GSEB Class 9 Science Solutions Chapter 10 Gravitation

Gujarat Board Textbook Solutions Class 9 Science Chapter 10 Gravitation

Gujarat Board Class 9 Science Gravitation InText Questions and Answers

Question 1. State the universal law of gravitation.
Answer: The universal law of gravitation states that every particle in the universe pulls every other particle with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between the two masses. The force's direction is always along the line that joins the two particles.
In simple words: Everything in the universe attracts everything else. This pull gets stronger if things are heavier and weaker if they are farther apart. The force always acts in a straight line between the objects.

Exam Tip: Remember to clearly state both the direct proportionality to masses and the inverse square proportionality to distance for a complete answer.

Question 2. Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the Earth.
Answer: To find the magnitude of the gravitational force between the Earth and an object on its surface, the formula is:
Take, \(M\) = mass of Earth
\(m\) = mass of object
\(r\) = distance between the object and the center of the Earth
The gravitational force between the object and the Earth is given by:
\( F = \frac{GMm}{r^{2}} \)
In simple words: To calculate the pull between Earth and an object, you multiply a special number (G) by the Earth's mass and the object's mass, then divide all of that by the square of the distance from the object to Earth's center.

Exam Tip: Define all variables used in the formula to ensure clarity and full marks. Remember 'r' is the distance to the center, not just the surface.

Question 1. What do you mean by free fall?
Answer: Whenever objects fall towards the Earth under the effect of only gravitational force, we say that these objects are in free fall.
In simple words: When something drops and only gravity is pulling it down, we call that a free fall.

Exam Tip: The key idea in free fall is that air resistance or other forces are considered negligible or absent, with only gravity acting on the object.

Question 2. What do you mean by the acceleration due to gravity?
Answer: The acceleration with which a body falls towards the Earth, caused by the Earth's gravitational pull, is known as acceleration due to gravity.
In simple words: It's how quickly things speed up when gravity pulls them down.

Exam Tip: Note that acceleration due to gravity is a specific type of acceleration caused by gravitational force, and it has a standard value on Earth's surface (approximately 9.8 m/s²).

Question 1. What are the differences between the mass of an object and its weight?
Answer:
Mass:

• The mass of a body is the quantity of matter contained within it.
• The mass of an object is constant and does not change from one place to another.
• The SI unit of mass is the kilogram (kg).

• The weight of an object is the force with which it is pulled towards the Earth.
• The weight of an object changes from place to place because the acceleration due to gravity varies with position and location.
• The SI unit of weight is Newton (N).

Exam Tip: Clearly differentiate between mass as an intrinsic property of matter and weight as a force dependent on gravity. Use bullet points for clear comparisons.

Question 2. Why is the weight of an object on the moon \( \frac {1}{6} \)th its weight on the Earth?
Answer: The weight of an object is given by the formula, \( w = mg \). The acceleration due to gravity on the moon is approximately \( \frac {1}{6} \)th of that on the Earth. Consequently, an object's weight on the moon is also \( \frac {1}{6} \)th of its weight on Earth.
In simple words: Your weight is how much gravity pulls you. Since the moon's gravity is about one-sixth as strong as Earth's, you would weigh one-sixth as much on the moon.

Exam Tip: Explain the relationship \( w = mg \) and then specifically mention the difference in gravitational acceleration (g) between the Earth and the Moon to justify the weight difference.

Question 1. Why it is difficult to hold a school bag having a strap made of a thin and strong string?
Answer: For a thin strap, the contact area is smaller, which results in more pressure being exerted on the shoulder. This happens because pressure is inversely proportional to the area. Therefore, it becomes difficult to hold a bag with a thin strap.
In simple words: A thin strap hurts your shoulder because it puts all the bag's weight on a tiny spot, creating a lot of pressure.

Exam Tip: This question relates to the concept of pressure. Clearly state the relationship between pressure, force, and area to earn full marks.

Question 2. What do you mean by buoyancy?
Answer: When a body is partially or entirely immersed in a fluid, an upward force acts upon it, which is called upthrust or buoyant force.
In simple words: Buoyancy is the upward push a liquid gives to anything placed in it.

Exam Tip: Use the terms "upthrust" and "buoyant force" interchangeably and emphasize that it's an upward force from a fluid.

Question 3. Why does an object float or sink when placed on the surface of the water?
Answer: When an object is placed in water, two main forces act on it: the object's weight and the buoyant force. If the object's weight is balanced by the buoyant force, then the object floats. However, if the object's weight is greater than the buoyant force, then the object sinks.
In simple words: An object floats if the upward push from the water (buoyant force) is strong enough to hold up its weight. If its weight is too heavy for the water's push, it sinks.

Exam Tip: Focus on the comparison between the object's weight and the buoyant force. This is the primary determinant of floating or sinking.

Question 1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer: If the measured mass is 42 kg, then your true mass is slightly greater than 42 kg. The measured value is less than your true value because of the effect of buoyant force which acts upwards, slightly reducing the apparent weight. Therefore, the true mass is actually a little bit more than 42 kg.
In simple words: Your weighing machine shows 42 kg, but your real mass is actually a little more. This is because the air pushes you up slightly, making you seem lighter than you actually are.

Exam Tip: Remember that air also exerts a buoyant force. Weighing scales measure apparent weight, not true mass, and air buoyancy slightly reduces this apparent weight.

Question 2. 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 than the iron bar. The volume of cotton is much larger than the volume of iron, so the buoyant force of air on cotton is much greater than that on the iron bar. This happens because the measured weight equals the actual weight minus the buoyant force, or conversely, the actual weight equals the measured weight plus the buoyant force.
In simple words: The cotton bag is actually heavier. Because cotton takes up more space, the air pushes up on it more, making it seem lighter on the scale. To get the true weight, you have to add back the air's push.

Exam Tip: This question tests understanding of buoyancy in air. The object with greater volume experiences a greater buoyant force from the air, meaning its measured weight is proportionally more reduced compared to its actual weight.

In-Text Activities Solved

Activity 10.1 Discussion and conclusion
Answer:

• The stone moves in a circular path.
• The direction of motion after release is tangent to the circular path at that specific point.

Exam Tip: This activity demonstrates centripetal force and inertia. The stone's natural tendency is to move in a straight line, but the string provides the force needed for circular motion.

Activity 10.3 Discussion and conclusion
Answer:

• Air resistance on paper is greater than that on a stone.
• In a vacuum, both a paper and a stone will reach the floor of the glass jar at the same time because air resistance is zero in a vacuum.

Exam Tip: This illustrates that in the absence of air resistance, all objects fall at the same rate, regardless of their mass or shape. Air resistance is the main factor causing different fall rates in everyday conditions.

Activity 10.4 Discussion and conclusion
Answer:

• Gravitational force acts downward, and buoyant force acts upwards.
• As we push the bottle downward, the immersed volume increases, and more water is displaced. Consequently, the buoyant force also increases.
• When the bottle is completely immersed in water, the buoyant force becomes greater than the gravitational force, which causes the bottle to bounce back to the surface.

Exam Tip: Remember Archimedes' principle: the buoyant force is equal to the weight of the fluid displaced. This explains why pushing the bottle deeper increases the buoyant force as more water is displaced.

Activity 10.5 Discussion and conclusion
Answer: An iron nail sinks. The weight of the iron nail is greater than the upthrust on the iron nail, or simply, the density of the iron nail is greater than the density of water.
In simple words: An iron nail sinks because it's heavier for its size than water, or simply, it's denser than water.

Exam Tip: For floating and sinking questions, always relate the object's density to the fluid's density. If the object's density is greater, it sinks; if less, it floats.

Activity 10.6 Discussion and conclusion
Answer: The iron nail will sink into the water, while the cork floats on its surface. This occurs because iron's density is greater than water's density, whereas cork's density is less than water's density.
In simple words: The iron nail sinks because it's denser than water, but the cork floats because it's less dense than water.

Exam Tip: This activity reinforces the concept of density as a primary factor in whether an object floats or sinks in a given fluid.

Activity 10.7 Discussion and conclusion
Answer: As the stone is slowly lowered into the water, the string's elongation or the balance reading will decrease. This reduction in elongation or balance reading will continue as more and more of the stone is dipped into the water. The elongation or reading will become minimal and will stay constant if the stone is lowered even deeper. The decrease in elongation is due to an increase in the upthrust. As the immersed volume grows, the upthrust also grows.
In simple words: When you put a stone slowly into water, the string holding it stretches less, or the scale shows a lighter weight. This happens because the water pushes up on the stone. The more stone you put in, the more the water pushes, making the stone feel lighter.

Exam Tip: This experiment demonstrates the direct relationship between the volume of an object immersed in a fluid and the buoyant force (upthrust) it experiences.

Gujarat Board Class 9 Science Gravitation Textbook Questions and Answers

Question 1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer: According to Newton's law of gravitation, the force between masses is given by \( F = \frac{Gm_1m_2}{r^{2}} \).
If the distance between the two objects, \(r\), is reduced to half, then the new distance becomes \( r' = \frac{r}{2} \).
The new force \( F' \) will be:
\( F' = \frac{Gm_1m_2}{(r/2)^2} \)
\( F' = \frac{Gm_1m_2}{r^2/4} \)
\( F' = 4 \frac{Gm_1m_2}{r^2} \)
\( F' = 4F \)
Therefore, when the distance between two objects is reduced to half, the gravitational force between them becomes four times its original value.
In simple words: If two things are pulled together by gravity, and you move them twice as close, the gravitational pull between them becomes four times stronger.

Exam Tip: Remember the inverse square law; if the distance changes by a factor, the force changes by the inverse square of that factor. For half the distance, the force is \( (1/ (1/2)^2) = 4 \) times.

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: All objects fall to the ground with constant acceleration, which is called acceleration due to gravity. This acceleration is constant and does not depend on an object's mass.
In simple words: Everything falls at the same speed, regardless of how heavy it is, because gravity makes all objects accelerate the same way, not caring about their mass.

Exam Tip: The key here is the acceleration due to gravity (g). Emphasize that 'g' is constant for all objects near Earth's surface and does not depend on their individual masses.

Question 3. What is the magnitude of the gravitational force between the Earth and a 1 kg object on its surface?
[Mass of the Earth is \( 6 \times 10^{24} \) kg and radius of the Earth is \( 6.4 \times 10^6 \) m]
Answer:
Given,
Mass of object, \( m = 1 \) kg
Mass of Earth, \( M = 6 \times 10^{24} \) kg
Radius of Earth, \( r = 6.4 \times 10^6 \) m
Gravitational constant, \( G = 6.67 \times 10^{-11} \) N m\(^2\) kg\(^{-2}\)
Force, \( F = ? \)
The gravitational force between two bodies is given by:
\( F = \frac{GMm}{r^{2}} \)
\( F = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24}) \times (1)}{(6.4 \times 10^6)^2} \)
\( F = \frac{40.02 \times 10^{13}}{40.96 \times 10^{12}} \)
\( F \approx 0.977 \times 10^1 \)
\( F \approx 9.77 \) N
\( F \approx 9.8 \) N
In simple words: We use a formula that multiplies a special number (G), the Earth's mass, and the object's mass, then divides by the square of the Earth's radius. When we do the calculation, the pull on a 1 kg object is about 9.8 Newtons.

Exam Tip: Ensure you use the correct values for G, Earth's mass, and radius. Pay close attention to the powers of 10 in the scientific notation during calculations to prevent errors.

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 or smaller or the same as the force with which the moon attracts the Earth? Why?
Answer: According to the universal law of gravitation, two objects always attract each other with an equal force, but in opposite directions. Therefore, the Earth attracts the moon with the same force with which the moon attracts the Earth.
In simple words: The Earth pulls the moon with the exact same strength that the moon pulls the Earth. It's an equal and opposite pull between them.

Exam Tip: This question relates to Newton's Third Law of Motion applied to gravitation. Emphasize that forces always occur in equal and opposite pairs.

Question 5. If the moon attracts the Earth, why does the Earth not move towards the moon?
Answer: The force exerted by the Earth on the moon is used to change the moon's path from a straight line into a circular orbit. The moon attracts the Earth with the same force. However, the Earth is far too massive to noticeably move towards the moon due to this force, as its inertia is much greater.
In simple words: Even though the moon pulls the Earth with the same force, the Earth is too big and heavy to move much. The moon's pull mostly affects the moon's own orbit around the Earth.

Exam Tip: Explain that while the forces are equal, the resulting acceleration (and thus movement) is inversely proportional to mass (\( F=ma \Rightarrow a=F/m \)). Earth's huge mass results in a negligible acceleration.

Question 6. What happens to the force between two objects, if –
1. the mass of one object is doubled?
2. the distance between the objects is doubled and tripled?
3. the masses of both objects are doubled?
Answer: From the universal law of gravitation, the force between two masses is given by \( F = \frac{Gm_1m_2}{r^{2}} \).
1. When the mass of one object is doubled, the force becomes twice.
Let the new mass be \( m'_1 = 2m_1 \).
The new force \( F' \) will be:
\( F' = \frac{G(2m_1)m_2}{r^2} \)
\( F' = 2 \frac{Gm_1m_2}{r^2} \)
\( F' = 2F \)
2. When the distance between the objects is doubled, the force becomes one-fourth of its previous value.
Initially, \( F = \frac{Gm_1m_2}{r^{2}} \).
If the distance (separation) is doubled, \( r' = 2r \).
The new force \( F' \) will be:
\( F' = \frac{Gm_1m_2}{(2r)^2} \)
\( F' = \frac{Gm_1m_2}{4r^2} \)
\( F' = \frac{1}{4} \frac{Gm_1m_2}{r^2} \)
\( F' = \frac{F}{4} \)
When the distance between the objects is tripled, the force becomes one-ninth of its previous value.
If the distance (separation) is tripled, \( r'' = 3r \).
The new force \( F'' \) will be:
\( F'' = \frac{Gm_1m_2}{(3r)^2} \)
\( F'' = \frac{Gm_1m_2}{9r^2} \)
\( F'' = \frac{1}{9} \frac{Gm_1m_2}{r^2} \)
\( F'' = \frac{F}{9} \)
3. When the masses of both objects are doubled, the force becomes four times.
Initially, \( F = \frac{Gm_1m_2}{r^{2}} \).
Now, let the new masses be \( m'_1 = 2m_1 \) and \( m'_2 = 2m_2 \).
The new force \( F' \) will be:
\( F' = \frac{G(2m_1)(2m_2)}{r^2} \)
\( F' = \frac{4Gm_1m_2}{r^2} \)
\( F' = 4F \)
In simple words: The gravitational pull changes a lot based on mass and distance. If you double one object's mass, the pull doubles. If you double the distance, the pull drops to one-fourth. If you triple the distance, the pull drops to one-ninth. But if you double both objects' masses, the pull becomes four times stronger.

Exam Tip: This question is a direct application of Newton's law of gravitation. For each scenario, clearly show the setup with new variables, substitute into the formula, and simplify to find the new force in terms of the original force.

Question 7. What is the importance of the universal law of gravitation?
Answer: The universal law of gravitation is important because it explains several key phenomena:

• 1. It explains the motion of the planets around the Sun.
• 2. It accounts for the motion of the moon and other artificial satellites around the Earth.
• 3. It clarifies how the tides are formed due to the gravitational pull of the moon.
• 4. It describes the force of gravity of the Earth and other planets.

Exam Tip: For importance questions, list distinct applications or consequences of the law. Use clear and concise points.

Question 8. What is the acceleration of free fall?
Answer: Acceleration of free fall is the acceleration generated when a body falls purely under the influence of the Earth's gravitational force. It is represented by 'g', and its value on the Earth's surface is approximately \( 9.8 \text{ ms}^{-2} \).
In simple words: It's how fast an object speeds up when only gravity is pulling it down. On Earth, this speed-up rate is about 9.8 meters per second every second.

Exam Tip: Be sure to provide both the definition and the standard numerical value for 'g' on Earth's surface, including its units.

Question 9. What do we call the gravitational force between the Earth and an object?
Answer: The gravitational force between the Earth and an object is called the Earth's gravity. The force exerted on an object due to Earth's gravity is also referred to as its weight.
In simple words: The pull between Earth and an object is called Earth's gravity, or simply, the object's weight.

Exam Tip: Differentiate between the general term "gravitational force" and the specific term "weight" when referring to the force between an object and the Earth.

Question 10. Amit buys some grains 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: Amit's friend will not agree with the weight of the gold bought. As the weight of an object on Earth is given by the formula \( w = mg \), and the acceleration due to gravity (\(g\)) at the pole is greater than at the equator, the gold will weigh more at the poles. Therefore, when Amit gives the gold to his friend at the equator, it will weigh less than what was measured at the poles, causing the friend not to agree.
In simple words: No, Amit's friend won't agree. Gold weighs more at the Earth's poles than at the equator because gravity is slightly stronger at the poles. So, the gold will feel lighter at the equator than when Amit bought it.

Exam Tip: Clearly explain that 'g' (acceleration due to gravity) varies with latitude, being highest at the poles and lowest at the equator, and how this directly affects the weight (\(w=mg\)) of an object.

Question 11. Why does a sheet of paper fall slower than one that is crumpled into a ball?
Answer: A flat sheet of paper will fall slower than one crumpled into a ball. This happens because air offers more resistance due to friction to the motion of the falling object. The air resistance against the flat sheet of paper is much greater than the resistance against the paper ball because the sheet has a significantly larger surface area.
In simple words: A flat piece of paper falls slowly because air pushes against its large surface. A crumpled paper ball falls faster because it's smaller and the air push is less.

Exam Tip: The crucial factor here is air resistance. Explain that a larger surface area leads to greater air resistance, which opposes the motion and slows down the fall.

Question 12. The gravitational force on the surface of the moon is approximately one-sixth as strong as the gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth?
Answer:
Given:
Mass of object, \( m = 10 \) kg
Acceleration due to gravity on Earth, \( g_e = 9.8 \) m/s\(^2\)
Acceleration due to gravity on Moon, \( g_m = \frac{1}{6} g_e \)
Weight of object on Earth, \( w_e \):
\( w_e = m \times g_e = 10 \text{ kg} \times 9.8 \text{ m/s}^2 = 98 \) N
Weight of object on moon, \( w_m \):
\( w_m = m \times g_m = 10 \text{ kg} \times \frac{9.8}{6} \text{ m/s}^2 \)
\( w_m = 10 \times 1.633... = 16.33 \text{ N} \)
\( w_m \approx 16.3 \) N
In simple words: On Earth, a 10 kg object weighs 98 Newtons. On the moon, because gravity is weaker (about one-sixth of Earth's), the same 10 kg object would only weigh about 16.3 Newtons.

Exam Tip: Clearly state the formula \( w = mg \). Calculate the weight on Earth first, then use the given ratio to find the weight on the Moon. Show all steps and units clearly.

Question 13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
1. the maximum height to which it rises,
2. the total time it takes to return to the surface of the Earth.
Answer:
Given:
Initial velocity, \( u = 49 \) m/s
Acceleration due to gravity, \( a = -g = -9.8 \) m/s\(^2\) (taking upward direction as positive)
1. For maximum height:
At the maximum height, the final velocity \( v = 0 \) m/s.
Applying the equation of motion: \( v^2 - u^2 = 2as \)
\( 0^2 - (49)^2 = 2(-9.8)h \)
\( -2401 = -19.6h \)
\( h = \frac{-2401}{-19.6} \)
\( h = 122.5 \) m
2. For the total time of flight:
The total time is the time of ascent plus the time of descent, so \( T_{total} = t_{ascent} + t_{descent} = 2t_{ascent} \).
To find the time of ascent, we consider the velocity at maximum height to be \( v = 0 \).
Applying the equation of motion: \( v = u + at \)
\( 0 = 49 + (-9.8)t \)
\( -49 = -9.8t \)
\( t = \frac{-49}{-9.8} \)
\( t = 5 \) s
The total time of flight is \( 2t = 2 \times 5 = 10 \) s.
In simple words: A ball thrown straight up with a speed of 49 m/s will reach a maximum height of 122.5 meters. It will take 5 seconds to go up and 5 seconds to come back down, making a total trip of 10 seconds.

Exam Tip: For problems involving vertical motion under gravity, always note the sign convention for acceleration (g) based on the direction of motion. At maximum height, the final velocity is always zero.

Question 14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answer:
Given:
Height of the tower, \( h = s = 19.6 \) m
Initial velocity, \( u = 0 \) m/s (since the stone is released)
Acceleration due to gravity, \( a = g = 9.8 \) m/s\(^2\)
We need to find the final velocity, \( v \).
Applying the equation of motion: \( v^2 - u^2 = 2as \)
\( v^2 - 0^2 = 2 \times 9.8 \times 19.6 \)
\( v^2 = 384.16 \)
\( v = \sqrt{384.16} \)
\( v = 19.6 \) m/s
The final velocity of the stone just before touching the ground is \( 19.6 \) m/s.
In simple words: If you drop a stone from a 19.6-meter tower, it will hit the ground with a speed of 19.6 meters per second.

Exam Tip: For objects released from rest, the initial velocity \( u \) is 0. Be careful with calculations involving decimals and squares/square roots.

Question 15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking \( g = 10 \text{ m/s}^2 \), 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 at maximum height, \( v = 0 \) m/s
Acceleration, \( a = -g = -10 \) m/s\(^2\) (taking upward direction as positive)
For maximum height reached by the stone:
Applying the equation of motion: \( v^2 - u^2 = 2as \)
\( 0^2 - (40)^2 = 2(-10)h \)
\( -1600 = -20h \)
\( h = \frac{-1600}{-20} \)
\( h = 80 \) m
The maximum height reached by the stone is \( 80 \) m.
For total distance covered:
The stone goes up \( 80 \) m and then falls back down \( 80 \) m.
Total distance covered = Upward journey + Downward journey \( = 80 \text{ m} + 80 \text{ m} = 160 \) m.
For net displacement:
As the final position coincides with its initial position (it returns to the starting point), the net displacement is \( 0 \).
In simple words: A stone thrown up at 40 m/s will go up 80 meters. It travels 80 meters up and 80 meters down, so the total distance is 160 meters. But because it ends up where it started, its overall change in position (displacement) is zero.

Exam Tip: Clearly distinguish between distance (total path length) and displacement (change in position). For an object returning to its starting point, displacement is zero, but distance is twice the maximum height.

Question 16. Calculate the force of gravitation between the Earth and the Sun, given that the mass of the Earth is \( 6 \times 10^{24} \) kg and of the Sun is \( 2 \times 10^{30} \) kg. The average distance between the two is \( 1.5 \times 10^{11} \) m.
Answer:
Given:
Mass of the Sun, \( M_S = 2 \times 10^{30} \) kg
Mass of the Earth, \( M_E = 6 \times 10^{24} \) kg
Average distance between the Earth and the Sun, \( r = 1.5 \times 10^{11} \) m
Gravitational constant, \( G = 6.67 \times 10^{-11} \) N m\(^2\) kg\(^{-2}\)
Applying Newton's law of gravitation, the force between the Earth and the Sun is given by:
\( F = \frac{GM_E M_S}{r^{2}} \)
\( F = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24}) \times (2 \times 10^{30})}{(1.5 \times 10^{11})^2} \)
\( F = \frac{6.67 \times 6 \times 2 \times 10^{-11 + 24 + 30}}{ (1.5)^2 \times (10^{11})^2 } \)
\( F = \frac{80.04 \times 10^{43}}{2.25 \times 10^{22}} \)
\( F = 35.573... \times 10^{43 - 22} \)
\( F \approx 35.57 \times 10^{21} \)
\( F \approx 3.56 \times 10^{22} \) N
In simple words: To find the gravitational pull between the Earth and the Sun, we use the gravitation formula with their masses and the distance between them. The calculation shows a very large force, about 3.56 multiplied by 10 to the power of 22 Newtons.

Exam Tip: Carefully handle the scientific notation (powers of 10) in both the numerator and denominator during calculations. Double-check the exponent rules for multiplication and division.

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:
Given:
Height of the tower, \( h = 100 \) m
Let the two stones meet after \( t \) seconds at a point P, which is at a height \( x \) above the ground.
For stone 1 (falling from the tower):
Initial velocity, \( u_1 = 0 \) m/s (released)
Displacement, \( s_1 = (100 - x) \) m (distance fallen)
Acceleration, \( a_1 = g = 9.8 \) m/s\(^2\)
Using \( s = ut + \frac{1}{2}at^2 \):
\( 100 - x = (0)t + \frac{1}{2}(9.8)t^2 \)
\( 100 - x = 4.9t^2 \)........(i)
For stone 2 (projected upwards from the ground):
Initial velocity, \( u_2 = 25 \) m/s
Displacement, \( s_2 = x \) m (height reached)
Acceleration, \( a_2 = -g = -9.8 \) m/s\(^2\) (taking upward direction as positive)
Using \( s = ut + \frac{1}{2}at^2 \):
\( x = 25t + \frac{1}{2}(-9.8)t^2 \)
\( x = 25t - 4.9t^2 \)........(ii)
From equation (i), we have \( x = 100 - 4.9t^2 \).
Substitute this value of \( x \) into equation (ii):
\( 100 - 4.9t^2 = 25t - 4.9t^2 \)
\( 100 = 25t \)
\( t = \frac{100}{25} \)
\( t = 4 \) s
Now, substitute the value of \( t \) back into equation (i) to find \( x \):
\( 100 - x = 4.9 \times (4)^2 \)
\( 100 - x = 4.9 \times 16 \)
\( 100 - x = 78.4 \)
\( x = 100 - 78.4 \)
\( x = 21.6 \) m
Therefore, the two stones will meet after \( 4 \) seconds at a height of \( 21.6 \) m from the base of the tower (or \( 100 - 21.6 = 78.4 \) m from the top of the tower).
In simple words: One stone is dropped from a 100m tower, and another is thrown up from the ground at the same time. They will pass each other after 4 seconds. This meeting point will be 21.6 meters above the ground.

Exam Tip: For problems involving two objects in motion, write down the equations of motion for each object separately. Choose a consistent coordinate system (e.g., upward positive, downward negative) and set up simultaneous equations to solve for time and position.

Question 18. A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up.
(b) the maximum height it reaches, and
(c) its position after 4s.
Answer:
Given:
Total time of flight, \( T_{total} = 6 \) s
Acceleration due to gravity, \( a = -g = -9.8 \) m/s\(^2\) (taking upward direction as negative)
(a) The velocity with which it was thrown up:
Since the ball returns to its initial position, its total displacement is zero, \( s = 0 \).
Applying the equation of motion: \( s = ut + \frac{1}{2}at^2 \)
\( 0 = u \times 6 + \frac{1}{2}(-9.8)(6)^2 \)
\( 0 = 6u - \frac{1}{2}(9.8)(36) \)
\( 0 = 6u - 4.9 \times 36 \)
\( 0 = 6u - 176.4 \)
\( 6u = 176.4 \)
\( u = \frac{176.4}{6} \)
\( u = 29.4 \) m/s
The initial velocity with which the ball was thrown up is \( 29.4 \) m/s.
(b) The maximum height it reaches:
The total time of flight is \( 6 \) s. Since the time of ascent equals the time of descent, the time to reach maximum height is \( t_{ascent} = \frac{T_{total}}{2} = \frac{6}{2} = 3 \) s.
At maximum height, the final velocity \( v = 0 \).
Using initial velocity \( u = 29.4 \) m/s, \( a = -9.8 \) m/s\(^2\), and \( t = 3 \) s.
Applying the equation of motion: \( s = ut + \frac{1}{2}at^2 \)
\( h = (29.4)(3) + \frac{1}{2}(-9.8)(3)^2 \)
\( h = 88.2 - \frac{1}{2}(9.8)(9) \)
\( h = 88.2 - 4.9 \times 9 \)
\( h = 88.2 - 44.1 \)
\( h = 44.1 \) m
The maximum height reached by the ball is \( 44.1 \) m.
(c) Its position after 4s:
In the first 3 seconds, the ball rises vertically upward to its maximum height. In the next 1 second (from \( t=3 \)s to \( t=4 \)s), it will fall towards the Earth from the maximum height.
We need to find the distance fallen in 1 second from rest (at max height).
Initial velocity from max height, \( u = 0 \).
Time, \( t = 4 - 3 = 1 \) s.
Acceleration, \( a = g = 9.8 \) m/s\(^2\).
Applying the equation of motion: \( s = ut + \frac{1}{2}at^2 \)
\( s = (0)(1) + \frac{1}{2}(9.8)(1)^2 \)
\( s = 0 + \frac{1}{2}(9.8) \times 1 \)
\( s = 4.9 \) m
So, after 4 seconds, the ball is \( 4.9 \) m below its maximum height.
Its position relative to the ground is \( h_{max} - s = 44.1 - 4.9 = 39.2 \) m from the ground.
In simple words: If a ball is thrown up and comes back in 6 seconds: (a) It was thrown up with a speed of 29.4 meters per second. (b) It went up to a highest point of 44.1 meters. (c) After 4 seconds, it will be 39.2 meters above the ground, having already passed its highest point and started falling down.

Exam Tip: Break down complex motion problems into segments. Use the appropriate equations of motion, paying careful attention to initial/final velocities and acceleration signs for each segment (upward vs. downward motion).

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 vertically upwards.
In simple words: The push from a liquid on an object always goes straight up.

Exam Tip: Always remember that buoyant force is an upward force, opposing gravity, which is why objects can float or feel lighter in a fluid.

Question 20. Why does a block of plastic released under water come up to the surface of the water?
Answer: The buoyant force or upthrust acting on the plastic block is greater than the object's weight. Because this upward force is stronger, the plastic block moves up to the water's surface.
In simple words: A plastic block floats up because the water pushes it up with more force than its own weight pulls it down.

Exam Tip: Remember that buoyancy is the upward force exerted by a fluid that opposes the weight of an immersed object. An object floats if buoyant force > weight, and sinks if buoyant force < weight.

Question 21. The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm, will the substance float or sink?
Answer: Given, the volume of the substance \( (u) = 20 \text{ cm}^3 \), and its mass \( (m) = 50 \text{ g} \). The density of water \( (d_w) = 1 \text{ g/cm}^3 \).
First, we calculate the substance's density \( (d_s) \):
\( d_s = \frac{m}{u} = \frac{50 \text{ g}}{20 \text{ cm}^3} = 2.5 \text{ g/cm}^3 \)
Since the substance's density \( (2.5 \text{ g/cm}^3) \) is greater than the density of water \( (1 \text{ g/cm}^3) \), the substance will sink.
In simple words: We find the substance's density by dividing its mass by its volume. If this density is more than water's density, it will sink; if it's less, it will float. In this case, it sinks.

Exam Tip: To predict if an object floats or sinks, always compare its density to the fluid's density. If the object is denser, it sinks; if less dense, it floats.

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 cm³? What will be the mass of the water displaced by this packet?
Answer: Given, the volume of the packet \( (\upsilon) = 350 \text{ cm}^3 \), and its mass \( (m) = 500 \text{ g} \). The density of water \( (d_w) = 1 \text{ g/cm}^3 \).
First, we calculate the packet's density \( (d_p) \):
\( d_p = \frac{m}{\upsilon} = \frac{500 \text{ g}}{350 \text{ cm}^3} \approx 1.428 \text{ g/cm}^3 \)
Since the packet's density \( (1.428 \text{ g/cm}^3) \) is greater than the density of water \( (1 \text{ g/cm}^3) \), the packet will sink.
As the packet is completely submerged in water, the mass of water displaced by the packet is:
Mass of water displaced = volume of packet \( \times \) density of water
Mass of water displaced = \( 350 \text{ cm}^3 \times 1 \text{ g/cm}^3 = 350 \text{ g} \)
In simple words: The packet is denser than water, so it sinks. When it sinks, it pushes aside an amount of water equal to its own volume, which is 350 grams of water.

Exam Tip: For submerged objects, the volume of fluid displaced is equal to the object's volume. Use this to calculate the mass or weight of displaced fluid.