Frank Brothers Solutions for ICSE Class 9 Physics Chapter 4 Fluids

Page No: 157

Question 1.
Answer: The thrust on the unit surface is known as pressure. The SI unit of pressure is \( \text{Nm}^{-2} \).
In simple words: Pressure is basically how hard you are pushing on a specific spot. Imagine pushing your thumb against a wall; the pressure is the force of your thumb divided by the size of your thumb's tip.

📝 Teacher's Note: Use the "needle vs. thumb" analogy. Even with the same force, a needle creates much higher pressure because its area is tiny.

🎯 Exam Tip: Always remember that pressure is force per unit area. Use the formula \( P = F/A \).

Question 2.
Answer: Pressure is given by
\( P = h \times \rho \times g \).
Where \( h \) is height of liquid column, \( \rho \) is density of liquid, \( g \) is acceleration due to gravity.
Density of mercury is \( = 1.36 \times 10^4 \text{ kg/m}^3 \).
\( h = \) height of mercury column which is given \( = 75 \text{ cm} = 0.75 \text{ m} \).
So pressure \( = 0.75 \times 1.36 \times 10^4 \times 9.8 = 9.996 \times 10^4 \text{ Nm}^{-2} \).
In simple words: The weight of a liquid column creates pressure at the bottom. By multiplying how tall the column is, how heavy the liquid is, and gravity, we find the total squeeze at the base.

📝 Teacher's Note: Make sure students convert centimeters to meters before plugging them into the formula to keep units consistent.

🎯 Exam Tip: If the question provides density in \( \text{g/cm}^3 \), convert it to \( \text{kg/m}^3 \) by multiplying by \( 1000 \).

Question 3.
Answer: Pressure is a scalar physical quantity.
In simple words: Scalar means it only has a size (like "how much pressure") but doesn't point in one specific direction.

📝 Teacher's Note: Explain that at any point in a fluid, pressure acts in all directions equally, which is why we don't assign it a single direction vector.

🎯 Exam Tip: Standard MCQ trap: Pressure is scalar, while Force (Thrust) is a vector.

Question 4.
Answer: One pascal is defined as the pressure exerted on a surface of area \( 1 \text{ m}^2 \) by a force of \( 1 \) Newton acting normally on the surface.
In simple words: If you take a giant square that is 1 meter long on each side and push on it with a force of 1 Newton (about the weight of a small apple), that "spread out" push is exactly one Pascal of pressure.

📝 Teacher's Note: 1 Pascal is actually a very small amount of pressure. We often use kilopascals (kPa) in real life.

🎯 Exam Tip: The word "normally" is key here—it means the force is hitting the surface at a perfect \( 90^{\circ} \) angle.

Question 5.
Answer: The force acting normally on a surface is known as thrust.
SI unit of thrust is \( \text{N} \).
In simple words: Thrust is just the total "push" force. If you push a door open, the force you use is called thrust.

📝 Teacher's Note: Distinguish between general force and thrust. Thrust specifically refers to force acting perpendicular to a surface.

🎯 Exam Tip: The SI unit for thrust is the same as the unit for force—the Newton (\( \text{N} \)).

Question 6.
Answer: We know pressure exerted by a liquid column of height \( h \), density \( \rho \) is
\( P = h \times \rho \times g \).

Pressure exerted by mercury column of height \( 76 \text{ cm} \).
Density of mercury \( = 13.6 \text{ g/cc} = 1.36 \times 10^4 \text{ kg/m}^3 \).
\( P_{\text{mercury}} = 0.76 \times 1.36 \times 10^4 \times 9.8 = 10.12 \times 10^4 \text{ Nm}^{-2} \).

Let height of water column \( = h \text{ m} \).
Density of water \( = 1 \text{ g/cc} = 10^3 \text{ kg/m}^3 \).
\( P_{\text{water}} = h \times 10^3 \times 9.8 = 9.8h \times 10^3 \text{ Nm}^{-2} \).

Now put \( P_{\text{mercury}} = P_{\text{water}} \)
\( 9.8 \text{ h} \times 10^3 = 10.12 \times 10^4 \)

\( \implies h = 10.12 / 9.8 = 10.34 \text{ m} \).

So, \( 10.34 \text{ m} \) height of water column would exert same pressure on its base as \( 76 \text{ cm} \) column of mercury.
In simple words: Mercury is much heavier than water. To create the same amount of pressure as a short \( 76 \text{ cm} \) mercury tube, you would need a water pipe over \( 10 \text{ meters} \) tall!

📝 Teacher's Note: This explains why we use mercury in barometers instead of water—a water barometer would be too tall to fit inside a normal room.

🎯 Exam Tip: When comparing two columns, you can simplify the math to \( h_1 \rho_1 = h_2 \rho_2 \) because \( g \) cancels out on both sides.

Question 7.
Answer: Water can’t be used in place of mercury in a barometer because of its low density. It would require \( 10.34 \text{ m} \) long tube to measure \( 1 \) atmospheric pressure which is not practically possible while mercury having high density (\( 13.6 \text{ g/cc} \)) would require only \( 0.76 \text{ m} \) long pipe which is practically possible.
In simple words: Using water for a barometer is like trying to use a giant \( 33 \text{-foot} \) straw to measure air pressure. Mercury is much more convenient because it is heavy and only needs a small tube.

📝 Teacher's Note: Also mention that mercury does not wet the glass and is easily visible because of its silvery color.

🎯 Exam Tip: Give density as the primary reason, and "practicality" or "size" as the secondary reason.

Question 8.
Answer: Pressure is the physical quantity which is measured in bar.
In simple words: "Bar" is just another unit for pressure, like how "inches" and "meters" are units for length. Weather forecasters use it a lot.

📝 Teacher's Note: 1 bar is approximately equal to standard atmospheric pressure at sea level (\( 10^5 \text{ Pascals} \)).

🎯 Exam Tip: Remember: \( 1 \text{ bar} = 10^5 \text{ Pa} \).

Question 9.
Answer: Thrust is a vector quantity.
In simple words: Since thrust is a force, it has a specific direction (like "pushing down"), so it is a vector.

📝 Teacher's Note: Reiterate the difference between vector (Thrust) and scalar (Pressure) to avoid confusion.

🎯 Exam Tip: Vector quantities require both magnitude and direction to be fully described.

Question 10.
Answer: Thrust on a surface is the force acting normally on a surface while pressure on a surface is thrust acting on the unit area of a surface.
In simple words: Thrust is the total "whole" force, while pressure is how that force is divided up across every tiny square inch of the surface.

📝 Teacher's Note: Use the example of standing on one foot vs. two feet. The thrust (your weight) is the same, but the pressure is higher when you stand on one foot.

🎯 Exam Tip: Always define pressure in terms of "unit area."

Question 11.
Answer:
(i) we know pressure exerted by a liquid column of height \( h \), density \( \rho \) is
\( P = h \times \rho \times g \).

Pressure exerted by mercury column of height \( 70 \text{ cm} \).
Density of mercury \( = 13.6 \text{ g/cc} = 1.36 \times 10^4 \text{kg/m}^3 \).
\( P_{\text{mercury}} = 0.7 \times 1.36 \times 10^4 \times 9.8 = 9.32 \times 10^4 \text{ Nm}^{-2} \).

Let height of water column \( = h \text{ m} \).
Density of water \( = 1 \text{g/cc} = 10^3 \text{ kg/m}^3 \).
\( P_{\text{water}} = h \times 10^3 \times 9.8 = 9.8h \times 10^3 \text{ Nm}^{-2} \).

Now put \( P_{\text{mercury}} = P_{\text{water}} \)
\( 9.8 \text{ h} \times 10^3 = 9.32 \times 10^4 \)

\( \implies h = 93.2 / 9.8 = 9.52 \text{ m} \).

So, \( 9.52 \text{ m} \) height of water column would exert same pressure on its base as \( 70 \text{ cm} \) column of mercury.
(ii) Height of the water column would not change if the cross section of the water column is made wider.
In simple words: (i) A \( 70 \text{ cm} \) mercury tube has the same pressure as a \( 9.5 \text{-meter} \) water tube. (ii) It doesn't matter if the tube is skinny or wide; the pressure only depends on how tall the liquid is.

📝 Teacher's Note: This is a crucial concept: Liquid pressure is independent of the shape or width of the container. It only cares about depth.

🎯 Exam Tip: Part (ii) is a common conceptual question. State clearly that pressure depends only on depth (\( h \)), density (\( \rho \)), and gravity (\( g \)).

Question 12.
Answer: Lake has greater pressure at the bottom than the surface as pressure increases with depth. So when gas bubble is released at the bottom of the lake it experiences more pressure and is small in size but as it rises upwards the pressure experienced by it decreases. So it grows in size as it moves towards the surface from bottom.
In simple words: Deep underwater, the water squeezes gas bubbles very hard, making them tiny. As they float up, the squeeze gets weaker, and the bubbles expand like a balloon.

📝 Teacher's Note: This is a real-world application of Boyle's Law—as pressure goes down, volume goes up.

🎯 Exam Tip: Explain the reason (depth) before describing the effect (size change) for a complete answer.

Question 13.
Answer: A dam has broader walls at the bottom than at the top because the pressure exerted by a liquid increases with its depth, and at any point at a particular depth liquid pressure is same in all directions. Now as more pressure is exerted by water on the wall of the dam as depth increases. Hence a thick wall is constructed at the bottom of dam to withstand greater pressure.
In simple words: The water at the bottom of a lake is pushing much harder than the water at the top. To keep the dam from breaking, engineers make the bottom much thicker and stronger.

📝 Teacher's Note: Show a picture of a dam cross-section. It looks like a triangle to match the pressure distribution.

🎯 Exam Tip: Mention "pressure increases with depth" as the fundamental physical law justifying this design.

Question 14.
Answer: Depth of water \( = 6 \text{ cm} = 0.06 \text{ m} \).
Density of water \( = 1000 \text{ kgm}^{-3} \).
Acceleration due to gravity \( = 10 \text{ ms}^{-2} \).

We know pressure
\( P = h \times \rho \times g \).
\( P = 0.06 \times 1000 \times 10 = 600 \text{ Nm}^{-2} = 600 \text{ Pa} \).

Area of base of cylindrical vessel \( = 300 \text{ cm}^2 = 300 \times 10^{-4} \text{ m}^2 = 0.03 \text{ m}^2 \).

We know
Thrust \( = \text{pressure} \times \text{area} \).
Thrust \( = 600 \times 0.03 = 18 \text{ N} \).

So thrust acting on base of vessel is \( 18 \text{ N} \).
In simple words: First, we find out how much the water is squeezing the bottom (pressure). Then we multiply that squeeze by the size of the bottom area to find the total push (thrust) in Newtons.

📝 Teacher's Note: Be careful with the area conversion: \( 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 \). This is a common place where students make decimal errors.

🎯 Exam Tip: Always show both formulas used: \( P = h \rho g \) and \( F = P \times A \).

Question 15.
Answer: The pressure at a point in a liquid depends upon on the following three factors:

• It depends on the point below the free surface (\( h \)).
• It depends on density of liquid (\( \rho \)).
• It depends upon acceleration due to gravity (\( g \)) of the place.

In simple words: Liquid pressure changes if you go deeper, if you use a heavier liquid like oil instead of water, or if you were on a different planet with stronger gravity.

📝 Teacher's Note: Students can remember the acronym "HDG" (Height, Density, Gravity) to keep these factors in mind.

🎯 Exam Tip: Do not include the "area of the container" or "shape" in this list—they do NOT affect pressure.

Question 16.
Answer: For calculating the total pressure in a liquid at a depth, we have to add the atmospheric pressure that acts on the free surface of the liquid.

Total pressure inside a liquid at depth '\( h \)' \( = \text{atmospheric pressure} + \text{pressure due to liquid column} \).

So Total pressure inside a liquid at depth \( = P_0 + h\rho g \).

Here \( P_0 \) is atmospheric pressure acting on the free surface of liquid.
In simple words: When you are underwater, the air above the water is already pushing down on the surface. So the total squeeze you feel is the air's push PLUS the water's push.

📝 Teacher's Note: This is why submarine hulls must be so strong; they have to withstand both the weight of the ocean and the air above it.

🎯 Exam Tip: Use the term "Absolute Pressure" if the question asks for the "Total" pressure including the atmosphere.

Question 17.
Answer: A substance having a tendency to flow is called fluid.
A fluid exerts pressure on the bottom due to its weight and on the walls of the container in which it is enclosed by virtue of its ability to flow. This is called fluid pressure.
In simple words: Liquids and gases are both "fluids" because they can flow. Because they are always moving and have weight, they push against every side of their container.

📝 Teacher's Note: Clarify that "fluid" is a category that includes both liquids and gases, not just water.

🎯 Exam Tip: Define fluids based on their "ability to flow" to get full credit.

Question 18.
Answer: The laws of liquid pressure are

• Pressure inside the liquid increases with the depth from the free surface of the liquid.
• Pressure is same at all points on a horizontal plane, in case of a stationary liquid.
• Pressure is same in all directions about a point inside the liquid.
• Pressure at same depth is different in different liquids. It increases with the increase in the density of the liquid.
• A liquid will always seek its own level.

In simple words: These are the rules water follows: it pushes harder as you go deep, it pushes equally in all directions at the same level, and it always tries to flatten out at the top.

📝 Teacher's Note: Use a "liquid levels" apparatus (connected tubes of different shapes) to prove that a liquid always seeks its own level regardless of shape.

🎯 Exam Tip: These five points are excellent for "long-form" descriptive answers on fluid mechanics.

Question 19.
Answer:
(i) When the diver moves horizontally the depth from the surface remains same and as we know that pressure at depth \( h \) is given by
\( P = h \times \rho \times g \) so as depth remains same pressure would remains same while moving horizontally. (Density of liquid, \( g \) also remains same).
(ii) When diver moves to depth, depth \( h \) increases and thus pressure increases as diver moves to depth.
In simple words: (i) If a diver swims sideways at the same depth, the pressure doesn't change. (ii) If the diver swims straight down, the pressure gets stronger.

📝 Teacher's Note: This reinforces that pressure is a function of depth (\( h \)), not horizontal distance.

🎯 Exam Tip: Mention the formula \( P = h \rho g \) to prove why moving horizontally doesn't change pressure.

Question 20.
Answer: A diving suit is a garment or device designed to protect a diver from the underwater environment.
In simple words: It's a special heavy-duty outfit that keeps divers dry and protects their bodies from the intense pressure of deep water.

📝 Teacher's Note: Explain that "Atmospheric Diving Suits" are like small submarines you wear to keep your body at normal air pressure even deep down.

🎯 Exam Tip: Focus on "protection" and "pressure management" as the main roles of the suit.

Question 21.
Answer: There are five main types of ambient pressure suits. These are wetsuits, drysuits, semidry suits, dive skins etc.
In simple words: Different suits are used for different jobs. Wetsuits keep you warm, while drysuits keep all the water out completely.

📝 Teacher's Note: Mention that wetsuits work by trapping a thin layer of water next to the skin, which the body then warms up.

🎯 Exam Tip: You usually only need to name two or three types to answer this question.

Question 22.
Answer: This diagram illustrates that pressure of liquid increases with depth. Hole at greater depth has more pressure and hence flow of liquid from there is more. Other hole is at less depth so the pressure of liquid there is less and hence flow of liquid from there is less. So the diagram shows that pressure increases with increase in depth.
In simple words: If you poke holes in a water jug at different heights, the water shoots out much further from the bottom hole because the pressure down there is much stronger.

📝 Teacher's Note: This is a very easy classroom experiment. Use a plastic bottle and tape over the holes before filling it with water.

🎯 Exam Tip: In your diagram, make sure the water stream from the lowest hole travels the horizontal distance furthest.

Question 23.
Answer: Manometer is a simple pressure gauge that measures differences in pressure at the two ends of the apparatus. Manometer is a U shaped tube containing water whose one limb is dipped in vessel and vessel is tightly covered with plastic sheet. U shaped tube has two limbs one towards the vessel and other is opened to atmosphere. Now if level of water toward atmospheric open limb is more than level of water in limb towards apparatus end then liquid is said to be at higher pressure than atmosphere. And if level of water toward atmospheric open limb is less than level of water in limb towards apparatus end then liquid is said to be at lower pressure than atmospheric pressure.
In simple words: A manometer is a U-shaped pipe with water in it. If you push on one side, the water rises on the other side. By looking at which side is higher, you can tell if the pressure you're measuring is stronger or weaker than the air around us.

📝 Teacher's Note: Use a clear plastic tube and colored water to make the "level difference" easier for students to see.

🎯 Exam Tip: Remember: Higher liquid level on the open side means the measured pressure is greater than atmospheric pressure.

Page No : 158

Question 24.
Answer:
(i) The pressure on the liquid surface in the limb Q is equal to atmospheric pressure i.e \( P_0 \).
(ii) According to manometer principle, difference in atmospheric pressure in two limbs is equal to difference in height of liquid in two limbs.
So pressure at \( P = \text{pressure at } Q + h\rho g \).
\( P_P = P_Q + h\rho g \).
In simple words: This math formula shows that the pressure inside the container (P) is just the outside air pressure plus the extra "weight" of the liquid height difference (h).

📝 Teacher's Note: Point out that the liquid in the U-tube acts like a balancing scale for pressure.

🎯 Exam Tip: Label the atmospheric pressure (\( P_0 \)) on the open limb of your manometer diagrams.

Question 25.
Answer: A hydraulic press works on the principle of pascal’s law. A hydraulic press can be used for extracting juice of sugarcane, sugar beet etc.
In simple words: A hydraulic press uses liquid pressure to create a massive squeeze. It's so powerful it can crush sugarcane or metal easily.

📝 Teacher's Note: Pascal's Law states that pressure applied to an enclosed fluid is transmitted equally in all directions.

🎯 Exam Tip: Always associate "hydraulic" devices with Pascal's Law.

Question 28.
Answer: Pascal’s law states that pressure applied to an enclosed liquid, is transmitted equally to every part of the liquid or in other words when pressure is applied at a point in a confined fluid, it is transmitted undiminished and equally in all directions throughout the liquid. Hydraulic machines such as hydraulic press, hydraulic brakes and hydraulic jack are application of pascal’s law.
In simple words: If you squeeze a closed balloon filled with water, the pressure spreads out and pushes against the whole balloon equally. This is why car brakes work so well!

📝 Teacher's Note: This law is the foundation of all hydraulic engineering. Without it, we couldn't lift cars or stop heavy trucks.

🎯 Exam Tip: Keywords are "enclosed/confined fluid" and "transmitted undiminished."

Question 29.
Answer: Altimeter is a device which is used in an aircraft to measure its altitude.
In simple words: It’s a special gadget in a plane that tells the pilot how high up they are by checking the air pressure.

📝 Teacher's Note: An altimeter is basically just a barometer that has a scale for height instead of pressure.

🎯 Exam Tip: Remember: Higher altitude \( = \) Lower air pressure.

Question 30.
Answer: Atmospheric pressure decreases with increase in height. our atmosphere comprises of a large number of parallel layers. The pressure on a layer is equal to the thrust or weight of the gaseous column on the unit area of that layer. Hence, as we go up, the weight of the gaseous column decreases, which decrease the pressure of the gaseous column.
In simple words: Think of the air like a stack of blankets. If you're at the bottom, all the blankets are on top of you (high pressure). If you climb higher, there are fewer blankets above you, so there is less weight pushing down.

📝 Teacher's Note: This is why people get out of breath on high mountains—the air is "thinner" or at lower pressure.

🎯 Exam Tip: Explain pressure as the "weight of the air column" above a point.

Question 31.
Answer: Aneroid means containing no liquid and aneroid barometer is evacuated so it tends to collapse under the pressure of air. The stout spring balances the thrust on the metal box due to normal air pressure and prevents the box from collapsing. As this type of barometer doesn’t contain any liquid so it got its name aneroid barometer.
In simple words: An aneroid barometer is a "dry" weather-checker. It uses a little metal box that is slightly crushed by air. It doesn't need messy liquids like mercury to work.

📝 Teacher's Note: These are more common today than mercury barometers because they are safer and more portable.

🎯 Exam Tip: The word "Aneroid" literally means "without liquid."

Question 32.
Answer: Barometer is a device which is used for measuring atmospheric pressure. Barometers are used in weather forecasting and in measuring altitudes.
In simple words: Barometers help us guess if it's going to rain (low pressure) or be sunny (high pressure), and they also help us know how high up a mountain we've climbed.

📝 Teacher's Note: Low pressure often brings storms because it allows air to rise and form clouds.

🎯 Exam Tip: Name "Weather forecasting" and "Altitude measurement" as the two main uses.

Question 33.
Answer: Mercury is used in barometer because

• It can be obtained in pure form.
• It does not vaporize at ordinary temperatures.
• Its density is high and hence the length of the mercury column supported by atmospheric pressure is \( 76 \text{ cm} \) which is practically possible.

In simple words: Mercury is perfect for barometers because it's super heavy, doesn't evaporate easily, and fits in a small, convenient tube.

📝 Teacher's Note: Mercury's high density is the most important scientific reason for its use in this instrument.

🎯 Exam Tip: Memorize these three points as they are frequently asked as a "Give Reason" question.

Page No: 173

Question 1.
Answer: All liquid exerts an upward force on the body placed in it. This Phenomenon is called buoyancy.
In simple words: Buoyancy is the "floating power" of water that pushes things up when they are submerged.

📝 Teacher's Note: Demonstrate this by trying to hold a floating ball under water; the force you feel pushing back is buoyancy.

🎯 Exam Tip: "Upward force" is the defining keyword here.

Question 2.
Answer: The upward force which any liquid exerts upon a body placed in it is called the upthrust. The SI unit of upthrust is \( \text{N} \).
In simple words: Upthrust is just another name for the buoyant force. It's measured in Newtons, just like any other force.

📝 Teacher's Note: Clarify that Buoyancy is the name of the effect, while Upthrust is the actual force value.

🎯 Exam Tip: Remember: Upthrust is a force, so its SI unit is Newtons (\( \text{N} \)).

Question 3.
Answer: Buoyant force act on a body in upward direction.
In simple words: Buoyancy always pushes straight up, trying to lift things to the surface.

📝 Teacher's Note: This force acts opposite to gravity (which pulls things down).

🎯 Exam Tip: Always draw the buoyant force arrow pointing vertically upward from the center of the submerged part.

Question 4.
Answer: Upthrust is defined as the upward force on the object provided by the liquid because the object has displaced some of the fluid.
In simple words: When you get into a bathtub, the water level rises because you pushed some water out of the way. That water you pushed aside is what pushes back up on you!

📝 Teacher's Note: This is a simple bridge to understanding Archimedes' Principle.

🎯 Exam Tip: Mention "displacement of fluid" to make your definition more scientific.

Question 5.
Answer: When block of cork is immersed in water buoyant force acts on it in upward direction so to overcome this force we have to apply an equal force in downward direction to keep block of cork inside water.
In simple words: Cork loves to float. To keep it underwater, you have to keep pushing it down because the water is constantly trying to shove it back up.

📝 Teacher's Note: This shows that upthrust can be stronger than the weight of an object.

🎯 Exam Tip: Use the term "resultant force" to explain why the cork rises if you let go.

Question 6.
Answer: Wood has density less than water so volume of water displaced by it is more than the volume of wooden block submerged so force of upthrust is greater than the weight of wood which pushes wooden block on the surface. Hence, a piece of wood when left under water again comes to the surface.
In simple words: Wood is very light for its size. The water it pushes aside weighs more than the wood itself, so the water wins the "pushing match" and shoves the wood back to the top.

📝 Teacher's Note: This is why wood floats even if it is a huge log.

🎯 Exam Tip: For any object that floats, the upthrust is equal to its total weight at the surface.

Question 7.
Answer: A body will weigh more in air as weight of body acts in downward direction and there is no force in upward direction while body submerged in water weigh less because an upthrust act on the body in upward direction so the resultant weight of the body decreases.
In simple words: You feel lighter in a pool than on land because the water is helping you carry your own weight by pushing up on you.

📝 Teacher's Note: Use a spring balance to show that a heavy stone has a lower reading when dipped in water.

🎯 Exam Tip: Use the term "Apparent Weight" to describe how heavy an object feels when it is submerged.

Question 8.
Answer: Upthrust or buoyant force depends on the following factors:

• Volume of body submerged in the liquid.
• Density of the liquid.
• Acceleration due to gravity.

In simple words: You get more upthrust if you're bigger, if the liquid is heavier (like salt water vs. fresh water), or if gravity is stronger.

📝 Teacher's Note: This is why it's easier to float in the salty Dead Sea than in a freshwater swimming pool—the salt water is denser.

🎯 Exam Tip: Memorize the formula for upthrust: \( F = V \times \rho \times g \).

Question 9.
Answer:
(i) \( F_2 \) represent the buoyant force acting on the stone in upward direction.
(ii) \( m_1 \) represent the apparent mass of the stone during motion through water.
(iii) Net force acting on the stone \( = F_1 - F_2 = (m - m_1)g \).
(iv) Mass of stone \( = 200 \text{ gm} = 0.2 \text{ kg} \).
Volume of stone \( = 80 \text{ cm}^3 = 80 \times 10^{-6} \text{ m}^3 \).
Density of water \( = 1 \text{ gcm}^{-3} = 1000 \text{ kgm}^3 \).
Acceleration due to gravity \( = 10 \text{ ms}^{-2} \).

Upthrust \( = V \times \rho \times g \).
Upthrust \( = 80 \times 10^{-6} \times 1000 \times 10 = 0.8 \text{ N} \).
Weight of the stone \( = \text{mass} \times \text{gravity} = 0.2 \times 10 = 2 \text{ N} \).
Resultant weight of the stone \( = \text{weight} - \text{upthrust} = 2 \text{ N} - 0.8 \text{ N} = 1.2 \text{ N} \).
Resultant acceleration of the gravity \( = \text{m} \times \text{a'} \)

\( \implies m \times a' = 1.2 \text{ N} \)
\( a' = 1.2 / 0.2 = 6 \text{ ms}^{-2} \).

Resultant acceleration of the stone as it falls through the water is \( 6 \text{ ms}^{-2} \).
In simple words: Normally things fall at \( 10 \text{ m/s}^2 \) because of gravity. But in water, buoyancy pushes up and slows the fall down. For this stone, it falls at only \( 6 \text{ m/s}^2 \).

📝 Teacher's Note: This calculation explains why things "sink slowly" instead of just crashing to the bottom like they do in air.

🎯 Exam Tip: The net force is Weight minus Upthrust. Use \( F = m \times a \) to find the new slower acceleration.

Question 10.
Answer: Weight of the body in air \( = 300 \text{ gf} \).
Apparent Weight of the completely immersed body in water \( = 280 \text{ gf} \).

• Loss in weight of the body \( = \text{Weight of body in air} - \text{apparent weight of immersed body} \).


• \( \implies \text{Loss in weight} = 300 \text{ gf} - 280 \text{ gf} = 20 \text{ gf} \).


• As upthrust on the body \( = \text{loss in weight} \).


• \( \implies \text{So upthrust} = 20 \text{ gf} \).

In simple words: If an object weighs \( 300 \) grams on a scale but feels like \( 280 \) grams in water, the "missing" \( 20 \) grams is exactly how much the water is pushing it up.

📝 Teacher's Note: The unit "gf" stands for gram-force. It is a convenient unit for these experiments.

🎯 Exam Tip: Archimedes' Principle can be simplified to: Upthrust \( = \) Weight in Air \( - \) Weight in Liquid.

Question 11.
Answer: Edge of metal cube \( = 5 \text{ cm} \).
Density of the metal cube \( = 9 \text{ gcm}^{-3} = 9 \times 10^3 \text{ kgm}^{-3} \).
Volume of the metal cube \( = 125 \text{ cm}^3 = 125 \times 10^{-6} \text{ m}^3 \).
Mass of the metal cube \( = 9 \times 10^3 \times 125 \times 10^{-6} = 1125 \times 10^{-3} = 1.125 \text{ kg} \).
Weight of the liquid \( = \text{mass} \times \text{gravity} = 1.125 \times 10 = 11.25 \text{ N} \).
Density of liquid \( = 1.2 \text{ gcm}^{-3} = 1.2 \times 10^3 \text{ kgm}^{-3} \).
Upthrust of the liquid \( = V \times \rho \times g \).

\( \implies \text{Upthrust} = 125 \times 10^{-6} \times 1.2 \times 10^3 \times 10 = 1.5 \text{ N} \).
Apparent weight of the body \( = \text{weight of liquid} - \text{upthrust} \)
Apparent weight \( = 11.25 \text{ N} - 1.5 \text{ N} = 9.75 \text{ N} \).
Tension in the string is equal to the apparent weight of the body. So, tension in string would be \( 9.75 \text{ N} \).
In simple words: If you hang a heavy metal block by a string into a bucket of liquid, the liquid helps hold the block up. The "pull" on the string (tension) becomes less because of this help from the liquid.

📝 Teacher's Note: Emphasize that "tension" is just the force measured by the string. If the object gets lighter due to upthrust, the tension goes down.

🎯 Exam Tip: Tension \( = \) Apparent Weight. This is a very common question type in physics finals.

Question 12.
Answer: It is easier to lift a heavy stone under water because in water an upthrust acts on the upward direction which reduces the apparent weight of the stone and makes it easy to lift.
In simple words: Picking up a heavy rock in a lake is much easier than picking it up on the beach because the water is pushing the rock up for you!

📝 Teacher's Note: This is why astronauts train in giant swimming pools—it simulates a low-gravity environment.

🎯 Exam Tip: Use the term "Apparent Weight" in your explanation to sound more precise.

Question 13.
Answer: Principle of Archimedes’ states that when a body is totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced by it.
In simple words: The "lift" an object gets from water is exactly the same as the weight of the water it pushed out of the way to make room for itself.

📝 Teacher's Note: This is the most important rule in all of fluid mechanics. Help students visualize the "displaced" water as the water that overflows from a full cup.

🎯 Exam Tip: Be sure to say "weight of fluid displaced," not just "volume."

Question 14.
Answer: We can verify archimedis’ principle experimentally by doing this experiment.
The stone weighed \( 0.67 \text{ N} \) in air and \( 0.40 \text{ N} \) when immersed in water. The displaced water weighed is \( 0.27 \text{ N} \) (\( = 0.67 - 0.40 \)).
Pour water into eureka can till the water starts overflowing through the spout. When the water stops dripping replace the beaker by another one of known weight. Suspend a stone with the help of a string from the hook of a spring balance and record the weight of the stone. Now, gradually lower the body into eureka can containing water and record its new weight in water when it is fully submerged in water. When no more water drips from the spout, weigh the beaker containing water.

After observation, we can conclude that apparent loss of weight of stone (calculated by differences in weight measured by spring balance) \( = \) weight of water displaced (weight of water in beaker). This proves the archimedis’ principle.
In simple words: To prove Archimedes was right, we weighed a stone in air and then in water. The weight it "lost" in water was exactly the same as the weight of the water that spilled out into the little beaker.

📝 Teacher's Note: Using a "Eureka can" (a can with a spout) is the classic way to demonstrate this in class.

🎯 Exam Tip: Describe the steps clearly: 1. Weigh in air, 2. Weigh in water, 3. Weigh displaced water, 4. Show they match.

Question 15.
Answer: The fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
\( (\text{Density of floating body / Density of liquid}) = \text{fraction submerged} \).
Density of liquid \( = (\text{Density of floating body / fraction submerged}) \).

Now in this case as density of floating body is same in all three cases. So, density of liquid is maximum when fraction submerged of wooden block is minimum. As liquid R has least fraction submerged of wooden block in it So, it has maximum density.
In simple words: If you put the same block in different liquids, it will float higher in the heavier (denser) liquids. Since the block floats highest in Liquid R, Liquid R must be the densest.

📝 Teacher's Note: This is a great conceptual question. Students often think the deeper it sinks, the denser the liquid is, but it's the opposite.

🎯 Exam Tip: Higher floating \( = \) Higher density of the liquid.

Page No : 174

Question 16.
Answer: Wood has density less than water so volume of water displaced by it is more than the volume of wooden block submerged so force of upthrust is greater than the weight of wood which makes it float on the water surface. And the apparent weight of the piece of the wood would be zero.
In simple words: When something floats perfectly on the surface, it feels weightless! The upward push from the water exactly balances its entire weight, so its "apparent weight" is zero.

📝 Teacher's Note: This "Zero weight" concept is key to the Law of Flotation.

🎯 Exam Tip: For any floating body, Apparent Weight is always \( 0 \).

Question 17.
Answer: Density of iron is less than the density of mercury so it will float on the surface of the mercury. Apparent weight of the floating iron ball is zero.
In simple words: Mercury is a liquid metal that is much heavier than iron. Because of this, iron balls actually float on mercury like wood floats on water!

📝 Teacher's Note: This often surprises students because they expect metal to always sink. It depends on the liquid!

🎯 Exam Tip: Always compare densities: If \( \rho_{\text{object}} < \rho_{\text{liquid}} \), it floats.

Question 18.
Answer: Iron nail has density less than that of mercury so it will float on the surface of mercury but in the case of water it will sink because the density of iron nail is more than that of water.
In simple words: An iron nail sinks in water because it's denser than water. But that same nail floats on top of mercury because mercury is even denser than iron!

📝 Teacher's Note: Use this to show that "sinking" isn't a permanent property of an object; it depends on the liquid you put it in.

🎯 Exam Tip: Sinking vs. Floating depends on the ratio of densities.

Question 19.
Answer: No, the relative density of a substance is the ratio of the density of the substance to the density of water at \( 4^{\circ}\text{C} \).
In simple words: Relative density isn't just about regular water; we specifically use water at \( 4^{\circ}\text{C} \) as the standard because that's when water is at its densest.

📝 Teacher's Note: This is a definition check. Ensure students understand why a specific temperature is mentioned.

🎯 Exam Tip: Always include "\( 4^{\circ}\text{C} \)" in your definition of Relative Density.

Question 20.
Answer:

• SI unit of buoyant force is \( \text{N} \).
• SI unit of density is \( \text{Kgm}^{-3} \).
• SI unit of weight of body is \( \text{N} \).
• Relative density is a pure ratio it has no dimension.

In simple words: These are the standard units. One important thing: Relative Density is just a number (like \( 2 \) or \( 5.5 \)) and doesn't have any units like kilograms or meters.

📝 Teacher's Note: Explain that since Relative Density is a ratio of two densities, the units cancel out.

🎯 Exam Tip: Never write a unit after a Relative Density value in your calculations.

Question 21.
Answer: Density of iron is more than the density of water so it sinks down in the water but in case of ship, it is design in such a manner that it encloses large quantity of air in air tight bags and in rooms and corridors which makes the average density of ship less than that of water and ship floats on the surface of the water.
In simple words: A solid chunk of iron sinks. But a ship is hollow and full of air. The air and iron together act like one giant, light "object" that is less dense than water, so it floats.

📝 Teacher's Note: This is why a metal bowl floats but a metal spoon sinks. The bowl traps air, lowering its overall density.

🎯 Exam Tip: The key concept is "Average Density" or "Effective Density" of the hollow object.

Question 22.
Answer: The fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
\( (\text{Density of floating body / Density of liquid}) = \text{fraction submerged} \).
The Fraction of ice submerged in water remain same as density of ice and water remain same during melting. As ice melts some volume of ice decrease and convert into water and volume of water increase by same amount. So, level of water remains same during melting.
In simple words: When ice cubes melt in your glass of water, the water level doesn't go up! This is because the ice was already pushing aside exactly as much water as it will turn into when it melts.

📝 Teacher's Note: This is a very common "trick" question. Students usually assume the water level will rise.

🎯 Exam Tip: State clearly that the "volume of displaced water" equals the "volume of melted ice."

Question 23.
Answer: The fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
\( (\text{Density of floating body / Density of liquid}) = \text{fraction submerged} \).
Height of wooden piece \( = 15 \text{ cm} \).
Height of wooden piece sinks in water \( = 10 \text{ cm} \).
Fraction of wooden piece submerged in water \( = 10/15 = 0.67 \).
As liquid is water so ratio of Density of wooden by density of water gives relative density of floating wooden piece. So, relative density of wooden block is \( 0.67 \).
Height of wooden piece \( = 15 \text{ cm} \).
Height of wooden piece sinks in spirit \( = 12 \text{ cm} \).
Fraction of wooden piece submerged in water \( = 12/15 = 0.8 \).
We know density of wooden piece \( = 0.67 \)
\( (\text{Density of floating body / Density of liquid}) = \text{fraction submerged} \).
Density of liquid/spirit \( = (\text{Density of floating body / fraction submerged}) \)
Density of liquid/spirit \( = 0.67/0.8 = 0.83 \).
Relative density of spirit is \( 0.83 \).
In simple words: By seeing how deep the wood sinks in water (\( 10 \text{cm} \)) and then in spirit (\( 12 \text{cm} \)), we can calculate how heavy spirit is compared to water. Spirit is slightly lighter than water (\( 0.83 \)).

📝 Teacher's Note: This is a two-step problem. First find the object's RD, then use it to find the second liquid's RD.

🎯 Exam Tip: RD of object \( = \) Fraction submerged in water. This is a very useful shortcut!

Question 24.
Answer:

• When a body is completely immersed in water then it displaces equal volume of water to its own weight. Volume of body of man is same in both river and sea so weight of water of sea displaced by him is equal to the weight of water of river displaced by him. And ratio of weights would be \( 1:1 \).
• Sea water contains mineral salts and density of sea water increase due to presence of these. As density of sea water is more than the normal water so it apply more buoyant force than usual one and a person find it easy to swim in sea water.

In simple words: (i) A person takes up the same amount of space in both river and sea water. (ii) However, sea water is "heavier" because of salt, so it pushes up on the swimmer harder, making it easier to stay afloat.

📝 Teacher's Note: Use the "Egg in saltwater" experiment to show how adding salt makes objects float better.

🎯 Exam Tip: Mention "presence of mineral salts" to explain the higher density of sea water.

Question 25.
Answer: The fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
\( (\text{Density of floating body / Density of liquid}) = \text{fraction submerged} \).
Fraction of wooden piece submerged in water \( = 2/3 = 0.67 \).
As liquid is water so ratio of Density of wooden by density of water gives relative density of floating wooden piece.
So, relative density of wooden block is \( 0.67 \).
Density of water in SI system \( = 1000 \text{ Kg m}^{-3} \).
Density of wood \( = \text{relative density} \times \text{density of water} = 0.67 \times 1000 \text{ Kg m}^{-3} = 670 \text{ kgm}^{-3} \).
Fraction of wooden piece submerged in oil \( = 3/4 = 0.75 \).
We know density of wooden piece \( = 0.67 \)
\( (\text{Density of floating body / Density of liquid}) = \text{fraction submerged} \).
Relative Density of oil \( = (\text{Relative Density of wooden block / fraction submerged}) \)
Density of oil \( = 0.67/0.75 = 0.893 \).
Density of water in SI system \( = 1000 \text{ Kg m}^{-3} \).
Density of oil \( = \text{relative density} \times \text{density of water} = 0.893 \times 1000 \text{ Kg m}^{-3} = 893 \text{ kgm}^{-3} \).
In simple words: We used the floating height to find out that wood is \( 670 \text{ kg/m}^3 \). Then, by seeing how it floats in oil, we calculated that the oil's density is \( 893 \text{ kg/m}^3 \).

📝 Teacher's Note: This is a comprehensive numerical. Remind students to always convert Relative Density to absolute density by multiplying by \( 1000 \).

🎯 Exam Tip: Keep your decimals up to 3 places (\( 0.893 \)) for better accuracy in physics marks.

Question 26.
Answer: Relative density of Ice \( = 0.92 \)
Relative density of sea water \( = 1.025 \)
Let total volume of iceberg \( = X \text{ cm}^3 \).
Volume of iceberg above water \( = 800 \text{ cm}^3 \).
Volume of iceberg in submerged in the water \( = (X - 800) \text{ cm}^3 \).
Fraction of iceberg submerged \( = (X - 800)/X \)
Now we know that fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
\( (\text{Density of ice / Density of sea water}) = \text{fraction submerged} \)
\( 0.92/1.025 = (X - 800)/X \)
\( 0.8975 \text{ X} = X - 800 \)
\( X - 0.8975 \text{ X} = 800 \)
\( 0.1025 \text{ X} = 800 \)
\( X = 800 / 0.1025 = 7804.8 \text{ cm}^3 \).
Total volume of iceberg \( = 7804.8 \text{ cm}^3 \).
In simple words: Icebergs are mostly underwater. For this specific iceberg, only \( 800 \text{ cm}^3 \) is showing above the waves, but its total size is over \( 7800 \text{ cm}^3 \)!

📝 Teacher's Note: This is the science behind the phrase "tip of the iceberg." Most of it is hidden from view.

🎯 Exam Tip: Set up the equation \( \text{RD}_1 / \text{RD}_2 = V_{\text{submerged}} / V_{\text{total}} \) carefully.

Question 27.
Answer: Relative density of wax \( = 0.95 \)
Relative density of brine \( = 1.1 \)
\( (\text{Density of wax / Density of brine}) = \text{fraction submerged} \)
\( 0.95/1.1 = \text{fraction of volume submerged} \)
Fraction of volume submerged \( = 0.86 \)
In simple words: Brine is very salty water. When wax is put into it, about \( 86\% \) of the wax sits underwater and \( 14\% \) floats above.

📝 Teacher's Note: Brine is much denser than regular water (\( 1.1 \) vs \( 1.0 \)), so things float better in it.

🎯 Exam Tip: Convert fractions to two decimal places for your final answer.

Question 28.
Answer: Relative density of Ice \( = 0.9 \text{ cm} \)
Relative density of sea water \( = 1.1 \text{ cm} \)
\( (\text{Density of ice / Density of sea water}) = \text{fraction submerged of iceberg} \)
\( 0.9/1.1 = \text{fraction of iceberg submerged} \)
Fraction of iceberg submerged \( = 9/11 \).
In simple words: In this ocean, nearly \( 82\% \) (\( 9/11 \)) of an iceberg would be hidden underwater.

📝 Teacher's Note: The source text mistakenly puts "cm" after relative density. Correct them in class: RD has no units.

🎯 Exam Tip: If the question asks for a fraction, leaving it as \( 9/11 \) is usually acceptable.

Question 30.
Answer: Lactometer is commonly used for testing the purity of milk.
In simple words: It’s a tool that floats in milk to tell you if it's pure or if it has been watered down.

📝 Teacher's Note: A lactometer works on the principle of buoyancy; pure milk is denser than watered-down milk.

🎯 Exam Tip: This is a classic "Name the instrument" question. Match "milk" with "lactometer."

Question 31.
Answer: Density of water at \( 4^{\circ}\text{c} \) in SI system is \( = 1000 \text{ Kgm}^{-3} \).
In simple words: One cubic meter of water weighs exactly \( 1000 \text{ kilograms} \). This is the standard measurement used in science.

📝 Teacher's Note: This is a fundamental constant in physics that students must memorize.

🎯 Exam Tip: Always specify the temperature (\( 4^{\circ}\text{C} \)) as density changes with temperature.

Question 32.
Answer: Side of wooden cube \( = 10 \text{ cm} \).
Volume of wooden cube \( = 10 \times 10 \times 10 = 1000 \text{ cm}^3 \).
Mass of wooden cube \( = 700 \text{ g} \).
Density of wooden cube \( = \text{mass / volume} = 700 / 1000 = 0.7 \text{ gcm}^{-3} \).
Density of water \( = 1 \text{ gcm}^{-3} \).
\( (\text{Density of floating body / Density of liquid}) = \text{fraction submerged} \)

\( \implies 0.7/1 = \text{fraction submerged} \)
Fraction of wooden cube submerged in water \( = 0.7 \)
Height of wooden cube \( = 10 \text{ cm} \)
Part of wooden cube which is submerged \( = 10 \times 0.7 = 7 \text{ cm} \)
So, wooden cube will float in water with \( 3 \text{ cm} \) height above the water surface.
In simple words: The block weighs \( 700 \text{ grams} \). When it's in water, \( 7 \text{ cm} \) of it sinks and \( 3 \text{ cm} \) stays above the water.

📝 Teacher's Note: This is a practical example showing how mass and volume determine the floating position.

🎯 Exam Tip: Calculate density first, then the fraction, and finally the height. Step-by-step logic scores full marks.

Question 33.
Answer: Volume of wooden block \( = 0.032 \text{ m}^3 \).
Mass of wooden block \( = 24 \text{ Kg} \).
Density of wooden block \( = \text{mass / volume} = 24 / 0.032 = 750 \text{ Kgm}^{-3} \).
Density of water \( = 1000 \text{ Kgm}^{-3} \).
\( (\text{Density of floating body / Density of liquid}) = \text{fraction submerged} \)
\( 750 / 1000 = \text{fraction submerged} \)
Fraction of wooden block submerged in water \( = 0.75 \)
Total volume of wooden block \( = 0.032 \text{ m}^3 \).
Part of volume of wooden block which is submerged \( = 0.032 \times 0.75 = 0.024 \text{ m}^3 \).
In simple words: About \( 75\% \) of this wooden block is underwater. We calculated its total volume submerged as \( 0.024 \) cubic meters.

📝 Teacher's Note: Remind students that "fraction submerged" refers to volume, but for uniform blocks, it also corresponds to height.

🎯 Exam Tip: Double-check your division: \( 24 / 0.032 \) should result in \( 750 \).

Question 34.
Answer: Relative density \( = \text{density of substance / density of water at } 4^{\circ}\text{C} \).
As relative density of platinum is \( 21.50 \), this means platinum is \( 21.5 \) times denser than water at \( 4^{\circ}\text{C} \).
In simple words: Relative density is a way of saying "how much heavier is this than water?" Platinum is very heavy—it’s \( 21.5 \) times heavier than the same amount of water!

📝 Teacher's Note: Use this to compare the "heaviness" of different materials intuitively.

🎯 Exam Tip: Platinum is one of the densest metals known; the high RD value reflects this.

Page No : 175

Question 35.
Answer: Density of mercury \( = 13600 \text{ Kgm}^{-3} \).
Density of water at \( 4^{\circ}\text{C} = 1000 \text{ kg m}^{-3} \).
Relative density \( = \text{density of substance / density of water at } 4^{\circ}\text{C} \).
Relative density of mercury \( = 13600 \text{ Kgm}^{-3} / 1000 \text{ kg m}^{-3} = 13.6 \).
In simple words: This confirms that mercury is \( 13.6 \) times heavier than water.

📝 Teacher's Note: This RD value is a standard constant used in many fluid pressure problems.

🎯 Exam Tip: In SI units, just divide density by \( 1000 \) to get the RD.

Question 36.
Answer: volume of body \( = 100 \text{ cm}^3 \).
Weight of body \( = 1 \text{ kgf} = 1000 \text{ gf} \)
Mass of body \( = 1000 \text{ gm} \).
Density of liquid \( = 1000 \text{ gm} / 100 \text{ cm}^3 = 10 \text{ gcm}^{-3} \).
Density of water at \( 4^{\circ} = 1 \text{ gcm}^{-3} \).
Relative density \( = \text{density of substance / density of water at } 4^{\circ} \text{ C} \)
Relative density \( = 10 \text{ gcm}^{-3} / 1 \text{ gcm}^{-3} = 10 \)
Mass of body \( = 1000 \text{ gm} \).
Density of water \( = 1 \text{ gcm}^{-3} \)
Acceleration due to gravity \( = 10 \text{ ms}^{-2} \).
Upthrust \( = V \times \rho \times g \).
Upthrust \( = 100 \times 1 \text{ gf} = 100 \text{ gf} \).
Resultant weight of the body \( = \text{weight} - \text{upthrust} = 1000 \text{ gf} - 100 \text{ gf} = 900 \text{ gf} \).
In simple words: This body starts at \( 1000 \text{ grams} \). Once it's in water, the water pushes up with \( 100 \text{ grams} \) of force, so the body feels like it only weighs \( 900 \text{ grams} \).

📝 Teacher's Note: This problem uses "gf" (gram-force) units. Ensure students understand that \( 100 \text{ cm}^3 \) of water weighs \( 100 \text{ grams} \), which is why the upthrust is \( 100 \text{ gf} \).

🎯 Exam Tip: Always subtract Upthrust from real weight to get the final "Resultant" weight.

Question 37.
Answer: When a body is completely immersed in water then it displaces equal volume of water to its own weight.

So, volume of body \( = 20000 \text{ cm}^3 \).
Mass of body \( = 70 \text{ kg} = 70000 \text{ gm} \)
Density of body \( = \text{mass / volume} = 70000 / 20000 = 3.5 \text{ gm cm}^{-3} \).
Density of water in C.G.S system \( = 1 \text{ g cm}^{-3} \).
Relative density of body \( = \text{density of body / density of water} = 3.5 \text{ gm cm}^{-3} / 1 \text{ g cm}^{-3} \).
Relative density \( = 3.5 \).
In simple words: This object is \( 3.5 \) times heavier than water. We found this by dividing its total mass (\( 70 \text{ kg} \)) by its volume.

📝 Teacher's Note: Remind students that \( 1 \text{ kg} = 1000 \text{ grams} \). Conversion errors are common here.

🎯 Exam Tip: In CGS units, Density and Relative Density have the same number because water density is \( 1 \).

Question 38.
Answer: Relative density \( = \text{density of mercury / density of water} \).
Density of mercury \( = \text{relative density} \times \text{density of water} \).
Relative density \( = 13.6 \).
Density of water in C.G.S system \( = 1 \text{ g cm}^{-3} \).
So, density of mercury in C.G.S system \( = 13.6 \times 1 = 13.6 \text{ gcm}^{-3} \).
Density of water in SI system \( = 1000 \text{ Kg m}^{-3} \).
So, density of mercury in SI system \( = 13.6 \times 1000 = 13.6 \times 10^3 \text{ Kgm}^{-3} \).
In simple words: Depending on the units we use, mercury’s density is either \( 13.6 \) (CGS) or \( 13,600 \) (SI).

📝 Teacher's Note: This highlights the importance of units in science. The number changes, but the material stays the same.

🎯 Exam Tip: Multiply CGS density by \( 1000 \) to convert it to SI units (\( \text{kg/m}^3 \)).

Question 39.
Answer: Density of iron is \( = 7.8 \times 10^3 \text{ Kg m}^{-3} \).
Density of water at \( 4^{\circ}\text{C} = 10^3 \text{ Kg m}^{-3} \).
Relative density of a substance is the ratio of the density of the substance to the density of water at \( 4^{\circ}\text{C} \).
So, relative density of iron \( = 7.8 \times 10^3 \text{ Kg m}^{-3} / 10^3 \text{ Kg m}^{-3} = 7.8 \)
In simple words: Iron is \( 7.8 \) times denser than water.

📝 Teacher's Note: Since both values are multiplied by \( 10^3 \), they cancel out in the division, leaving just the \( 7.8 \).

🎯 Exam Tip: Relative density is always unitless.

Question 40.
Answer:

• Mass of a metallic piece remains unchanged with increase in temperature.
• Volume of metallic piece increases with increase in temperature.
• Density of metallic piece decreases with increases in temperature.

In simple words: When you heat metal, it doesn't get heavier, but it does expand (get bigger). Because it’s bigger but weighs the same, it’s now "less dense."

📝 Teacher's Note: Use the "Hot Air Balloon" analogy—heated air expands and becomes less dense, causing it to float.

🎯 Exam Tip: Mass is constant regardless of temperature changes in a closed system.

Question 41.
Answer: Density of water decreases with the increase in temperature and increases with decreases in temperature.
In simple words: Hot water is lighter and floats on top of cold water. As water gets colder, it gets more "packed" and heavy.

📝 Teacher's Note: Note the exception: between \( 0^{\circ}\text{C} \) and \( 4^{\circ}\text{C} \), water actually gets *denser* as it warms up. This is called anomalous expansion.

🎯 Exam Tip: Standard density rules usually apply only above \( 4^{\circ}\text{C} \).

Question 42.
Answer:
(i) Mass \( = \text{VOLUME} \times \text{density} \).
(ii) SI unit of density is \( \text{Kgm}^{-3} \).
(iii) Density of water is \( 1000 \text{ Kgm}^{-3} \).
(iv) Density in \( \text{Kgm}^{-3} = 1000 \times \text{ density in gcm}^{-3} \).
In simple words: These are the four basic rules for calculating and measuring density.

📝 Teacher's Note: These are the building blocks for every numerical in this chapter.

🎯 Exam Tip: Memorize the conversion factor in part (iv); it is used in almost every multi-step problem.

Page No: 177

Question 1.
Answer: All liquid exerts a upward force on the body placed in it. This Phenomenon is called buoyancy.
In simple words: Buoyancy is the "invisible lift" water gives to anything you put in it.

📝 Teacher's Note: Mention that gases also exert buoyancy (e.g., helium balloons).

🎯 Exam Tip: Define it as an "upward force" for a complete answer.

Question 2.
Answer: The upward force which any liquid exerts upon a body placed in it is called the upthrust.
In simple words: Upthrust is the actual force pushing up against an object in a fluid.

📝 Teacher's Note: Reiterate that "Upthrust" and "Buoyant force" mean the same thing.

🎯 Exam Tip: Mention both terms if asked to explain how things float.

Question 3.
Answer: Pressure is a scalar quantity.
In simple words: Pressure doesn't have a direction; it just tells you how much squeeze there is.

📝 Teacher's Note: Scalar quantities like pressure can be added together easily because they don't have direction.

🎯 Exam Tip: Always list Pressure as scalar in multiple-choice questions.

Question 4.
Answer: Thrust is a vector quantity.
In simple words: Thrust is a force that acts in a straight line, so it has direction.

📝 Teacher's Note: This is a standard contrast to Pressure (scalar).

🎯 Exam Tip: Force, Weight, and Thrust are all vectors.

Question 5.
Answer: SI unit of density is \( \text{Kgm}^{-3} \).
In simple words: It’s kilograms per cubic meter.

📝 Teacher's Note: Remind students that \( \text{m}^{-3} \) means "per cubic meter."

🎯 Exam Tip: Always use the SI unit unless the question specifically asks for CGS.

Question 6.
Answer: The relative density of a substance is the ratio of the density of the substance to the density of water at \( 4^{\circ}\text{C} \).
In simple words: It tells us how much more or less dense something is compared to water.

📝 Teacher's Note: At \( 4^{\circ}\text{C} \), water density is exactly \( 1 \text{ g/cm}^3 \), which makes it a perfect standard.

🎯 Exam Tip: Always specify water at \( 4^{\circ}\text{C} \) in your definition.

Question 7.
Answer: The pressure at a point in a liquid depends upon the following three factors:
(i) It depends on the point below the free surface (\( h \)).
(ii) It depends on density of liquid (\( \rho \)).
(iii) It depends upon acceleration due to gravity (\( g \)) of the place.
In simple words: Depth, the type of liquid, and gravity are the only things that change liquid pressure.

📝 Teacher's Note: This is a repeat of a previous question, showing its importance in the curriculum.

🎯 Exam Tip: Use the formula \( P = h \rho g \) to remember these three factors.

Question 8.
Answer: Principle of Archimedes’ states that when a body is totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced. Yes, it applies to gases also.
In simple words: The "push up" from any fluid (like air or water) is the same weight as the fluid you moved out of the way.

📝 Teacher's Note: Archimedes' Principle explains why ships float AND why helium balloons rise.

🎯 Exam Tip: The "Yes, it applies to gases" is a common bonus point part of the question.

Question 9.
Answer: Thrust on a surface is the force acting normally on a surface while pressure on a surface is thrust acting on the unit area of a surface.
In simple words: Thrust is the whole push; pressure is the push on each little bit of space.

📝 Teacher's Note: This is a core comparison between force and its concentration.

🎯 Exam Tip: "Normal force" means perpendicular force.

Question 10.
Answer: Pascal’s law states that pressure applied to an enclosed liquid, is transmitted equally to every part of the liquid or in other words when pressure is applied at a point in a confined fluid, it is transmitted undiminished and equally in all directions throughout the liquid.
In simple words: Squeeze a liquid in a closed box, and every single part of that liquid feels the exact same extra squeeze.

📝 Teacher's Note: Use a syringe filled with water to show how the pressure at the plunger is felt at the tip.

🎯 Exam Tip: Use the word "undiminished" to show the pressure doesn't get weaker as it travels.

Question 11.
Answer: Yes, all liquid exert pressure.
In simple words: Whether it's oil, water, or honey, any liquid will push against its container.

📝 Teacher's Note: Liquid pressure is due to the weight of the liquid and the movement of its molecules.

🎯 Exam Tip: A simple "Yes" with a brief explanation of weight-based pressure is usually enough.

Question 12.
Answer: Hydraulic machines such as hydraulic press, hydraulic brakes and hydraulic jack are application of pascal’s law.
In simple words: These machines use liquid to turn a small push into a huge lifting force.

📝 Teacher's Note: This is "Force Multiplication"—using physics to do heavy lifting easily.

🎯 Exam Tip: Be ready to name these three examples: press, brakes, and jack.

Question 13.
Answer: Pascal’s law is principle of hydraulic machines.
In simple words: Every hydraulic machine works because liquid pressure spreads out evenly.

📝 Teacher's Note: This reinforces the link between theory and practical engineering.

🎯 Exam Tip: This is a standard fill-in-the-blank question.

Question 14.
Answer: Brahma press depends upon Pascal’s law.
In simple words: Even specific industrial machines like the Brahma press follow the basic rules of liquid pressure.

📝 Teacher's Note: The Bramah press was the first practical hydraulic press invented.

🎯 Exam Tip: Don't get confused by the name "Brahma"—it’s just a type of hydraulic press.

Question 15.
Answer:

• A hydraulic press can be used for extracting juice of sugarcane, sugar beet etc.
• A hydraulic press can be used for pressing cotton bales, quilts, books etc.

In simple words: Hydraulic presses are like giant nutcrackers—they can squeeze liquid out of plants or pack huge bundles of clothes into small boxes.

📝 Teacher's Note: Discuss why using liquid is better than using manual labor for these massive squeezing tasks.

🎯 Exam Tip: Provide one "extraction" and one "compression" use to show full understanding.

Question 16.
Answer: Atmospheric point at any point in air at rest is equal to the weight of a vertical column of air on a unit area surrounding the point, the column extending to the top of atmosphere.
In simple words: Air pressure is just the weight of all the air sitting on your head! Imagine a giant invisible pipe of air that goes from you all the way to outer space; that weight is what creates pressure.

📝 Teacher's Note: This weight is surprisingly high—about the weight of a small car on every square meter of your body!

🎯 Exam Tip: Define it as the "weight of a vertical air column" for full marks.

Question 17.
Answer: Atmospheric pressure at sea level is about \( 10^5 \text{ N/m}^2 \).
In simple words: At the beach, the air is pushing on everything with a force of \( 100,000 \) Newtons for every square meter.

📝 Teacher's Note: This is a standard reference value in physics (\( 1 \text{ atmosphere} \)).

🎯 Exam Tip: Remember this number: \( 10^5 \text{ Pa} \).

Question 18.
Answer: Barometer is used for measuring the atmospheric pressure.
In simple words: It’s a tool that lets us see if the air's "push" is getting stronger or weaker.

📝 Teacher's Note: Most wall barometers you see in homes are the "aneroid" (liquid-free) type.

🎯 Exam Tip: Do not confuse "Barometer" (Pressure) with "Thermometer" (Temperature).

Question 19.
Answer: Altimeter is a device which is used in an aircraft to measure its altitude.
In simple words: It uses air pressure to tell the pilot how many thousands of feet high the plane is flying.

📝 Teacher's Note: An altimeter is a modified barometer. As the plane goes higher, pressure drops, and the needle moves.

🎯 Exam Tip: Altitude measurement is the primary use of an altimeter.

Question 20.
Answer: A falling barometer indicates the approach of rain or storm or both.
In simple words: When air pressure drops quickly, it usually means bad weather is coming soon.

📝 Teacher's Note: This is because low pressure allows warm, moist air to rise and condense into clouds.

🎯 Exam Tip: "Falling barometer" \( = \) "Bad weather."

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Question 21.
Answer: A atmospheric pressure diving suit is a garment or device designed to protect a diver from the underwater environment. Yes, diving suits create buoyancy.
In simple words: A diving suit keeps you safe from the cold and pressure. Because it contains air, it also helps push you up towards the surface.

📝 Teacher's Note: This "buoyancy" can sometimes be a problem for divers, so they wear heavy lead weights to help them sink.

🎯 Exam Tip: Focus on "protection" and "buoyancy" as the two key features.

Question 23.
Answer: A fluid exerts pressure on the bottom due to its weight and on the walls of the container in which it is enclosed by virtue of its ability to flow. This is called fluid pressure.
In simple words: Because water can flow, it doesn't just push down; it pushes out against the sides of the bucket too!

📝 Teacher's Note: Contrast this with a solid block, which mostly just pushes down on its base.

🎯 Exam Tip: Mention "walls" and "bottom" to describe where fluid pressure acts.

Question 24.
Answer: A dam has broader walls at the bottom than at the top because the pressure exerted by a liquid increases with its depth, and at any point at a particular depth liquid pressure is same in all directions. Now as more pressure is exerted by water on the wall of the dam as depth increases. Hence a thick wall is constructed at the bottom of dam to withstand greater pressure.
In simple words: Deep water pushes harder than shallow water. Dams are built with extra-thick bases so they don't burst under all that pressure.

📝 Teacher's Note: This design is a perfect real-world illustration of \( P = h \rho g \).

🎯 Exam Tip: Clearly state that "pressure increases with depth" to justify the thick bottom.

Question 25.
Answer: Pascal’s law states that pressure applied to an enclosed liquid, is transmitted equally to every part of the liquid.
Means when pressure is applied at a point in a confined fluid, it is transmitted undiminished and equally in all directions throughout the liquid.
In simple words: Inside a liquid container, every single drop feels any extra push you give it, no matter where they are.

📝 Teacher's Note: This is why hydraulic brakes stop all four car wheels at the same time with one pedal push.

🎯 Exam Tip: "Enclosed liquid" is a critical condition for Pascal's law to work.

Question 27.
Answer: Factors which affect the atmospheric pressure as we go up are

• Weight of gaseous column.
• Density of gaseous column.
  In simple words: As you climb a mountain, there is less air above you (less weight) and the air is less "crowded" (lower density), so the pressure drops.

📝 Teacher's Note: Both factors contribute to why it’s hard to breathe at very high altitudes.

🎯 Exam Tip: List both "weight" and "density" to get full marks on this question.

Question 28.
Answer: Atmospheric pressure decreases with increase in height. Our atmosphere comprises of a large number of parallel layers. The pressure on a layer is equal to the thrust or weight of the gaseous column on the unit area of that layer. Hence, as we go up, the weight of the gaseous column decreases, which decrease the pressure of the gaseous column.
In simple words: Gravity pulls most of the air down near the ground. As you go higher, there's less air left above you to push down on you.

📝 Teacher's Note: Use the "stack of people" analogy—the person at the bottom feels the most pressure because everyone else is on top of them.

🎯 Exam Tip: Relate pressure directly to the "weight of the overhead air column."

Question 29.
Answer: Barometer is a device used for measuring atmospheric pressure.
In simple words: It’s the standard tool for measuring air pressure in a lab or at home.

📝 Teacher's Note: Mention that there are different types, like the Fortin and Aneroid barometers.

🎯 Exam Tip: Keep it simple: Barometer \( = \) Atmospheric Pressure.

Simple Barometer Has Two Main Defects

• It is not suitable for making accurate measurement of atmospheric pressure as any change in the level of mercury in the tube changes the level of the free surface of mercury is trough and fixed scale cannot be used with it.
• Simple barometer is not portable. So, it cannot be used by airmen, navigators, mountaineers, who need a portable barometer.

In simple words: The basic barometer is hard to read accurately and way too big and messy to carry around in a backpack or airplane.
📝 Teacher's Note: This is why the "Aneroid" barometer was invented—to solve these two problems.

🎯 Exam Tip: "Accuracy" and "Portability" are the two keywords for this answer.

Question 31.
Answer: We don’t feel uneasy even under enormous pressure of the atmosphere above us because our blood also exerts a pressure called blood pressure, which is greater than atmospheric pressure. So, there is balance between our blood pressure and atmospheric pressure.
In simple words: The air is pushing in on us, but our bodies are pushing back out with blood pressure. It’s like two people pushing on opposite sides of a door with the same strength—the door doesn't move, and we don't feel "crushed."

📝 Teacher's Note: This is a great way to explain why deep-sea fish explode if brought to the surface too quickly—their internal pressure is built for the deep ocean and becomes too much when the external pressure drops.

🎯 Exam Tip: The key term here is "balance" or "equilibrium" between internal and external pressure.

Question 32.
Answer: Reading of a barometer would rise if it is taken to the mine as pressure increases with depth.
Reading of a barometer would fall if it is taken to a hill as pressure decreases with increase in height.
In simple words: Going down into a mine is like going to the bottom of an "air ocean"—there is more air above you, so the pressure goes up. Going up a hill is the opposite; there is less air above you, so the pressure drops.

📝 Teacher's Note: Remind students that the "zero point" for height in atmospheric pressure is the top of the atmosphere, not the ground.

🎯 Exam Tip: Remember: Downward movement (depth) \( \implies \) Pressure Increase; Upward movement (altitude) \( \implies \) Pressure Decrease.

Question 33.
Answer: Weight of solid in air \( = 2.10 \text{ N} \)
Relative density of solid \( = 8.4 \)
Now, Relative density \( = \text{weight of solid in air / loss of weight of solid in water} \).
Loss of weight of solid in water \( = \text{weight of solid in air / Relative density} \).

\( \implies \text{Loss of weight of solid in water} = 2.10 / 8.4 = 0.25 \text{ N} \).
Weight of solid in water \( = \text{weight in air} - \text{loss of weight in water} \)

\( \implies \text{Weight of solid in water} = 2.10 - 0.25 = 1.85 \text{ N} \).
Relative density of liquid \( = 1.2 \)
We know
Relative density of liquid \( = \text{Loss of weight of solid in liquid / loss of weight of solid in water} \).
Loss of weight of solid in liquid \( = \text{Relative density} \times \text{loss of weight of solid in water} \).

\( \implies \text{Loss of weight of solid in liquid} = 1.2 \times 0.25 = 0.3 \text{ N} \).
Weight of solid in liquid \( = \text{weight of solid in air} - \text{loss of weight of solid in liquid} \).

\( \implies \text{Weight of solid in liquid} = 2.10 - 0.3 = 1.8 \text{ N} \).
In simple words: We used the "heaviness" of the solid to figure out how much weight it loses in different liquids. It loses \( 0.25 \text{ N} \) in water and \( 0.3 \text{ N} \) in the other liquid.

📝 Teacher's Note: This problem teaches students how to use the "Loss of Weight" method to find both object density and liquid density.

🎯 Exam Tip: Always calculate the loss of weight in water first, as it is the standard reference for finding the volume/relative density of the object.

Question 34.
Answer: Density of iron is \( 7800 \text{ Kgm}^{-3} \).
This means a cube of iron having side \( 1 \text{ m} \) would weigh \( 7800 \text{ Kg} \).
Density of water at \( 4^{\circ}\text{C} \) is \( 1000 \text{ Kgm}^{-3} \).
In simple words: Iron is quite dense. If you had a box of iron the size of a washing machine (\( 1 \text{ cubic meter} \)), it would weigh as much as \( 5 \) or \( 6 \) cars!

📝 Teacher's Note: Using a \( 1 \text{ m}^3 \) cube is the standard way to help students visualize what density numbers actually mean.

🎯 Exam Tip: When explaining density, always mention the mass and the specific volume it occupies.

Question 35.
Answer: Relative density of body \( = 0.52 \)
Density of water at \( 4^{\circ}\text{C} = 1000 \text{ Kgm}^{-3} \).
Density of body \( = 0.52 \times 1000 \text{ Kgm}^{-3} = 520 \text{ Kgm}^{-3} \)
We know density \( = \text{mass / volume} \).
Mass \( = \text{density} \times \text{volume} \)

\( \implies \text{Mass} = 520 \times 2 = 1040 \text{ Kg} \).
Mass of given body is \( 1040 \text{ Kg} \).
In simple words: Since the body's relative density is \( 0.52 \), it is about half as heavy as water. We found its total mass by multiplying its density by its size (\( 2 \text{ cubic meters} \)).

📝 Teacher's Note: Ensure students realize that since the RD is less than \( 1 \), this object will float in water.

🎯 Exam Tip: Always write the formula \( \text{Mass} = \text{Density} \times \text{Volume} \) before doing the calculation.

Question 36.
Answer: Piece of metal weighs in air \( = 44.5 \text{ f} \)
Piece of metal weighs in liquid \( = 39.5 \text{ f} \).
Loss of weight of metal in liquid \( = 44.5 - 39.5 = 5 \text{ f} \).
Relative density \( = \text{weight of solid in air / loss of weight of solid in water} \).
Relative density of liquid \( = 44.5 \text{ f} / 5 \text{ f} = 8.9 \)
Relative density of liquid \( = 8.9 \)
In simple words: By comparing how much weight the metal lost when dipped in the liquid, we calculated that the liquid is \( 8.9 \) times denser than the reference.

📝 Teacher's Note: The "f" in the source likely stands for gram-force (\( \text{gf} \)). Treat it as a unit of force/weight.

🎯 Exam Tip: When dividing two values with the same unit (like \( 44.5 \text{ f} / 5 \text{ f} \)), the units cancel out, leaving a pure number for Relative Density.

Question 37.
Answer: Volume of body \( = 100 \text{ cm}^3 \).
Weight of body \( = 1 \text{ kgf} = 1000 \text{ gf} \)
Mass of body \( = 1000 \text{ gm} \).
Density of liquid \( = 1000 \text{ gm} / 100 \text{ cm}^3 = 10 \text{ gcm}^{-3} \).
Density of water at \( 4^{\circ} = 1 \text{ gcm}^{-3} \).
Relative density \( = \text{density of substance / density of water at } 4^{\circ}\text{ C} \)

\( \implies \text{Relative density} = 10 \text{ gcm}^{-3} / 1 \text{ gcm}^{-3} = 10 \)
Mass of body \( = 1000 \text{ gm} \).
Density of water \( = 1 \text{ gcm}^{-3} \)
Acceleration due to gravity \( = 10 \text{ ms}^{-2} \).
Upthrust \( = V \times \rho \times g \).
Upthrust \( = 100 \times 1 \text{ xf} = 100 \text{ gf} \).
Resultant weight of the body \( = \text{weight} - \text{upthrust} = 1000 \text{ gf} - 100 \text{ gf} = 900 \text{ gf} \).
In simple words: This is a repeat calculation showing that a body in water feels an upward push (upthrust) equal to the weight of water it displaces, making it feel lighter.

📝 Teacher's Note: Remind students that in CGS units, the volume in \( \text{cm}^3 \) is numerically equal to the upthrust in \( \text{gf} \) for water.

🎯 Exam Tip: "Resultant weight" is the weight measured while the object is submerged.

Question 38.
Answer: Principle of floatation states that a floatating body displaces an amount of fluid equal to its own weight.
Ship is designed in such a manner that it encloses large quantity of air in air tight bags and in rooms and corridors which makes the average density of ship less than that of water.
In simple words: A ship stays afloat because it is hollow and full of air, making it "lighter" for its size than the water it displaces. It follows the rule that it must push aside its own weight in water to float.

📝 Teacher's Note: Explain that "average density" includes the heavy metal hull AND the light air inside the ship.

🎯 Exam Tip: The condition for floating is: Weight of body \( = \) Weight of liquid displaced by submerged part.

Question 40.
Answer: Acid battery hydrometer is used to check the concentration of sulphuric acid in an acid battery.
In simple words: This is a tool that floats in battery acid. If it floats high, the acid is strong; if it sinks low, the acid is weak and needs charging.

📝 Teacher's Note: Car mechanics use this to check if a car battery is still "good" or "dead."

🎯 Exam Tip: Match "Hydrometer" with "Density" or "Acid Concentration" in matching columns.

Question 41.
Answer: Iron nail has density less than that of mercury so it will float on the surface of mercury but in the case of water it will sink because the density of iron nail is more than that of water.
In simple words: Mercury is a super-heavy liquid metal. Because it is much denser than iron, the nail floats on it, even though that same nail sinks like a stone in water.

📝 Teacher's Note: This is a classic paradox used to test if students understand that "sinking" depends on the liquid, not just the object.

🎯 Exam Tip: Always mention the comparison of densities (\( \rho_{\text{nail}} \) vs \( \rho_{\text{mercury}} \)).

Question 43.
Answer: Principle of floatation states that a floatation body displaces an amount of fluid equal to its own weight.
Hydrometer is based on principle of floatation.
A hydrometer is a device used for measuring the relative density of a liquid directly.
It usually consists of a glass float with a long thin stem which is graduated. The glass float is a large hollow bulb which increases with buoyancy. The narrow stem increases the sensitivity of hydrometer. The bottom of hydrometer is made heavier by loading it with lead shots so that it floats vertically.
Hydrometer works on the principle of floatation. Consider a thin walled and flat bottomed test tube. Add some lead shots in the test tube and place it in jar containing water. Adjust the number of lead shots such that it floats vertically with some of its portion outside the surface of water in jar.
If \( l \) is the length of the test tube inside water, \( a \) is area of cross section of the test tube, \( d \) is density of water in which it floats, then the weight of water displaced \( = aldg \) and it is equal to the weight of loaded test tube.
Now allow the test tube to float in the other jar filled with a liquid of density \( d_1 \) and note the level \( l_1 \) at which it floats in that liquid.
Weight of the liquid displaced \( = al_1d_1g \)
Using law of floatation
\( aldg = al_1d_1g \)

\( \implies ld = l_1d_1 \)

\( \implies l / l_1 = d_1 / d \)
or \( l \) is inversely proportional to the density.
It sinks more in a lighter liquid so as to displace more volume of the lighter liquid whose weight is equal to the weight of Hydrometer. Hence, it will sink less in a denser liquid so that it has displace less volume of the denser liquid whose weight will be equal to the hydrometer. In this way this measures relative density of a liquid.
Barometer can be used as lactometer which can be used to check the purity of milk.
Barometer can be used as acid battery hydrometer which can be used to check the concentration of sulphuric acid in an acid battery.
In simple words: A hydrometer is like a floating ruler. It sinks deep into light liquids and floats high in heavy liquids. By checking where the liquid surface hits the ruler, we can tell the density.

📝 Teacher's Note: This is a long-answer question. Make sure students understand the inverse relationship: deeper immersion means lower density.

🎯 Exam Tip: Draw the diagram and mention that the stem is "graduated" (has marks) and the bottom is "weighted" with lead shots.

Question 44.
Answer:

• A balloon filled with hydrogen has low density than air so it rises over the air but as height increases density of air decreases and at a certain height the density of hydrogen in balloon and density of air become equal. And as there is no density difference there is no pressure difference also and hence balloon stops rising further.
• Density of egg is greater than fresh water so it sinks in fresh water but due to addition of salt density of water increases which makes the density of salt water greater than egg and hence floats in a strong solution of salt.
• The bottom of the hydrometer is made heavier by loading it with lead shots so that it floats vertically with some of its portion outside the surface of water in the jar.
• Relative density of Ice is \( = 0.9 \text{ cm}^{-3} \)
  Relative density of sea water is \( = 1 \text{ cm}^{-3} \)
  \( (\text{Density of ice / Density of sea water}) = \text{fraction submerged of iceberg} \)


• \( \implies 0.9/1 = \text{fraction of iceberg submerged} \)
  Fraction of iceberg submerged \( = 9/10 \).
  Thus in colder countries where there are icebergs in oceans, only about \( 1/10 \) is seen above water and the remaining water \( 9/10 \) remain submerged. Hence, there is danger of these icebergs to the ships sailing in these oceans.
  In simple words: Hydrogen balloons stop rising when they meet air that is just as light as they are. Salt water is heavier than an egg, so an egg floats in it. For icebergs, \( 90\% \) of the ice is actually underwater, which is why ships have to be very careful!

📝 Teacher's Note: The source text again adds "cm-3" to relative density incorrectly. Clarify that RD has no units.

🎯 Exam Tip: Explain the "floating egg" experiment using the concept of density change in the liquid.

Question 45.
Answer: We can find the relative density of a solid which floats on water by following experiment.
Choose a sinker and find its weight in water by suspending it in water.
Tie the solid to the string attached to the sinker and find its weight in air but sinker in water.
Remove the solid and tie it together with the sinker and suspend it in water and find the weight of the solid together with the sinker in water.
Record your observation as shown below:
Weight of sinker in water \( = x \text{ gf} \)
Weight of sinker in water + solid in air \( = y \text{ gf} \)
Weight of solid in air \( = (y - x) \text{ gf} \)
Weight of solid + sinker in water \( = z \text{ gf} \)
upthrust on solid in water \( = (y - z) \text{ gf} \).
The upthrust on in water also represents the weight of the water displaced by the solid.
Relative density of solid \( = (\text{weight of cork in air}) / (\text{weight of equal volume of water}) \)
Relative density \( = (y - x) / (y - z) \).
In simple words: Since cork won't sink on its own, we use a heavy "sinker" to pull it underwater. By weighing them in air and then underwater, we can calculate how much water the cork displaces and find its density.

📝 Teacher's Note: This is a sophisticated lab experiment. The "sinker" is needed because cork is less dense than water and cannot be submerged without help.

🎯 Exam Tip: Memorize the three-step weighing process: (a) Sinker in water, (b) Sinker in water + solid in air, (c) Both in water.

Question 46.
Answer: We can find the relative density of a solid denser than water by following experiment.
Find the weight (\( W_1 \text{ gf} \)) of a solid in air using a hydrostatic balance.
Tie the solid firmly with a thread and suspend it from the hook. Lower the solid in water and find its weight.
Record the result as shown below:
Weight of solid in air \( = W_1 \text{ gf} \).
Weight of solid in water \( = W_2 \text{ gf} \).
Apparent loss of weight of solid \( = (W_1 - W_2) \text{ gf} \).
Relative density \( = W_1 / (W_1 - W_2) \)
Relative density of the solid \( = (\text{weight of solid in air}) / (\text{Apparent loss of weight of solid in water}) \).
In simple words: This is the simplest way to find density. Weigh it in air, weigh it in water. The weight it "lost" is divided into its original air weight to get the answer.

📝 Teacher's Note: Hydrostatic balance is a special scale designed to allow a sample to be hung underneath it into a beaker of water.

🎯 Exam Tip: The denominator (\( W_1 - W_2 \)) is equal to the weight of the water displaced by the object.

Question 47.
Answer: Principle of Archimedes’ states that when a body is totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced.
We can verify archimedis’ principle experimentally by doing this experiment.
The stone weighed \( 0.67 \text{ N} \) in air and \( 0.40 \text{ N} \) when immersed in water. The displaced water weighed is \( 0.27 \text{ N} \) (\( = 0.67 - 0.40 \)).
Pour water into eureka can till the water starts overflowing through the spout. When the water stops dripping replace the beaker by another one of known weight. Suspend a stone with the help of a string from the hook of a spring balance and record the weight of the stone.
Now, gradually lower the body into eureka can containing water and record its new weight in water when it is fully submerged in water. When no more water drips from the spout, weigh the beaker containing water.
After observation, we can conclude that apparent loss of weight of stone (calculated by differences in weight measured by spring balance) \( = \) weight of water displaced (weight of water in beaker).
This proves the archimedis’ principle that when a body is totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced.
In simple words: This experiment is a classic. You show that the "weight loss" of an object in water is exactly the same as the weight of the water that spills into a side cup.

📝 Teacher's Note: This is a repeat of Solution 14 from earlier, which emphasizes that this is a core practical skill for the students.

🎯 Exam Tip: Be ready to draw and label the Eureka can and the overflow beaker.

Question 48.
Answer: Barometer is a device for measuring atmospheric pressure.
It consists of a long glass tube and of bore \( 1 \text{ cm}^2 \) having thick walls and closed at one end. The tube is made wide to minimize the depression of mercury in the tube due to surface tension. The tube contains pure and dry mercury and stands inverted with its open end Immersed deep in mercury contained in trough called reservoir. The lower end of the tube is drawn out to avoid oscillations of mercury in the tube while making adjustment.
The barometer tube is enclosed for most of its length in a tube of brass. Near the upper end of the brass tube, there are two vertical rectangular slits cut diametrically opposite to each other. They enable us see the upper level of mercury through them. Two scales are engraved on the brass tube along the edges of front slit. The scale along one edge is graduated in centimeters and along the other in inches. The scales have graduations from \( 68 \text{ cm} \) to \( 80 \text{ cm} \) and \( 27 \text{ inch} \) to \( 32 \text{ inch} \). It is so done as ordinarily the pressure varies only between these limits. The zero of both the scale is at the tip of an ivory pointer which projects downward from the ceiling of the reservoir. The pressure is read with a vernier scale.
In simple words: This is a high-tech version of a mercury tube. It has protective brass casing, windows to see the mercury, and two different rulers (inches and cm) to read the pressure very accurately.

📝 Teacher's Note: This describes a Fortin Barometer. Point out the "ivory pointer"—it's the most famous feature of this design used to calibrate the zero point.

🎯 Exam Tip: Describe the "brass tube" and "rectangular slits" as key parts for accuracy and protection.

Question 49.
Answer: Aneroid barometer is a portable type of barometer.
In this no liquid is used. It consists of a metal box corrugated to make it flexible. The lower side of the box is fixed to the base of the instrument and the upper side is supported by a stout spring S. The box is partially evacuated. So it tends to collapse under the pressure of air. The stout spring balances the thrust on the metal box due to normal air pressure and prevents the box from collapsing.
When atmospheric pressure changes, the surface supported by the spring S moves in if pressure increases or out if pressure decreases slightly. This small movement is increased considerably by a system of levers one end of which is connected to the spring S. The other end of the lever system is connected to a chain which is wrapped round a spindle carrying a pointer. The pointer moves over a scale which is graduated to read the pressure in centimeters or inches.
In simple words: This barometer uses a squishy metal box instead of liquid. When the air pushes on the box, a system of tiny gears and levers moves a needle on a dial to show the pressure.

📝 Teacher's Note: This is the type of barometer found in aircraft (altimeters) because it doesn't spill and is very small.

🎯 Exam Tip: "No liquid," "Evacuated metal box," and "System of levers" are the three points you must include in your description.

Question 50.
Answer: Fortin barometer is used to measure the atmospheric pressure.

Before reading the fortin barometer, the free surface of mercury in the reservoir is made to touch the tip of ivory pointer by raising or lowering the level with the help of the screw \( S_2 \). The vernier scale is then adjusted with the screw \( S_1 \) till its lower edge touches the upper meniscus of mercury in the tube. While making this adjustment the eye is kept in level with the mercury meniscus. For this adjustment, the vernier is moved till the wall of glass plate just ceases to be visible through the slits. The vernier can read to \( 1/20 \text{ mm} \) or \( 1/500 \text{ inch} \). The reading of the barometer then gives the true atmospheric pressure.
A thermometer is usually provided with the barometer which is also read and correction is made for the expansion of scale with rise in temperature and also for the change in density of mercury with temperature. The scale graduations are correct only for the temperature at which the graduations are made.
In simple words: To use this barometer, you first turn a screw until the mercury just barely touches a little ivory tip. Then you move a special sliding ruler (vernier) to see exactly where the top of the mercury is. It even has a thermometer because heat can make the mercury expand slightly and mess up the reading!

📝 Teacher's Note: The ivory pointer is essential to ensure the "zero" of the ruler is exactly at the surface of the mercury in the bowl.

🎯 Exam Tip: Explain the role of the "ivory pointer" and the "vernier scale" for high-precision measurements.

Question 51.
Answer: Air exerts pressure and this can be demonstrated by this experiment.
Take a glass tumbler. Fill it to the brim with water, with water overflowing the rim of the tumbler. Cover it with a piece of cardboard. Press the cardboard so that it touches the rim of the tumbler at all points. This will ensure that no air is left inside.
With the hand on the cardboard, invert the tumbler gently and remove the hand. Both cardboard and water shall not fall. This shows that air exerts pressure on the lower surface of the cardboard in the upward as shown in figure. The upward thrust acting on the cardboard due to air pressure supports the combined weight of cardboard and water in the tumbler.
So, this shows that air exert pressure.
In simple words: If you fill a cup with water and cover it with cardboard, then flip it upside down, the water stays inside! This is because the air outside is pushing up against the cardboard harder than the water is pushing down.

📝 Teacher's Note: This is a "magic trick" experiment that always engages students. Remind them that the cup *must* be full so there is no air inside to push down.

🎯 Exam Tip: State that the "upward thrust of air" is greater than the "weight of water + cardboard."