These AP 9th Class Physics Important Questions 9th Lesson Gravitation will help students prepare well for the exams.
AP 9th Class Physics Science 9th Lesson Gravitation Important Questions
Class 9 Physical Science Chapter 9 Important Questions – 2 Marks
Question 1.
What will happen if no centripetal force is acting on the stone ?
Answer:
In the absence of this force, the stone flies off along a straight line.
This straight line will be a tangent to the circular path.
Question 2.
What is the Universal Law of Gravitation ?
Answer:
- Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects.
- F = G\(\frac{\mathrm{Mm}}{\mathrm{~d}^2}\)
Question 3.
What is G ? What is the value of G ? Who found it ?
Answer:
- G is a universal gravitation constant.
- Henry Cavendish (1731 – 1810) found the value of G using a sensitive balance.
- The accepted value of G is 6.673 × 10-11 Nm2 kg-2.
Question 4.
What is the acceleration due to the earth’s gravitational force ?
Answer:
- The acceleration due to the earth’s gravitational force is denoted by ’g’.
- It is the acceleration experienced by objects falling towards the earth under the influence of gravity alone.
- Its value is approximately 9.8 m/s2.
Question 5.
Is there any change in the velocity of falling objects ?
Answer:
- While falling, there is no change in the direction of motion of the objects.
- But due to the earth’s attraction, there will be a change in the magnitude of the velocity.
Question 6.
What is the relation between the gravitational force, mass and acceleration due to gravity ?
Answer:
The magnitude of the gravitational force on an object is equal to the product of its mass and acceleration due to the earth’s gravitational force. Mathematically, F = mg.
Question 7.
How Galileo proved that all objects hollow or solid, big or small, should fall at the same rate ?
Answer:
Galileo dropped different objects from the top of the Leaning Tower of Pisa in Italy to prove that all objects hollow or solid, big or small, should fall at the same rate.
Question 8.
What is mass ? Does it change from place to place ?
Answer:
- The mass of an object is the measure of its inertia.
- It remains the same whether the object is on the earth, the moon or even in outer space.
- Thus, the mass of an object is constant and does not change from place to place.
Question 9.
What is the weight of an object ?
Answer:
- The force of attraction of the earth on an object is known as the weight of the object.
- It is denoted by W.
- W = m × g
Question 10.
Compare mass and radius of the earth and the moon.
Answer:
Celestial body | Mass (kg) | Radius (m) |
Earth | 5.98 × 1024 | 6.37 × 106 |
Moon | 7.36 × 1022 | 1.74 × 106 |
Question 11.
Define pressure. Write its units.
Answer:
- The thrust on unit area is called Pressure.
Pressure = \(\frac{\text { Thrust }}{\text { Area }}\) - The S.I. unit of pressure as N/m2 or Nm-2.
Question 12.
What are pascal units ?
Answer:
- S.I. unit of pressure is called pascal, denoted as Pa.
- N/m2 or Nm-2 is also denoted by Pascal.
Question 13.
Why does a nail have a pointed tip ?
Answer:
A nail has a pointed tip because it allows the same force to be applied over a smaller area, resulting in a larger pressure that can penetrate materials more easily.
Question 14.
Why do buildings have wide foundations ?
Answer:
- Buildings have wide foundations because they need to support the weight of the structure and distribute it evenly over the ground.
- By spreading the weight over a larger area, the pressure on the ground is reduced, which prevents the building from sinking or collapsing.
Question 15.
What is the meaning of buoyancy ?
Answer:
- Buoyancy is the upward force exerted by a fluid on an object immersed in it.
- This force is equal to the weight of jhe fluid displaced by the object.
Question 16.
Why does a ship made of iron and steel not sink in seawater, but the same amount of iron and steel in the form of a sheet would sink ?
Answer:
- The ship made of iron and steel does not sink in seawater because it displaces a large amount of water and the buoyant force acting on the ship is greater than the weight of the ship.
- On the other hand, a sheet of iron and steel would sink because it does not displace enough water to create a buoyant force that is greater than its weight.
Question 17.
Write the applications of Archimedes Principle.
Answer:
- It is used in designing ships and submarines.
- Lactometer which checks purity of milk is designed on this principle.
- Hydrometer used to determine density of liquids also designed based on this principle.
Question 18.
What is Relative density?
Answer:
It is the ratio of density of a substance by density of water.
Relative density = \(\frac{\text { Density of substance }}{\text { Density of water }}\)
Note : Relative density has no units.
Question 19.
What do you mean by acceleration due to gravity ?
Answer:
- Acceleration due to gravity (g) refers to the acceleration experienced by an object when it falls freely under the influence of gravity.
- It is a constant acceleration that is always directed towards the centre of the Earth.
- It has a value of approximately 9.81 m/s2 near the Earth’s surface.
Question 20.
Why is it easy to cut with a sharp edge knife than with a blunt one ?
Answer:
- Sharp knife exert more pressure than a blunt knife.
- Lesser the area greater the pressure.
- Knife with sharp edge has greater area than blunt knife.
Question 21.
Why is it easy to walk on a grass than on a gravel ?
Answer:
- When we are walking on a grass the feet cover more are means less pressure makes our walking more comfortable.
- When we are walking on a gravel feet cover lesser area means more pressure this gives more pain.
Question 22.
Why egg sinks in fresh water and floats in sea water ?
Answer:
Sea water contains dissolved minerals which makes it more denser than fresh water due this egg sinks in sea water & floats in fresh water.
Question 23.
Write the equations of motion for a projected body.
Answer:
The equations of motion for a body projected vertically upwards are given by
v = u – gt ; s = ut – \(\frac{1}{2}\)gt2 ; v2 – u2 = -2gh
Question 24.
Why do cutting tools have sharp edges ?
Answer:
- Sharp edges on cutting tools reduce the surface area of contact between the tool and the material being cut, which concentrates the force of the tool onto a smaller area.
- This increases the pressure exerted by the tool, making it easier to cut through the material.
Question 25.
What is buoyancy ? On which factor does it depend ?
Answer:
- The upward force exerted by the water on the bottle is known as upthrust or buoyant force. In fact, all objects experience a force of buoyancy when they are immersed in a fluid.
- The magnitude of this buoyant force depends on the density of the fluid.
Question 26.
a) What keeps the moon in uniform circular motion around the earth ?
b) Why do astronauts in space feel weightless ?
Answer:
a) Gravitational force between the moon and the earth keeps moon in uniform circular motion around the earth.
b) They do not exert any force / weight on their spaceship in the absence of gravity in space.
Question 27.
What is meant by free fall ? A ball is dropped from the roof of a building. It takes 10 seconds to reach the ground. Find the height of the building.
Answer:
The fallings of a body from a height towards earth under the gravitational force of earth (with no other force acting on it) is called “FREE FALL”.
H = ut + \(\frac{1}{2}\)gt2 = 0 × 10 + 1/2 × 9.8 × 102 = 490 m
Question 28.
Gravitational force on an object of an imaginary planet is half of that of the gravitational force of same body on earth. Find (i) the value of acceleration due to gravity on this planet and (ii) weight of an object of mass 50kg on that planet.
Answer:
gplanet = \(\frac{1}{2}\) × gearth = \(\frac{1}{2}\) × 10 m/s2 = 5 m/s2
W = mg ⇒ Wplanet = 50 × 5 = 250N
Question 29.
What happens to the force between two objects if
i) the mass of one object is doubled ?
ii) the distance between the objects is tripled ?
Answer:
F = \(\frac{G m_1 m}{d^2}\)
i) If m1 = 2m, then F it becomes twice.
ii) If d = 3d, then F becomes one – nineth.
Question 30.
State two factors on which gravitational force depends and also mention how it depends on these factors?
Answer:
- Masses of object.
- Distance between them as F ∝ m1 × m2 and F ∝ 1/d2
Gravitation Class 9 Important Questions – 3 Marks
Question 1.
What causes to the motion of the moon around the earth ?
(OR)
If there is no centripetal force, what could happen to the moon ?
Answer:
- The motion of the moon around the earth is due to the centripetal force.
- The centripetal force is provided by the force of attraction of the earth.
- If there were no such force, the moon would pursue a uniform straight-line motion.
Question 2.
It is seen that a foiling apple is attracted towards the earth. Does the apple attract the earth ? If so, we do not see the earth moving towards an apple. Why ?
Answer:
- According to the third law of motion, the apple does attract the earth.
- But according to the second law of motion, for a given force, acceleration is inversely proportional to the mass of an object.
- The mass of an apple is negligibly small compared to that of the earth.
- So, we do not see the earth moving towards the apple.
Question 3.
Why does the earth not move around the moon ?
Answer:
- According to the second law of motion, for a given force, acceleration is inversely proportional to the mass of an object.
- The mass of the moon is small compared to that of the earth.
- So, we do not see the earth moving towards the moon.
Question 4.
Write the importance of the universal law of gravitation.
Answer:
The universal law of gravitation successfully explained several phenomena which were believed to be unconnected :
- the force that binds us to the earth.
- the motion of the moon around the earth.
- the motion of planets around the Sun and
- the tides due to the moon and the Sun.
Question 5.
Why does the value of g vary at different locations on the earth ?
Answer:
- The value of g varies at different locations on the earth due to the uneven distribution of mass in the earth’s interior and the earth’s rotation.
- The earth is not a perfect sphere and its shape is slightly flattened at the poles and bulges at the equator resulting in a slightly lower value of g at the equator compared to the poles.
Question 6.
Write the equations for the uniformly accelerated motion of objects due to gravity.
Answer:
- As ‘g’ is constant near the earth, all the equations for the uniformly accelerated motion of objects become valid with acceleration ‘a’ replaced by ‘g’.
- The equations are : v = u + at ; s = ut + \(\frac{1}{2}\)at2 ; v2 = u2 + 2as
Where ‘u’ and ‘v’ are the initial and final velocities and ‘s’ is the distance covered in time ‘t’.
Question 7.
Why do we use mass to determine our weight ?
(OR)
What is the relation between mass and weight ?
Answer:
We know that the value of g is constant at a given place.
- Therefore, at a given place, the weight of an object is directly proportional to the mass, say m, of the Object, that is, W ∝ m.
- It is due to this reason that at a given place, we can use the weight of an object as a measure of its mass.
- The mass of an object remains the same everywhere, that is, on the earth and on any planet whereas its weight depends on its location because ‘g’ depends on location.
Question 8.
Write the equations of motion for a free-falling body.
Answer:
Imagine an object is falling freely for time t seconds, with final velocity v, from a height . h, due to gravity g. It will follow the following equations of motion as :
- h = \(\frac{1}{2}\) gt2
- v2 = 2gh
- v = gt
Where, ’
h = height, v = final velocity, g = acceleration due to gravity, t = time taken for a free-falling body, the initial velocity (u) is zero.
Question 9.
Why can a camel run easily in a desert ?
Answer:
- Camels have evolved to survive in desert environments and their physical adaptations allow them to move efficiently over sandy terrain.
- Camels have large, padded feet that distribute their weight over a larger surface area, reducing the pressure they exert on the ground.
Question 10.
Why do army tanks rest upon a continuous chain ?
Answer:
- The weight of an army tank can be enormous and the tank needs to be able to move over rough terrain without sinking.
- The continuous chain distributes the weight of the tank over a larger area, reducing the pressure on the ground and providing better traction.
- This helps the tank to move forward even in difficult conditions, such as muddy or sandy terrain.
Question 11.
Why does a cork float while a nail sinks in water ?
Answer:
- A cork floats while a nail sinks in water because the density of cork is less than the density of water, whereas the density of an iron nail is more than the density of water. The upthrust of water on the cOrk is greater than its weight, so it floats.
- On the other* hand, the upthrust of water on the iron nail is less than its weight, so it sinks.
Question 12.
How is the density of a substance defined and how is it related to an object’s ability to float or sink in a liquid ?
Answer:
- The density of a substance is defined as its mass per unit volume.
- The density of an object determines its ability to float or sink in a liquid.
- Objects of density less than that of a liquid float on the liquid because the upthrust of water on the object is greater than its weight.
- Objects of density greater than that of a liquid sink in the liquid because the upthrust of water on the object is less than its weight.
Question 13.
How do you show a free falling object with an activity ?
Answer:
- Take a stone.
- Throw it upwards.
- It reaches a certain height and then it starts falling down.
- This is due to the gravitational force.
- Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall.
Question 14.
Why objects float or sink when placed on the surface of water? Explain with an activity.
Answer:
- Take a beaker filled with water.
- Take an iron nail and place it on the surface of the water.
- Observe what happens. The nail sinks.
- 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.
- But the downward force acting on the nail is greater than the upthrust of water on the nail. So, it sinks.
Question 15.
Why does an object float or sink when placed on the surface of water ?
Answer:
- An object floats or sinks when placed on the surface of water due to its density relative to the density of water.
- If the object is less dense than water, it will float because the buoyant force acting on it will be greater than its weight, causing it to rise to the surface.
- On the other hand, if the object is denser than water, it will sink because the weight of the object is greater than the buoyant force acting on it.
- The buoyant force is determined by the volume of the fluid displaced by the object, which is why larger objects will displace more water and experience a greater buoyant force.
Question 16.
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 other. Can you say which one is heavier alnd why ?
Answer:
- But in reality, the iron bar is heavier than the bag of cotton. This is because the iron bar has a higher density than the bag of cotton. To calculate the density by taking equal volumes of the iron and ‘cotton and compare their masses.
- According to given data,
The weight of 100 kg of iron bar = mg = 100 × 9.8 = 98 N
The weight of 100 kg of cotton bag = mg = 100 × 9.8 = 98 N
Both are same in weights (heaviness) (Here their volumes are different).
Question 17.
Why is it difficult to hold a school bag having a strap made of a 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 thin string does not distribute the weight of the bag evenly on the shoulders, which can cause discomfort and pain.
- Additionally, the thin string can cut into the skin or cause strain on the muscles due to increased pressure.
- Therefore, it is better to have wider straps that can distribute the weight evenly and provide better support.
Question 18.
Differentiate between Gravitational acceleration (g) and Gravitational constant (G).
Answer:
Acceleration due to gravity (g) | Universal gravitational constant (G) |
1) Acceleration due to gravity is the acceleration acquired by a body due to the earth’s gravitational pull on it. | Gravitational constant is numerically equal to the force of attraction between two masses of 1 kg that are separated by a distance of 1m. |
2) ‘g’ is a vector quantity. | ’G’ is a scalar quantity. |
3) Its value also varies from one celestial body to another. | Its value is the same everywhere in the universe. |
Question 19.
A stone is released from the top of a tower of height 19.6 m. Calculate its velocity just before touching the ground.
Solution:
Height of tower h = 19.6 m ; Initial velocity u = 0 m/s; Final velocity v = ?
Acceleration due to gravity g = 9.8 m/s2
v2 = u2 + 2gh
v2 = 0 + 2 × 9.8 × 19.6
v2 = 19.6 × 19.6m
v = 19.6 m/s
Question 20.
A body of weight 600N rests on the floor of a lift. If the lift begins to fall freely under the gravity. What is the force with which the body presses on the floor ?
Answer:
When the lift is falling freely under the gravity, then the body of weight 600N kept in it also falls freely under the action of gravity. In this case the reaction (force) of the floor of the lift on the body is zero, so the action (force) of the body on the floor of the lift should also be zero (By Newton’s 3rd Law of Motion). Hence, no force is exerted by the body on the floor of the lift when the lift is falling freely under the gravity. The body is weightless under such situations.
Important Questions on Gravitation Class 9 – 5 Marks
Question 1.
Write the differences between gravity (g) and universal gravitational constant (G).
Answer:
Parameter | Gravity (g) | Universal Gravitational Constant (G) |
Definition | Acceleration due to gravity at a specific location on Earth’s surface | Constant of proportionality that determines the strength of the gravitational force between objects |
Symbol | g | G |
Units | m/s2 | N.m2/kg2 (S.I. units) |
Formula | g = GM/r2 | F = Gm1m2/r2 |
Value | g = 9.81 m/s2 (approximate) | G = 6.6743 × 10-11 N.m2/kg2 |
Importance | Determines the weight of objects on Earth | Determines the strength of the gravitational force between objects in the universe. Used in many physics calculations, such as orbits, gravitational waves and cosmology |
Question 2.
Write the differences between mass and weight.
Answer:
Mass | Weight | |
Definition | The amount of matter in an object | The force exerted on an object due to gravity |
S.I. Unit | Kilogram (kg) | Newton (N) |
Measuring Instrument | Balance | Spring scale |
Effect of Gravity | Has no effect on mass | Affects weight ; weight changes depending on the strength of gravity |
Symbol | m | w or W |
Formula | Not applicable | w = mg (where g is the acceleration due to gravity) |
Dependence on Location | Does not change from place to place. | Weight changes depending on location, since gravity varies with location |
Question 3.
What is Universal Law of Gravitation ? Derive a formula for the force between two objects. (OR) Derive F = G\(\frac{\mathrm{Mm}}{\mathrm{~d}^2}\).
Answer:
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects.
Let two objects A and B of masses M and m lie at a distance d from each other as shown in above figure. Let the force of attraction between two objects be F. According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses. That is,
F ∝ M × m ………….. (1)
And the force between two objects is inversely proportional to the square of the distance between them, that is,
F ∝ \(\frac{1}{\mathrm{~d}^2}\) …………….. (2)
Combining Eqs. (1) arrd (2), we get
F ∝ \(\frac{\mathrm{M} \times \mathrm{m}}{\mathrm{~d}^2}\) ……….. (3) or, F = G\(\frac{\mathrm{M} \times \mathrm{m}}{\mathrm{~d}^2}\) ……………… (4)
where G is the constant of proportionality and is called the universal gravitation constant.
By multiplying crosswise, Eq. (4) gives
F × d2 = G M × m or G = \(\frac{\mathrm{Fd}^2}{\mathrm{M} \times \mathrm{m}}\) ……………. (5)
Question 4.
Derive a relation between G and g. (OR) Find the g value.
Answer:
- We know from the second law of motion that force is the product of mass and acceleration.
- We already know that there is acceleration involved in falling objects due to the gravitational force and is denoted by g.
- Therefore the magnitude of the gravitational force F will be equal to the product of mass and acceleration due to the gravitational force, that is, F = m g …………….. (1)
- mg = G\(\frac{\mathrm{Mm}}{\mathrm{~d}^2}\)
- g = G\(\frac{\mathrm{M}}{\mathrm{~d}^2}\) ………….. (2)
- Where M is the mass of the earth and d is the distance between the object and the earth.
- Let an object be on or near the surface of the earth. The distanced in the above equation will be equal to R, the radius of the earth. Thus, for objects on or near the surface of the earth.
- mg = G\(\frac{\mathrm{Mm}}{\mathrm{R}^2}\) ………….. (3)
- g = G\(\frac{\mathrm{M}}{\mathrm{R}^2}\) ………… (4)
- To calculate the value of g, we should put the values of G, M and Rin Eq. (4) namely, universal gravitational constant, G = 6.7 × 10-1 Nm2 kg2
Mass of the earth, M = 6 × 1024 kg and
Radius of the earth, R = 6.4 × 106 m.
g = G\(\frac{\mathrm{M}}{\mathrm{R}^2}\) = \(\frac{6.7 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2} \times 6 \times 10^{24} \mathrm{~kg}}{\left(6.4 \times 10^6 \mathrm{~m}\right)^2}\) = 9.8 m s-2
Thus, the value of acceleration due to gravity of the earth, g = 9.8 ms-2.
Question 5.
Find the weight of the object on the moon.
Answer:
Let the mass of an object be m. Let its weight on the moon be Wm. Let the mass of the moon be Mm and its radius be Rm.
By applying the universal law of gravitation, the weight of the object on the moon will be
Wm = G\(\frac{M_m \times m}{R_m^2}\) ………… (1)
Let the weight of the same object on the earth be We. The mass of the earth is M and its radius is R.
Celestial body | Mass (kg) | Radius (m) |
Earth | 5.98 × 1024 | 6.37 × 106 |
Moon | 7.36 × 1022 | 1.74 × 106 |
From Eqs. g = G\(\frac{\mathrm{M}}{\mathrm{R}^2}\) and We = m × g we have, W = G\(\frac{\mathrm{M} \times \mathrm{m}}{\mathrm{R}^2}\) …………. (2)
Substituting the values from above table in Eqs. (1) and (2), we get
Wm = G\(\frac{7.36 \times 10^{22} \mathrm{~kg} \times \mathrm{m}}{\left(1.74 \times 10^6 \mathrm{~m}\right)^2}\) ⇒ Wm = 2.431 × 1010 G × m …………… (3)
and We = 1.474 × 1011 G × m …………… (4)
Dividing Eq. (3) by Eq. (4), we get
\(\frac{W_m}{W_e}\) = \(\frac{2.431 \times 10^{10}}{1.474 \times 10^{11}}\) or \(\frac{W_m}{W_e}\) = 0.165 ≈ \(\frac{1}{6}\) …………. (5)
\frac{\text { Weight of the object on the moon }}{\text { Weight of the object on the earth }}=\frac{1}{6}
Weight of the object on the moon = (1/6) x its weight on the earth.
Question 6.
Write an activity to prove that the effect of thrust depends on the area on which it acts. (OR) Explain the difference between thrust and a pressure with an activity.
Answer:
Activity:
Case 1 : You stand on loose sand’.
Your feet go deep into the sand.
Case 2 : Now, lie down on the sand.
You will find that your body will not go that deep in the sand.
Reason:
- In both cases the force exerted on the sand is the weight of your body.
- Here the force is acting perpendicular to the surface of the sand.
- The force acting on an object perpendicular to the surface is called thrust.
- When you stand on loose sand, the force, that is, the weight of your body is acting on an area equal to area of your feet.
- When you lie down, the same force acts on an area equal to the contact area of your whole body, which is larger than the area of your feet.
- Thus, the effects of forces of the same magnitude on different areas are different.
- In the above cases, thrust is the same. But effects are different.
- Therefore, the effect of thrust depends on the area on which it acts.
Question 7.
Write an activity to demonstrate the centripetal force of a body.
(OR)
What is centripetal force ? Explain with an activity.
Answer:
- Take a piece of thread.
- Tie a small stone at one end. Hold the other end of the thread and whirl it round, as shown in the adjacent figure.
- Note the motion of the stone.
- Release the thread.
- Again, note the direction of motion of the stone.
- Before the thread is released, the stone moves in a circular path with a certain speed and changes direction at every point.
- The change in direction involves 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 the centripetal (meaning ‘centre-seeking’) force. v
- In the absence of this force, the stone flies off along a straight line.
- This straight line will be a tangent to the circular path.
Question 8.
How do you verify that acceleration experienced by an object is independent of its mass ?
(OR)
Which one has a greater acceleration due to gravity, a sheet of paper or a stone ? Justify your answer. (OR)
Show that all objects hollow or solid, big or small, should fall at the same rate.
Answer:
- 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.
- We see that paper reaches the ground a little later than the stone.
- This happens because of 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.
- If we do the experiment in a glass jar from which air has been sucked out, the paper and the stone would fall at the same rate.
- We know that an object experiences acceleration during free fall.
- From Equation g = G\(\frac{\mathrm{M}}{\mathrm{R}^2}\) this acceleration experienced by an object is independent of its mass.
- This means that all objects hollow or solid, big or small, should fall at the same rate.
Question 9.
Write an activity to show the buoyancy.
Answer:
- Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper.
- Put it in a bucket filled with water.
- You see that the bottle floats.
- Push the bottle into the water.
- You 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 bottle is pushed deeper till it is completely immersed.
- Now, release the bottle.
- It bounces back to the surface.
- The force due to the gravitational attraction of the earth acts on the bottle in the downward direction.
- So, the bottle is pulled downwards.
- But the water exerts an upward force on the bottle.
- The upward force exerted by the water on the bottle is known as upthrust or buoyant force.
Question 10.
Write an activity to show that sink or float depends on the density of an object.
Answer:
- Take a beaker filled with water.
- Take a piece of cork and an iron nail of equal mass.
- Place them on the surface of water.
- Observe what happens. The cork floats while the nail sinks.
- This happens because of the difference in their densities.
- The density of a substance is defined as the mass per unit volume.
- 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.
- Therefore objects of density less than that of a liquid float on the liquid.
- The objects of density greater than that of a liquid sink in the liquid.
Question 11.
Write an activity to explain the concept of Archimedes’ principle.
(OR)
Observe the activity given in the figure and answer the following questions.
a) Why does the extension of the spring balance or elongation of the string decrease once the stoifie is lowered in water ?
b) i) What is the magnitude of the buoyant force experienced by a body ?
ii) Is it the same in all fluids for a given body ?
iii) Do all bodies in a given fluid experience the sarnie buoyant force ?
Answer:
- Take a piece of stone and tie it to one end of a rubber string or a spring balance.
- Suspend the stone by holding the balance or the string as shown in the figure (a).
- Note the elongation of the string or the reading on the spring balance due to the weight of the stone.
- Now, slowly dip the stone in the water in a container as shown in the figure (b).
- Observe what happens to the elongation of the string or the reading on the balance.
- You will find that the elongation of the string or the reading of the balance decreases as the stone is gradually lowered in the water.
- However, no further change is observed once the stone gets fully immersed in the water.
a) Explanation:
- We know that the elongation produced in the string or the spring balance is due to the weight of the stone.
- Since the extension decreases once the stone is lowered in the water, it means that some force acts on the stone in an upward direction.
- As a result, the net force on the string decreases and hence the elongation also decreases.
b) Reasons for the given questions :
i) The magnitude of the buoyant force experienced by a body is equal to the weight of the fluid displaced by the body.
ii) The buoyant force experienced by a body is not necessarily the same in all fluids for a given body. The magnitude of the buoyant force depends on the density of the fluid the body is immersed in,.as well as the volume of the fluid displaced by the body. Therefore, if a body is submerged in a less dense fluid, it will experience a smaller buoyant force than if it were submerged in a more dense fluid.
iii) Similarly, not all bodies in a given fluid experience the same buoyant force. The buoyant force experienced by a body depends.on its volume and shape, as well as the density of the fluid it is submerged in.
Question 12.
What is weight ? Derive a formula for weight. Is weight a scalar or a vector ?
Answer:
- We know that the earth attracts every object with a certain force and this force depends on the mass (m) of the object and the acceleration due to gravity (g).
- The weight of an object is the force with which it is attracted towards the earth.
- We know that F = m × a, that is, F = m × g ……………. (1)
- The force of attraction of the earth on an object is known as the weight of the object.
- It is denoted by W.
- Substituting the same in Eq.(1), we have W = m × g
- As the weight of an object is the force with which it is attracted towards the earth, the S.I. unit 6f weight is the same as that of force, that is, newton (N).
- The weight is a force acting vertically downwards; it has both magnitude and direction. Hence, it is a vector.
Question 13.
Write the formula to find the magnitude of the gravitational, force between the earth and an object on the surface of the earth.
Answer:
1) The formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth is F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}\).
2) Where,
F is the magnitude of the gravitational force between the earth and the object.
G is the universal gravitational constant (6.674 × 10-11 N-m2 /kg2)
m1 is the mass of the earth (5.97 × 1024 kg)
m2 is the mass of the object
r is the distance between the centres of mass of the earth and the object (equal to . the radius of the earth, which is approximately 6,371 km).
3) If we take the mass of the earth (M) is too bigger when compared to the mass of the object, then F = \(\frac{\mathrm{GM}}{\mathrm{r}^2}\).
Question 14.
State Universal Law of Gravitation. Derive an expression for gravitational force between two bodies.
Answer:
Universal law of gravitation states that the force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Let the two bodies ‘A’ and ‘B’ be of masses ‘M’ and ‘m’ respectively, which are separated by a distance ‘r’.
According to Universal Law of Gravitation,
then F ∝ M × m ……………… (1) ; F ∝ l/r2 ………….. (2)
Combining (1) & (2)
F = G\(\frac{\mathrm{M} \times \mathrm{m}}{\mathrm{r}^2}\)
Where ‘G’ is called Universal gravitation constant.
The numerical value of G = 6.67 × 10-11 Nm2kg-2
Question 15.
To estimate the height of a bridge over a river, a stone is dropped freely in the river from the bridge. The stone takes 2s to touch the water surface in the river. Calculate the height of the bridge from the water level.
Answer:
The stone is being dropped freely from rest, so the initial velocity of the stone, u = 0. Again the velocity of the stone is increasing as it comes down, so the acceleration due to gravity ‘g’ is to be taken as positive.
Initial velocity u = 0 m/s ; Time taken t = 2 seconds
Let h be the height of the bridge.
Apply 2nd kinematic equation, h = ut + \(\frac{1}{2}\) gt2 = [0 + \(\frac{1}{2}\) × 9.8(2.0)2] m = 19.6 m
∴ The height ofjthe bridge from the water level is 19.6 m.
Question 16.
What happens to the magnitude of the force of gravitation between two objects if
a) mass of one of the object is tripled ?
b) distance between the objects is doubled ?
Answer:
The force between two objects is given by ‘Universal Gravitational Law’. It is numerically stated as F = G\(\frac{m_1 m_2}{d^2}\)
a) Mass of one object is tripled :
F = \(\frac{G\left(3 m_1\right) m_2}{d^2}\)
F = \(\frac{3 G\left(m_1 m_2\right)}{d^2}\)
Force will be tripled.
b) Distance between the objects is doubled :
F = G\(\frac{\left(\mathrm{m}_1 \mathrm{~m}_2\right)}{(2 \mathrm{~d})^2}\)
F = G\(\frac{\left(m_1 m_2\right)}{4 d^2}\) F = \(\frac{\frac{1}{4}\left\{G\left(m_1 m_2\right)\right\}}{d^2}\)
Question 17.
A ball is thrown upwards from the ground at a tower with a speed of 20 m/s. There is a window in the tower at the height of 15 m from the ground. How many times and when will the ball passthe window ? (Take g = 10 m/s-2)
Answer:
Initial velocity, u = 20 m/s ; Maximum height the ball will reach (h) = ?
Using equation v2 = u2 + 2gh (∴ at maximum height v = 0)
h = \(\frac{-u^2}{2 g}\) = \(\frac{-(20)^2}{2(-10)}\) = \(\frac{400}{20}\) ∴ h = 20 m
This means ball will reach the height of 20 m and come back. It will pass tHe window two times. Now to calculate the time ball will take to reach 15 m height.
h = ut – \(\frac{1}{2}\)gt2
25 = 20t – \(\frac{1}{2}\)(10)t2
5t2 – 20t + 15 = 0
t2 – 4t + 3 = 0
t2 – 3t – t + 3 = 0
t(t – 3) – 1(t – 3) = 0
(t – 1) (t – 3) = 0
⇒ t = 1, 3
Thus, ball will pass the window at 1 second and 3 seconds respectively.
Extra Questions on Gravitation Class 9 – 4 Marks
Question 1.
You wish to fix a poster on a bulletin board. To do this you have to press drawing pins with your thumb. You will apply force on the surface area of the head of the pin. Based on the above information answer the following questions.
a) Why drawing pins have sharp edges ?
Answer:
Sharp edge exerts more pressure so, it is easy to insert into drawing board.
b) In which direction force of the pin acts on board ?
Answer:
Force acts in perpendicular direction to the surface
Question 2.
You stand on loose sand your feet go deep into sand now, lie down on the sand.
a) You will find that your body will not go that deep in the sand.
Answer:
Weight of the body acting on an area equal to your feet in downward which is very small due to this the feet sink in sand.
b) Give the reason for the above situation.
Answer:
When you lay down the same weight acting downward equal to contact of. your whole body which is larger than area of your feet that’s why it won’t sink.