Maharashtra Board Class 8 Science Chapter 3 Force and Pressure Solutions

Std 8 Science Chapter 3 Force and Pressure Question Answer Maharashtra Board

Class 8 Science Chapter 3 Force and Pressure Question Answer Maharashtra Board

Exercise 1. Write Proper Word In The Blank Space:

Question a.The Sl unit of force is the ............... (dyne, newton, joule)
Answer:The Sl unit of force is the newton.
In simple words: The standard international unit for measuring force is the newton.

🎯 Exam Tip: Remember the correct SI units for fundamental physical quantities to avoid common mistakes.

 

Question b.The air pressure on our body is equal to the ............. pressure. (atmospheric, sea bottom, space)
Answer:The air pressure on our body is equal to the atmospheric pressure.
In simple words: Our bodies experience pressure from the air around us, which is called atmospheric pressure.

🎯 Exam Tip: Understanding atmospheric pressure is key to explaining many natural phenomena. Focus on its definition and effects.

 

Question c.For a given object, the buoyant force in liquids of different ............... is ............... (the same, density, different, area)
Answer:For a given object, the buoyant force in liquids of different density is the same.
In simple words: The upward force a liquid exerts on an object, known as buoyant force, depends on the liquid's density; different densities result in different buoyant forces for the same object.

🎯 Exam Tip: Buoyant force is directly proportional to the density of the fluid. Keep this relationship clear for problem-solving.

 

Question d.The Sl unit of pressure is ............... (N/m³, N/m², kg/m², Pa/m²)
Answer:The Sl unit of pressure is N/m².
In simple words: Pressure is measured in Newtons per square meter (N/m²), which is also known as Pascal.

🎯 Exam Tip: Knowing the SI unit of pressure (N/m² or Pascal) is fundamental. Understand that it relates force and area.

 

Exercise 2. Make A Match.

Question a.

Group 'A'	Group 'B'
1. Fluid	a. Higher pressure
2. Blunt knife	b. Atmospheric pressure
3. Sharp needle	c. Specific gravity
4. Relative density	d. Lower pressure
5. Hectopascal	e. Same pressure in all directions

Answer:

Group 'A'	Group 'B'
1. Fluid	e. Same pressure in all directions
2. Blunt knife	d. Lower pressure
3. Sharp needle	a. Higher pressure
4. Relative density	c. Specific gravity
5. Hectopascal	b. Atmospheric pressure

In simple words: This match-the-following exercise connects various physical concepts like fluid behavior, pressure applications, density, and atmospheric pressure units.

🎯 Exam Tip: Practice matching key terms with their definitions or associated phenomena to strengthen your understanding of concepts.

 

Exercise 3. Answer The Following Questions In Brief.

Question a.A plastic cube is released in water. Will it sink or come to the surface of water?
Answer:It will come to the surface of water. [Note: This is because its density is less than that of water. When it floats, the unbalanced force acting on it is zero.]
In simple words: A plastic cube floats because plastic is less dense than water, meaning it displaces enough water to balance its weight.

🎯 Exam Tip: The principle of buoyancy and density is crucial here: an object floats if its density is less than the fluid's density, resulting in an upward buoyant force that balances its weight.

 

Question b.Why do the load carrying heavy vehicles have large number of wheels?
Answer:The pressure produced by a given force depends on the area of the surface on which the force acts. The greater the surface area, the less is the pressure produced. Load carrying heavy vehicles have large number of wheels so that the load (weight, force) is distributed over large surface area of the wheels in contact with the road. Hence, the pressure decreases and the tyres do not burst.
In simple words: Heavy vehicles use many wheels to spread their weight over a larger road area, which reduces the pressure on the tires and prevents them from bursting.

🎯 Exam Tip: Remember the inverse relationship between pressure and area (P = F/A). Increasing the contact area reduces pressure, a common application in engineering.

 

Try this : Pressure of a liquid: Activity 1: Take a plastic bottle. Take a 10 cm long piece of a glass tube on which a rubber balloon can be fitted. Warm up one end of the glass tube and gently push it into the bottle at about balloon inflates. What is observed? The pressure of water acts on the side of the bottle as well.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक प्लास्टिक की बोतल दिखाता है जिसमें एक कांच की ट्यूब फिट की गई है, जिसके एक सिरे पर एक रबर का गुब्बारा लगा है और बोतल में पानी भरा है। गुब्बारे का फुलाव यह दर्शाता है कि बोतल के अंदर पानी द्वारा लगाए गए दबाव के कारण यह बाहर की ओर धकेला जा रहा है, जिससे तरल पदार्थों में पार्श्व दबाव का सिद्धांत स्पष्ट होता है। [Note: Here, the area of cross section of the tube remains the same. As the level of the water in the bottle rises, the mass of the water increases resulting in increase in the weight. As the applied force increases, the pressure increases. Therefore, the balloon increases in size.]
In simple words: When water is poured into a bottle with a balloon attached to a side tube, the balloon inflates, showing that water exerts pressure sideways, and this pressure increases with the water level.

🎯 Exam Tip: This activity demonstrates that liquids exert pressure in all directions and that pressure increases with depth. Focus on how this observation supports the concept of liquid pressure.

 

Activity 2: Take a plastic bottle. Pierce it with a thick needle (or with a hot nail) at the points 1, 2, 3 as shown in the Fig. Fill water in the bottle up to full height. A shown in the figure, water jets will be seen emerging and projecting out. The water jet emerging from the hole at the top will fall closest to the bottle. The jet from the lowest hole falls farthest from the bottle. Also, jets coming out from the two holes at the same level fall at the same distance from the bottle. What is understood from this? At any one level, the liquid pressure is the same. Also, the pressure increases as the depth of the liquid increases.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक प्लास्टिक की बोतल दिखाता है जिसमें तीन अलग-अलग ऊँचाइयों पर छेद (1, 2, 3) किए गए हैं, और बोतल में पानी भरा है। पानी सबसे ऊपरी छेद से सबसे कम दूरी पर और सबसे निचले छेद से सबसे अधिक दूरी पर निकलता है, यह दर्शाता है कि तरल का दबाव गहराई के साथ बढ़ता है, जबकि समान स्तर के छेद से पानी समान दूरी पर गिरता है।
In simple words: This experiment shows that liquid pressure increases with depth, as water from lower holes sprays farther, and pressure is equal at the same depth, as water from holes at the same level sprays the same distance.

🎯 Exam Tip: This activity visually confirms two key properties of liquid pressure: it increases with depth and acts equally at the same horizontal level. These are fundamental principles for understanding fluid dynamics.

 

Question c.How much pressure do we carry on our heads? Why don't we feel it?
Answer:The air pressure at the sea level is about \(101 \times 10^3\) Pa. This is the pressure that we carry on our heads. The cavities in our body are filled with air, and arteries and veins are filled with blood. Their pressure balances the pressure due to the atmosphere. Hence, we don't feel the atmospheric pressure.
In simple words: We carry about \(101 \times 10^3\) Pascals of atmospheric pressure on our heads, but we don't feel it because the internal pressure from air and blood in our bodies balances this external atmospheric pressure.

🎯 Exam Tip: Emphasize the concept of internal pressure balancing external atmospheric pressure. This is a common question testing understanding of pressure in living systems.

 

Exercise 4. Why Does It Happen?

Question a.Why does it happen? A ship dips to a larger depth in freshwater as compared to marine water.
Answer:The buoyant force acting on a body is proportional to the density of the fluid in which the body is immersed. The density of freshwater is less than that of marine water. Hence, the buoyant force on a body in freshwater is less than that in marine water. Therefore, a ship dips to a larger depth in freshwater as compared to marine water.
In simple words: A ship sinks deeper in freshwater because freshwater is less dense than saltwater, so it provides less buoyant force, requiring the ship to displace more water to float.

🎯 Exam Tip: This question highlights Archimedes' principle and the role of fluid density in buoyancy. Objects float higher in denser liquids as less volume needs to be displaced.

 

Question b.Why does it happen? Fruits can easily he cut with a sharp knife.
Answer:
1. It is easy to cut vegetables. fruits with a sharp knife. A blunt knife does not work here.
2. The force exerted perpendicularly on a unit area is called 'pressure'
\[ \text{Pressure} = \frac{\text{Force}}{\text{Area on which the force in applied}} \]
3. Presently we are considering only the force acting on an area in a direction perpendicular to it.
In simple words: A sharp knife cuts easily because its small edge area concentrates the applied force into high pressure, while a blunt knife spreads the force over a larger area, resulting in lower pressure.

🎯 Exam Tip: This is a direct application of the pressure formula (P=F/A). A smaller area for the same force leads to higher pressure, making cutting easier.

 

Question c.Why does it happen? The wall of a dam is broad at its base.
Answer:
1. The pressure at a point in a liquid is proportional to the height of the liquid column above it. Hence, the pressure of water in a dam is much greater at the bottom of the dam than at the top.
2. To withstand this high pressure, the wall of a dam is made stronger and thicker (broad) at the base than at the top.
In simple words: Dams are built wider at the bottom because water pressure increases with depth, requiring a stronger, thicker base to withstand the immense force at lower levels.

🎯 Exam Tip: This question assesses understanding of liquid pressure and its variation with depth. A strong answer connects the increasing pressure to the structural design of the dam.

 

Question d.Why does it happen? If a stationary bus suddenly speeds up, passengers are thrown in the backward direction.
Answer:
1. When passengers sit or stand in a stationary bus, they are in a state of rest. When the bus suddenly speeds up, the lower (parts of their body in contact with the bus acquire the speed of the bus.
2. The upper parts of their body, however, continue to be in the state of rest due to inertia. Hence, they are thrown in the backward direction.
In simple words: Passengers in a bus lurch backward when it accelerates suddenly because their bodies, due to inertia, tend to remain at rest even as the bus moves forward.

🎯 Exam Tip: This scenario is a classic example of Newton's First Law of Motion, specifically inertia of rest. Clearly explain how the body's tendency to maintain its state of rest causes the backward motion.

 

Exercise 5. Complete The Following Tables.

Question a.Mass (kg) Volume (m³) Density (kg/m³)

Mass (kg)	Volume (m³)	Density (kg/m³)
350	175	-
-	190	4

Answer:Using the formula, density = mass/volume:

Mass (Kg)	Volume (m³)	Density (Kg/m³)
350	175	2
760	190	4

In simple words: To complete the table, calculate density by dividing mass by volume, or calculate mass by multiplying density by volume.

🎯 Exam Tip: Ensure you correctly apply the density formula (\(\text{density} = \text{mass}/\text{volume}\)) for each row. Pay attention to unit consistency (kg and m³). The calculations are: Row 1: \(350 \text{ kg} / 175 \text{ m}^3 = 2 \text{ kg/m}^3\). Row 2: \(190 \text{ m}^3 \times 4 \text{ kg/m}^3 = 760 \text{ kg}\).

 

Question b.Density of Metal (kg/m³) Density of water (kg/m³) Relative Density

Density of Metal (kg/m³)	Density of water (kg/m³)	Relative Density
\(5 \times 10^3\)	\(10^3\)	5
\(8.5 \times 10^3\)	\(10^3\)	-

Answer:Using the formula, relative density = density of a metal/density of water:

Density of Metal (Kg/m³)	Density of water (Kg/m³)	Relative Density
\(5 \times 10^3\)	\(10^3\)	5
\(8.5 \times 10^3\)	\(10^3\)	8.5

In simple words: To find the relative density, divide the density of the metal by the density of water.

🎯 Exam Tip: Relative density is a dimensionless quantity. It's calculated by dividing the density of a substance by the density of water (usually \(1000 \text{ kg/m}^3\) or \(1 \text{ g/cm}^3\)). For the second row, \(8.5 \times 10^3 \text{ kg/m}^3 / 10^3 \text{ kg/m}^3 = 8.5\).

 

Question c.Weight (N) Area (m²) Pressure (Nm-²)

Weight (N)	Area (m²)	Pressure (N.m-²)
-	0.04	20,000
1500	500	-

Answer:Using the formula, pressure = weight/area:

Weight (N)	Area (m²)	Pressure (N.m-²)
800	0.04	20000
1500	500	3

In simple words: Use the formula pressure = weight/area to complete the table; if pressure and area are given, find weight by multiplying them. If weight and area are given, divide to find pressure.

🎯 Exam Tip: Ensure correct unit conversion if necessary. Pressure is in N/m² (Pascals). For the first row, \(20000 \text{ N/m}^2 \times 0.04 \text{ m}^2 = 800 \text{ N}\). For the second row, \(1500 \text{ N} / 500 \text{ m}^2 = 3 \text{ N/m}^2\).

 

Question 6. The density of a metal is \(10.8 \times 10^3 \text{ kg/m}^3\). Find the relative density of the metal.
Answer:Question a.The density of a metal is \(10.8 \times 10^3 \text{ kg/m}^3\). Find the relative density of the metal. Solution: Data: Density of the metal = \(10.8 \times 10^3 \text{ kg/m}^3\) density of water = \(10^3 \text{ kg/m}^3\) relative density of the metal = ? \[ \text{Relative density of a substance} = \frac{\text{density of the substance}}{\text{density of water}} \] \[ = \frac{10.8 \times 10^3 \text{ kg/m}^3}{10^3 \text{ kg/m}^3} = 10.8 \] The relative density of the metal = 10.8.
In simple words: The relative density of a substance is found by dividing its density by the density of water. In this case, the metal's relative density is 10.8.

🎯 Exam Tip: Relative density is a ratio of densities, so it is a dimensionless quantity (has no units). Always remember the density of water is \(10^3 \text{ kg/m}^3\) or \(1 \text{ g/cm}^3\).

 

Question 7. The volume of an object is 20 cm³ and the mass is 50 g. The density of water is 1 gm³. Will the object float on water or sink in water?
Answer:Question a.The volume of an object is 20 cm³ and the mass is 50 g. The density of water is 1 gm³. Will the object float on water or sink in water? Solution: \[ \rho \text{ (object)} = \frac{50 \text{ g}}{20 \text{ cm}^3} = 2.5 \text{ g/cm}^3 \] It is greater than the density of water. Hence, the object will sink in water.
In simple words: To determine if an object floats or sinks, compare its density to water's density; if the object is denser, it sinks. Here, \(2.5 \text{ g/cm}^3\) is greater than \(1 \text{ g/cm}^3\), so it sinks.

🎯 Exam Tip: The core concept here is density comparison. If an object's density is greater than the fluid's density, it sinks; otherwise, it floats. Make sure units are consistent (g/cm³ in this case).

 

Question 8. The volume of a plastic-covered sealed box is 350 cm³ and the box has a mass 500 g. Will the box float on water or sink in water? What will be the mass of water displaced by the box?
Answer:Question a.The volume of a plastic-covered sealed box is 350 cm³ and the box has a mass 500 g. Will the box float on water or sink in water? What will be the mass of water displaced by the box? Solution: \[ \rho \text{ (box)} = \frac{500 \text{ g}}{350 \text{ cm}^3} = \frac{10}{7} \text{ g/cm}^3 \] It is greater than that of water. Hence, the box will sink in water. The volume of water displaced by the box (V) = the volume of the box = 350 cm³ Now, density \( (\rho) = \frac{\text{mass (m)}}{\text{volume (V)}} \)
\( \implies m = \rho V \)
\( \implies \) The mass of water displaced by the box = \(1 \text{ g.cm}^{-3} \times 350 \text{ cm}^3 = 350 \text{ g}. \)
In simple words: The box sinks because its density (\(10/7 \text{ g/cm}^3\)) is greater than water's density (\(1 \text{ g/cm}^3\)); when it sinks, it displaces water equal to its own volume, which is 350 g.

🎯 Exam Tip: To determine floating/sinking, calculate the object's density and compare it to water's density. If it sinks, the volume of water displaced equals the object's volume, which can then be used to find the mass of displaced water.

 

Project:

Question a.Video record all the experiments (Try it) in this chapter with the help of mobile phone and send to others.
Answer:[This is a project-based task and does not require a specific answer to be digitized. It encourages practical application.]
In simple words: The project involves recording the experiments from this chapter using a mobile phone to share with others, promoting practical learning.

🎯 Exam Tip: While this is a project, understanding the principles behind the experiments will be directly beneficial for exam questions on force and pressure.

 

Class 8 Science Chapter 3 Force And Pressure Additional Important Questions And Answers

Exercise 1. Rewrite The Sentences After Filling The Blanks

Question 1.The tendency of an object to remain in its existing state is called its ...............
Answer:The tendency of an object to remain in its existing state is called its inertia.
In simple words: Inertia is the natural tendency of an object to resist changes in its state of motion or rest.

🎯 Exam Tip: Inertia is a foundational concept in physics. Understand that it's a property of matter and directly related to an object's mass.

 

Question 2.Pressure = ...............
Answer:Pressure = \[ \frac{\text{Force}}{\text{Area on which the force in applied}} \]
In simple words: Pressure is defined as the force applied perpendicularly to a surface divided by the area over which the force is distributed.

🎯 Exam Tip: This is the fundamental formula for pressure. Memorize it and understand the inverse relationship between pressure and area for a constant force.

 

Question 3.1 bar = ............... N/m².
Answer:1 bar = \(10^5\) N/m².
In simple words: One bar, a unit of pressure, is equivalent to 100,000 Newtons per square meter.

🎯 Exam Tip: Be familiar with common pressure unit conversions, especially between bar and Pascal (N/m²), as they are frequently used in atmospheric and engineering contexts.

 

Question 4.1 atmosphere = ............... Pa.
Answer:1 atmosphere = \(101 \times 10^3\) Pa.
In simple words: One standard atmosphere of pressure is approximately equal to 101,000 Pascals.

🎯 Exam Tip: Knowing the value of standard atmospheric pressure in Pascals is important for problems involving atmospheric pressure calculations.

 

Question 5.The Sl unit of density is ...............
Answer:The Sl unit of density is kg/m³
In simple words: The standard international unit for measuring density is kilograms per cubic meter.

🎯 Exam Tip: Understand the SI units for basic physical quantities like density (kg/m³), force (Newton), and pressure (Pascal) as they are foundational for calculations.

 

Exercise. Write Proper Word In The Blank Space:

Question 1.According to Archimedes' principle, the magnitude of the force of buoyancy acting on a body is ............... (Vpg, Vp/g, Vpm, mpg)
Answer:According to Archimedes' principle, the magnitude of the force of buoyancy acting on a body is Vpg.
In simple words: Archimedes' principle states that the buoyant force on an object is equal to the weight of the fluid it displaces, often expressed as Vpg, where V is displaced volume, p is fluid density, and g is gravity.

🎯 Exam Tip: Memorize Archimedes' principle and its formula (Vpg) as it's a critical concept for understanding flotation and sinking. Pay attention to what each variable represents.

 

Question 2.The pascal is the unit of ............... (velocity, pressure, mass, force)
Answer:The pascal is the unit of pressure.
In simple words: Pascal is the scientific unit used to measure pressure.

🎯 Exam Tip: Ensure you correctly associate Pascal with pressure, as it's the SI unit and frequently appears in physics problems.

 

Question 3.Keeping the surface area constant, if the applied force is doubled, the pressure ............... (becomes double, remains the same, becomes four times, becomes half)
Answer:Keeping the surface area constant, if the applied force is doubled, the pressure becomes double.
In simple words: If you double the force while keeping the area the same, the pressure will also double because pressure is directly proportional to force.

🎯 Exam Tip: This question tests the direct proportionality between force and pressure (P = F/A). If one doubles while the other is constant, the result also doubles.

 

Exercise. State Whether The Following Statements Are True Or False:

Question 1.The density of water is 1000 g/cm³.
Answer:False. [The density of water is \(1000 \text{ kg/m}^3\) (or \(1 \text{ g/cm}^3\))]
In simple words: The statement is false because the density of water is \(1 \text{ g/cm}^3\) or \(1000 \text{ kg/m}^3\), not \(1000 \text{ g/cm}^3\).

🎯 Exam Tip: Be precise with units for density values. \(1000 \text{ g/cm}^3\) would be \(1,000,000 \text{ kg/m}^3\), which is incorrect for water. The correct value is \(1 \text{ g/cm}^3\).

 

Question 2.Force and weight have the same units.
Answer:True.
In simple words: Both force and weight are measured in Newtons because weight is a type of force caused by gravity.

🎯 Exam Tip: Understand that weight is the gravitational force acting on an object. Therefore, both force and weight are measured in Newtons (N) in the SI system.

 

Question 3.Atmospheric pressure at sea level is about \(10^6\) dynes/cm².
Answer:True.
In simple words: The statement is true; atmospheric pressure at sea level is approximately \(10^6\) dynes per square centimeter.

🎯 Exam Tip: While Pascals are the SI unit, understanding atmospheric pressure in CGS units (dynes/cm²) can be useful. Remember the approximate value for sea-level pressure in various units.

 

Question 4.The buoyant force due to a liquid is proportional to the acceleration due to gravity.
Answer:True.
In simple words: Buoyant force is directly proportional to the acceleration due to gravity, meaning it would be stronger on planets with higher gravity and weaker on those with lower gravity.

🎯 Exam Tip: The formula for buoyant force, \(F_b = V\rho g\), clearly shows its dependence on \(g\) (acceleration due to gravity). This is a direct application of Archimedes' principle.

 

Question 5.Atmospheric pressure increases with altitude.
Answer:False. (Atmospheric pressure decreases with altitude.)
In simple words: The statement is false; atmospheric pressure actually decreases as you go higher in altitude because there is less air above you.

🎯 Exam Tip: This is a key concept related to atmospheric pressure. Higher altitude means a thinner column of air above, resulting in less pressure. This has implications for phenomena like ear popping and breathlessness.

 

Question 6.Pressure due to a given force is directly proportional to the area on which the force acts.
Answer:False. (Pressure due to a given force is inversely proportional to the area on which the force acts.)
In simple words: The statement is false; pressure is inversely proportional to the area, meaning a smaller area experiences more pressure from the same force.

🎯 Exam Tip: This directly relates to the pressure formula \(P=F/A\). Understanding the inverse relationship between pressure and area (for a constant force) is crucial for many applications.

 

Question 7.When a body is completely immersed in a liquid, the buoyant force acting on it due to the liquid is proportional to the volume of the liquid displaced by the body.
Answer:True.
In simple words: The statement is true; when an object is fully submerged, the buoyant force it experiences is directly proportional to the volume of the liquid it pushes aside.

🎯 Exam Tip: This is a direct statement of Archimedes' principle: the buoyant force equals the weight of the fluid displaced, and the weight of the fluid is proportional to its volume (\(W = V\rho g\)).

 

Question 8.The density of a material is useful to determine its purity.
Answer:True.
In simple words: The statement is true; by comparing a material's measured density to its known pure density, one can assess its purity.

🎯 Exam Tip: Pure substances have characteristic densities. Deviations indicate impurities, making density a practical tool for quality control and material identification.

 

Question 9.One tends to slip over a banana peel on the street and one can slip due to mud are events that occur due to reduced friction.
Answer:True.
In simple words: Slipping on a banana peel or mud happens because both significantly reduce the friction between your feet and the ground, making it hard to maintain grip.

🎯 Exam Tip: This question highlights the practical importance of friction. Understand that friction is a force opposing motion and its reduction can lead to instability.

 

Question 10.Frictional force is electromagnetic in origin.
Answer:True.
In simple words: The statement is true; frictional forces arise from the electromagnetic interactions between the atoms and molecules at the surfaces in contact.

🎯 Exam Tip: While often simplified as a macroscopic force, frictional force ultimately stems from the fundamental electromagnetic interactions at the atomic level between contacting surfaces.

 

Exercise. Identify The Odd Term:

Question 1.Density, Pressure exerted by a gas, Mass, Force.
Answer:Force. (Force is a vector quantity; other quantities are scalar quantities.)
In simple words: "Force" is the odd one out because it is a vector quantity (has direction), while density, pressure, and mass are scalar quantities (only have magnitude).

🎯 Exam Tip: Clearly distinguish between scalar (magnitude only) and vector (magnitude and direction) quantities. This fundamental distinction is important in all areas of physics.

 

Question 2.Lactometer, Hydrometer, Voltmeter, Submarine.
Answer:Voltmeter. (Its working is not based on Archimedes' principle. The working of a lactometer, hydrometer and submarine is based on Archimedes' principle.)
In simple words: "Voltmeter" is the odd term because it measures electrical voltage, whereas a lactometer, hydrometer, and submarine all operate based on Archimedes' principle of buoyancy.

🎯 Exam Tip: Understand the working principles of common scientific instruments. Lactometers and hydrometers measure density/purity using buoyancy, and submarines adjust buoyancy to dive or surface.

 

Exercise. Rewrite The Following Table In Such A Way That Column 2 And Column 3 Will Match With Column 1:

Question 1.

Column 1	Column 2	Column 3
1. Pressure	Mass/volume	Specific gravity
2. Density	Force/area	Decreases with increase in height above the sea level
3. Atmospheric pressure	No unit	Useful to determine the purity of a substance
4. Relative density	The pascal	Decreases with increase in area

Answer:

Column 1	Column 2	Column 3
1. Pressure	Force/area	Decreases with increase in area
2. Density	Mass/volume	Useful to determine the purity of a substance
3. Atmospheric pressure	The pascal	Decreases with increase in height above the sea level
4. Relative density	No unit	Specific gravity

In simple words: This table correctly matches physical quantities (Column 1) with their definitions/formulas (Column 2) and related characteristics/units (Column 3).

🎯 Exam Tip: Practice matching definitions, formulas, and properties for core physics concepts like pressure, density, atmospheric pressure, and relative density. This table is an excellent summary for review.

 

Exercise. Answer The Following Questions In One Sentence:

Question 1.Which of the following has more inertia? A Rs. 10 coin and a Rs. 1 coin.
Answer:A Rs. 10 coin has more inertia than a Rs. 1 coin.
In simple words: The Rs. 10 coin has more inertia because it typically has more mass than a Rs. 1 coin, and inertia is directly proportional to mass.

🎯 Exam Tip: Inertia is directly proportional to mass; the more massive an object is, the greater its inertia, meaning it's harder to change its state of motion or rest.

 

Question 2.Name the physical quantity expressed in pascal.
Answer:Pressure is expressed in pascal.
In simple words: The physical quantity measured in pascals is pressure.

🎯 Exam Tip: It is essential to know the SI units for various physical quantities. Pascal is the standard unit for pressure.

 

Question 3. State the Sl unit of pressure.
Answer: The Sl unit of pressure is N/m², also called the pascal.
In simple words: The standard unit for measuring pressure is the Pascal (Pa), which is equivalent to one Newton per square meter (N/m²).

🎯 Exam Tip: Remember that Pascal (Pa) and N/m² are interchangeable and both are SI units for pressure. Knowing this helps in unit conversions and calculations.

 

Question 4. Name the property of a liquid due to which it exerts an upward force on an object immersed in it.
Answer: Buoyancy is the property of a liquid due to which it exerts an upward force on an object immersed in it.
In simple words: The upward push a liquid gives to an object placed in it is called buoyancy.

🎯 Exam Tip: Understanding buoyancy is crucial for explaining why objects float or sink. It's an essential concept in fluid mechanics.

 

Question 5. Name the principle used in designing ships and submarines.
Answer: Archimedes' principle is used in designing ships and submarines.
In simple words: Archimedes' principle explains how ships and submarines float or submerge by relating buoyant force to displaced fluid.

🎯 Exam Tip: Archimedes' principle is a foundational concept in physics with real-world applications in marine engineering. Be prepared to state and explain it.

 

Question 6. What is specific gravity?
Answer: The specific gravity of a substance is another name used for relative density, i. e., the ratio of the density of the substance to the density of water.
In simple words: Specific gravity, also known as relative density, compares how dense a substance is to the density of water.

🎯 Exam Tip: Specific gravity is a dimensionless quantity because it's a ratio of two densities. It helps in quickly determining if a substance will float or sink in water.

 

Question 7. State any one factor on which the pressure exerted by a liquid at a point inside the liquid depends.
Answer: The pressure exerted by a liquid at a point inside the liquid depends on the density of the liquid.
In simple words: The pressure from a liquid at a certain depth depends on how dense that liquid is.

🎯 Exam Tip: Other factors affecting liquid pressure include depth and acceleration due to gravity. Focusing on these relationships is key for problem-solving.

 

Question 8. State any one factor on which the buoyant force due to a liquid depends.
Answer: The buoyant force due to a liquid depends on the density of the liquid.
In simple words: How much an object is pushed up by a liquid (buoyant force) depends on the liquid's density.

🎯 Exam Tip: Buoyant force is also proportional to the volume of the immersed object and acceleration due to gravity. Keep all these factors in mind.

 

Question 9. Name the device used to determine the purity of a sample of milk.
Answer: The lactometer is used to determine the purity of a sample of milk.
In simple words: A lactometer is a tool that helps check how pure milk is.

🎯 Exam Tip: The lactometer works on the principle of buoyancy, relating the density of milk to its purity. This is a common application question.

 

Question 10. Name the device used to determine the density of a liquid.
Answer: The hydrometer is used to determine the density of a liquid.
In simple words: A hydrometer is a device used to measure the density of liquids.

🎯 Exam Tip: Hydrometers, like lactometers, operate on Archimedes' principle, measuring liquid density based on how deeply they float.

 

Question 11. Name two instruments whose working is based on Archimedes' principle.
Answer: Working of the lactometer and hydrometer is based on Archimedes' principle.
In simple words: Both lactometers and hydrometers use Archimedes' principle to measure properties of liquids.

🎯 Exam Tip: Knowing practical applications like these demonstrates a deeper understanding of scientific principles. Be ready to give examples for various principles.

 

Answer The Following Questions:

 

Question 1. Give three examples to show that a force acts on two bodies through an interaction between them.
Answer:
1. Consider a car at rest on a level (plane) road. If it is pushed from behind, it moves in the forward direction.
2. Iron nails get attracted to the poles of a magnet and stick to the magnet.
3. The moon revolves around the earth.
In simple words: Force often happens when two things interact; for example, pushing a car, a magnet pulling nails, or the Earth's gravity pulling the moon.

🎯 Exam Tip: These examples illustrate both contact (pushing a car) and non-contact forces (magnetic attraction, gravitational pull), emphasizing the concept of interaction.

 

Question 2. What is a contact force? Give one example.
Answer: A force that acts through a direct contact of two objects or via one more object, is called a contact force.
Example: If a ball at rest on the ground is kicked, it starts moving.
In simple words: A contact force occurs when two objects physically touch each other to apply a force.

🎯 Exam Tip: Examples of contact forces are pushing, pulling, friction, and normal force. Be able to differentiate between contact and non-contact forces.

 

Question 3. What is a non contact force? Give one example.
Answer: A force that acts between two objects even if the two objects are not in contact, is called a non contact force.
Example: The earth revolves around the Sun.
In simple words: A non-contact force acts between objects without them physically touching, like gravity between Earth and the Sun.

🎯 Exam Tip: Gravitational, magnetic, and electrostatic forces are primary examples of non-contact forces. Understanding these fundamental forces is crucial.

 

Use Your Brainpower!

 

Question. Make a list of some more examples in which contact and non contact forces are applied. Write the types of force.
Answer:
1. Some examples in which contact forces are applied:
• to cut an apple with a knife (muscular force, frictional force)
• to lift a ball lying on the ground (muscular force, frictional force)
2. Some examples in which non contact forces are applied:
• the motion of the earth around the Sun (gravitational force)
• the motion of an electron around the nucleus of an atom (mainly the electric force).
In simple words: Contact forces like cutting with a knife require touch, while non-contact forces like gravity (Earth-Sun) or electric force (electron-nucleus) act from a distance.

🎯 Exam Tip: Being able to identify and categorize various forces helps consolidate understanding of their underlying mechanisms and impacts.

 

Question 4. In the following examples, state whether the force is a contact force or non contact force:
Answer:
1. a reluctant dog is being pulled by his master: contact force
2. a boy playing football is kicking the ball away: contact force
3. when iron nails are brought near a magnet, they are attracted to the poles of the magnet and stick to the magnet: non contact force
4. a coconut is falling from the coconut tree: non contact force
5. when a comb is rubbed against hair, small pieces of paper kept on a table get attracted to the comb: non contact force
6. when brakes are applied to a moving bicycle, it stops after some time: contact force.
In simple words: Forces that require touch (like pulling a dog or kicking a ball) are contact forces, while forces acting over a distance (like magnetic attraction or gravity) are non-contact forces.

🎯 Exam Tip: Practice classifying forces as contact or non-contact. This skill tests your fundamental understanding of how forces are transmitted and interact.

 

Question 5. Give one example in which frictional force is useful.
Answer: While walking, we push the ground behind with our feet. In the absence of friction between the ground and the lower surface of our feet, we will slip and will not be able to walk.
[Note: Frictional force is electromagnetic in origin.]
In simple words: Friction is useful because it allows us to walk without slipping by providing grip between our feet and the ground.

🎯 Exam Tip: Frictional force is often seen as a hindrance, but it's vital for many daily activities like walking, writing, and braking. Be ready to explain its importance.

 

Try This:

 

Take two plastic bottles with rectangular shape. Close their openings by fitting the lids tightly. Keep two small bar magnets on them and fix them neatly using a sticking tape.
Fill a big plastic tray with water and leave the two bottles floating with magnets at the top. Take one bottle near the other. If the north pole of the magnet is near the south pole of the other magnet, the bottles will head towards each other, because unlike poles attract each other.
Observe what will happen when the directions of the bottles are changed. We can observe change in the motion of the bottles without any direct contact. This means that there exists a non contact force between the two magnets.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो आयताकार प्लास्टिक की बोतलों को दिखाता है जो पानी में तैर रही हैं, प्रत्येक पर एक छोटा बार चुंबक लगा हुआ है। एक चुंबक का उत्तरी ध्रुव (N) दूसरे के दक्षिणी ध्रुव (S) के पास है, जिससे उनके बीच आकर्षण बल (गैर-संपर्क बल) काम कर रहा है।
In simple words: This activity demonstrates how magnets exert a non-contact force; floating bottles with magnets either attract or repel depending on the pole orientation, moving without physical touch.

🎯 Exam Tip: Experiments like this vividly illustrate abstract concepts like non-contact forces. Understanding the setup and expected observations is key.

 

magnet is near the north pole of the other magnet (or the south pole of one magnet is near the south pole of the other magnet), the bottles will move away from each other because like poles repel each other.

 

Use Your Brainpower!

 

Question. You have learnt about static electricity in the previous standard. Electrostatic force is a non contact force. To verify this, which experiment will you perform?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक प्लास्टिक की कंघी को दिखाता है जिसे बालों से रगड़ने के बाद कागज के छोटे टुकड़ों के पास लाया जा रहा है। कंघी कागज के टुकड़ों को बिना छुए अपनी ओर आकर्षित कर रही है, जो स्थिरविद्युत बल के कारण होता है।
Very small pieces of paper being attracted towards the comb
Do not switch on the fan in the room. Keep very small pieces of paper on the table. Rub a plastic comb against hair and bring it near the bits of paper. You will find that the bits of paper are attracted by the comb. The comb, on rubbing acquires electrostatic charge. It induces opposite charges on the bits of paper. Hence, the bits of paper are attracted by the comb.
In simple words: To show electrostatic force, rub a comb on hair and bring it near small paper pieces; the comb attracts the paper without touching, demonstrating a non-contact force.

🎯 Exam Tip: This classic experiment clearly shows electrostatic attraction, a non-contact force. Explaining the charging and induction process adds depth to your answer.

 

Try This:

 

Balanced forces and unbalanced force:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मेज पर रखे कार्डबोर्ड बॉक्स को दिखाता है जिसके दोनों ओर से तार लटके हुए हैं। इन तारों से वजन के पलड़े (weighing pans) जुड़े हैं। इस व्यवस्था का उपयोग संतुलित और असंतुलित बलों को प्रदर्शित करने के लिए किया जाता है।
Take a cardboard box, tie thick string to its two sides and keep it on a smooth table as shown in Fig. Take the strings on both sides of the table. Tie weighing pans to the two ends. Keep equal masses in both the pans. The box does not move on the table.
If more mass is kept in one of the pans than in the other, the box starts moving in the direction of that pan. Equal gravitational force acts on both the pans when equal masses are kept in them. This means balanced forces act on the box, with effective force equal to zero as these are acting in opposite directions.
On the contrary, if more mass is kept in one pan than in the other, the box starts moving in the direction of the pan with more mass. When unequal forces are applied to the box on the two sides, an unbalanced force acts on the box resulting in imparting motion to the box.
Children playing tug of war pull the rope in their respective directions. If the pull of the force is equal on the two sides, the rope does not move. If the force is more on one side, the rope moves in that direction. This means that initially, the two forces are balanced; the rope moves in the direction of higher force when the forces become unbalanced.
Let us see one more example. When a big grain storage container is required to slide on the ground, it becomes easier if two persons push it rather than one person. When the force is applied by both in the same direction, the movement is easy. You may have experienced this. What do we understand from this example?
1. If several forces are applied on an object in the same direction, a force equal to their addition acts on that object.
2. If two forces are applied on one object in directions opposite to each other, a force equal to their difference acts on the object.
3. A force is expressed in magnitude and direction.
Force is a vector quantity. If more than one force are acting on a body, then the effect on the body is due to the net force. When a force is applied on a stationary object it moves, its speed and direction change. Similarly, a force is required to stop an object in motion.
An object can change its shape due to force. While kneading a dough made from flour, the dough changes its shape when a force is applied. A potter applies a force in a specific direction while shaping the pot. Rubber, when stretched, expands. There are many such examples.,
In simple words: This experiment shows that when equal forces pull a box in opposite directions, it stays still (balanced forces), but if one force is stronger, the box moves (unbalanced forces).

🎯 Exam Tip: Understanding balanced and unbalanced forces is fundamental to Newton's laws of motion. Remember that unbalanced forces cause a change in motion or shape.

 

Question 6. What are balanced forces?
Answer: If a body is acted upon by two forces, equal in magnitude, opposite in direction and having the same line of action, the forces are called balanced forces. Here, the net force acting on the body is zero.
In simple words: Balanced forces are equal and opposite forces acting on an object, resulting in no change in its motion or zero net force.

🎯 Exam Tip: When forces are balanced, an object at rest stays at rest, and an object in motion continues at a constant velocity. No acceleration occurs.

 

Question 7. What is an unbalanced force?
Answer: If two or more forces act on a body such that their resultant is not zero, the resultant is an unbalanced force.
[Note: Unbalanced force acting on a body = mass of the body × acceleration of the body.]
In simple words: Unbalanced forces are unequal forces on an object that cause it to accelerate or change its motion.

🎯 Exam Tip: Unbalanced forces are directly responsible for causing acceleration (change in speed or direction) in an object, as described by Newton's second law.

 

Question 8. Explain: Force has magnitude as well as direction. OR Force is a vector quantity.
Answer: The effect of force applied to a body depends upon how much force we supply, i.e., the magnitude of the force, and the direction in which the force is applied. Consider a ball at rest on the ground. When- we push it, it starts rolling. The greater the applied force, the greater is the speed acquired by the ball.
Consider a body moving in a straight line, If we apply a force in the direction of motion of the body, the speed of the body increases. On the contrary, if we apply a force in the direction opposite to that of motion of the body, the speed of the body decreases. These (examples show that force has magnitude as well as direction, i.e., force is a vector quantity.
In simple words: Force is a vector because its effect depends on both how strong it is (magnitude) and the way it's applied (direction); pushing a ball harder or in a different direction changes its movement.

🎯 Exam Tip: Always remember that force is a vector quantity. This implies that when dealing with multiple forces, both their magnitudes and directions must be considered for vector addition.

 

Question 9. Explain the term balanced forces.
Answer: Consider a rigid body acted upon by two forces, equal in magnitude, opposite in . direction and having the same line of action. These forces are called balanced forces as their net effect on the body is zero.
Example: A glass slab kept on a table is acted upon by two balanced forces: (i) the weight of the slab acting downward and (ii) the upward force on the slab due to the table. Their net effect on the slab being zero, the slab remains at rest.
In simple words: Balanced forces are equal in strength and opposite in direction, acting on the same line, causing no change in an object's motion, like a book resting on a table.

🎯 Exam Tip: Ensure you include the "same line of action" detail when defining balanced forces, as forces of equal magnitude and opposite direction but different lines of action can cause rotation.

 

Question 10. Explain the term unbalanced force.
Answer: A single force acting on a body is an unbalanced force. It produces acceleration in the body. If two or more forces act on a body such that their resultant is not zero, the resultant is an unbalanced force responsible for accelerating the body.
Example: When a ball lying on the ground is hit with a bat, the ball is set in motion by the applied force.
In simple words: An unbalanced force is a net force on an object that is not zero, causing it to change its speed or direction, like a bat hitting a ball.

🎯 Exam Tip: The presence of an unbalanced force is the direct cause of acceleration, a core concept of Newton's second law. Be prepared to provide real-world examples.

 

Question 11. What will happen if the force is removed completely when an object acquires a certain speed?
Answer: If the force is removed completely when an object acquires a certain speed, the object will move with the velocity it has at the instant the force is removed.
For example, a body moving with constant speed along a circular path in a horizontal plane will fly tangentially in the sense of motion if the centripetal force (the force directed towards the centre of the circle) is removed completely.
In simple words: If a force is entirely removed from a moving object, it will continue to move at that exact speed and in the same direction, unless another force acts on it.

🎯 Exam Tip: This question relates to Newton's First Law (Law of Inertia). In an ideal, frictionless environment, an object in motion stays in motion at constant velocity if no net force acts on it.

 

Always Remember:

 

1. The tendency of an object to remain in its existing state is called its inertia.
2. This is why an object in stationary state remains in the same state and an object in motion remain in the state of motion in the absence of an external force.
In simple words: Inertia is an object's natural resistance to changes in its state of motion, meaning it tends to stay still if still, or keep moving if moving, without outside forces.

🎯 Exam Tip: Inertia is a fundamental concept in physics, directly related to an object's mass. The greater the mass, the greater its inertia.

 

Types Of Inertia:

 

1. Inertia of the state of rest: An object in the state of rest cannot change its state of rest due to its inherent property. This property is called the inertia of the state of rest.
2. Inertia of motion: The inherent property of an object due to which its state of motion cannot change, is called its inertia of motion. For example, a revolving' electric fan continues to revolve even after it is switched off, passengers sitting in the running bus get aerk in the forward direction if the bus suddenly stops.
3. Directional inertia: The inherent property of an object due to which the object cannot change the direction of its motion, is called directional inertia. For example, if a vehicle in motion along a straight line suddenly turns, the passengers sitting in it are thrown opposite to the direction of turning.
In simple words: There are three types of inertia: inertia of rest (staying still), inertia of motion (keeping moving), and directional inertia (maintaining direction), all resisting changes to an object's current state.

🎯 Exam Tip: Be sure to provide clear examples for each type of inertia (rest, motion, direction) to demonstrate your understanding of how inertia manifests in different situations.

 

Try This:

 

Activity 1:
Take a postcard and keep it on a glass. Keep a 5 Rupee coin on it. Now skilfully push the card. The coin straight away falls in the glass. Have you ever done this?
Answer: Yes. (Explanation: The postcard moves forward due to the applied force and then falls due to the earth's gravitational force. In the absence of adequate frictional force between the coin and the postcard, the coin does not move forward with the postcard, but straightaway falls in the glass due to the earth's gravitational force.)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक कांच के ऊपर एक पोस्टकार्ड और उस पर एक सिक्का रखा हुआ दिखाता है। जब पोस्टकार्ड को अचानक धकेला जाता है, तो सिक्का जड़त्व के कारण अपनी जगह पर रहता है और फिर गुरुत्वाकर्षण बल के कारण सीधे कांच में गिर जाता है।
In simple words: Yes, this experiment demonstrates inertia of rest; the coin stays put due to inertia while the card moves, then gravity pulls the coin into the glass.

🎯 Exam Tip: This is a classic demonstration of inertia of rest. Students should be able to explain why the coin falls into the glass rather than moving with the card.

 

Activity 2: Hang a half a kg mass from a stand, with a string 1. Tie another string 2 to the mass and keep it hanging. Now pull the string 2 with a jerk. The string 2 breaks but the mass does not fall. Heavy mass does not move. Now pull the string 2 slowly. The string 1 breaks and the mass fall down. This is because of the tension developed in the string 2 due to the mass.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक स्टैंड से लटके 0.5 किलोग्राम द्रव्यमान को दिखाता है, जो दो तार (स्ट्रिंग 1 और स्ट्रिंग 2) से जुड़ा है। स्ट्रिंग 1 द्रव्यमान को ऊपर से पकड़े हुए है, और स्ट्रिंग 2 द्रव्यमान के नीचे से लटकी है। यह प्रयोग जड़त्व (inertia) और बल के संचरण (transmission of force) को दर्शाता है।
(Explanation: (1) As no force acts on the mass, it remains at rest due to inertia. (2) The transmission of force results in the tension in the string 1. As the string 1 cannot withstand it, it breaks and the mass falls down.)
In simple words: This activity shows that a sudden jerk on the bottom string breaks it because of the mass's inertia, but a slow pull breaks the top string due to steady tension.

🎯 Exam Tip: This experiment highlights how inertia resists sudden changes in motion. The difference in outcomes (which string breaks) depends on the speed of the applied force.

 

Question 12. What are the three types of inertia?
Answer: Types of inertia:
• inertia of rest
• inertia of motion
• inertia of direction.
In simple words: The three types of inertia are inertia of rest (resisting change from stillness), inertia of motion (resisting change from movement), and inertia of direction (resisting change in direction).

🎯 Exam Tip: Being able to list and briefly define the three types of inertia is a basic requirement for understanding the concept of inertia.

 

Question 13. What is inertia of rest ? Give two examples of inertia of rest.
Answer: The inherent property of a body by virtue of which it cannot change its state of rest is called the inertia of rest.
Examples:
1. When we dust a carpet, the carpet moves but the dust particles in it remain at rest due to inertia and hence get separated from the carpet. Hence, the carpet becomes clean.
2. When a bus starts suddenly, the passengers experience a backwarderk due to inertia.
In simple words: Inertia of rest means an object at rest tends to stay at rest; for example, dust stays on a carpet when shaken, or passengers lean back when a bus starts.

🎯 Exam Tip: Provide clear and distinct examples for inertia of rest. The carpet and bus passenger examples are common and effective for illustration.

 

Question 14. What is inertia of motion? Give two examples of inertia of motion.
Answer: The inherent property of a body by virtue of which it cannot change its state of motion is called the inertia of motion.
Examples:
1. When a fan is switched off, its blades continue to rotate for some time. Due to internal friction and friction with air, the blades of the fan stop rotating after some time.
2. Passengers in a bus experience a forwarderk when the bus stops suddenly due to application of brakes.
In simple words: Inertia of motion means an object in motion tends to stay in motion; for example, a fan keeps spinning after being turned off, or bus passengers lurch forward when the bus brakes.

🎯 Exam Tip: When explaining inertia of motion, clarify that objects eventually stop due to opposing forces like friction, not because inertia ceases.

 

Question 15. What is inertia of direction? Give two examples of inertia of direction.
Answer: The inherent property of a body by virtue of which it cannot change its direction of motion is called the inertia of direction.
Examples:
1. While sharpening a knife, sparks fly off tangentially in the sense of motion from the grinding stone.
2. When a vehicle moves, the mud particles sticking to its wheels fly off tangentially in the sense of motion. Hence, mudguards are fitted to vehicles.
In simple words: Inertia of direction means an object moving in a certain way tends to keep going that way; like sparks flying straight off a grinder, or mud flying tangentially from a spinning tire.

🎯 Exam Tip: Use examples that clearly show objects resisting a change in their path of motion, not just speed. Mudguards are a great practical illustration.

 

Question 16. Why do we fall sideways when we are sitting in a bus and it takes a sharp turn?
Answer: When we sit in a bus and the bus is in motion, we are in a state of motion in the same direction. When the bus takes a sharp turn, our body tends to maintain the state of motion in the straight line due to inertia. The portion of our body in firm contact with the seat acquires the motion along the curved path, but the upper portion of our body, tends to move in the initial direction of motion. Hence, we fall sideways.
In simple words: When a bus turns sharply, our body wants to keep moving straight due to directional inertia, while the lower part moves with the bus, causing us to lean sideways.

🎯 Exam Tip: This question is an excellent application of directional inertia. Clearly explain how different parts of the body react differently to the sudden change in motion.

 

Question 17. What happens when you shake a wet piece of cloth? Explain your observation.
Answer:
1. When we shake a wet piece of cloth, water droplets come out.
2. Initially, the wet piece of cloth is at rest. When the cloth is shaken, it is accelerated, but the water droplets in it, due to inertia, tend to maintain the state of rest. Hence, the droplets come out.
In simple words: When you shake a wet cloth, water droplets fly off because they tend to stay at rest due to inertia even as the cloth moves quickly.

🎯 Exam Tip: This demonstrates inertia of rest. The cloth moves, but the water tries to remain in its initial state, leading to its separation from the cloth.

 

Question 18. If brakes are suddenly applied to a moving car, the passengers in the car are pushed in the forward direction. Explain why.
Answer:
1. The passengers in a moving car have the same velocity as that of the car. When brakes are suddenly applied to the car, it stops suddenly and the lower parts of the passengers' bodies in contact with the seats, come to rest.
2. The upper parts of their bodies, however, continue to be in a state of motion due to inertia. Hence, the passengers are pushed in the forward direction.
In simple words: When a car brakes suddenly, passengers lurch forward because their upper body, due to inertia of motion, tries to keep moving at the car's original speed.

🎯 Exam Tip: This is a classic example of inertia of motion. Distinguish between the motion of the car and the tendency of the passengers' bodies to maintain their original state.

 

Question 19. A person alighting from a moving train is likely to fall in the direction of motion of the train. Explain why.
Answer:
1. A person in a moving train has the same velocity as that of the train. After alighting from the moving train his feet come to rest on the platform.
2. However, due to inertia, the upper part of his body continues to be in a state of motion in the direction of motion of the train. Hence, he is likely to fall in the direction of motion of the train.
In simple words: A person jumping off a moving train tends to fall forward because their body, due to inertia of motion, wants to continue moving at the train's speed, even if their feet stop on the platform.

🎯 Exam Tip: This is another strong example of inertia of motion. Emphasize the difference in state of motion between the feet (at rest) and the upper body (still in motion).

 

Try This:

 

Activity 3: Take some sharp pointed nails and push them into a wooden plank by hammering on their heads. Now take- a nail and hold it with its head on the plank and hammer it down from the pointed end. When pressing the drawing pins into a drawing board, they get into the board easily. By applying a force using the thumb one can push the pins into the boards. On the contrary, while pressing ordinary pins into the board with a thumb, the thumb may get hurt.
What does this simple experiment tell?
The nail easily penetrates into wood from its pointed end. From this you will notice that when a force is applied on the head of the nail, it is easy to hammer it into the plank.
Explanation: The less the area of the surface on which the force is applied, the greater is the effect of the force.
In simple words: This experiment shows that a smaller contact area results in greater pressure; pointed nails or pins penetrate easily because the same force is concentrated over a tiny surface.

🎯 Exam Tip: This activity vividly illustrates the inverse relationship between pressure and area (P=F/A). Explain how a small area concentrates force, leading to high pressure.

 

Question 20. Define pressure.
Answer: The force exerted perpendicularly on a unit area is called pressure.
In simple words: Pressure is the amount of force pushing down on a specific unit of surface area.

🎯 Exam Tip: The key here is "perpendicularly" - force normal to the surface. Also, remember the formula P = F/A.

 

Use Your Brain Power!

 

Question 1. It is easy to cut vegetables, fruits with a sharp knife. A blunt knife does not work here. Why does this happen?
Answer: The effect of a given force varies - inversely as the area of the surface on (which the force is applied. The less the surface area, the greater is the effect of the force. The cutting edge of a sharp- knife has less cross sectional area relative to that of a blunt knife. Hence, it is easy to cut vegetables, fruits with a sharp knife, rather than with a blunt knife. For a given force, pressure is inversely proportional to the area of the surface on which the force acts.
In simple words: A sharp knife cuts easily because it concentrates the force over a very small area, creating high pressure, unlike a blunt knife which spreads the force out.

🎯 Exam Tip: This question is a practical application of the pressure formula (P=F/A). Focus on how reducing the area of contact dramatically increases pressure, making cutting easier.

 

Question 21. State the formula for pressure, Hence, determine the unit of pressure.
Answer: Pressure = \( \frac{\text{force}}{\text{area on which the force is applied}} \)
\( \implies \) The unit of pressure = \( \frac{\text{the unit of force}}{\text{the unit of area}} \)
The SI unit of force is the newton (N) and that of area is m². Therefore, the SI unit of pressure is N/m². It is called the pascal (Pa). \( 1 \text{ Pa} = 1 \text{ N/m}^2 \).
[Note: The unit pascal is named in honour of Blaise Pascal (1623-62), French mathematician, physicist and philosopher.]
In simple words: Pressure is calculated by dividing force by area, and its SI unit is the Pascal (Pa), which is equal to one Newton per square meter (N/m²).

🎯 Exam Tip: Memorize the pressure formula (P=F/A) and its SI unit (Pascal or N/m²). Understanding the derivation of the unit is also important.

 

Question 22. State the CGS unit of pressure. State the relation betweeen the Sl and CGS units of pressure.
Answer: The CGS unit of pressure is the dyne/cm².
\( 1 \text{N} = 10^5 \text{ dynes} \), \( 1\text{m}^2 = (10^2 \text{ cm})^2 = 10^4 \text{ cm}^2 \)
\( \implies 1 \text{ Pa} = 1 \text{ N/m}^2 = \frac{10^5 \text{ dynes}}{10^4 \text{ cm}^2} = 10 \text{ dynes/cm}^2 \).
In simple words: The CGS unit for pressure is dyne/cm²; one Pascal (SI unit) is equivalent to 10 dyne/cm².

🎯 Exam Tip: Be able to convert between SI (Pascal) and CGS (dyne/cm²) units for pressure. This requires knowing the conversion factors for force (Newton to dyne) and area (m² to cm²).

 

Question 23. State the factors on which pressure depends.
Answer: Pressure depends on the applied force and the area of the surface on which the force is applied.
In simple words: Pressure depends on two main things: how much force is applied and how big the area over which that force is spread.

🎯 Exam Tip: Clearly state the direct relationship with force and the inverse relationship with area. This fundamental understanding is crucial for all pressure-related problems.

 

Question 25.With neat diagrams, describe an experiment to show that pressure increases if the surface area is decreased, keeping the applied force the same.
Answer:Take a brick measuring 20 cm x 10 cm x 5 cm. Take some clay in a glass trough. Add water to it and knead it into a soft dough. Place the brick on the dough with one of its faces measuring 20 cm x 10 cm in contact with the dough. Observe how deep the brick penetrates into the dough.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक ईंट को नरम मिट्टी के आटे पर अलग-अलग तरीकों से रखा हुआ दिखाता है। चित्र (a) में ईंट को उसके 20 सेमी x 10 सेमी वाले सबसे बड़े फलक पर रखा गया है, जबकि चित्र (b) में इसे उसके 10 सेमी x 5 सेमी वाले छोटे फलक पर रखा गया है, यह दर्शाता है कि छोटे क्षेत्र पर अधिक दबाव पड़ता है। Clean the brick and place it on the dough with one of its faces measuring 10 cm x 5 cm in contact with the dough. Observe how deep the brick penetrates into the dough. You will find that the brick penetrates deeper in this case than that in the first case.
- In the first case, the weight of the brick acts on a surface area of 200 cm².
- In the second case, the weight of the brick acts on a surface area of 50 cm².
- This shows that pressure increases if the surface area is decreased, keeping the applied force the same.In simple words: This experiment demonstrates that for the same force (weight of the brick), a smaller contact area results in greater pressure, causing the brick to sink deeper into the soft dough.

🎯 Exam Tip: When describing experiments, clearly state the objective, materials, procedure, and observation, followed by a conclusion based on scientific principles.

 

Question 26.State the unit for pressure used in atmospheric science. How is it related to the unit pascal?
Answer:In atmospheric science, the unit used for pressure is the bar. 1 bar = 10 Pa (pascal).In simple words: In atmospheric science, pressure is measured in bars, and one bar is equal to 10 pascals.

🎯 Exam Tip: Remember the specific units used in different scientific contexts, especially for pressure (Pascal, bar, N/m²), and their conversion factors.

 

Try this:

Question 1.Do the activity as depicted in Figure What is seen?
Answer:In Fig.(a), the plank is horizontal. In Fig.(b), four books are placed side by side on the plank. The plank bends slightly due to the pressure produced by the weight of the books.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक क्षैतिज तख्ते पर पुस्तकों के वजन के कारण लगने वाले दबाव को दर्शाता है। चित्र (a) में तख्ता खाली है, (b) में चार पुस्तकें अगल-बगल रखी हैं जिससे तख्ता थोड़ा झुकता है, और (c) में चार पुस्तकें एक के ऊपर एक रखी हैं, जिससे दबाव छोटे क्षेत्र पर केंद्रित होता है और तख्ता अधिक झुकता है। In Fig.(c), the four books are placed one above the other in the middle of the plank. Here the area of the surface of the plank on which the force acts is reduced by a factor of four relative to the earlier case. Hence, the plank bends considerably. [Note: This shows that for a given force, pressure varies inversely as the area of the surface on which the force acts.]In simple words: When books are placed on a plank, the plank bends more when the books are stacked (smaller area of contact) compared to when they are spread out (larger area of contact), demonstrating that pressure increases as the contact area decreases.

🎯 Exam Tip: Visualizing experiments helps understand abstract concepts. Focus on the relationship between force, area, and pressure shown in such activities.

 

Use your brainpower!

Question 1.You must have seen a vegetable vendor carrying a basket on her head. She keeps a twisted piece of cloth on the head, below the basket. How does it help?
Answer:Keeping a twisted piece of cloth on the head increases the area of the surface on which the weight of the basket containing vegetables acts. Hence, the pressure produced by the force (weight) is reduced and it becomes easier to carry the basket.In simple words: A twisted cloth increases the contact area between the basket and the head, distributing the weight over a larger area and thus reducing the pressure, making it more comfortable to carry the load.

🎯 Exam Tip: Real-life examples effectively illustrate scientific principles. Relate this to the inverse relationship between pressure and area.

 

Question 2.We cannot stand at one place for a long time. How then can we sleep on a place for 8 and odd hours?
Answer:When we stand, our weight acts on relatively small surface area, resulting in increased tension on the muscles of our legs. Hence, we cannot stand at one place for a long time. When we sleep, our weight acts on relatively large surface area, resulting in comparatively reduced tension. Therefore, we can sleep on a place for 8 and odd hours.In simple words: Standing concentrates body weight on a small area, causing higher pressure and muscle strain, while sleeping distributes weight over a much larger area, reducing pressure and allowing for longer rest.

🎯 Exam Tip: Practical applications of physics concepts like pressure distribution are important. Consider how surface area impacts comfort and endurance.

 

Question 3.For skiing on ice, why are long flat skis used?
Answer:The pressure produced by a given force depends on the area of the surface on which the force acts. The greater the surface area, the less is the pressure produced. The skis used to slide over snow are long and flat so that the area is increased and hence the pressure is decreased. This makes it easier to slide over snow.In simple words: Long, flat skis increase the contact area with the snow, which reduces the pressure exerted by the skier's weight. This reduced pressure prevents the skier from sinking deep into the snow, making it easier to glide.

🎯 Exam Tip: Understanding how design (like ski shape) is optimized based on physical principles (pressure and area) is a common exam focus.

 

Question 27.Why do the blades of a pair of scissors have sharp edges?
Answer:The pressure produced by a given force depends on the area of the surface on which the force acts. The less the surface area, the greater is the pressure produced. The blades of a pair of scissors have sharp edges so that the area is decreased and hence the pressure is increased. This makes cutting - an object such as cloth easier. [Note: Answers to questions such as why is the blade of an axe sharp? or why is the blade of a saw sharp? can be written on the basis of the answer given above.]In simple words: Scissors blades are sharp to minimize the contact area, which increases the pressure for a given force, making it easier to cut through materials.

🎯 Exam Tip: This question is a classic example of pressure application. Remember that sharp objects work by concentrating force onto a tiny area.

 

Question 28.Why does a needle have s sharp point?
Answer:The pressure produced by a given force depends on the areas of the surface on which the force acts. The less the surface area, the greater is the pressure produced. A needle has a sharp point so that the area is decreased and hence the pressure is increased. This makes the action of piercing easier.In simple words: A needle has a sharp point to create a very small contact area. This small area, under applied force, generates high pressure, allowing it to pierce materials easily.

🎯 Exam Tip: Similar to scissors, the sharp point of a needle illustrates the principle of pressure being inversely proportional to the area. Always explain this relationship.

 

Question 29.Why do school bags have broad shoulder straps?
Answer:The pressure produced by a given force depends on the area of the surface on which the force acts. The greater the surface area, the less is the pressure produced. School bags have broad shoulder straps so that the weight of the bag is distributed over a large surface area thereby decreasing the pressure on the shoulders of the student carrying the bag.In simple words: School bags have broad shoulder straps to spread the weight of the bag over a larger area of the shoulders. This reduces the pressure, making the bag more comfortable to carry.

🎯 Exam Tip: This is another practical application of pressure. Always link broader straps to increased area and decreased pressure for comfort.

 

Question 30.How will the pressure change if the area is doubled keeping the force the same?
Answer:If the area is doubled, keeping the force the same, the pressure will become half the initial pressure.In simple words: Since pressure is inversely proportional to area, doubling the area while keeping the force constant will halve the pressure.

🎯 Exam Tip: Remember the formula \( P = F/A \). If A doubles, P becomes \( F/(2A) \), which is half of \( F/A \).

 

Question 31.How will the pressure change if the force is doubled, keeping the area the same?
Answer:If the force is doubled, keeping the area the same, the pressure will become double the initial pressure.In simple words: Since pressure is directly proportional to force, doubling the force while keeping the area constant will double the pressure.

🎯 Exam Tip: Remember the formula \( P = F/A \). If F doubles, P becomes \( (2F)/A \), which is double of \( F/A \).

 

Question 32.State the characteristics of the pressure due to a liquid (or a fluid in general). OR Write a short note on the pressure due to a liquid (a fluid in general).
Answer:Characteristics of the pressure due to a liquid (or a fluid):
1. The pressure at a point in a liquid (or a fluid) is due to the weight of the liquid (fluid) column above that point.
2. It acts on all sides of the container.
3. At a given depth it is the same in all directions.
4. It is independent of the size and shape of the container.
5. It is proportional to the height of the liquid (fluid) column above the given point.
6. It is proportional to the density of the liquid (fluid).
7. It is proportional to the acceleration due to gravity at the given place. [Note: The pressure exerted by a liquid (or gas or fluid) at a depth h below the free surface of the liquid = hpg, where p is the density of the liquid (or gas or fluid) and g is the acceleration due to gravity.]In simple words: Liquid pressure depends on depth, density, and gravity, acts equally in all directions at a given depth, and is independent of the container's shape.

🎯 Exam Tip: Memorize the formula \( P = h\rho g \) and each characteristic point. This is a fundamental concept for understanding fluid mechanics.

 

Question 33.Give two examples to show that air exerts equal pressure in all directions.
Answer:
1. When air is filled in a balloon, it acquires its characteristic shape such as round or oval.
2. When a bicycle tube is filled with air, it acquires its characteristic (tube-like) shape throughout. This shows that air exerts equal pressure in all directions.In simple words: Air-filled objects like balloons and bicycle tubes take a uniform shape because air pressure acts equally from all internal directions on their surfaces.

🎯 Exam Tip: Visualizing how objects inflate uniformly is a good way to remember that fluid pressure (like air) acts isotropically.

 

Question 34.Whatisafluid?Givetwoeamples.
Answer:A fluid is a substance which can flow. Examples:
1. Water (liquid)
2. Air (gas) [Note: Liquids and gases together are called fluids. Gases have very low viscosity compared to liquids. A liquid with low viscosity flows easily. A liquid with high viscosity does not flow easily.]In simple words: A fluid is any substance that can flow, including both liquids (like water) and gases (like air).

🎯 Exam Tip: Distinguish between solids, liquids, and gases based on their ability to flow. Fluids encompass both liquids and gases.

 

Question 35.Take two rubber balloons. Fill one with water and blow air into the other. Now prick both balloons with a pin. What do you observe?
Answer:Water and air both come out of the balloons. Air escapes quickly compared to water and produces a loud sound.In simple words: Both water and air escape when their respective balloons are pricked, but air escapes faster and with an audible sound due to its lower density and higher compressibility compared to water.

🎯 Exam Tip: This experiment highlights differences in fluid properties (compressibility, viscosity) between gases and liquids when under pressure.

 

Question 36.State the characteristics of pressure exerted by a fluid. OR Write a short note on pressure exerted by a fluid.
Answer:Characteristics of pressure exerted by a fluid:
1. A fluid due to its weight, exerts pressure on the base as well as the walls of the container that holds it.
2. A fluid exerts pressure on a body immersed in it.
3. The pressure exerted on any confined mass of fluid is transmitted undiminished in all directions. [Note: The pressure exerted by a fluid is a scalar quantity.]In simple words: Fluid pressure acts on all container surfaces and immersed bodies, is transmitted equally throughout a confined fluid, and its magnitude is a scalar quantity.

🎯 Exam Tip: This question repeats aspects of Question 32, emphasizing the key properties of fluid pressure, especially Pascal's principle (point 3).

 

Question 37.Take an empty can. Pour small quantity of water in it. Boil this water for a few minutes until the steam has driven out most of the air. Now close the can with the stopper tightly. Allow it to cool by pouring cold water over it. What do you observe?
Answer:The can gets gradually crushed. [Note: The steam inside the can condenses to form water as the can cools. Therefore, the pressure inside the can becomes much less than the external pressure of the air. Hence, the can gets crushed.]In simple words: Boiling water in a sealed can and then cooling it causes the can to crush because the internal steam condenses, creating a low-pressure vacuum that cannot withstand the greater external atmospheric pressure.

🎯 Exam Tip: This is a classic demonstration of atmospheric pressure. Ensure you explain both the internal pressure drop (condensation) and the external atmospheric pressure.

 

Question 38.Put a folded newspaper on a plastic bag. Blow air into the bag. What do you observe?
Answer:The plastic bag inflates as air is blown into it. This raises the folded news-paper put on the bag.In simple words: Blowing air into the bag increases the internal pressure, causing it to inflate and lift the newspaper placed on top.

🎯 Exam Tip: This simple activity illustrates how increased internal air pressure can exert an upward force, lifting objects.

 

Question 39.What is meant by atmospheric pressure?
Answer:The earth is surrounded by air from all sides. This layer of air is called the atmosphere. Its density is high up to about 16 km from the earth's surface. Beyond that, up to about 400 km, its density is very low. Air, due to its weight, exerts pressure on the surface of the earth. The pressure exerted by air or the atmosphere surrounding the earth is known as the atmospheric pressure. It is the ratio of the weight of the air to the area of the surface of the earth. It decreases with altitude as the density of air decreases with altitude and also the weight of the air column above a given place. [Note: At sea level the atmospheric pressure is about 105 Pa. We do not feel it because the pressure of blood and other fluids in our body balances it.]In simple words: Atmospheric pressure is the force exerted by the weight of the air column above a given area on the Earth's surface, decreasing with altitude due to less air above.

🎯 Exam Tip: Define atmospheric pressure clearly, relating it to the weight of the air column. Also, mention its variation with altitude and why humans don't "feel" it constantly.

 

Question 40.State the relation between 1 atmosphere and the pascal.
Answer:1 atmosphere = (about) \( 101 \times 10^3 \) Pa (pascal) [Note: The air pressure at the sea level is (about) 1 atmosphere. 1 atmosphere = 101325 Pa. 1 bar = 103 mbar (millibar) 1 mbar w 103 Pa (hectopascal) Atmospheric pressure is expressed in mbar or hectopascal (hPa).]In simple words: One atmosphere, a standard unit for pressure, is approximately equal to 101,000 pascals, which is the SI unit of pressure.

🎯 Exam Tip: Know the conversion factor between atmospheres and pascals. Also, be aware of other units like bar, mbar, and hPa for atmospheric pressure.

 

Figures for reference :
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र वायुमंडलीय दबाव को दो तरीकों से दर्शाता है। पहला चित्र एक काल्पनिक वायु स्तंभ (Air column) दिखाता है जो पृथ्वी की सतह पर दबाव डालता है। दूसरा ग्राफ (Atmospheric pressure (kPa) vs Height (m) from the earth's surface) दर्शाता है कि ऊंचाई बढ़ने के साथ वायुमंडलीय दबाव (किलोपास्कल में) कैसे घटता जाता है, जो समुद्र तल पर अधिकतम और अधिक ऊंचाई पर न्यूनतम होता है।

 

Use your brain power!

Question 1.At the sea level the atmospheric pressure \( 101 \times 10^3 \) Pa is acting on a table top of size 1 m². Under such a heavy pressure, why doesn't the table top crumble down?
Answer:The air below the table top exerts pressure \( 101 \times 10^3 \) Pa on it in the upward direction. Hence, the table top doesn't crumble down.In simple words: The table top doesn't crumble because atmospheric pressure acts equally on both its top and bottom surfaces, resulting in balanced forces that prevent collapse.

🎯 Exam Tip: This question highlights the balanced nature of atmospheric pressure. Ensure you explain that pressure acts from all directions, including upwards.

 

Question 41.Think - Why?
(i) Some people feel their ears popping at the top of a mountain.
Answer:Atmospheric pressure decreases with altitude. At the top of a mountain, it becomes less than the internal pressure in the ear. Hence, some people feel their ears popping at the top of a mountain.
(ii) Some people feel breathless as they climb higher and higher on a mountain.
Answer:Atmospheric pressure decreases with altitude. Hence, some people feel breathless as they climb higher and higher on a mountain.
(iii) We can enjoy a cold-drink or fruit juice with the help of a straw but can we imagine drinking a cold-drink or fruit juice on the moon using a straw?
Answer:When we suck air in the straw, the pressure of the air in the straw becomes less than that of the outside air on the cold drink or fruit juice in the bottle (or the glass). Hence, the cold drink or fruit juice rises in the straw and enters our mouth. We can then drink it. There is no atmosphere on the moon. Hence, it is not possible to enjoy a cold drink or fruituice on the moon by using a straw.
(iv) People are often advised not to carry fountain pens while travelling by air.
Answer:The ink in a fountain pen (filled at sea level at atmosphere pressure) may come out through its mouth while travelling by air if the outside pressure becomes less than the pressure in the ink holder of the pen. This can spoil the clothes/purse/bag. Hence, people are often advised not to carry fountain pens while travelling by air.In simple words: These phenomena (ear popping, breathlessness, straw function, leaking pens) all occur due to changes in external atmospheric pressure, which decreases with increasing altitude.

🎯 Exam Tip: These are all direct consequences of changes in atmospheric pressure. Understand that pressure differences drive these effects; a lower external pressure leads to expansion or outward flow from a higher internal pressure zone.

 

Question 42.Why do some people have earache when they travel by an aeroplane?
Answer:When an aeroplane descends at a high speed, there is an increase in air pressure. This increases the pressure on the ear drum. Hence, some people have earache when they travel by an aeroplane.In simple words: During an airplane's descent, the external air pressure rapidly increases, pushing on the eardrum, which can cause earache due to the pressure difference.

🎯 Exam Tip: Rapid pressure changes, especially increases during descent, can affect the body. Focus on the pressure difference across the eardrum.

 

Question 43.Explain why a person may bleed from the nose when at a great height above the sea level.
Answer:The pressure exerted by the blood in blood capillaries is slightly more than the atmospheric pressure and acts in a direction opposite to that of the atmospheric pressure. Atmospheric pressure decreases with height and at a great height above the sea level, it is very low. As a result, there arises a difference in the internal and external pressures on the walls of the cells and blood capillaries. If the difference is large, it may cause the cell wall and the blood capillaries to burst. Thus, the capillaries in the nose (and ear) may burst causing bleeding.In simple words: At high altitudes, the external atmospheric pressure drops significantly. Since internal blood pressure remains relatively constant and higher, the large pressure difference can cause fragile capillaries in the nose to burst, leading to bleeding.

🎯 Exam Tip: This question requires explaining the pressure difference between internal body fluids and external atmosphere. Highlight the fragility of capillaries at high pressure differentials.

 

Question 44.When a rubber sucker is pressed onto a flat glass surface, it sticks tightly on the surface. Why? You need a large force to separate it from the surface. Why?
Answer:When a rubber sucker is pressed onto a flat glass surface, practically all the air between the surfaces of the sucker and the glass is pushed out. The air pressure there becomes much less than the atmospheric pressure. Hence, the sucker sticks to the glass due to the external atmospheric pressure. The atmospheric pressure is about 105 Pa. It is very large. Hence, to work against it to separate the sucker from the glass, a large force is needed.In simple words: A rubber sucker sticks because pressing it removes most of the air underneath, creating a partial vacuum. The higher external atmospheric pressure then pushes the sucker tightly against the surface, requiring a strong force to overcome it.

🎯 Exam Tip: This is a classic demonstration of atmospheric pressure. Explain the creation of a pressure differential and how the greater external pressure holds the sucker in place.

 

Question 45.Describe a simple experiment to demonstrate atmospheric pressure.
Answer:Fill a glass completely with water (to its brim) and cover it with a flat and stiff card paper (or a piece of glass). Holding your palm on the card, turn the glass upside down and take the palm away from the card.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक प्रयोग को दर्शाता है जिसमें एक पानी से भरा गिलास एक कार्डबोर्ड से ढका हुआ है और उल्टा पकड़ा गया है। निचले तीरों से दर्शाया गया वायुमंडलीय दबाव ऊपर की ओर कार्य कर रहा है, जो पानी के वजन से अधिक है, जिससे कार्डबोर्ड अपनी जगह पर बना रहता है और पानी नहीं गिरता। You will find that the water does not spill. The atmospheric pressure on the card (acting upward) is greater than the pressure of the water in the glass (acting downward). Hence, the water in the glass does not spill.In simple words: Filling a glass with water, covering it with a card, and inverting it demonstrates atmospheric pressure. The upward atmospheric pressure on the card is greater than the downward pressure of the water, preventing the water from spilling.

🎯 Exam Tip: This is a common and effective experiment to show atmospheric pressure. Clearly explain the opposing forces (water's weight vs. atmospheric pressure) and why the water stays in.

 

Question 46.Explain the working of an ink dropper. [Application]
Answer:An ink dropper consists of a tube of glass or plastic, with one end tapering to a narrow opening and the other end fitted with a small rubber bulb. When the narrow open end is dipped into the ink and the rubber bulb is pressed, some air in the tube escapes through the open end. This reduces the air pressure inside the dropper.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक स्याही ड्रॉपपर की संरचना को दर्शाता है, जिसमें एक रबर का बल्ब, एक ट्यूब और एक संकीर्ण निचला सिरा है। यह उपकरण वायुमंडलीय दबाव का उपयोग करके स्याही को ट्यूब में खींचता है जब बल्ब दबाकर हवा बाहर निकाली जाती है, और फिर स्याही को बाहर निकालने के लिए फिर से दबाया जाता है। On releasing the bulb, the atmospheric pressure on the ink pushes the ink into the dropper. The dropper is then taken out and its open end is held over the open barrel of the pen. The bulb is then pressed so that the ink in the dropper enters the pen. [Note: The medicine dropper works in the same manner.]In simple words: An ink dropper works by using atmospheric pressure. Squeezing the bulb expels air, creating a lower pressure inside. Releasing the bulb allows the higher external atmospheric pressure to push ink up into the dropper.

🎯 Exam Tip: The ink dropper's function is a direct application of atmospheric pressure. Focus on the creation of a pressure differential by squeezing and releasing the bulb.

 

Question 47.Why is the opening of a dropper very narrow?
Answer:The pressure produced by a given force is inversely proportional to the area of the surface on which the force acts. The opening of a dropper is very narrow. Hence, its area of cross-section is very small. As a result, even if the dropper has a small amount of ink it, its pressure can equal the atmospheric pressure. As the opening is narrow, it is easier to transfer the ink to the pen. The possibility of ink spilling is very low.In simple words: The dropper's narrow opening ensures that a small force generates high pressure, making it easy to release ink, while also preventing spills due to the small surface area.

🎯 Exam Tip: Relate the narrow opening to a small area, which, by the pressure formula (\( P = F/A \)), results in high pressure for easy ink transfer and reduced spillage.

 

Question 48.What is the characteristic of the cap of eye drop bottles?
Answer:The cap of an eye drop bottle is fitted with a dropper.In simple words: Eye drop bottles often have caps integrated with droppers for precise and controlled dispensing of the liquid.

🎯 Exam Tip: This question highlights a common design feature. Recognize how practical tools incorporate basic physics principles (like droppers and pressure) for specific functions.

 

Question 49.Explain the working of a syringe used by children when they play with coloured water. [Application]
Answer:As shown in the figure, a syringe used by children when they play with coloured water consists of a cylinder made of plastic or metal fitted with a piston. One end of the cylinder is in the form of a narrow
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सिरिंज की संरचना को दर्शाता है, जिसमें एक हैंडल, पिस्टन रॉड के साथ पिस्टन, ढक्कन, सिलेंडर और एक संकीर्ण नली शामिल हैं। यह वायुमंडलीय दबाव के सिद्धांत पर कार्य करता है, जहां पिस्टन को खींचने या धकेलने से सिलेंडर के अंदर दबाव बदलता है, जिससे तरल पदार्थ अंदर या बाहर खींचता है। tube. The snugly fitting piston can slide in, and out smoothly. The rod connected to the piston passes through a hole in the centre of the lid and has a handle at the other end. When the tip of the narrow tube is dipped in coloured water (or any other liquid) and the piston is pushed towards the tip, up to the bottom, most of the air in the cylinder escapes through the tube, reducing the pressure. When the piston is moved up, the coloured water rises in the part of the cylinder below the piston due to the atmospheric pressure. Finally, the inner pressure equals the atmospheric pressure and no more coloured water enters in or comes out. To spray the coloured water, the tube is taken out and the piston is moved towards the opening of the tube. As the inner pressure is now greater than the atmospheric pressure, the coloured water gushes out of the narrow opening of the tube.In simple words: A syringe works by creating pressure differences. Pushing the piston expels air, then pulling it up reduces internal pressure, allowing external atmospheric pressure to push water in. Pushing the piston again increases internal pressure, expelling the water.

🎯 Exam Tip: Focus on how the piston movement creates areas of high and low pressure, enabling the suction and expulsion of liquid due to atmospheric pressure.

**Question 50. How does the doctor's syringe work?**
Answer: As shown in the figure, a syringe used by children when they play with coloured water consists of a cylinder made of plastic or metal fitted with a piston. One end of the cylinder is in the form of a narrow
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सिरिंज के घटकों को दर्शाता है, जिसमें हैंडल, पिस्टन रॉड के साथ पिस्टन, ढक्कन, सिलेंडर और एक संकीर्ण नली शामिल हैं। यह एक सामान्य सिरिंज की संरचना को दर्शाता है जिसका उपयोग द्रव को भरने या निकालने के लिए किया जाता है। tube. The snugly fitting piston can slide in, and out smoothly. The rod connected to the piston passes through a hole in the centre of the lid and has a handle at the other end. When the tip of the narrow tube is dipped in coloured water (or any other liquid) and the piston is pushed towards the tip, up to the bottom, most of the air in the cylinder escapes through the tube, reducing the pressure. When the piston is moved up, the coloured water rises in the part of the cylinder below the piston due to the atmospheric pressure. Finally, the inner pressure equals the atmospheric pressure and no more coloured water enters in or comes out. To spray the coloured water, the tube is taken out and the piston is moved towards the opening of the tube. As the inner pressure is now greater than the atmospheric pressure, the coloured water gushes out of the narrow opening of the tube. The tip of a syringe is fitted with a very fine and hollow needle. The required quantity of medicine can be taken in the syringe with the help of the piston. The medicine can then be injected into the body of a patient using the needle and the piston.
In simple words: A doctor's syringe works by creating a pressure difference. Pushing the piston forces air out, creating low pressure inside; pulling it back allows atmospheric pressure to push liquid in. For injection, the piston pushes the liquid out through a fine needle.

🎯 Exam Tip: Understanding the role of atmospheric pressure and piston movement in creating pressure differences is key to explaining syringe functionality. Focus on how pressure changes facilitate fluid movement.

**Question 51. Explain the origin of pressure due to a gas enclosed in a container.**
Answer: Molecules of a gas are in a state of continuous random motion. These molecules possess energy due to motion. When the molecules collide with a wall of the container, they rebound. A large number of molecules collide with the wall every second. Hence, a significant force is exerted on the wall. Pressure is the force per unit surface area. Thus, a pressure is exerted by the gas on a wall of the container.
In simple words: Gas pressure originates from countless gas molecules constantly moving and colliding with the container walls. Each collision exerts a tiny force, and the cumulative effect of these many collisions over a unit area creates the observable pressure.

🎯 Exam Tip: When explaining gas pressure, remember to mention the random motion of molecules, their collisions with the container walls, and how these collisions result in a force per unit area. This directly links microscopic behavior to macroscopic pressure.

Try This: Buoyant Force

Take a plastic bottle and fix the lid tightly. Now place it in wafer and see. It will float on water. Try and push it into the water. Even when pushed, it continues to float. This experiment can also be done with a plastic hollow ball. Now fill the bottle with water to the fullest capacity and close the lid, and release in water. The bottle will float inside the water. Why does this happen?
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक प्लास्टिक की बोतल को पानी में तैरते हुए दिखाया गया है, जिस पर दो बल कार्य कर रहे हैं - नीचे की ओर गुरुत्वाकर्षण बल (\(f_g\)) और ऊपर की ओर उत्प्लावन बल (\(f_b\))। यह बोतल के पानी में तैरने की स्थिति को दर्शाता है जब दोनों बल संतुलित होते हैं। Balanced and unbalanced buoyant force
Answer: The empty plastic bottle floats on the surface of water. On the contrary, the bottle full of water floats inside water but does not go to the bottom. The weight of the empty bottle is negligible as compared with the weight of the water inside. Such a bottle with water neither floats on the surface, nor does it go to the bottom. This means the force due to gravity acting downwards (\(f_g\)), must have been balanced by an opposing force in the upward direction (\(f_b\)) on the bottle filled with water. This force must have originated from the water surrounding the bottle. The upward force acting on the object in water or other fluid or gas is called buoyant force (\(f_g\)).
In simple words: An empty bottle floats on water because its weight is less than the buoyant force. A water-filled bottle floats inside the water because its weight (including water) is balanced by the buoyant force from the displaced water, allowing it to remain suspended without sinking or fully surfacing.

🎯 Exam Tip: For problems involving buoyancy, always compare the object's weight with the buoyant force. If weight is less than buoyant force, it floats; if greater, it sinks; if equal, it floats submerged.

**Question 52. Define buoyant force.**
Answer: The upward force acting on the object in water or other fluid or gas is called the buoyant force.
In simple words: Buoyant force is the upward push exerted by a fluid on an object immersed in it, which counteracts the object's weight.

🎯 Exam Tip: A clear, concise definition of buoyant force as the upward force exerted by a fluid on an immersed object is essential. Emphasize its upward direction.

**Question 53. State the factors on which the buoyant force depends.**
Answer: The buoyant force depends upon the volume of the object immersed in the fluid (V), the density of the fluid (\(p_1\)) and the acceleration due to gravity (g) at that place. [Note: Magnitude of the buoyant force = \(Vp_1 g\).]
In simple words: Buoyant force depends on three key factors: the volume of the object submerged in the fluid, the density of that fluid, and the acceleration due to gravity.

🎯 Exam Tip: Remember the formula for buoyant force, \(F_b = V\rho g\), to easily recall the three factors it depends on: Volume of displaced fluid, density of fluid, and acceleration due to gravity.

Use Your Brain Power!

**Question. While pulling a bucket from a well, the bucket full of water immersed fully in water appears to weigh less than when it has been pulled out of water. Why?**
Answer:1. When a bucket full of water is immersed in water, the net force acting on the bucket = weight of the bucket full of water-the buoyant force exerted by the water on the bucket. The buoyant force due to a fluid is proportional to the density of the fluid. 2. The density of water is much greater than that of air. Therefore, the buoyant force acting on a bucket full of water while it is in water is much greater than that when it is out of water, i.e., in air. Hence, it appears to weigh less, while it is in water than when it has been pulled out of water.
In simple words: A bucket of water feels lighter when submerged because the water exerts an upward buoyant force, reducing the effective downward weight. Once out of the water, this buoyant support from the surrounding fluid (water) is lost, and it feels its full weight.

🎯 Exam Tip: This question tests understanding of buoyant force and apparent weight. Highlight that the buoyant force opposes gravity and is significantly greater in water than in air due to water's higher density.

Try This:

**Question 1. Take a piece of thin aluminum sheet and dip it in water in a bucket. What do you observe?**
Answer: It sinks in the water.
In simple words: A flat aluminum sheet sinks because its density is greater than water's, and the shape doesn't displace enough water to create a buoyant force equal to its weight.

🎯 Exam Tip: Connect the observation directly to density. If an object's density is greater than the fluid's, it will sink.

**Question 2. Now shape the same piece of aluminium into a small boat and place it on the surface of water. It floats, isn't it?**
Answer: Yes, the boat floats on the surface of water.
In simple words: Yes, shaping the aluminum into a boat makes it float because it now displaces a larger volume of water, increasing the buoyant force to balance its weight.

🎯 Exam Tip: This illustrates the principle that an object's shape can significantly affect the volume of fluid it displaces, thereby altering the buoyant force and its ability to float.

**Question 54. A lemon sinks in a glass filled with water but it floats when we stir in two spoons of salt in the water. Explain why.**
Answer: When salt is dissolved in water, the density of the water increases. The buoyant force is proportional to the density of the liquid. Hence, when the lemon is immersed in the water containing salt, the magnitude of the buoyant force acting on the lemon becomes greater than the magnitude of the weight of the lemon. Therefore, it floats in water.
In simple words: A lemon floats in salt water because adding salt increases the water's density. This denser fluid provides a greater upward buoyant force, which can then counteract the lemon's weight, causing it to float.

🎯 Exam Tip: Focus on the relationship between fluid density and buoyant force. An increase in fluid density directly leads to an increase in buoyant force for the same submerged volume, which can cause an object to float.

Do You Know?

**How is it decided that an object left in a liquid will get sink in the liquid, will float on the surface, or will float inside the liquid?**
Answer:1. The object floats if the buoyant force is larger than its weight. 2. The object sinks if the buoyant force is smaller than its weight. 3. The object floats inside the liquid if the buoyant force is equal to its weight. Which forces are unbalanced in the above cases? Unbalanced force: 1. Magnitude of the unbalanced force acting on the object - magnitude of the buoyant force on the object-magnitude of the weight of the object. The direction of the unbalanced force is the direction of the buoyant force. 2. Magnitude of the unbalanced force on the object - magnitude of the weight of the object - magnitude of the buoyant force on the object. The direction of the unbalanced force is the direction of the weight of the object. 3. Here, the unbalanced force is zero.
In simple words: An object's fate in a liquid (sink, float on surface, or float submerged) is determined by comparing its weight to the buoyant force. If buoyant force is greater, it floats on the surface; if less, it sinks; if equal, it floats submerged within the liquid. Unbalanced forces are present when sinking or floating to the surface, but balanced when floating submerged.

🎯 Exam Tip: Clearly differentiate between floating on the surface (buoyant force > weight, but eventually balances) and floating inside (buoyant force = weight). Emphasize that sinking implies weight > buoyant force, leading to a net downward force.

Try This:

Take a long rubberband and cut it at one point. At one of its ends tie a clean washed stone or a 50 g weight as shown in Figure Now hold the other end of the rubberband and make a mark there. Keep the stone hanging in air and measure the length of the rubberband from the stone to the mark made earlier. Now take water in a pot and hold the rubberband at such a height that the stone sinks in it. Again measure the length of the rubberband up to the mark. What is observed? This length is shorter than the earlier length. While dipping the stone in water, length of the stretched rubber gets slowly reduced and is minimum when it sinks completely. What could be the reason for a shorter length of the rubberband in water?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो स्थितियों को दर्शाता है: पहली में एक पत्थर एक रबरबैंड से हवा में लटका हुआ है, जिससे रबरबैंड खिंचा हुआ है। दूसरी स्थिति में, वही पत्थर पानी में डूबा हुआ है, और रबरबैंड कम खिंचा हुआ दिखाई दे रहा है, जो पानी के उत्प्लावन बल के प्रभाव को दर्शाता है। Buoyant force
Answer: When the stone is sunk in water, a buoyant force acts on it in the upward direction. The weight of the stone acts downwards. Therefore, the force which acts on it in the downward direction is effectively reduced. (This decreases the downard force on the rubberband. Hence, its length decreases.)
In simple words: The rubberband is shorter in water because the water exerts an upward buoyant force on the stone. This buoyant force reduces the stone's effective weight, thus decreasing the downward pull on the rubberband.

🎯 Exam Tip: This experiment demonstrates apparent weight loss due to buoyant force. Explain that the buoyant force directly counteracts a portion of the object's actual weight, making it seem lighter.

**Question 55. State Archimedes' principle.**
Answer: Archimedes' principle: When an object is partially or fully immersed in a fluid, a force of buoyancy acts on it in the upward direction. This force is equal to the weight of the fluid displaced by the object. [Note: The two forces mentioned here are equal in magnitude and opposite in direction.]
In simple words: Archimedes' principle states that when an object is placed in a fluid, it experiences an upward push (buoyant force) that is exactly equal to the weight of the fluid it pushes aside.

🎯 Exam Tip: The core of Archimedes' principle is that buoyant force equals the weight of the displaced fluid. Be precise in stating this relationship and the upward direction of the force.

**Question 56. Using Archimedes' principle, determine the magnitude of the buoyant force.**
Answer: Let m = mass of the body (object) immersed in the fluid, V = volume of the body, \( \rho \) = density of the body, \(p_1\) = density of the fluid, g = magnitude of the acceleration due to gravity. Suppose that the body is completely immersed in the fluid. Then the volume of the fluid displaced by the body = V. According to Archimedes' principle, magnitude of the buoyant force = magnitude of the weight of the fluid displaced by the body = mass of the displaced fluid \( \times \) g = volume of the displaced fluid \( \times \) density of the fluid \( \times \) g (as density = mass/volume) \( =Vp_1 g \)
\( \implies Vp_1 g = (\frac{m}{\rho}) p_1 g \)
\( \implies (\frac{m}{\rho}) p_1 g = mg (\frac{p_1}{\rho}) \)
\( \implies (\therefore \rho = \frac{m}{V}) \) If the body is partially immersed in the fluid, the volume of the fluid displaced by the immersed part of the body = xV ; here \( 0 < x < 1 \). In this case, the magnitude of the buoyant force \( =xVp_1g \)
\( \implies xVp_1g = x(\frac{m}{\rho}) p_1g \)
\( \implies x(\frac{m}{\rho}) p_1g = xmg (\frac{p_1}{\rho}) \)
In simple words: The buoyant force is calculated as the volume of the displaced fluid multiplied by the fluid's density and the acceleration due to gravity (\(F_b = V_{displaced} \times \rho_{fluid} \times g\)). If an object is partially immersed, only the submerged volume is used in this calculation.

🎯 Exam Tip: Remember the formula \(F_b = V\rho g\) and understand how to apply it for both fully and partially immersed objects. Clearly define each variable used in the formula.

**Question 57. Using the formula for the magnitude of the buoyant force, state under what conditions the body will**
(i) sink in the fluid
(ii) float in the fluid
(iii) remain in equilibrium anywhere within the fluid.
Answer:Magnitude of the buoyant force on the body = \( mg (\frac{p_1}{p}) \) \( = \) magnitude of the weight of the body \( \times (\frac{p_1}{p}) \) 1. If the density of the fluid (\(p_1\)) is less than the density of the body (\(p\)), the magnitude of the buoyant force on the body will be less than the magnitude of the weight of the body. Therefore, the body will sink in the fluid. 2. If the density of the fluid is greater than that of the body, the magnitude of the buoyant force on the body will be greater than that of the weight of the body. Therefore, the body will float in the fluid. 3. If the density of the fluid is equal to that of the body, the magnitude of the buoyant force on the body will be equal to that of the weight of the body. Therefore, the body will remain in equilibrium anywhere within the fluid.
In simple words: An object sinks if its density is greater than the fluid's, floats if its density is less than the fluid's, and remains suspended within the fluid if their densities are equal. These outcomes depend on the balance between the object's weight and the buoyant force.

🎯 Exam Tip: The critical comparison is between the object's density and the fluid's density. This determines whether the buoyant force is sufficient to support the object against its weight. Clearly link each condition (sink, float, equilibrium) to the relative densities.

Use Your Brain Power!

**Question. Explain the observations in the earlier experiments according to the Archimedes' principle.**
Answer: The increase in the length of the rubberband (y) is proportional to the downward force (f) acting on it. 1. When the stone tied to the rubberband is hanging in air, the magnitude of the buoyant force exerted by air on the stone is negligible compared to that of the weight of the stone. Hence, ignoring it, we have \( f = f_g = mg \), where m is the mass of the stone and g is the acceleration due to gravity. 2. When the stone is immersed partially in water, \( f = f_g - f_b \), where \(f_b\) is the magnitude of the buoyant force exerted by water on the stone. Now, \( f_g = V\rho g \) and \( f_b = xV\rho_w g \), where V = volume of the stone, \( \rho \) = density of the stone, xV= volume of the water displaced by the part of the stone immersed in water = volume of the part of the stone immersed in water (\( x < 1 \)) and \( \rho_w \) = density of water. \( \therefore f = V\rho g - xV\rho_w g = V\rho g (1 - x \frac{\rho_w}{\rho}) \)
\( \implies f = f_g (1 - x \frac{\rho_w}{\rho}) \) This shows that as the stone is gradually lowered in water, x goes on increasing and hence \( \frac{f_g}{f} \) goes on decreasing. Therefore, elongation (y) of the rubberband goes on decreasing, i.e., the length of the rubberband goes on decreasing.
In simple words: The rubberband experiment shows that when the stone is immersed in water, the buoyant force from the water reduces the stone's effective weight. This reduced downward force causes the rubberband to stretch less, making its length shorter compared to when the stone is in air.

🎯 Exam Tip: To answer this, clearly state that buoyant force acts opposite to weight. Explain how the *apparent* weight reduction in water (due to buoyant force) leads to less stretching of the rubberband, demonstrating Archimedes' principle in action.

3. When the stone is completely immersed in water, x becomes maximum, equal to 1. Here, \( f = f_g (1 - \frac{\rho_w}{\rho}) \). This is the minimum value of f. Here, the elongation of the rubberband is minimum, i.e., the length of the rubberband is minimum.
In simple words: When the stone is fully submerged, the buoyant force is at its maximum, leading to the greatest reduction in its apparent weight and thus the minimum stretching of the rubberband.

🎯 Exam Tip: Note that for full immersion, the maximum buoyant force is achieved, leading to the smallest apparent weight and minimal stretching of the rubberband.

**Question 58. State the applications of Archimedes' principle.**
Answer: Archimedes' principle is used in the construction of ships and submarines. The working of the lactometer and hydrometer is based on Archimedes' principle. [Note: The hydrometer is used to determine the density or relative density of a liquid.]
In simple words: Archimedes' principle is used to design ships and submarines, ensuring they float or submerge as intended, and also forms the basis for instruments like lactometers and hydrometers that measure liquid densities.

🎯 Exam Tip: Remember common applications like ships, submarines, lactometers, and hydrometers. For each, briefly explain *how* Archimedes' principle applies (e.g., controlling buoyancy for ships/submarines, density measurement for instruments).

**Question 59. If a spring balance is used to weigh a body, will the weight of the body be the same in vacuum and air? Explain why.**
Answer: When a body is suspended in air, the buoyant force acts on the body. Hence, the magnitude of the net downward force on the body = the magnitude of the weight of the body - the magnitude of the buoyant force on the body. Hence, when a spring balance is used to weigh a body, the weight of the body in air is less than that in vacuum.
In simple words: The weight of a body measured by a spring balance will be slightly less in air than in a vacuum because air, being a fluid, exerts a small upward buoyant force that reduces the apparent weight. In a vacuum, there is no fluid, so no buoyant force.

🎯 Exam Tip: Key to this explanation is the presence of buoyant force in air and its absence in a vacuum. Emphasize that the buoyant force in air makes the object *appear* lighter.

**Question 60. What is density of a substance? Obtain its SI unit.**
Answer: The density of a substance is the ratio of its mass to volume. The SI unit of density = \( \frac{\text{the SI unit of mass}}{\text{the SI unit of volume}} = \text{kg/m}^3 \) [Note: Density is useful in determining the purity of a substance. The CGS unit of density is g/cm³. \( 1 \text{ kg/m}^3 = 10^3 \text{g}/(100 \text{ cm})^3 = 10^{-3} \text{g/cm}^3 \)
\( \therefore 1 \text{g/cm}^3 = 1000 \text{ kg/m}^3 \)]
In simple words: Density is a measure of how much mass is packed into a given volume of a substance, calculated as mass divided by volume. Its SI unit is kilograms per cubic meter (\(\text{kg/m}^3\)).

🎯 Exam Tip: Define density precisely (mass per unit volume) and correctly state its SI unit (\(\text{kg/m}^3\)). Mentioning its use in purity determination is a good additional point.

**Question 61. What is relative density of a substance?**
Answer: The relative density of a substance is the ratio of its density to the density of water. [Note: Relative density is also called specific gravity. It is a ratio of two equal (same) physical quantities. Hence, it has no unit.]
In simple words: Relative density, also known as specific gravity, compares a substance's density to the density of water. It's a unitless ratio showing how much denser or less dense a substance is compared to water.

🎯 Exam Tip: Define relative density as the ratio of an object's density to water's density. Crucially, state that it is a unitless quantity because it's a ratio of two identical units.

**Question 62. A piece of wood floats both in water and kerosine. In which liquid does it sink more during floating? Why?**
Answer: The piece of wood sinks more in kerosine than in water during floating. The density of kerosine is less than that of water. The buoyant force on a body is proportional to the density of the liquid in which the body is immersed. When a body floats, the magnitude of the buoyant force acting on the body is equal to that of the weight of the body. Hence, the volume of the liquid displaced by a floating body is inversely proportional to the density of the liquid. As a result, when a piece of wood floats in kerosine, it displaces greater volume of kerosine compared to the volume of water displaced when the piece of wood floats in water. Hence, it sinks more in kerosine than in water. [Note: When a body floats in a liquid \( f_b = f_g \)
\( \implies V p_1 g = V_b \rho_b g \)
\( \implies V = V_b \frac{\rho_b}{\rho_1} \), where \(V_b\) is the volume of the body, \( \rho_b \) is the density of the material of the body and \( \rho_1 \) is the density of the liquid. Thus, \( V \propto \frac{1}{\rho_1} \)]
In simple words: Wood sinks more in kerosene because kerosene is less dense than water. To generate the same buoyant force (equal to the wood's weight), the wood must displace a larger volume of the less dense kerosene, thus submerging deeper.

🎯 Exam Tip: This question highlights the inverse relationship between the volume of displaced fluid and the fluid's density for a floating object. Less dense fluid requires a larger displaced volume (deeper sinking) to match the object's weight.

**Question 63. State whether the body will float or sink in a liquid if the density of the body is**
(i) greater than that of the liquid
(ii) less than that of the liquid
(iii) equal to that of the liquid.
Answer:1. The body will sink in the liquid if the density of the body is greater than that of the liquid. 2. The body will float in the liquid if the density of the body is less than that of the liquid. 3. The body will float inside the liquid if the density of the body is equal to that of the liquid.
In simple words: An object sinks if its density is greater than the liquid's, floats if its density is less than the liquid's, and floats completely submerged within the liquid if their densities are equal.

🎯 Exam Tip: This is a fundamental concept of buoyancy. Clearly stating the three density-based conditions for sinking, floating, and neutral buoyancy (floating inside) is crucial for a complete answer.

**Question 64. If the relative density of a body is greater than 1, will it float in water?**
Answer: If the relative density of a body is greater than 1, it will not float in water. [Note: The relative density of a pin is much greater than 1. But when kept gently on the surface of water, it floats. This is due to the surface tension of water.]
In simple words: No, if a body's relative density is greater than 1, it means it is denser than water and will sink, not float. Surface tension can temporarily support very small, denser objects, but not true floating.

🎯 Exam Tip: Relative density greater than 1 signifies higher density than water, leading to sinking. Briefly mentioning surface tension as an exception for small objects demonstrates deeper understanding.

**Question 65. A glass of water has an ice cube floating in water. The water level must touches the rim of the glass. Will the water overflow when the ice melts? Give the reason.**
Answer: The water will not overflow when the ice melts. The water level will remain the same. Ice floats on water because its density is less than that of water. When ice melts, the volume of the water formed is less than the volume of the ice which has melted. When ice in water melts, this difference equals the volume of the water formed when the part of ice above the surface of water melts. Therefore, the water level remains the same. Hence, there is no overflow of water when the ice melts.
In simple words: The water will not overflow because the floating ice cube already displaces a volume of water equal to the weight of the ice. When the ice melts, the resulting water volume is exactly equal to the volume it previously displaced, so the total water level remains unchanged.

🎯 Exam Tip: This is a classic application of Archimedes' principle. The key insight is that a floating object displaces its own weight in fluid, and ice displaces a volume of water exactly equal to the volume of water it will become when melted.

**Question 66. A plastic ball is released underwater. State whether it will sink or come up to the surface of water. Give the reason.**
Answer: A plastic ball released under water will come up to the surface of water. The density of water is greater than that of plastic. Hence, when a plastic ball is under water, the magnitude of the buoyant force exerted by water on the ball is greater than the magnitude of the weight of the ball. Therefore, the ball will start moving upward. As it comes up with part of the ball above the water surface, the volume of the water displaced by the ball becomes less and hence at a certain stage, the buoyant force and the weight balance each other. Then the ball continues to remain in that state, as the net force on the ball becomes zero. [Note: Initially, the ball moves slightly up and down near the water surface. The force due to friction with water, opposing the motion of the ball, finally makes the ball steady.]
In simple words: A plastic ball released underwater will rise to the surface because plastic is less dense than water. This means the buoyant force acting on the submerged ball is greater than its weight, pushing it upwards until it floats partly above the surface.

🎯 Exam Tip: The core reason is the density difference: plastic is less dense than water. Therefore, the buoyant force (from displaced water) will be greater than the plastic ball's weight, causing it to accelerate upwards.

Write Short Notes:

**Question 1. Buoyant force.**
Answer:1. When a body is immersed partially or completely in a liquid, the liquid exerts forces on all sides of the body. This force is perpendicular to the surface of the body and equals the product of pressure and area at that point. 2. The resultant of all these forces acts upward. It is called the upthrust or buoyant force. 3. The buoyant force is proportional to (i) the volume of the liquid displaced by the body (ii) the density of the liquid (iii) the acceleration due to gravity. Its magnitude equals the magnitude of the weight of the liquid displaced by the body.
In simple words: Buoyant force is the net upward force exerted by a fluid on an immersed object. It's the sum of all pressure forces on the object, acting upwards, and its magnitude is equal to the weight of the fluid displaced by the object.

🎯 Exam Tip: For short notes on buoyant force, cover its definition as an upward force, its origin from fluid pressure, its proportionality to displaced volume and fluid density, and its magnitude as per Archimedes' principle (weight of displaced fluid).

**Question 2. Applications of Archimedes' principle.**
Answer:1. The working of a lactometer, a device used to determine the purity of a sample of milk, and a hydrometer, a device used to determine the density of a liquid, is based on Archimedes' principle. The extent to which a lactometer floats (or sinks) depends on the density (and hence purity) of the milk. The same thing is true for a hydrometer. The greater the density of a liquid, the less is the extent to which a body sinks in it. 2. Archimedes' principle is used in design of ships and submarines. A submarine is provided with large tanks at the front and the back. Its weight can be increased by filling the tanks with sea water or air from compressed air reservoirs. The weight can be decreased by pumping out water from the tanks by forcing compressed air in them. By controlling the weight, it can be made to sink or rise to the surface as desired. 3. The density of a body that floats or sinks in water or kerosine can be determined by. Archimedes' principle. 4. The density of kerosine can be determined by Archimedes' principle, using a body of material that is not affected by water and kerosine.
In simple words: Archimedes' principle is applied in designing ships and submarines to control their buoyancy, allowing them to float or submerge. It also underpins the function of lactometers (for milk purity) and hydrometers (for liquid density), as these devices float at different levels based on the fluid's density. Additionally, it helps determine the density of various substances and liquids.

🎯 Exam Tip: When listing applications, provide a brief explanation for each. For ships/submarines, mention buoyancy control via weight/volume adjustment. For lactometers/hydrometers, explain how their depth of immersion relates to fluid density. Also, include its use in determining densities of objects and liquids.

Give Scientific Reasons:

**Question 1. The tiles are placed over a slushy patch of ground to help cross It.**
Answer:1. Tiles have greater area than the area of our feet. 2. The weight of the person crossing the slushy patch is exerted over a large area of the tiles. 3. Therefore, there is a decrease in the pressure and hence the tiles do not sink much in the slushy patch of ground. This helps to cross the slushy patch of ground. [Note: If there were no tiles, the feet will come in direct contact with the slushy ground. The area of the feet being less, the weight of the person will act over a smaller area. Therefore. there will be more pressure and hence the feet will sink into the slushy ground.]
In simple words: Tiles are placed on slushy ground because they spread the person's weight over a larger area. This significantly reduces the pressure exerted on the ground, preventing the tiles and the person from sinking deep into the mud.

🎯 Exam Tip: This question is about the inverse relationship between pressure and area (P=F/A). Emphasize that increasing the contact area reduces pressure, preventing sinking in soft surfaces.

**Question 2. Drawing pins have flattened heads.**
Answer:1. The head of a drawing pin is flattened and the other end is pointed. 2. When enough force is applied to the head of the pin, the pressure due to the force on the pointed end increases tremendously and the pin can be easily inserted in the drawing board. 3. When we press the flattened end, the force applied spreads over a larger area. This reduces the pressure of the reaction force acting on the thumb. Hence, the thumb is not injured. 4. If the head of the pin is sharp, then the pressure due to the force would be more and hence the pressure of the reaction force would also be more and the sharp end would prick the thumb causing injury.
In simple words: Drawing pins have flattened heads to distribute the force from your thumb over a larger area, reducing pressure and preventing injury. The pointed tip, however, concentrates the force onto a tiny area, creating high pressure for easy insertion into a surface.

🎯 Exam Tip: Explain both aspects: the large area of the head reduces pressure on the thumb (for safety and comfort), while the small area of the point increases pressure for effective penetration. This demonstrates the practical application of pressure concepts.

**Question 3. An iron nail sinks in water but a ship made from iron floats on water.**
Answer:1. An iron nail sinks in water because its density is more than that of water. 2. A ship made from iron, due to the particular shape given to it, displaces a large amount of water so that the buoyant force acting on the ship due to water balances the weight of the ship. Hence, the ship floats on water.
In simple words: An iron nail sinks because iron's density is greater than water's. A ship made of iron floats because its hollow design allows it to displace a much larger volume of water. This large displaced volume creates a buoyant force equal to the ship's total weight, making it float even though the iron itself is denser than water.

🎯 Exam Tip: This classic example illustrates how average density, not just material density, determines floating. Emphasize that a ship's design (displacing a large volume) reduces its *average* density below that of water.

**Question 4. A piece of iron sinks in water but floats on mercury.**
Answer: The density of iron is more than that of water but less than that of mercury. Hence, a piece of iron sinks in water but floats on mercury.
In simple words: Iron sinks in water because it is denser than water. However, it floats on mercury because mercury is significantly denser than iron, meaning the buoyant force from displaced mercury is greater than the iron's weight.

🎯 Exam Tip: This question reinforces the concept of relative density. An object floats if its density is less than the fluid's density and sinks if greater. Compare the density of iron with both water and mercury.

**Question 5. A sheet of metal that sinks in water can float if shaped like a pan.**
Answer:1. A sheet of metal sinks in water because its density is more than that of water. 2. If the sheet is shaped like a pan, it can displace a large amount of water such that the buoyant force on the pan due to water balances the weight of the pan. Hence, it can float on water.
In simple words: A flat metal sheet sinks because it's denser than water. However, if shaped into a pan, it encloses a volume of air. This larger overall volume, when submerged, displaces enough water to create a buoyant force that balances the pan's weight, allowing it to float.

🎯 Exam Tip: Emphasize that shaping the metal into a pan increases the *volume of water displaced* for the same mass of metal. This increased displaced volume generates a greater buoyant force, which can then overcome the metal's weight.

Solve The Following Examples:

**Problem 1.**
(i) Calculate the pressure exerted by the wooden block when it is kept in the vertical position. Given: The length of the wooden block is 80 cm, the breadth is 50 cm, the thickness is 20 cm and the weight is 500 N
(ii) Also calculate the pressure when the wooden block is kept in the horizontal position with its surface 80 cm \( \times \) 50 cm touching the floor.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक लकड़ी के गुटके को दो अलग-अलग स्थितियों में दिखाता है: (a) लंबवत स्थिति में, जहाँ 20 cm x 50 cm का छोटा क्षेत्र सतह को छू रहा है, और (b) क्षैतिज स्थिति में, जहाँ 80 cm x 50 cm का बड़ा क्षेत्र सतह को छू रहा है। यह दर्शाता है कि सतह पर गुटके के उन्मुखीकरण के आधार पर संपर्क क्षेत्र कैसे बदलता है।
Answer:Solution: Data: F = W = 500 N, l = 80 cm = 0.8 m, b = 50 cm = 0.5 m. h = 20 cm = 0.2 m (i) A = bh = \( 0.5 \text{ m} \times 0.2 \text{ m} = 0.1 \text{ m}^2 \)
\( P = \frac{F}{A} = \frac{500 \text{ N}}{0.1 \text{ m}^2} = 5000 \text{ N/m}^2 \) or \( 5000 \text{ Pa} \) The pressure exerted in the vertical position of the block = \( 5000 \text{ N/m}^2 \) or \( 5000 \text{ Pa} \).
(ii) A = lb = \( 0.8 \text{ m} \times 0.5 \text{ m} = 0.4 \text{ m}^2 \)
\( P = \frac{F}{A} = \frac{500 \text{ N}}{0.4 \text{ m}^2} = 1250 \text{ N/m}^2 \) or \( 1250 \text{ Pa} \) The pressure exerted in the horizontal position of the block = \( 1250 \text{ N/m}^2 \) or \( 1250 \text{ Pa} \).
In simple words: To calculate pressure, we divide the force (weight of the block, 500 N) by the area of contact. In the vertical position, the contact area is smaller (0.1 m²), resulting in a pressure of 5000 Pa. When placed horizontally, the contact area is larger (0.4 m²), which reduces the pressure to 1250 Pa for the same weight.

🎯 Exam Tip: Remember the formula \(P = \frac{F}{A}\). Ensure all units are consistent (SI units are preferred). For pressure calculations, always identify the force and the *exact area over which that force is distributed*. This problem clearly shows the inverse relationship between pressure and area: a larger contact area for the same force leads to less pressure.

**Problem 2. Measure the length, breadth, height and mass of a rectangular tiffin box. Find the weight of the box and calculate the pressure in two different positions as in Ex. (1) above.**
Answer:Solution: Let l = 0.25 m, b = 0.1 m, h = 0.05 m, F = W= 0.5 N (i) A = bh = \( 0.1 \text{ m} \times 0.05 \text{ m} = 0.005 \text{ m}^2 \)
\( \therefore \text{Pressure, } P = \frac{F}{A} = \frac{0.5 \text{ N}}{0.005 \text{ m}^2} = 100 \text{ Pa} \). (ii) A = lb = \( 0.25 \text{ m} \times 0.1 \text{ m} = 0.025 \text{ m}^2 \)
\( \therefore \text{Pressure, } P = \frac{F}{A} = \frac{0.5 \text{ N}}{0.025 \text{ m}^2} = 20 \text{ Pa} \).
In simple words: By taking sample dimensions and weight for a tiffin box, we calculated pressure in two orientations. In the first position (smallest area), the pressure was 100 Pa. In the second position (larger area), the pressure was 20 Pa, demonstrating how pressure changes with contact area for a constant force.

🎯 Exam Tip: This problem requires students to apply the pressure formula with their own measured values. Emphasize showing the calculation steps clearly, including unit conversions and final units for pressure.

Question 3.A force of 1000 N is applied over an area 50 cm × 20 cm. Find the corresponding pressure.
Answer:Solution: Data: F = 1000 N, A = 50 cm x 20 cm = 0.5 m x 0.2 m = 0.1 m², pressure = ? P = \( \frac{F}{A} = \frac{1000 \, N}{0.1 \, m^2} \) = 10\(^4\) N/m² The pressure = 10 N/m².
In simple words: To find pressure, convert the area to square meters, then divide the force (N) by the area (m²) to get the pressure in N/m².

🎯 Exam Tip: Remember to always convert units to SI units (meters, kg, N) before performing calculations for pressure to avoid errors.

Question 4.A metal block has dimensions 10 cm × 5 cm × 2 cm and the density of the metal is 8 × 10\(^3\) kg/m³. It is kept on a table with the face 10 cm x 5 cm in contact with the table. Find the force and pressure exerted by the block on the table. (g = 9.8 m/s²)
Answer:Solution: Data: l = 10 cm, b = 5 cm, h = 2 cm, \( \rho \) = 8 × 10\(^3\) kg/m³, g = 9.8 m/s², A = lb = 10cm × 5cm = 50cm² = 50 × 10\(^{-4}\)m² = 5 × 10\(^{-3}\)m², force =?, pressure = ? Volume of the block = lbh = 10 cm x 5 cm × 2 cm = 100 cm³ = 100 × 10\(^{-6}\) m³ = 1 × 10\(^{-4}\)m³ Mass of the block = volume × density (
\( \implies \) density = mass/volume)
\( \implies \) Mass of the block, m = 1 × 10\(^{-4}\) m³× 8 × 10\(^3\) kg/m³ = 0.8 kg Weight of the block = mg = 0.8 kg × 9.8 m/s² = 7.84 N
\( \implies \) The force exerted by the block on the table = 7.84 N. Pressure = \( \frac{\text{force}}{\text{area}} = \frac{7.84 \, N}{5 \times 10^{-3} \, m^2} \) = 1.568 × 10\(^3\) N/m² or 1.568 × 10\(^3\) Pa The pressure exerted by the block on the table = 1.568 × 10\(^3\) N/m² or 1.568 × 10\(^3\) Pa.
In simple words: First calculate the mass of the block using its volume and density, then find its weight (force) using g. Finally, divide this force by the contact area to determine the pressure exerted on the table.

🎯 Exam Tip: Pay close attention to unit conversions (cm to m) and ensure all values are in consistent units (SI units are preferred) before starting calculations.

Question 5.A body of volume 100 cm³ is immersed completely in water. Find the weight of the water displaced by the body. [g = 9.8 m/s², \( \rho \) (water) = 10\(^3\) kg/m³]
Answer:Solution: Data: V = 100 cm³ = 100 × 10\(^{-6}\) m³ = 1 x 10\(^{-4}\) m³, \( \rho \)(water) = 10\(^3\) kg/m³ g = 9.8 m/s², weight of the displaced water ? Density = \( \frac{\text{mass}}{\text{volume}} \)
\( \implies \) Mass = volume × density Volume of the water displaced by the body = 1 × 10\(^{-4}\) m³
\( \implies \) Mass of the water displaced, m = 1 x 10\(^{-4}\) m³ x 10\(^3\) kg/m³ = 0.1 kg
\( \implies \) Weight of the water displaced = mg = 0.1 kg × 9.8 m/s² = 0.98 N. The weight of the water displaced by the body = 0.98 N.
In simple words: When a body is completely immersed, the volume of water displaced is equal to the body's volume. Multiply this volume by the density of water and then by 'g' to find the weight of the displaced water.

🎯 Exam Tip: Archimedes' principle is key here: the buoyant force (equal to the weight of displaced fluid) is directly related to the volume of the submerged object and the fluid's density.

Question 6.A body of mass 200 g and volume 50 cm³ is put in a bucket containing water. Will it float or sink? [ \( \rho \)(water) = 1 g/cm³]
Answer:Solution: Data: m = 200 g, V= 50 cm³, \( \rho \)(water) = 1 g/cm³ Density(\( \rho \)) = \( \frac{\text{mass}}{\text{volume}} \)
\( \implies \) \( \rho \)(body) = \( \frac{200}{50 \, \text{cm}^3} \) = 4 g/cm³ It is greater than the density of water. Hence, the body will sink in water.
In simple words: To determine if an object floats or sinks, compare its density to the density of the fluid. If the object's density is higher, it sinks; if lower, it floats.

🎯 Exam Tip: Remember the basic rule of buoyancy: an object sinks if its density is greater than the fluid it's in, and floats if its density is less.

Question 7.A body of mass 200 g and volume 400 cm³ is put in a bucket containing water. Will it float or sink? [ \( \rho \)(water)=1 g/cm³]
Answer:Solution: Proceed as above. \( \rho \)(body) = \( \frac{200 \, \text{g}}{400 \, \text{cm}^3} \) = 0.5 g/cm³ It is less than the density of water.
\( \implies \) The body will float in water.
In simple words: Calculate the object's density. If it's less than water's density (1 g/cm³), the object will float.

🎯 Exam Tip: This question reinforces the density-based rule for floating and sinking. A clear calculation of the object's density is crucial for full marks.

Question 8.The mass of a tile is 500 g. If the density of the tile is 2.5 g/cm³, what will be the weight of the tile when it is completely immersed in water? (g = 9.8 m/s², \( \rho \)(water) = 1000 kg/m³)
Answer:Solution: Data: m = 500 g = 0.5 kg, \( \rho \)(tile) = 2.5 g/cm³ = 2500 kg/m³, g = 9.8 m/s², \( \rho \)(water) = 1000 kg/m³, weight of the tile when completely immersed in water (also called the apparent weight) = ?
\( \rho \) = \( \frac{m}{V} \)
\( \implies \) Volume of the tile, V= \( \frac{m}{\rho} = \frac{0.5 \, \text{kg}}{2500 \, \text{kg/m}^3} \) = \( \frac{0.5}{2500} \) m³ = 2 × 10\(^{-4}\) m³
\( \implies \) Volume of water displaced by the tile = 2 × 10\(^{-4}\) m³
\( \implies \) Mass of water displaced by the tile m' = \( \rho \)(water) V = 1000 kg/m³ × 2 × 10\(^{-4}\) m³ = 0.2kg
\( \implies \) Magnitude of the weight of this water = mg = 0.2 kg × 9.8 m/s² = 1.96 N
\( \implies \) Buoyant force exerted on the tile = 1.96 N Magnitude of the weight of the tile = mg = 0.5 kg × 9.8 m/s² = 4.9 N
\( \implies \) Weight of the tile when completely immersed in water (apparent weight) = weight of the tile in air-buoyant force on the tile = 4.9 N – 1.96 N = 2.94 N(downward)
In simple words: First, find the volume of the tile. Then, calculate the buoyant force (weight of displaced water) using this volume and water's density. The apparent weight in water is the tile's actual weight minus the buoyant force.

🎯 Exam Tip: Convert all given quantities to SI units (kg, m, N) at the start of the problem to ensure consistency and accuracy in calculations for apparent weight and buoyant force.

Examples For Practice:

Question 1.Calculate the relative density of a metal having density 7.5 g/cm³.
Answer:7.5
In simple words: Relative density is the ratio of the substance's density to water's density (1 g/cm³). So, 7.5 g/cm³ divided by 1 g/cm³ gives 7.5.

🎯 Exam Tip: Remember that relative density is a unitless quantity, as it's a ratio of two densities, and the density of water is commonly 1 g/cm³ or 1000 kg/m³.

Question 2.Find the density of steel if its relative density is 8 and the density of water is 10\(^3\) kg/m³.
Answer:8 × 10\(^3\) kg/m³
In simple words: To find the density of steel, multiply its relative density by the density of water. Since relative density is 8 and water density is 10\(^3\) kg/m³, steel's density is 8 times that.

🎯 Exam Tip: Relative density is a multiplier for the density of water; always multiply the relative density by the given density of water to find the material's actual density.

Question 3.A body has mass 200 g and volume 100 cm³. Find its density and relative density.
Answer:2 g/cm³, 2
In simple words: Divide the mass (200 g) by the volume (100 cm³) to get the density (2 g/cm³). Since water's density is 1 g/cm³, the relative density is 2.

🎯 Exam Tip: Density is mass per unit volume. Relative density compares an object's density to water's density, making it easy to calculate if the units are consistent.

Question 4.If the relative density of a material is 2.5, find its density.
Answer:2.5 × kg/m³ or 2.5 g/cm³
In simple words: The density of the material is 2.5 times the density of water. So, if water's density is 1000 kg/m³, the material's density is 2500 kg/m³ (or 2.5 g/cm³ if using 1 g/cm³ for water).

🎯 Exam Tip: When converting relative density to actual density, always specify the units being used (e.g., kg/m³ or g/cm³) as the numerical value will change based on the unit chosen for water's density.

Question 5.A force of 100 N is applied on an area 40 cm × 25cm. Find the corresponding pressure.
Answer:10\(^3\) N/m²
In simple words: First, convert the area from cm² to m² (40x25 = 1000 cm² = 0.1 m²). Then, divide the force (100 N) by the area (0.1 m²) to get the pressure (1000 N/m² or 10\(^3\) N/m²).

🎯 Exam Tip: Always remember to convert the area to square meters (m²) when calculating pressure in N/m² (Pascals), as 1 m² = 10,000 cm².

Question 6.If the pressure exerted on an area 10 cm × 10 cm is 1000 dynes/cm², find the applied force.
Answer:10\(^5\) dynes
In simple words: To find the force, multiply the pressure (1000 dynes/cm²) by the area (10 cm x 10 cm = 100 cm²). This gives 100,000 dynes, or 10\(^5\) dynes.

🎯 Exam Tip: In CGS units, force is in dynes, area in cm², and pressure in dynes/cm². Ensure consistency in these units when solving problems.

Question 7.A metal block of mass 10 kg is kept on a table. If the contact surface area Is 100 cm², find the pressure on the table.
Answer:9.8 × 10\(^3\) N/m² or 9.8 × 10\(^3\) Pa
In simple words: Calculate the weight (force) of the block by multiplying mass (10 kg) by acceleration due to gravity (9.8 m/s²), which is 98 N. Convert the area (100 cm²) to m² (0.01 m²). Divide force (98 N) by area (0.01 m²) to get pressure (9800 N/m² or 9.8 × 10\(^3\) Pa).

🎯 Exam Tip: For objects resting on a surface, the force causing pressure is its weight (mass x g). Don't forget to convert area to m² for SI pressure units.

Question 8.A body of volume 50 cm³ is immersed completely in water. Find the weight of the water displaced by the body.
Answer:0.49 N
In simple words: The volume of displaced water is 50 cm³ (or 50 × 10\(^{-6}\) m³). Multiply this by water's density (1000 kg/m³) and 'g' (9.8 m/s²) to find the weight of displaced water, which is 0.49 N.

🎯 Exam Tip: The weight of the displaced fluid is equivalent to the buoyant force. Always ensure you use the density of the *fluid* and the *volume displaced* for this calculation.

Question 9.A block of mass 100 g and volume 20 cm³ is put in a bucket filled with water. Will it float or sink?
Answer:The body will sink in water.
In simple words: The block's density is 100 g / 20 cm³ = 5 g/cm³. Since this is greater than water's density (1 g/cm³), the block will sink.

🎯 Exam Tip: Comparing the object's density to the fluid's density is the quickest way to determine if it will float or sink. Density greater than water means it sinks.

Question 10.Will a block of mass 100g and volume 400 cm³ float or sink in water?
Answer:The block will float in water.
In simple words: The block's density is 100 g / 400 cm³ = 0.25 g/cm³. As this is less than water's density (1 g/cm³), the block will float.

🎯 Exam Tip: A density less than water means the object will float. Always calculate the density first for floating/sinking problems.

Question 11.The volume of a cube is 125 cm³ and its mass is 250 g. It is put in a tub containing water. Will it float or sink?
Answer:It will sink in water.
In simple words: The cube's density is 250 g / 125 cm³ = 2 g/cm³. Since 2 g/cm³ is greater than water's density (1 g/cm³), the cube will sink.

🎯 Exam Tip: The principle remains consistent: an object with a density greater than the liquid it is placed in will always sink.