Class 9 Science Notes For Chapter 9 Gravitation
It has been observed that an object dropped from a height falls towards the earth. Newton generalised this idea and said that not only the earth but every object in the universe attracts every other object.
This force of attraction between two objects is called the force of gravitation or gravitational force. In this chapter, we shall learn about gravitation and the universal law of gravitation.
Gravitation
Gravitation is defined as the force of attraction between any two bodies in the universe. The earth attracts (or pulls) all objects lying on or near its surface towards its centre. The force with which the earth pulls the objects towards its centre is called the gravitational force of the earth or gravity of the earth.
Universal Law of Gravitation
The universal law of gravitation was given by Isaac Newton. According to this law, the attractive force between any two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
The direction of the force is along the line joining the centres of two objects. Consider two bodies A and B having masses m1 and m2, whose centres are at a distance d from each other.
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The gravitational force between two objects is directed along the line joining their centres Then, the force between two bodies is directly proportional to the product of their masses,
⇒ \(F \propto m_1 m_2\) → Equation 1
and the force between two bodies is inversely proportional to the square of the distance between them, i.e.
F \(\propto \frac{1}{d^2}\)
F \(\propto \frac{m_1 m_2}{d^2}\) or F=G\( \frac{m_1 m_2}{d^2}\)
where, G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\) is called the universal
Combining Eqs. (1) and (2), we get
where G= 6.67 x 10_11N-m2/kg² is called the universal gravitational constant.
Its value does not depend on the medium between the two bodies and the masses of the bodies or the distance between them. Suppose the masses of two bodies are 1 kg each and the distance d between them is 1 m, then
F = G [m1 = m2 = 1 kg and d = m]
Hence, the universal gravitational constant is defined as the gravitational force between two objects of unit masses separated by a unit distance from each other placed anywhere in space. The SI unit of G is N-m2 kg. The value of G was found out by Henry Cavendish (173T1810) by using a sensitive balance.
Importance of Universal Law of Gravitation
- The universal law of gravitation successfully explained several phenomena given below:
- The force that binds us to the earth.
- The motion of the moon around the earth.
- The motion of planets around the sun.
- The occurrence of tides is due to the gravitational force of attraction of the moon.
- The flow of water in rivers is also due to the gravitational force of the earth on water.
Motion of Moon around Earth and Centripetal Force
The force that keeps a body moving along the circular path acting towards the centre is called centripetal (centre-seeking) force. The motion of the moon around the Earth is due to the centripetal force. The centripetal force is provided by the gravitational force of attraction of the earth. If there were no such force, then the moon would pursue a uniform straight-line motion.
Example 1. Find the gravitational force between the earth and an object of 2 kg mass placed on its surface. (Givenr mass of the earth = 6 x \(10^24\) kg and radius of the earth = 6.4 x \(10^6\) m)
Answer:
Given, mass of the earth, \(m_e=6 \times 10^{24} \mathrm{~kg}\)
Mass of an object, \(m_o=2 \mathrm{~kg}\)
Radius of the earth, R=6.4 \(\times 10^6 \mathrm{~m}\)
Gravitational force, F=?
F=G \(\frac{m_e m_o}{d^2}\) [by universal law of gravitation]
F=\(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2}{\left(6.4 \times 10^6\right)^2}=19.5 \mathrm{~N}\)
Example 2. The mass of Mars is 639 x 1023 kg and that of Jupiter is 189 x 1027 kg. If the distance between Mars and Jupiter is 749 x 123 m. Calculate the force exerted by the Jupiter on the mars. (G =6.7 x 10~n Nm2 kg2)
Answer:
Given, the mass of the mars \(M_m=639 \times 10^{23} \mathrm{~kg}\)
The mass of the Jupiter \(M_j=1.89 \times 10^{27} \mathrm{~kg}\)
The distance between Mars and Jupiter
d=7.49 \(\times 10^5 \mathrm{~m}\)
G=6.7 \(\times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2\)
The force exerted by Jupiter on Mars
F =G \(\frac{M_m \times M_j}{d^2}\)
= \(\frac{6.7 \times 10^{-11} \times 639 \times 10^{23} \times 1.89 \times 10^{27}}{\left(7.49 \times 10^5\right)^2}\)
=1.44 \(\times 10^{29} \mathrm{~N}\)
Thus, the force exerted by the Jupiter on the mars is 1.44 \(\times 10^{29} N\).
Free Fall
When objects fall towards the earth under the influence of the earth’s gravitational force alone, then these are called freely falling objects and such a motion is called free fall.
Acceleration due to Gravity [g]
Whenever an object falls towards the earth, acceleration is involved. This acceleration is due to the earth’s gravitational pull and is called acceleration due to gravity. It is denoted by g.
The SI unit of is the same as that of acceleration, i.e. m/s². Let the mass of the earth be M and an object falling freely towards it be m. The distance between the centres of the earth and the object is R.
From Newton’s law of gravitation,
Also, from the second law of motion, the force exerted on an object, F =ma
Since, a = g (i.e. acceleration due to gravity)
F = mg Equation (2)
Equating RHS of Equation. (1) and (2), we get GMm
m g=\(\frac{G M m}{R^2}\) or g=\(\frac{G M}{R^2}\)
From the formula, it is clear that acceleration due to gravity does not depend on the mass of a falling object. It depends only on the mass of the earth or celestial bodies.
Equations of Motion for Free Fall
The three equations of motion which we have derived earlier are for bodies under uniform acceleration. In the case of motion of bodies under free fall, there is a uniform acceleration, i.e. acceleration due to gravity (g) acting downward.
So, the previous three equations of motion can be applied to the motion of bodies under free fall as follows:
General equations Equations for body under free fall of motion
where h is the height from which the object falls, t is the time of fall, u is the initial velocity and v is the final velocity when the body accelerates at g.
In solving numerical problems, we should remember the following points:
- If an object falls vertically downwards, then acceleration due to gravity is taken as positive, since its velocity increases while falling.
- If an object is thrown vertically upwards, then acceleration due to gravity is taken as negative, since its velocity decreases as it moves upward.
Example 3. A car falls off a ledge and drops to the ground in 0.6 seconds. The value of g is 10 m/s2 (for simplifying the calculation).
- What is its speed on striking the ground?
- What is its average speed during the 0.6 s?
- How high is the ledge from the ground?
Answer:
Initial velocity u = 0
Acceleration due to gravity g=10 \(\mathrm{~m} / \mathrm{s}^2\)
(1) Speed v=u+g t
v=0+10 \(\times\) 0.6
v=6 \(\mathrm{~m} / \mathrm{s}\)
(2) Average speed =\(\frac{u+v}{2}=\left(\frac{0+6}{2}\right)=3.0 \mathrm{~m} / \mathrm{s}\)
(3) Distance travelled h=u t+\(\frac{1}{2} g t^2\)
h=0+\(\frac{1}{2} g t^2\)
h=\(\frac{1}{2} \times 10 \times(0.6)^2=5 \times 0.36=1.80 \mathrm{~m}\)
Example 4. An object is thrown vertically upwards and rises to a height of 13.07 m. Calculate
- The velocity with, which the object was thrown upwards
- The Time taken by the object to reach the highest point.
Answer:
Distance travelled, h =13.07 m
Final velocity v = 0
Acceleration due to gravity g=-9.8 \(\mathrm{~m} / \mathrm{s}^2\) (upward motion)
(1) \(v^2=u^2+2 g h\)
0=\(u^2+2 \times(-9.8) \times 13.07\)
⇒ \(u^2=256 \Rightarrow u=16 \mathrm{~m} / \mathrm{s}\)
(2) v=u+a t
0=16-9.8 \(\times t \Rightarrow t=1.63 \mathrm{~s}\)
Example 5. A ball is thrown vertically upwards with a velocity of 25 m/s. If g is 10 m/s², then calculate
- the height it reaches.
- time taken to return back.
Answer:
Given, initial velocity, u = 25 m/s, final velocity, v =0
If a body is thrown upwards, then its velocity becomes zero at the highest point, where it reaches, acceleration due to gravity,
g=-10 \(\mathrm{~m} / \mathrm{s}^2\)
(1) Height, h =\(\frac{v^2-u^2}{2 g}\)
=\(\frac{0-(25)^2}{2(-10)}=\frac{-625}{-20}=31.25 \mathrm{~m}\)
Height, h=\(\frac{v^2-u^2}{2 g}\)
(2) Time, t=\(\frac{v-u}{g}=\frac{0-25}{-10}=2.5 \mathrm{~s}\)
Time taken to return back,
T = time of ascent + time of descent =21 Time taken to return back, T = 2×2.5 = 5s
Weight
The weight of an object is the force with which it is attracted towards the earth.
Weight of an object, w = mg
where, m = mass, g = acceleration due to gravity or w=\(\frac{G M m}{R^2}\)
Here, M = mass of the earth and R = radius of the earth
Important points regarding weight are as follows:
- Weight is a vector quantity, it acts in a vertically downward direction, and its SI unit is newton (N). Weight of 1 kg mass is 9.8 N. (i.e. 1 kg-wt =9.8 N)
- The weight of an object is not constant, it changes from place to place.
- In the space, where g = 0, the weight of an object is zero.
- At the centre of the earth, weight becomes zero. This is due to the fact that on going down to the earth, the value of g decreases and at the centre of the earth, g = 0.
Weight of an Object on the Moon
Let the mass of an object be m and its weight on the moon be wm. Suppose the mass of the moon is Mm and its radius is Rm. According to the universal law of gravitation, the weight of an object on the moon will be
⇒ \(w_m=G \frac{M_m \times m}{R_m^2}\)
Let the weight of the same object on the earth be we. Let the mass of the earth be Me and the radius of the earth be Re.
⇒ \(\frac{w_m}{w_e} =\frac{\frac{G M_m m}{R_m^2}}{\frac{G M_e m}{R_e^2}}=\frac{M_m}{M_e} \times \frac{R_e^2}{R_m^2}\)
Now, \(M_m =5.98 \times 10^{24} \mathrm{~kg} ; M_e=736 \times 10^{22} \mathrm{~kg}\)
⇒ \(R_m =1.74 \times 10^6 ; R_e=6.37 \times 10^6 \)
⇒ \(\frac{w_m}{w_e} =\frac{5.98 \times 10^{24}}{736 \times 10^{22}} \times \frac{\left(637 \times 10^6\right)^2}{\left(1.74 \times 10^6\right)^2} \approx \frac{1}{6}\)
Thus, the weight of an object on the moon is one-sixth of its weight on the Earth.
Example 6. The mass of an object is 12 kg. Calculate
- Its weight on the earth.
- Its weight on the moon.
Answer:
Given, m= 12 kg
(1) Acceleration due to gravity on earth, ge =9.8 m/s² weight earth we = mg =12 x 9.8 = 117.6 N
(2) Acceleration due to gravity on the moon \(g_m=\frac{g_e}{6}\)
⇒ \(g_m=\frac{9.8}{6} \mathrm{~m} / \mathrm{s}^2\)
Weight on moon \(w_m=m g_m=12 \times \frac{9.8}{6}=9.8 \times 2=19.6 \mathrm{~N}\)
Example 7. A man weighs 600 N on the earth. What is his mass, if g is 10 m/s²? On the moon, his weight would be 100 N. What is the acceleration due to gravity on the moon?
Answer:
Given, the weight of a man on the earth, \(w_e=600 \mathrm{~N}\)
Acceleration due to gravity on the earth, \(g_e=10 \mathrm{~m} / \mathrm{s}^2\)
Weight of man on the moon, \(w_m=100 \mathrm{~N}\)
Acceleration due to gravity on the moon, \(g_m\)=?
As we know, mass of the man, m=\(\frac{w_e}{g_e}=\frac{600}{10}=60 \mathrm{~kg}\)
⇒ \(g_e=\frac{w_e}{m}\)
Similarly, for the moon \(g_m=\frac{w_m}{m}\)
⇒ \(g_m=\frac{100}{60}=1.66 \mathrm{~m} / \mathrm{s}^2\)
Thus, acceleration due to gravity on the moon is 1.66 \(\mathrm{~m} / \mathrm{s}^2\), i.e. \(g_m=\frac{g_e}{6}\).
Example 8. A particle weighs 120 N on the surface of the earth. At what height above the earth’s surface will its weight be 30 N? Radius of the earth = 6400 km.
Answer:
The weight of a particle on the surface of the earth is
w=m g=\(\frac{m M G}{R^2} [g=\frac{G M}{R^2}]\)
Let \(w_1\) be the weight of a particle at height h above the earth’s surface.
So, \(\frac{w}{w_1} =\frac{G \frac{M}{R^2}}{G \frac{M}{(R+h)^2}}=\frac{(R+h)^2}{R^2}\)
⇒ \(\frac{120}{30} =\left(\frac{R+h}{R}\right)^2\)
4=\(\left(\frac{R+h}{R}\right)^2 \Rightarrow \quad 2=\frac{R+h}{R}\)
2 R=R+h \(\Rightarrow\) R=h
Height of the particle, h= Radius of the earth,
b=6400 km.
Thrust and Pressure
- Thrust is the force acting on an object perpendicular to its surface. The effect of thrust depends on the area on which it acts.
- The unit of thrust is the same as that of force, i.e. the SI unit of thrust is Newton (N). It is a vector quantity.
- Pressure is the force acting perpendicularly on a unit area of an object.
- Pressure(p)=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{\text { Thrust }}{\text { Area }}\)
- The SI unit of pressure is Nm -2, which is also called Pascal (Pa) named after the scientist Blaise Pascal. It is a scalar quantity.
- \(1 \mathrm{~Pa}=1 \mathrm{Nm}^{-2}\)
- From the formula of pressure, it is clear that the same force can produce different pressures depending on the area over which it acts. A force acting on a smaller area exerts a large pressure while the same force acting on a larger area exerts small pressure.
Example 9. A Force of 200 N is applied to an object of an area of 4 m2. Find the pressure.
Answer:
Given, force, F=200 \(\mathrm{~N}\), area, A=4 \(\mathrm{~m}^2\)
Now, Pressure, p=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{200}{4}=50 \mathrm{Nm}^{-2}\)
Example 10. A woman is wearing sharp-heeled sandals (stilettos). If the mass of a woman is 60 kg and the area of one heel is 1 cm², find out the pressure exerted on the ground, when the woman stands on just one heel. (Given, g = 10ms-2)
Answer:
In the given case, the force will be the weight of the woman which is given by m x g [where m is the mass of the woman and g is the acceleration due to gravity].
Force, F = mx g [weight of woman]
= 60 x 10 N [given, m = 60 kg, A = 1 cm²]
= 600 N
And area, A=1 \(\mathrm{~cm}^2=\frac{1}{10000} \mathrm{~m}^2\)
Pressure, p=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{600 \times 10000}{1}\)
=6000000 \(\mathrm{Nm}^{-2}\) (or } 6000000Pa
Thus, the pressure exerted by the woman (of 60 kg) standing on only one heel of area 1 cm² is 6000000 Nm-2.
Example 11 A block of wood is kept on a tabletop. The mass of the wooden block is 6 kg and its dimensions are 50 cm x 30 cm x 20 cm.
Find the pressure exerted by the wooden block on the tabletop if it is made to lie on the tabletop with its sides of dimensions
- 30 cm x 20 cm
- 50 cm x 30 cm
Answer:
The mass of the wooden block = 6 kg
The dimensions = 50 cm x 30 cm x 20 cm Thrust F = mg = 6 x 9.8 = 58.8 N
(1) Area of a side = Length x Breadth
=30 \(\times 20=600 \mathrm{~cm}^2=0.06 \mathrm{~m}^2\)
Pressure \(p_1 =\frac{F}{A}=\frac{58.8}{0.06}\)
=980 \(\mathrm{~N} / \mathrm{m}^2\)
(2) When the block lies on its side of dimensions 50 \(\mathrm{~cm} \times 30 \mathrm{~cm}\), it exerts the same thrust
Area = Length x Breadth
=50 \(\times\) 30=1500 \(\mathrm{~cm}^2=0.15 \mathrm{~m}^2\)
Pressure \(p_2=\frac{F}{A}=\frac{58.8}{0.15}=392 \mathrm{~N} / \mathrm{m}^2\)
Some Daily Life Applications of Pressure
- The handles of bags, suitcases, etc., are made broad so that less pressure is exerted on the hand.
- Buildings are provided with broad foundations so that the pressure exerted on the ground becomes less.
- Railway tracks are laid on cement or iron sleepers so that the pressure exerted by the train could spread over a larger area and thus pressure decreases.
- Pins, needles and nails are provided with sharp pointed ends to reduce the area and hence to increase the pressure.
- Cutting tools have sharp edges to reduce the area so that with lesser force, more pressure can be exerted.
- Pressure on the ground is more when a man is walking than when he is standing because, in the case of walking, the effective area is less.
- Depression is much more common when a man stands on the cushion than when he lies down on it because in the standing case, the area is lesser than in the case of lying.
- The tractors have broad tyres, to create less pressure on the ground so that tyres do not sink into comparatively soft ground in the field.
Pressure in Fluids
All liquids and gases are together called fluids. Water and air are the two most common fluids. Solids exert pressure on a surface due to their weight.
Fluids also have weight, therefore fluids also exert pressure on the base and walls of the container in which they are enclosed. Fluids exert pressure in all directions.
Buoyancy
- The tendency of a liquid to exert an upward force on an object immersed in it is called buoyancy. Gases also exhibit this property of buoyancy.
- Buoyant Force is an upward force which acts on an object when it is immersed in a liquid. It is also called upthrust.
- It is the buoyant force due to which a heavy object seems to be lighter in water. As we lower an object into a liquid, the liquid underneath it provides an upward force.
- For Example, A piece of cork is held below the surface of water. When we apply a force by our thumb, the cork immediately rises to the surface. This is due to the fact that every liquid exerts an upward force on the objects immersed in it.
Factors Affecting Buoyant Force
The magnitude of buoyant force depends on the following factors:
(1) Density of the Fluid
The liquid having higher density exerts more upward buoyant force on an object than another liquid of lower density. This is the reason why it is easier to swim in seawater in comparison to normal water.
The Sea water has higher density and hence exerts a greater buoyant force on the swimmer than the freshwater having lower density.
(2) Volume of Object Immersed in the Liquid
As the volume of a solid object immersed inside the liquid increases, the upward buoyant force also increases. The magnitude of buoyant force acting on a solid object does not. depend on the nature of the solid object. It depends only on its volume.
For Example, When two balls made of different metals having different weights but equal volume are fully immersed in a liquid, they will experience an equal upward buoyant force as both the balls displace an equal amount of the liquid due to their equal volumes.
Floating or Sinking of Objects in Liquid
When an object is immersed in a liquid, the following two forces act on it:
Weight of the object which acts in a downward direction, i.e. it tends to pull down the object.
Buoyant force (upthrust) which acts in an upward direction, i.e. it tends to push up the object.
Whether an object will float or sink in a liquid, depends on the relative magnitudes of these two forces which act on the object in opposite directions. There are three conditions of floating and sinking of objects. These are:
- If the buoyant force or upthrust exerted by the liquid is less than the weight of the object, the object will sink into the liquid.
- If the buoyant force is equal to the weight of the object, the object will float in the liquid.
- If the buoyant force is more than the weight of the object, the object will rise in the liquid and then float.
Density
- The density of a substance is defined as mass per unit volume.
- Density =\(\frac{\text { Mass of the substance }}{\text { Volume of the substance }}\) or \(\rho=\frac{m}{V}\)
- The SI unit of density is kilogram per metre cube (kg/m³). It is a scalar quantity. The density of a substance under specified conditions always remains the same.
- Hence, the density of a substance is one of its characteristic properties. It can help us to determine its purity. It is different for different substances. The lightness and the heaviness of different substances can be described by using the word density.
- Objects having a density less than that of a liquid, float on the liquid. Objects having greater density than that of liquid, sink in the liquid. It decreases with an increase in temperature.
Example 12. A sealed can of mass 700 g has a volume of 500 cm³. Will this can sink in water? (The density of water is 1 g cm-3)
Answer:
Given, the mass of the can, m=700 g
Volume of can, V=500 \(\mathrm{~cm}^3\)
Density of can, \(\rho=\frac{m}{V}\)
=\(\frac{700}{500}=1.4 \mathrm{~g} \mathrm{~cm}^{-3}\)
Since the density of the can is greater than the density of the water, the can will sink into the water.
Archimedes’ Principle
The statement ‘When an object is fully or partially immersed in a liquid, it experiences a buoyant force or upthrust, which is equal to the weight of the liquid displaced by the object’, i.e.
Buoyant force or upthrust acting on an object = Weight of liquid displaced by the object
Even gases like air, exert an upward force or buoyant force on the objects placed in them. It is buoyant force or upthrust due to displaced air which makes a balloon rise in air.
Applications of Archimedes’ Principle
- Archimedes’ principle is used in:
- Designing ships and submarines.
- lactometer (a device used to determine the purity of milk). hydrometer (a device used for determining the density of liquid).
Example 13 If an iron object is immersed in water, it displaces 8 kg of water. How much is the buoyant force acting on the iron object in Newton? (Given, g = 10 ms-2)
Answer:
According to Archimedes’ principle, “the buoyant force acting on this iron object will be equal to the weight of water displaced by this iron object.”
We know that, weight, w = mx g. Here, the mass of water, m= 8kg
Acceleration due to gravity, g =10 ms-2
On putting values, w = 8 x 10 = 80 N
Since the weight of water displaced by the iron object is 80 N, therefore the buoyant force acting on the iron object (due to water) will also be 80 N.
Activity Zone
Activity 1
Objective: To describe a circular path in the motion of a stone.
Materials Required:
Thread and stone.
Procedure
- A stone describing a circular path with a velocity of constant magnitude
- Take a piece of thread.
- Tie a small stone at one end and hold the other end of the thread. Whirl it as shown in the figure.
- Observe the motion of the stone.
- Release the thread.
- Again note the direction of motion of the stone.
Observation
- The stone moves in a circular path with a certain speed and changes direction at every point.
- The change in direction involves a change in velocity or acceleration.
- The force that causes this acceleration and keeps the body moving along the circular path is acting towards the centre.
- This force is called centripetal force. When the thread is released, the stone makes a tangent to the circle and falls down.
Question 1. Name the motion of the stone.
Answer:
The motion of the stone is along a circular path.
Question 2. What happens to the stone if centripetal force vanishes?
Answer:
Due to the absence of centripetal force, the stone flies off along a tangent of a circle.
Question 3. Name the straight line that meets the circle at one point.
Answer:
The tangent is a straight line that meets the circle at one point.
Activity 2
Objective: To study free-falling or rising bodies.
Material Required: Stone
Procedure
1. Take a stone.
2. Throw it upwards.
3. It reaches a certain height and then it starts falling down.
Observations
- The stone is attracted by the earth hence it falls down due to the earth’s gravitational force.
- The stone gets accelerated, when it goes upwards, g is negative when it falls towards the earth, g is positive.
Question 1. Which force accelerates a body in free fall?
Answer:
The force of gravity on earth accelerates a body in free fall.
Question 2. What is the direction of motion of an object if g is taken negative?
Answer:
The direction of motion of an object is upwards if g is taken negative.
Question 3. What is the direction of acceleration due to gravity?
Answer:
The direction of acceleration due to gravity is always downward.
Activity 3
Objective: To study any object that may be hollow or solid, big or small should fall at the same rate.
Materials Required
- Glass jar
- One sheet of paper
Procedure
- Take a sheet of paper and a stone.
- Drop them simultaneously from the first floor of a building.
- Observe whether both of them reach the ground simultaneously.
- Also, repeat in a glass jar and then observe what happens.
Observation
- The paper reaches the ground a little later than the stone. This is due to air resistance. The air offers resistance due to friction to the motion of the falling objects.
- The resistance offered by air to the paper is more than the resistance offered to the stone.
- In a glass jar, the air is sucked out, the paper and the stone falls at the same rate. When the earth attracts the paper sheet and crumbled paper, they experience the same acceleration during free fall.
- This acceleration experienced by an object is independent of its mass. Thus, any object whether it be hollow or solid, big or small should fall at the same rate.
Question 1. Which one reaches the ground first?
Answer:
The stone reaches the ground first.
Question 2. The paper reaches the ground later, why?
Answer:
The paper reaches the ground later due to air resistance.
Question 3. What happens when air is sucked in along a glass jar?
Answer:
When air is sucked into a long glass jar both stone and paper reach at the bottom at the same time.
Activity 4
Objective: To understand buoyant force.
Materials Required: Plastic bottle, stopper, bucket and water.
Procedure
- Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper. Put it in a bucket filled with water. You will see that the bottle will float.
- Push the bottle into the water. You will feel an upward push. Try to push it further down.
- You will find it difficult to push deeper and deeper. This indicates that water exerts a force on the bottle in the upward direction. The upward force exerted by the water goes on increasing as the bride is pushed deeper till it is completely immersed.
- Now, release the bottle. It bounces back to the surface.
Observation
- The force due to the gravitational attraction of the Earth acts on this bottle in a downward direction. But the water exerts an upward force on the bottle. This upthrust is known as buoyant force.
- When the bottle is immersed, the upward force exerted by the water on the bottle is greater than its weight. Therefore, it rises up when released.
- To keep the bottle completely immersed, the upward force on the bottle due to water must be balanced. This can be achieved by an externally applied force acting downwards.
- This force must at least be equal to the difference between the upward force and the weight of the bottle.
Conclusion
It is the buoyant force which keeps the object float.
Question 1. When a body is fully immersed in a liquid, what happens to its weight?
Answer:
When a body is fully immersed in a liquid, its weight decreases due to buoyant force.
Question 2. What are the factors affecting buoyant force?
Answer:
The buoyant force depends on
- the density of the liquid
- the volume of the solid immersed in liquid.
Question 3. What happens, if the density of an object is more than the density of the liquid in which it is immersed?
Answer:
The object will sink into the liquid.
Question 4. Who exerts buoyant force?
Answer:
The buoyant force is exerted by the liquid molecules on the body when it is dipped in that liquid.
Question 5. By which force does a body start floating?
Answer:
Buoyant force helps the body to float in a liquid in which it is immersed.
Activity 5
Objective: To know why objects float or sink when placed on the surface of water.
Materials Required: Two beakers filled with water, an iron nail and a piece of cork.
Procedure
- Take a beaker filled with water.
- Take an iron nail.
- Place it on the surface of the water.
- Take another beaker filled with water.
- Take a piece of cork and an iron nail of equal mass.
- Place them on the surface of the water.
Observation
- The nail sinks.
- The cork floats (as shown alongside).
Reasons
- The force due to the gravitational attraction of the Earth on the iron nail, pulls it downwards. There is an upthrust of water on the nail, which pushes it upwards.
- However, the downward force acting on the nail is greater than the upthrust of water on the nail. So, it sinks. This happens because of the difference in their densities.
- The density of cork is less than the density of water. This means that the upthrust of water on the cork is greater than the weight of the cork. So, it floats.
- The density of an iron nail is more than the density of water. This means that the upthrust of water on the iron nail is less than the weight of the nail. So, it sinks.
Conclusion
The object of density less than that of a liquid floats on the surface of the liquid. Whereas, the object of density greater than that of a liquid sinks in the liquid.
Question 1. What happens to buoyant force, when we increase the density of the fluid?
Answer:
The buoyant force increases with an increase in the density of the fluid.
Question 2. What is the condition for floating of a body?
Answer:
If the density of a body is less than that of the liquid in which it is immersed, then the body floats in liquid.
Question 3. Give the condition for the sinking of a body.
Answer:
If the density of the body is greater than that of the liquid, it sinks into the liquid.
Question 4. If an object floats with \(\frac{1}{9}, \frac{2}{11}\) and \(\frac{3}{7}\) parts of its volume
Answer:
- Outside the surface of the liquid of densities \(d_1, d_2\) and \(d_3\). Then, arrange the densities in increasing order.
- Upthrust due to liquid on an object is directly proportional to the density of the liquid. Therefore, densities in increasing order are of \(d_1<d_2<d_3\)
Question 5. A nail sinks while a cork floats. Why?
Answer:
The density of the nail is greater than that of liquid, so it sinks. But the density of cork is less than that of water, so it floats.
Gravitation Question And Answer
Question 1. State the universal law of gravitation.
Answer:
The universal law of gravitation states that the force of attraction between any two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This law is applicable to any two objects anywhere in the universe.
Question 2. Write the formula for the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer:
The formula for the magnitude of the gravitational force between the earth and an object on the surface of the earth is given by F=G \(\frac{M m}{R^2}\)
where, M = mass of the earth
m = mass of an object
G = gravitational constant
= \(6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
and R = distance between the centres of the earth and an object.
Question 3. What do you mean by free fall?
Answer:
The falling of a body from a height towards the earth under the gravitational force of the earth is called free fall. Hence, the motion of a particle falling down or going up under the action of gravity means the body is in free fall.
Question 4. What do you mean by acceleration due to gravity?
Answer:
The acceleration of a body during free fall towards a celestial body is called acceleration due to gravity. Its value is 9.8 m/s².
For objects on or near the surface of the earth,
m g=G \(\frac{M \times m}{R^2}\)
where, g = acceleration due to gravity
M = mass of the earth
m = mass of an object
and R = radius of the earth
Hence, g=\(\frac{G M}{R^2}\)
Question 5. What are the differences between the mass of an object and its weight?
Answer:
Differences between the mass of an object and its weight are as follows:
Question 6. Why is the weight of an object on the moon 1/6 its weight on the earth?
Answer:
Weight of an object, w = mg [where, m = mass of an object and g — acceleration due to gravity]
The mass of an object m remains constant at all places. Acceleration due to gravity changes from place to place. So, we can say that the weight of an object depends on the acceleration due to gravity.
On the moon, the acceleration due to gravity is l/6th that of the earth, this is the reason why the weight of an object on the moon is 1 /6th its weight on the earth.
Question 7. Why is it difficult to hold a school bag having a strap made of thin and strong string?
Answer:
It is difficult to hold a school bag having a strap made of a thin and strong string because the area under the strap is small. Hence, a large pressure is exerted by the strap on the fingers or shoulders. Due to the large pressure, the strap tends to cut the skin and hence pain is caused.
Question 8. What do you mean by buoyancy?
Answer:
The upward force exerted by a liquid on any object immersed in it is called buoyancy or upthrust.
Question 9. Why does an object float or sink when placed on the surface of water?
Answer:
When an object is placed on the surface of water, two forces act on the object.
- The weight of the object acts vertically downwards.
- The upthrust of the water acts vertically upwards.
The object will float on the surface of the water if the upthrust is greater than the weight of the object. The object will sink if the weight of the object is more than the upthrust of the water.
Question 10. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
Mass is more than 42 kg. As the buoyant force due to air is acting on us, the reading of the weighing machine will be less than the actual mass of a person.
Question 11. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than the other. Can you say which one is heavier and why?
Answer:
The cotton bag is heavier as the buoyant force acting on the cotton bag is more as its surface area is more than that of the iron piece.
Exercises
Question 1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
The force of gravitation between two objects is given by
F=G \(\frac{m_1 m_2}{r^2}\)
If the distance is reduced to half, i.e. \(r^{\prime}=r / 2\). Then, a new force of gravitation,
⇒ \(F^{\prime}=\frac{G m_1 m_2}{r^{\prime 2}}=\frac{G m_1 m_2}{(r / 2)^2}=4 \times \frac{G m_1 m_2}{r^2}=4 F\)
i.e. The force of gravitation becomes 4 times the original value.
Question 2. Gravitational force acts on all objects in proportion- to their masses. Why, then a heavy object does not fall faster than a light object?
Answer:
Acceleration due to gravity (g) is independent of the mass of the falling object and is equal for all objects at a point. So, a heavy object falls with the same acceleration as a light object.
Question 3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Take, the mass of the earth is 6 \(\times 10^{24}\) kg and the radius of the earth is 6.4 \(\times 10^6 \mathrm{~m}\).)
Answer:
The gravitational force between the earth and an object is given by F=\(\frac{G M m}{R^2}\).
where, G = gravitational constant
=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
M = mass of the earth =6 \(\times 10^{24} \mathrm{~kg}\)
R = radius of the earth =\(6.4 \times 10^6 \mathrm{~m}\)
and m = mass of an object =1 kg
F =\(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^6\right)^2}\)
=\(9.77 \approx 9.8 \mathrm{~N}\)
Thus, the magnitude of the gravitational force between the earth and a 1 kg object is 9.8 N.
Question 4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater than smaller than or equal to the force with which the moon attracts the earth? Why?
Answer:
When two objects attract each other, then the gravitational force of attraction applied by the first object on the second object is the same as the force applied by the second object on the first object. So, both the earth and moon attract each other by the same gravitational force of attraction.
Question 5. If the moon attracts the earth, then why does the earth not move towards the moon?
Answer:
The earth does not move towards the moon in spite of the attraction by the moon because the mass of the earth is much greater than the mass of the moon and for a given force, acceleration is inversely proportional to the mass of the object.
Question 6. What happens to the force between two objects, if
- the mass of one object is doubled?
- the distance between the objects is doubled and tripled?
- the masses of both objects are doubled?
Answer:
The force of attraction between two objects is given by
F=\(\frac{G m_1 m_2}{r^2}\)
where, \(m_1\) and \(m_2\) = masses of the objects
r= distance between the objects
and
G= gravitational constant
(1) If the mass of one object is doubled, then the new force,
\(F^{\prime} =\frac{G\left(2 m_1\right) m_2}{r^2}\)=2 \(\times \frac{G m_1 m_2}{r^2}\)=2 F i.e. Force becomes double.
(2) If the distance between the objects is doubled, then the new force,
⇒ \(F^{\prime}=\frac{G m_1 m_2}{(2 r)^2}=\frac{G m_1 m_2}{4 r^2}=\frac{1}{4} \cdot \frac{G m_1 m_2}{r^2}\)
=\(\frac{F}{4}\)
i.e. Force becomes one-fourth. If the distance between the objects is tripled, then new force,
⇒ \(F^{\prime} =\frac{G m_1 m_2}{(3 r)^2}=\frac{G m_1 m_2}{9 r^2}\)
= \(\frac{1}{9}\left(\frac{G m_1 m_2}{r^2}\right)=\frac{F}{9}\)
i.e. Force becomes one-ninth.
(3) If the masses of both objects are doubled, then the new force,
⇒ \(F^{\prime} =\frac{G\left(2 m_1\right)\left(2 m_2\right)}{r^2}\)
=4 \(\times \frac{G m_1 m_2}{r^2}=4 F\)
i.e. Force becomes four times.
Question 7. Calculate the force of gravitation between the earth and the sun given that the mass of the earth = 6 \(\times 10^{24}\) kg and of the sun = 2 \(\times 10^{30}\) kg. The average distance between the two is 1.5 \(\times 10^{11}m\).
Answer:
The force of attraction between the Earth and the sun is given by
F=\(\frac{G M_s M_e}{r^2}\)
where, G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
Mass of the sun, \(M_s=2 \times 10^{30} \mathrm{~kg}\)
Mass of the earth, \(M_e=6 \times 10^{24} \mathrm{~kg}\)
Average distance between the Earth and the sun
F =\(\frac{6.67 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{\left(1.5 \times 10^{11}\right)^2}=1.5 \times 10^{11} \mathrm{~m}\)
=3.6 \(\times 10^{22} \mathrm{~N}\)
Thus, the force between the earth and the sun is 3.6 \(\times 10^{22} \mathrm{~N}\).
Question 8. What is the importance of the universal law of gravitation?
Answer:
- The universal law of gravitation successfully explained several phenomena given below:
- The force that binds us to the earth.
- The motion of the moon around the earth.
- The motion of planets around the sun.
- The tides are due to the moon and the sun.
- The flow of water in rivers is also due to the gravitational force of the earth on water.
Question 9. What is the acceleration of free fall?
Answer:
The acceleration of free fall is the acceleration produced in the motion of an object when it falls freely towards the Earth. It is also called acceleration due to gravity. Its value on the earth’s surface is 9.8 m/s².
Question 10. What do we call gravitational force between the earth and an object?
Answer:
The gravitational force between the earth and an object is called the force of gravity or gravity.
Question 11. Amit buys a few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why?
[Hint The value of g is greater at the poles than at the equator.]
Answer:
No, his friend will not agree with the weight of gold. w = mg
⇒ \(g \propto \frac{1}{R^2}\)
The value of g is greater at the poles than at the equator. Therefore, gold at the equator weighs less than that at the poles. Thus, Amit’s friend will not agree with the weight of the gold bought.
Question 12. Why does a sheet of paper fall slower than one that is crumpled into a ball?
Answer:
The sheet of paper falls slower than one that is crumpled into a ball because in the first case, the area of the sheet is more, so it experiences a large opposing force due to air. In contrast, the sheet crumpled into a ball experienced less opposing force due to the small area. This opposing force arises due to air resistance or air friction.
Question 13. The gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in Newton of a 10 kg object on the moon and on the earth?
Answer:
Given, the mass of the object, m=10 kg
Weight on the earth, w=m g=10 \(\times 9.8=98 \mathrm{~N}\)
Weight on the moon =\(\frac{1}{6}\) of the weight on the earth
=\(\frac{1}{6} \times\) 98=16.33 N
Question 14. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
- the maximum height to which it rises.
- the total time it takes to return to the surface of the earth.
Answer:
Given, initial velocity, u = 49 m/s
(1) At the maximum height velocity becomes zero.
Final velocity, v = 0
⇒ \(v^2 =u^2-2 g h\)
0 =\((49)^2-2 \times 9.8 \times h\)
h =\(\frac{(49)^2}{2 \times 9.8}=122.5\) m
Maximum height attained =122.5 m
(2) Time taken by the ball to reach the maximum height.
From the first equation of motion, v=u-g t or
0=49-9.8 \(\times\) t
t=\(\frac{49}{9.8}\)=5 s
For the motion against gravity, the time of descent is the same as the time of ascent. So, the time taken by the ball to fall from maximum height is 5 s
Total time taken by the ball to return to the surface of the earth =5+5=10 s.
Question 16. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g=10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answer:
Given, initial velocity, u = 40 m/s
Final velocity becomes zero, i.e. v = 0 [at maximum height]
From the third equation of upward motion,
⇒ \(v^2 =u^2-2 g h\)
⇒ \((0)^2 =(40)^2-2 \times 10 \times h\)
0 =1600-20 h
h =\(\frac{1600}{20}=80 \mathrm{~m}\)
The maximum height reached by the stone = 80 m.
After reaching the maximum height, the stone will fall towards the earth and will reach the earth’s surface covering the same distance.
So, the distance covered by the stone = 80 + 80 = 160 m. Displacement of the stone = 0.
Because the stone starts from the earth’s surface and finally reaches the earth’s surface again, i.e. the initial and final positions of the stone are the same.
Question 17. A stone is allowed to fall from the top of a tower 100 m high and at the same time, another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Answer:
Let after time t both stones meet and s be the distance travelled by the stone dropped from the top of the tower at which the stones will meet.
Distance travelled by the stone dropped = s
Distance travelled by the stone projected upwards = (100-5) m
For the stone dropped from the tower,
s=u t+\(\frac{1}{2} g t^2=0+\frac{1}{2}(10) t^2\)
[ u=0 because the stone is dropped, i.e. it starts from rest] s=5 \(t^2\)
For the stone projected upwards, \(s^{\prime}=u t-\frac{1}{2} g t^2\)
[due to upward motion, negative sign is taken] (100-s) =25 \(t-\frac{1}{2} \times 10 t^2\)
100-s = \(25 t-5 t^2\)
From Eq. (1), we get
⇒ \(100-5 t^2=25 t-5 t^2\)
25 t=100
t=4 s
So, the stones will meet after 4 s.
s=5 \(t^2=5 \times(4)^2=80 \mathrm{~m}\)
So, the stones will be at a distance of 80 m from the top of the tower or 20 m (100 m – 80 m) from the base of the tower.
Question 18. A ball thrown up vertically returns to the thrower after 6 seconds, find
- the velocity with which it was thrown up,
- the maximum height it reaches and
- its position after 4 s.
Answer:
Total time taken =6 s
Time taken to reach the maximum height = \(\frac{6}{2}\) = 3 s [ time of ascent = time of descent]
(1) From the first equation of motion, v = u = gt [negative sign is taken due to upward motion]
0 =U -9.8 x 3 [v at maximum height, v = 0]
⇒ u- 29.4 m/s
(2) From the third equation of motion,
⇒ \(v^2=u^2-2\) g h [negative sign is taken due to upward motion]
0 =\((29.4)^2-2 \times 9.8 \times h\)
h =\(\frac{(29.4)^2}{2 \times 9.8}=44.1 \mathrm{~m}\)
The maximum height attained by the ball is 44.1 m.
(3) In the initial 3 s, the ball will rise, and then in the next 3 s it falls toward the earth.
The position after 4 s
= Distance covered in Is in the downward motion
From the second equation of motion,
h=u t+\(\frac{1}{2} g t^2=0+\frac{1}{2} \times 9.8 \times(1)^2=4.9 \mathrm{~m}\)
i.e. The ball will be at 4.9 m below the top of the tower or the height of the ball from the ground will be at (44.1-4.9) =39.2
Question 19. In what direction, does the buoyant force on an object immersed in a liquid act?
Answer:
The buoyant force on an object immersed in a liquid always acts in the vertically upward direction.
Question 20. Why does a block of plastic released under water, come up to the surface of water?
Answer:
The upthrust or buoyant force acting on the block of plastic by the water is greater than the weight of the plastic block. So, the plastic block comes up to the surface of the water.
Question 21. The volume of 50 g of a substance is 20 \(\mathrm{~cm}^3\). If the density of water is 1 g \(\mathrm{cm}^{-3}\), will the substance float or sink?
Answer:
Given, the mass of the substance, m = 50 g
Volume of substance, V=20 \(\mathrm{~cm}^3\)
Density of substance,
⇒ \(\rho=\frac{\text { Mass }}{\text { Volume }}=\frac{50}{20}=2.5 \mathrm{~g} \mathrm{~cm}^{-3}\)
i.e. The density of the substance is greater than the density of water, so it will sink in water.
Question 22. The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water, if the density of water is 1 g \(\mathrm{~cm}^{-3}\)? What will be the mass of the water displaced by this packet?
Answer:
Given, the mass of the packet =500 g
Volume of packet =350 \(\mathrm{~cm}^3\)
Density of packet, \(\rho=\frac{\text { Mass }}{\text { Volume }}\)
=\(\frac{500 \mathrm{~g}}{350 \mathrm{~cm}^3}\)
=1.43 \(\mathrm{~g} \mathrm{~cm}^{-3}\)
i.e. The density of the packet is greater than the density of the water, so it will sink in the water.
The mass of water displaced by the packet
= Volume of packet x Density of water =350 x 1 = 350 g
Summary
- Gravitation is defined as the non-contact force of attraction between any two bodies in the universe.
- According to the universal law of gravitation, the attractive force between any two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
- Mathematically, F = \(\frac{G M m}{d^2}\) where G is called the universal gravitational constant and its value is 6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
- Free fall Whenever objects falls towards the earth under the earth’s gravitational force alone, then such a motion is called free fall.
- The acceleration with which an object falls towards the earth due to the earth’s gravitational pull is called acceleration due to gravity. It is denoted by g.
- At the surface of earth, g = \(\frac{G M}{R^2}\)
- The SI unit of g is \(\mathrm{ms}^{-2}\) and its value is 9.8 \(\mathrm{~m} / \mathrm{s}^2\).
Equations of motion for freely falling bodies
v =u+g t
h =\(u t+\frac{1}{2} g t^2\)
⇒ \(v^2 =u^2+2 g h\)
- where h is the height from which the object falls, t is the time of fall, u is the initial velocity and v is the final velocity when the body accelerates at g.
- The total amount of matter contained in an object is called its mass.
- The SI unit of mass is kilogram (kg). Mass is a scalar quantity.
- The weight of an object is the force with which it is attracted towards the earth i.e. weight of an object, w = mg The SI unit of weight is Newton (N) Weight is a vector quantity.
- The weight of an object on the moon is 1 / 6 th of its weight on the earth.
- Thrust is the force, acting on an object perpendicular to its surface.
- The SI unit of thrust is Newton (N). Thrust is a vector quantity.
- Pressure is the force acting perpendicularly on a unit area of an object.
- Pressure can be calculated as, (p)=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{\text { Thrust }}{\text { Area }}\)
- The SI unit of pressure is \(\mathrm{Nm}^{-2}\) or pascal(Pa).
- Pressure is a scalar quantity.
- All the liquids and gases are called fluids.
- Buoyant force is an upward force, which acts on an object when it is immersed in a liquid. It is also called upthrust. Factors Affecting the Buoyant Force Density of the Fluid.
- The volume of the object immersed in the liquid.
- If the buoyant force exerted by the liquid is less than the weight of the object, the object will sink in the liquid.
- If the buoyant force is equal to the weight of the object, the object will float in the liquid.
- If the buoyant force is more than the weight of the object, the object will rise in the liquid and then float.
- The density of a substance is defined as mass per unit volume. Density = \(\frac{\text { Mass of the substance }}{\text { Volume of the substance }}\)
- The SI unit of density is kilogram per metre cube (kg/m3) and it is a scalar quantity.
- According to Archimedes’ principle, when an object is fully or partially immersed in a liquid, it experiences a buoyant force or upthrust which is equal to the weight of liquid displaced by the object.