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**Page No. 134**

**Q.1.** **State the universal law of gravitation.****Ans.** Everybody in this universe attracts every other body with a force, which is directly proportional to the product of their masses and inversely proportional to the square of distance between their centres.**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 mass of earth = M,

mass of object = m.

If radius of object is comparatively negligible, r ≈ radius of earth = R

F = GMm/R^{2}**Page No. 136****Q.1. What do you mean by free fall?****Ans.** When an object falls with a constant acceleration, under the influence of force of gravitation of the earth, object is said to have free fall.**Q.2. What do you mean by acceleration due to gravity?****Ans.** During the course of its free fall, a body accelerates due to the force of gravity acting on it. This acceleration is known as acceleration due to gravity.**Page No. 138****Q.1. What are the differences between the mass of an object and its weight?**

**Ans.**

Mass | Weight |

The quantity of matter contained in a body is called the mass of the body. | The force with which the earth attracts a body towards its centre is the weight of the body. |

SI unit of mass is kilogram (kg). | SI unit of weight is newton (N). |

Mass of a body remains constant. | Weight of a body changes from place to place. |

Mass is a scalar quantity. | Weight is a vector quantity. |

Mass of a body is never zero. | Weight of a body at the centre of earth is zero. |

Mass is measured by a beam balance. | Weight is measured by a weighing machine or a spring balance. |

**Q.2. Why is the weight of an object on the moon 1/6th its weight on the earth?****Ans.** Mass of the moon (M) = 7.4 × 10^{22} kg

Radius of the moon (R) = 1.74 × 10^{6} m

Gravitational constant (G) = 6.7 × 10^{11} Nm^{2}kg^{2}

Acceleration due to gravity = on moon (g_{m})

g_{m} =

Gravitation =

g_{m} = 1.63 ms^{2}

g_{m} = g_{e}**Page No. 141****Q.1. ****Why is it difficult to hold a school bag having a strap made of a thin and strong string?Ans.** It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.

Ans.

When the distance between them is reduced to half, the gravitational force.

The gravitational force becomes four times.

Ans.

Ans.

Mass of the object, m = 1 kg, radius of the earth, R = 6.4 x 10

The magnitude of gravitational force between the earth and object is given by

Ans.

(i) the mass of one object is doubled?

(ii) the distance between the objects is doubled and tripled?

(iii) the masses of both objects are doubled?

According to universal law of gravitation F ∝ m

(i) If one mass, say m

(ii) If distance is doubled and F ∝ 1/r

If distance is tripled and F ∝ 1/r

(iii) If both masses are doubled and F ∝ m

(a) it explains about the force which holds us on earth.

(b) it explains the motion of planets around the sun.

(c) it explains the motion of moon around the earth.

(d) it explains about occurrence of tides in the ocean.

Ans.

Weight of the object on the earth’s surface = mg

The weight of object on moon is 1/6th that of object on earth, so the weight of the object on the moon

= 1/6 x 98 N= 16.3 N.

(i) the maximum height to which it rises.

(ii) the total time it takes to return to the surface of the earth.

Maximum height, H = ?

Acceleration due to gravity, g = + 9.8 ms

As object is moving upward, so a = -g

From relation

⇒

⇒ H = 122.5 m

(ii) When ball returns on the surface of earth, displacement, s = 0

∴

⇒ t (49 -4.9 t) = 0

t = 0 or 49 - 4.9 t = 0 ⇒

As t ≠ 0, so t = 10s

The maximum height is 122.5 m and total time is 10 s.

Final velocity,

= 19.6 ms

g = -10 ms

Final velocity, v = 0

During upward motion, g = -10 ms

Net displacement on returning back = zero

Total distance = 80 m + 80 m = 160 m

Force,

= 3.56 x 10

Let the stones meet at a height h above the ground. Stone 1 covers a distance (100 - h); g = 10 m/s

⇒ 100 - h = 5 t

u

Stone 2 covers a distance h

⇒ h = 25 t - 5 t

From (i) and (ii), we get

100 - 25 t + 5 t

⇒ 100 - 25 t = 0

or t = 4 s (Stones meet 4 s after they are thrown)

h = 25(4) - 5(4)

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

(c) its position after 4 s.

Ans.

⇒ Time taken to reach the top, t = 3 s.

Final velocity, v = 0, g = -9.8 ms

(a) Initial velocity, u = v - gt = 0 - (-9.8) (3)

= 29.4 ms

(b)

= 44.1 m

(c) After 4 s, ball has started falling and has fallen by some distance h' for 1 s.

Here, initial velocity u' = 0, t = 1 s,

g = +9.8 ms

The ball is at a height, (44.1 - 4.9) = 39.2 m above the ground.

Ans.

Here, density of the substance =

The density of the substance is more than the density of water (1 g cm

The density of the substance is more than the density of water (). Hence, it will sink in water.

The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 g.

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