Class 9 Science Chapter 9 Gravitation

Key Concept: Universal Law of Gravitation

a) $8.34 \times 10^{-10} \, \text{N}$

[Solution Description]

The gravitational force between two objects can be calculated using the formula:

$F = G \frac{m_1 \times m_2}{d^2}$

Here, $m_1 = 5 \, \text{kg}$, $m_2 = 10 \, \text{kg}$, $d = 2 \, \text{m}$, and $G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$. Plugging in the values:

$F = 6.67 \times 10^{-11} \times \frac{5 \times 10}{2^2}$

$F = 6.67 \times 10^{-11} \times \frac{50}{4}$

$F = 6.67 \times 10^{-11} \times 12.5$

$F = 8.3375 \times 10^{-10} \, \text{N}$

So, the gravitational force between the two objects is $8.34 \times 10^{-10} \, \text{N}$.

Key Concept: Gravitational Force

c) 49 N

[Solution Description]

To find the gravitational force acting on the stone, we use the formula:

$F = m \times g$

Here, $m = 5 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$.

Substituting the values:

$F = 5 \times 9.8 = 49 \, \text{N}.$

Therefore, the gravitational force acting on the stone is 49 N.

Key Concept: Archimedes’ Principle Application

d) 9,800,000 N

[Solution Description]

According to Archimedes’ principle, the buoyant force is equal to the weight of the fluid displaced. The mass of the displaced water can be calculated as: $\text{Mass} = \text{Volume} \times \text{Density} = 1000 \text{ m}^3 \times 1000 \text{ kg/m}^3 = 1,000,000 \text{ kg}$. The buoyant force is then calculated as: $\text{Buoyant Force} = \text{Mass} \times \text{Acceleration due to Gravity} = 1,000,000 \text{ kg} \times 9.8 \text{ m/s}^2 = 9,800,000 \text{ N}$.

Key Concept: Universal Law of Gravitation

c) The gravitational force exerted by the Earth keeps us bound to it.

[Solution Description]

The universal law of gravitation states that every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This gravitational force binds us to the Earth, as the Earth’s mass creates a strong gravitational pull on our bodies.

Key Concept: Gravitational Force

d) 49 N

[Solution Description]

The gravitational force acting on an object can be calculated using the formula $F = m g$, where $m$ is the mass of the object and $g$ is the acceleration due to gravity. Given that $m = 5 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$, we substitute these values into the formula:

$F = 5 \times 9.8 = 49 \, \text{N}$

Therefore, the gravitational force acting on the object is 49 N.

Key Concept: Equations of Motion

c) 20 m

[Solution Description]

To find the maximum height, we use the equation $v^2 = u^2 + 2as$. At the highest point, the final velocity $v = 0$, initial velocity $u = 20 \, \text{m/s}$, and acceleration $a = -g = -10 \, \text{m/s}^2$.

$0 = (20)^2 + 2(-10)s \\ \\ 0 = 400 – 20s \\ \\ 20s = 400 \\ \\ s = 20 \, \text{m}$

So, the maximum height reached by the ball is 20 meters.

Key Concept: Density, Upthrust

c) The wooden block has a lower density

[Solution Description]

The density of the wooden block is less than the density of water, so the upthrust exerted by water on the wooden block is greater than its weight, causing it to float. On the other hand, the density of the metal block is greater than the density of water, so the upthrust exerted by water on the metal block is less than its weight, causing it to sink.

Key Concept: Archimedes’ Principle Application

b) It displaces water equal to its weight

[Solution Description]

According to Archimedes’ principle, an object floats when the buoyant force acting on it is equal to the weight of the fluid it displaces. In the case of a ship, its design allows it to displace a large volume of water, creating a buoyant force that equals the ship’s weight. This enables the ship to float on the surface of the water.

Key Concept: Archimedes’ Principle, Buoyant Force, Density

b) $5,333.33 \, \text{kg/m}^3$

[Solution Description]

First, calculate the buoyant force acting on the metal block:

$\text{Buoyant Force} = (\text{Mass in air} – \text{Mass in liquid}) \times g = (10 \, \text{kg} – 8.5 \, \text{kg}) \times g = 1.5g \, \text{N}$

Next, calculate the volume of the metal block using the buoyant force:

$\text{Volume} = \frac{\text{Buoyant Force}}{\text{Density of liquid} \times g} = \frac{1.5g}{800 \times g} = 0.001875 \, \text{m}^3$

Finally, calculate the density of the metal:

$\text{Density of metal} = \frac{\text{Mass in air}}{\text{Volume}} = \frac{10 \, \text{kg}}{0.001875 \, \text{m}^3} = 5,333.33 \, \text{kg/m}^3$

Key Concept: Centripetal Force

c) Centripetal force

[Solution Description]

The force that causes an object to move in a circular path is called the centripetal force. This force acts towards the center of the circular path and keeps the object in its orbit.

Key Concept: Weight on Moon, Universal Law of Gravitation

b) 6.0 N

[Solution Description]

To find the weight of the object on the moon, we use the relationship between the weight on the moon and the weight on Earth. Given that the weight on the moon is $\frac{1}{6}$th the weight on Earth:

$W_m = \frac{W_e}{6} = \frac{36 \text{ N}}{6} = 6 \text{ N}$

Therefore, the weight of the object on the moon is 6 N.

Key Concept: Archimedes’ Principle, Buoyant Force

b) 5,022,500 N

[Solution Description]

To find the buoyant force, we use Archimedes’ principle: The buoyant force equals the weight of the fluid displaced by the object. The formula for buoyant force is $F_b = \rho \cdot V \cdot g$, where $\rho$ is the density of the fluid, $V$ is the volume of the fluid displaced, and $g$ is the acceleration due to gravity ($9.8 \text{ m/s}^2$). Substituting the given values:

$F_b = 1025 \cdot 500 \cdot 9.8 = 5,022,500 \text{ N}$

Key Concept: Density and Buoyancy

c) It will float on the surface

[Solution Description]

The density of the object ($0.8 \, \text{g/cm}^3$) is less than the density of water ($1.0 \, \text{g/cm}^3$). According to Archimedes’ principle, if the density of an object is less than the density of the liquid it is placed in, the object will float on the surface of the liquid.

Key Concept: Mass and Weight

b) 10 kg

[Solution Description]

The weight of an object is given by the formula $W = m \times g$. Rearranging the formula to find mass, we get $m = W / g$. Substituting the given values, $m = 98 / 9.8 = 10 \, \text{kg}$.

Key Concept: Weight on Earth

a) 49 N

[Solution Description]

We know that weight $W = m \times g$. Given, mass $m = 5$ kg and acceleration due to gravity $g = 9.8$ m/s$^2$. Substituting these values, we get:

$W = 5 \times 9.8 = 49 N$

Therefore, the weight of the object on the earth is 49 N.

Key Concept: Pressure in Fluids

b) Due to their weight

[Solution Description]

Fluids, like liquids and gases, have weight. This weight causes them to exert pressure on the base and walls of the container they are in. The pressure is transmitted equally in all directions within the fluid.

Key Concept: Equations of Motion

b) 10 m

[Solution Description]

At maximum height, the final velocity $v$ is 0. The initial velocity $u$ is 14 m/s, and the acceleration $a = -g = -9.8 \, \text{m s}^{-2}$. Using the equation $v^2 = u^2 + 2as$, we can solve for the distance $s$.

Plugging in the values:

$0 = (14 \, \text{m s}^{-1})^2 + 2 \times (-9.8 \, \text{m s}^{-2}) \times s$

$0 = 196 \, \text{m}^2 \text{s}^{-2} – 19.6 \, \text{m s}^{-2} \times s$

$19.6 \, \text{m s}^{-2} \times s = 196 \, \text{m}^2 \text{s}^{-2}$

$s = \frac{196 \, \text{m}^2 \text{s}^{-2}}{19.6 \, \text{m s}^{-2}} = 10 \, \text{m}$

Thus, the maximum height reached is 10 meters.

Key Concept: Archimedes’ Principle, Buoyancy Force, Floating and Sinking

c) $0.25$

[Solution Description] For the object to float, the buoyant force must equal the weight of the object. The weight of the object is $W = \rho_{\text{object}} V_{\text{object}} g = 0.9 \times 100 \times g = 90g$. The buoyant force is given by $F_{\text{buoyant}} = \rho_{\text{liquid}} V_{\text{submerged}} g$. At equilibrium, $W = F_{\text{buoyant}}$, so $90g = 1.2 \times V_{\text{submerged}} \times g$. Solving for $V_{\text{submerged}}$, we get $V_{\text{submerged}} = \frac{90}{1.2} = 75 \text{ cm}^3$. The fraction of the volume submerged is $\frac{75}{100} = 0.75$. Therefore, the fraction of the volume above the liquid surface is $1 – 0.75 = 0.25$.

Key Concept: Universal Law of Gravitation

a) $8.34 \times 10^{-10} \, \text{N}$

[Solution Description]

To find the gravitational force between two objects, we use the universal law of gravitation:

$F = G \frac{m_1 \times m_2}{d^2}$

Here, $m_1 = 5 \, \text{kg}$, $m_2 = 10 \, \text{kg}$, and $d = 2 \, \text{m}$. Plugging in the values:

$F = 6.67 \times 10^{-11} \times \frac{5 \times 10}{2^2}$

$F = 6.67 \times 10^{-11} \times \frac{50}{4}$

$F = 6.67 \times 10^{-11} \times 12.5$

$F = 8.3375 \times 10^{-10} \, \text{N}$

So, the gravitational force between the two objects is $8.34 \times 10^{-10} \, \text{N}$.

Key Concept: Buoyancy, Archimedes’ Principle

b) 39.2 N

[Solution Description]

To find the buoyant force, we use Archimedes’ principle. The buoyant force is equal to the weight of the fluid displaced by the object. The formula for buoyant force is:

$F_b = \rho_{\text{fluid}} \times V_{\text{displaced}} \times g$

Here, $\rho_{\text{fluid}}$ is the density of water (1000 kg/m$^3$), $V_{\text{displaced}}$ is the volume of the stone (0.004 m$^3$), and $g$ is the acceleration due to gravity (9.8 m/s$^2$). Substituting these values:

$F_b = 1000 \times 0.004 \times 9.8 = 39.2 \, \text{N}$

So, the buoyant force acting on the stone is 39.2 N.

Key Concept: Gravitational Constant

a) N m$^2$ kg$^{-2}$

[Solution Description]

The universal gravitational constant $G$ can be determined from the formula $F = G \frac{M \times m}{d^2}$. By substituting the units of force (N), mass (kg), and distance (m), we get:

$[G] = \frac{N \cdot m^2}{kg^2}$

Thus, the SI unit of $G$ is N m$^2$ kg$^{-2}$.

Key Concept: Weight of an Object

c) 98 N

[Solution Description]

The weight of an object is calculated using the formula $W = m \times g$
. Given the mass $m = 10 kg$ and acceleration due to gravity $g = 9.8 m s^{-2}$, we calculate the weight as follows: $W = 10 kg \times 9.8 m s^{-2} = 98 N$.

Key Concept: Thrust and Pressure

c) Pascal

[Solution Description]

The SI unit of pressure is Pascal (\$Pa\$). This was named in honour of the scientist Blaise Pascal. It is defined as \$1 Pa = 1 N/m^2\$.

Key Concept: Effect of air resistance

c) Both reach the ground simultaneously

[Solution Description]

In a vacuum, there is no air resistance. Both the stone and the paper will experience the same acceleration due to gravity ($g$) and will fall at the same rate.

Therefore, both objects will reach the ground simultaneously.

Key Concept: Archimedes’ Principle, Apparent Weight

b) $0.0020 \, \text{m}^3$

[Solution Description]

The apparent loss in weight when the block is submerged in water is due to the buoyant force acting on it. According to Archimedes’ principle, the buoyant force is equal to the weight of the fluid displaced.

The apparent weight is given by:

$W_{\text{apparent}} = W_{\text{air}} – F_{\text{buoyant}}$

Given $W_{\text{apparent}} = 30 \text{ N}$ and $W_{\text{air}} = 50 \text{ N}$, we can find the buoyant force:

$F_{\text{buoyant}} = W_{\text{air}} – W_{\text{apparent}} = 50 – 30 = 20 \text{ N}$

The buoyant force is also given by:

$F_{\text{buoyant}} = \rho_{\text{water}} \times g \times V$

where $\rho_{\text{water}} = 1000 \text{ kg/m}^3$ and $g = 9.81 \text{ m/s}^2$.

$20 = 1000 \times 9.81 \times V \implies V = \frac{20}{9810} \approx 0.00204 \text{ m}^3$

Key Concept: Universal Law of Gravitation

b) The force decreases by a factor of 4

[Solution Description]

According to the Universal Law of Gravitation, the force $F$ between two objects is inversely proportional to the square of the distance $d$ between them. Mathematically, this is expressed as $F \propto \frac{1}{d^2}$. If the distance is doubled ($2d$), the new force $F’$ will be $F’ \propto \frac{1}{(2d)^2} = \frac{1}{4d^2}$. Therefore, the force becomes one-fourth of its original value.

Key Concept: Effect of Radius on g

b) $2.45 \, \text{m s}^{-2}$

[Solution Description]

We know that $g = \frac{G M}{R^2}$. If the radius is doubled, the new radius $R’ = 2R$. The new value of $\textit{g}$ will be:

$g’ = \frac{G M}{(2R)^2} = \frac{G M}{4R^2} = \frac{g}{4}$

Substituting the original value of $g = 9.8 \, \text{m/s}^2$:

$g’ = \frac{9.8}{4} = 2.45 \, \text{m s}^{-2}$

Thus, the new value of $\textit{g}$ would be $2.45 \, \text{m s}^{-2}$.

Key Concept: Weight on Moon

b) 20 N

[Solution Description]

The weight of an object on the moon is approximately $\frac{1}{6}$ of its weight on Earth.

Therefore:

$W_m = \frac{W_e}{6} = \frac{120}{6} = 20 \, \text{N}$