF format. You can view them online or download PDF file for future use.

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### Download NCERT Solutions Class 9 Science chapter 10 Gravitation In PDF Format

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### Question & Answer

Q.1:State the universal law of gravitation.

Ans :The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square Of the distance between their centers. For two objects of masses m_{1}and m_{2}and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as: \(F = \frac { \mathrm { G } m _ { 1 } m _ { 2 } } { r ^ { 2 } }\) Where, G is the universal gravitation constant given by: \(\mathrm { G } = 6.67 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } \mathrm { kg } ^ { - 2 }\)

Q.2:Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

Ans :Let M_{E}be the mass of the Earth and m be the mass of an object on its surface. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation: \(F = \frac { G m _ { 1 } m _ { 2 } } { r ^ { 2 } }\)

Q.3:What do you mean by free fall?

Ans :Gravity of the Earth attracts every object towards its centre. When an object is released from a height, it falls towards the surface of the Earth under the influence of gravitational force. The motion of the object is said to have free fall.

Q.4:What do you mean by acceleration due to gravity?

Ans :When an object falls towards the ground from a height, then its velocity changes during the fall. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s^{2}.

Q.5:Why is the weight of an object on the moon 1/6 th its weight on the earth?

Ans :Let M_{E}be the mass of the Earth and m be an object on the surface of the Earth. Let R_{E}be the radius of the Earth. According to the universal law of gravitation, weight W_{E}of the object on the surface of the Earth is given by, \(W _ { \mathrm { E } } = \frac { G M _ { \mathrm { E } } m } { R _ { \mathrm { E } } ^ { 2 } }\) Let M_{M}and R_{M}the mass and radius of the moon. Then, according to the universal law of gravitation, weight WN of the object on the surface of the moon is given by: \(W _ { \mathrm { M } } = \frac { \mathrm { G } M _ { \mathrm { M } } m } { R _ { \mathrm { M } } ^ { 2 } }\) \(\frac { W _ { \mathrm { M } } } { W _ { \mathrm { E } } } = \frac { M _ { \mathrm { M } } R _ { \mathrm { E } } ^ { 2 } } { M _ { \mathrm { E } } R _ { \mathrm { M } } ^ { 2 } }\) \(M _ { \mathrm { E } } = 5.98 \times 10 ^ { 24 } \mathrm { kg } , M _ { \mathrm { M } } = 7.36 \times 10 ^ { 22 } \mathrm { kg }\) \(R _ { \mathrm { E } } = 6.4 \times 10 ^ { 6 } \mathrm { m } , R _ { \mathrm { M } } = 1.74 \times 10 ^ { 6 } \mathrm { m }\) \(\therefore \frac { W _ { \mathrm { M } } } { W _ { \mathrm { E } } } = \frac { 7.36 \times 10 ^ { 22 } \times \left( 6.37 \times 10 ^ { 6 } \right) ^ { 2 } } { 5.98 \times 10 ^ { 24 } \times \left( 1.74 \times 10 ^ { 6 } \right) ^ { 2 } } = 0.165 \approx \frac { 1 } { 6 }\) Therefore, weight of an object on the moon is ⅙ of its weight on the Earth.

## NCERT / CBSE Book for Class 9 Science

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### All NCERT Solutions Class 9

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