Selina Concise Solutions for ICSE Class 9 Physics Chapter 4 Pressure In Fluids And Atmospheric Pressure

Exercise 4(A)

Question 1S. What is thrust? State its S.I. unit.
Answer: Thrust is the force acting normally on a surface. Its S.I. unit is ‘newton’.
In simple words: Imagine pushing a pin straight into a board; that straight-down push is thrust. Just like normal weight or force, we measure it in Newtons.

📝 Teacher's Note: Emphasize that "normal" in physics means "perpendicular" or at a 90-degree angle to the surface.

🎯 Exam Tip: Always specify the direction ("normally" or "perpendicularly") when defining thrust to get full marks.

Question 2S. Define pressure and state its S.I. unit.
Answer: Pressure is the thrust per unit area of the surface. Its S.I. unit is ‘newton per metre\( ^2 \)’ or ‘pascal’.
In simple words: Pressure tells us how "spread out" a force is. If you push a pin with the same force as your thumb, the pin has more pressure because its tip is tiny.

📝 Teacher's Note: Use the formula \( P = \frac{F}{A} \) to show students that for the same force, a smaller area creates a much higher pressure.

🎯 Exam Tip: Note that \( 1 \text{ Pa} = 1 \text{ N/m}^2 \). Both units are acceptable, but 'Pascal' is the specific S.I. name.

Question 3S. (a) Name another unit of pressure. (b) Relate it to pascal.
Answer:
(a) Pressure is measured in ‘bar’.
(b) \( 1 \text{ bar} = 10^5 \text{ pascal} \).
In simple words: A 'bar' is a large unit used by meteorologists to measure air pressure, kind of like using kilometers instead of meters for long distances.

📝 Teacher's Note: Mention that atmospheric pressure at sea level is approximately 1 bar.

🎯 Exam Tip: Memorize the conversion \( 10^5 \); it is a common 1-mark objective question.

Question 4S. Define one pascal.
Answer: One pascal is the pressure exerted on a surface of area \( 1 \text{ m}^2 \) by a force of \( 1 \text{ N} \) acting normally on it.
In simple words: If you take a square table that is 1 meter long on each side and push straight down on it with exactly 1 Newton of force, you are applying 1 Pascal of pressure.

📝 Teacher's Note: Defining a unit is a standard question. Students should always use the "unit value" (1 Newton, 1 Square Meter) in their definition.

🎯 Exam Tip: Include the word "normally" or "perpendicularly" to indicate the direction of the force.

Question 5S. Is thrust a scalar or a vector quantity?
Answer: Thrust is a vector quantity.
In simple words: Since thrust is a type of force and force always has a direction (like pushing down or pulling up), it is a vector.

📝 Teacher's Note: Remind students that vectors have both magnitude and direction.

🎯 Exam Tip: If asked why, state that it is a force acting in a specific direction (perpendicular to the surface).

Question 6S. Is pressure a scalar or a vector quantity?
Answer: Pressure is a scalar quantity.
In simple words: Pressure doesn't point in one specific direction; in a fluid, it pushes equally in every direction at once, so we treat it as a scalar.

📝 Teacher's Note: This is a common trap. While force (thrust) is a vector, pressure (the ratio of force to area) is a scalar.

🎯 Exam Tip: Always label pressure as scalar in multiple-choice questions.

Question 8S. How does pressure depend on the area of the surface? Give an example.
Answer: Pressure exerted by thrust is inversely proportional to area of surface on which it acts. Thus, larger the area on which the thrust acts, lesser is the pressure exerted by it.
Example: If we stand on loose sand, our feet sink into the sand, but if we lie on that sand, our body does not sink into the sand. In both the cases, the thrust exerted on the sand is equal (equal to the weight of the body). However, when we lie on sand, the thrust acts on a large area and when we stand, the same thrust acts on a small area.
In simple words: Heavy trucks have many wheels so their weight is spread out. If the weight is spread over more ground (large area), the pressure on the road is low and they don't sink.

📝 Teacher's Note: Use the "High Heels vs. Flat Shoes" analogy to explain why heels sink into mud while flats don't.

🎯 Exam Tip: Use the keyword "inversely proportional" to describe the relationship between pressure and area.

Question 9S. Why is the tip of an allpin made sharp?
Answer: The tip of an allpin is made sharp so that large pressure is exerted at the sharp end and it can be driven into with less effort.
In simple words: A sharp point has a tiny area. Even a small push becomes a huge pressure on that tiny spot, making it easy to poke through paper.

📝 Teacher's Note: This is a practical application of increasing pressure by decreasing area.

🎯 Exam Tip: Relate the "sharpness" specifically to "small area" and "high pressure".

Question 10S. Explain why: (a) It is easier to cut with a sharp knife than a blunt one. (b) Wide wooden sleepers are placed below railway tracks.
Answer:
(a) It is easier to cut with a sharp knife because even a small thrust causes great pressure at the edges and cutting can be done with less effort.
(b) Wide wooden sleepers are placed below the railway tracks so that the pressure exerted by the rails on the ground becomes less.
In simple words: (a) A thin blade concentrates your strength. (b) The wide wood planks spread out the heavy train's weight so the tracks don't sink into the dirt.

📝 Teacher's Note: These are classic "Conceptual Reasoning" questions. Always focus on the Area-Pressure relationship.

🎯 Exam Tip: For part (b), mention that spreading the force over a "larger area" prevents the "ground from giving way."

Question 11S. What is a fluid?
Answer: A substance which can flow is called a fluid.
In simple words: Fluids are not just liquids like water; gases like air are also fluids because they can both move and flow from one place to another.

📝 Teacher's Note: Clarify that "Fluid" is an umbrella term covering both Liquids and Gases.

🎯 Exam Tip: Remember: Solids are NOT fluids because their particles are fixed in position.

Question 12S. What is fluid pressure?
Answer: Due to its weight, a fluid exerts pressure in all directions; the pressure exerted by the fluid is called fluid pressure.
In simple words: Just like you have weight, water and air have weight too. This weight presses against everything they touch.

📝 Teacher's Note: Emphasize that fluids push against the walls of their container, not just the bottom.

🎯 Exam Tip: Mention "in all directions" to distinguish fluid pressure from the pressure exerted by solids.

Question 13S. How does the pressure exerted by a solid differ from a fluid?
Answer: A solid exerts pressure only on the surface on which it is placed, i.e. at its bottom, but a fluid exerts pressure at all points in all directions.
In simple words: A brick only pushes down on the table. But water in a cup pushes down on the bottom AND sideways against the walls of the cup.

📝 Teacher's Note: Explain that fluids have no fixed shape, which allows them to transmit pressure to every part of their container.

🎯 Exam Tip: This is a frequent comparison question. Use the phrase "in all directions" for fluids.

Question 14S. Describe an experiment to show that a liquid exerts pressure at all points on the wall of its container.
Answer: Take a can or large plastic bottle filled with water. Place it on a horizontal surface. Make a series of holes in the wall of the vessel anywhere below the free surface of the liquid. The water spurts out through each hole. This shows that the liquid exerts pressure at each point on the wall of the bottle.
In simple words: If you poke holes in a bottle, water shoots out sideways. This proves the water is pushing against the walls of the bottle.

📝 Teacher's Note: Point out that the water from the lowest hole shoots out the furthest because pressure increases with depth.

🎯 Exam Tip: Drawing the streams of water correctly (the bottom one traveling the furthest) is important for diagram marks.

Question 15S. On what factors does the pressure at a point in a liquid depend?
Answer: Pressure at a point in a liquid depends upon the following three factors:
(1) Depth of the point below the free surface.
(2) Density of liquid.
(3) Acceleration due to gravity.
In simple words: Liquid pressure depends on how deep you are, how heavy the liquid is (like water vs. syrup), and the pull of gravity.

📝 Teacher's Note: Relate this to the formula \( P = h\rho g \). Students should see that the math reflects these physical factors.

🎯 Exam Tip: Listing all three factors is necessary for a full-mark descriptive answer.

Question 16S. Write the formula for the total pressure at a depth inside a liquid.
Answer: \( P = P_0 + h\rho g \)
Here, \( P = \text{Pressure exerted at a point in the liquid} \)
\( P_0 = \text{Atmospheric pressure} \)
\( h = \text{Depth of the point below the free surface} \)
\( \rho = \text{Density of the liquid} \)
\( g = \text{Acceleration due to gravity} \)
In simple words: The total pressure underwater is the weight of the water above you plus the weight of the air above the water.

📝 Teacher's Note: Clarify the difference between "gauge pressure" (\( h\rho g \)) and "total pressure" which includes the atmosphere.

🎯 Exam Tip: In numericals, read carefully to see if the question asks for "total pressure" or just "pressure due to liquid column."

Question 17S. Derive the expression for pressure at a depth h in a liquid of density \( \rho \).
Answer: Consider a vessel containing a liquid of density \( \rho \). Let the liquid be stationary. In order to calculate pressure at a depth, consider a horizontal circular surface PQ of area A at a depth h below the free surface XY of the liquid. The pressure on the surface PQ will be due to the thrust of the liquid contained in cylinder PQRS of height h with PQ as its base and top face RS lying on the frees surface XY of the liquid.
Total thrust exerted on the surface PQ
= Weight of the liquid column PQRS
= Volume of liquid column PQRS \( \times \) density \( \times \) g
= (Area of base PQ \( \times \) height) \( \times \) density \( \times \) g
= \( (A \times h) \times \rho \times g \)
This thrust is exerted on the surface PQ of area A. Therefore, pressure is given as shown below.
\( P = \frac{\text{Thrust on surface}}{\text{Area of surface}} \)

\( \implies P = \frac{Ah\rho g}{A} = h\rho g \)
Thus, Pressure = depth \( \times \) density of liquid \( \times \) acceleration due to gravity
In simple words: We calculate the weight of an imaginary column of water sitting on a small area. When we divide that weight by the area, the area cancels out, leaving us with depth times density times gravity.

📝 Teacher's Note: This is a core derivation. Walk through each step: Volume \( \rightarrow \) Mass \( \rightarrow \) Weight \( \rightarrow \) Pressure.

🎯 Exam Tip: Clearly show how the area 'A' cancels out in the final step to prove that pressure at a point doesn't depend on the size of the container.

Question 18S. Why is the pressure at the same depth in sea water more than in river water?
Answer: Due to dissolved salts, density of sea water is more than the density of river water, so pressure at a certain depth in sea water is more than that at the same depth in river water.
In simple words: Sea water is "heavier" (denser) because of the salt. Heavier liquid pushes down harder, creating more pressure.

📝 Teacher's Note: Remind students of the formula \( P = h\rho g \). If \( h \) and \( g \) are constant, pressure is directly proportional to density \( \rho \).

🎯 Exam Tip: Explicitly mention that "density of sea water is greater" to justify the difference.

Question 19S. Consider two points at depths \( h_1 \) and \( h_2 \) (\( h_2 > h_1 \)). (a) Write the relation between pressures \( P_1 \) and \( P_2 \) at these points. (b) Which pressure is greater?
Answer:
(a) \( P_2 = P_1 + h \rho g \),
(b) \( P_2 > P_1 \)
In simple words: The deeper you go, the higher the pressure gets.

📝 Teacher's Note: Here \( h \) represents the vertical distance between the two points.

🎯 Exam Tip: Deepest point always has highest pressure in a single fluid.

Question 20S. Explain why a gas bubble released at the bottom of a lake grows in size as it rises for the surface of lake.
Answer: The reason is that when the bubble is at the bottom of the lake, total pressure exerted on it is the atmospheric pressure plus the pressure due to water column. As the gas bubble rises, due to decrease in depth the pressure due to water column decreases. By Boyle’s law, PV = constant, so the volume of bubble increases due to decrease in pressure, i.e., the bubble grows in size.
In simple words: Deep down, the water squeezes the bubble tight. As it moves up, the water push gets weaker, so the air inside the bubble can expand and make it bigger.

📝 Teacher's Note: Connect this to Boyle's Law: Pressure and Volume are inverse. Less pressure means more space for the gas.

🎯 Exam Tip: Mention both "decrease in depth" and "Boyle's law" (or pressure-volume relationship) for a full answer.

Question 21S. Why is the wall of a dam made thicker at the bottom?
Answer: The pressure exerted by a liquid increases with its depth. Thus as depth increases, more and more pressure is exerted by water on wall of the dam. A thicker wall is required to withstand greater pressure, therefore, the thickness of the wall of dam increases towards the bottom.
In simple words: The water at the bottom of the lake pushes much harder than the water at the top. To stop that strong push from breaking the dam, the bottom has to be super thick and strong.

📝 Teacher's Note: This is a real-world engineering application of the principle \( P \propto h \).

🎯 Exam Tip: Keywords are "pressure increases with depth" and "withstand greater pressure."

Question 22S. Why do sea divers need a special protective suit?
Answer: The sea divers need special protective suit to wear because in deep sea, the total pressure exerted on the diver’s body is much more than his blood pressure. To withstand it, he needs to wear a special protective suit.
In simple words: Deep in the ocean, the water would try to squash a person flat because the pressure is so high. The suit protects them from being crushed.

📝 Teacher's Note: Explain that our bodies are used to 1 atmosphere of pressure. Divers go to places where it is dozens of times higher.

🎯 Exam Tip: Compare the "external water pressure" to the "internal blood pressure."

Question 23S. State the laws of liquid pressure.
Answer: Laws of liquid pressure:
(1) Pressure at a point inside the liquid increases with the depth from its free surface.
(2) In a stationary liquid, pressure is same at all points on a horizontal plane.
(3) Pressure is same in all directions about a point in the liquid.
(4) Pressure at same depth is different in different liquids. It increases with the increase in the density of liquid.
(5) A liquid seeks its own level.
In simple words: These are the rules water follows: it pushes harder deeper down, it pushes equally in all directions, and it always levels itself out.

📝 Teacher's Note: Use "liquid seeks its own level" to explain how water towers or teapots work.

🎯 Exam Tip: Memorize at least 3 to 4 points for a standard long-answer question.

Question 24S. A container has two holes A and B at different heights. Why does liquid from hole B reach further than hole A?
Answer: The liquid from hole B reaches a greater distance on the horizontal surface than that from hole A. This explains that liquid pressure at a point increases with the depth of point from the free surface.
In simple words: Hole B is lower down, so the water there is being pushed out much harder by the weight above it, making it spray further.

📝 Teacher's Note: This is the observational proof of \( P \propto h \). More pressure equals more velocity of efflux.

🎯 Exam Tip: Relate the horizontal distance directly to the "increase in pressure with depth."

Question 25S. What happens to pressure on a diver when he moves (i) deeper, (ii) horizontally?
Answer:
(i) As the diver moves to a greater depth, pressure exerted by sea water on him also increases.
(ii) When the diver moves horizontally, his depth from the free surface remains constant and hence the pressure on him remains unchanged.
In simple words: Diving down makes the squeeze stronger. Swimming left or right at the same level keeps the squeeze exactly the same.

📝 Teacher's Note: Use this to reinforce that pressure depends on vertical depth, not horizontal position.

🎯 Exam Tip: "Unchanged" is the keyword for horizontal movement in a stationary liquid.

Question 26S. State Pascal’s law.
Answer: Pascal’s law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.
In simple words: If you squeeze one end of a water balloon, the pressure travels through the water and pushes out on the other side instantly.

📝 Teacher's Note: Emphasize the word "confined"—the law only applies if the liquid is in a closed container.

🎯 Exam Tip: Use the words "equally" and "undiminished" to get full marks for this law's statement.

Question 27S. Give two applications of Pascal’s law.
Answer: Two applications of Pascal’s law:
(1) Hydraulic press
(2) Hydraulic jack
In simple words: We use this law to make machines that can lift heavy cars with just a small push.

📝 Teacher's Note: Mention that hydraulic brakes in cars also work on this same principle.

🎯 Exam Tip: Just naming the devices is usually enough for a 1 or 2 mark question.

Question 28S. State the principle of a hydraulic machine.
Answer: The principle of a hydraulic machine is that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston. Hydraulic press and hydraulic brakes work on this principle.
In simple words: It's like a force-multiplier. A small push on a tiny pipe becomes a giant push on a large pipe.

📝 Teacher's Note: Show the math: Since \( P_1 = P_2 \), then \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \). If \( A_2 \) is 100 times bigger than \( A_1 \), the force \( F_2 \) is 100 times stronger.

🎯 Exam Tip: Always mention the "small force on small piston" vs "large force on large piston" relationship.

Question 29S. Describe a hydraulic press and give one use.
Answer: Hydraulic press works on principle of hydraulic machine. It states that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.
Use: It is used for squeezing oil out of linseed and cotton seeds.
In simple words: It's a giant machine that uses water power to crush things or squeeze seeds to get oil out.

📝 Teacher's Note: Another common use is for compressing bales of cotton or paper for transport.

🎯 Exam Tip: Be ready to name at least one industrial use for the hydraulic press.

Question 30S. Name parts and describe the action of a hydraulic machine.
Answer:
(i) X : Press Plunger; Y: Pump Plunger
(ii) When the lever is moved down, valve B closes and valve A opens, so the water from cylinder P is forced into the cylinder Q.
(iii) Valve B closes due to an increase in pressure in cylinder P. This pressure is transmitted to the connecting pipe and when the pressure in connecting pipe becomes greater than the pressure in the cylinder Q, valve A opens up.
(iv) When the release valve is opened, the ram (or press) plunger Q gets lowered and water of the cylinder Q runs out in the reservoir.
In simple words: By moving a lever up and down, you pump liquid from a small pipe to a big one. One-way valves make sure the liquid only goes toward the big side, lifting the heavy load.

📝 Teacher's Note: Explain that valves act like "checkpoints" to ensure the fluid moves in a loop or specific direction without flowing backward.

🎯 Exam Tip: Study the diagram carefully to identify which valve opens during the "downstroke" vs "upstroke" of the pump plunger.

Question 31S. Describe the working of a hydraulic lift.
Answer: Working: When handle H of the lever is pressed down by applying an effort, the valve V opens because of increase in pressure in cylinder P. The liquid runs out from the cylinder P to the cylinder Q. As a result, the piston B rises up and it raises the car placed on the platform. When the car reaches the desired height, the handle H of the lever is no longer pressed. The valve V gets closed (since the pressure on the either side of the valve becomes same) so that the liquid may not run back from the cylinder Q to cylinder P.
In simple words: You pump a handle to push oil under a platform. The oil pushes the platform up to lift a car. When you stop, a valve locks the oil in place so the car stays up.

📝 Teacher's Note: This is basically the same mechanism as the hydraulic press but with a platform for lifting instead of a surface for crushing.

🎯 Exam Tip: Use the term "effort" for the input force on the handle and "load" for the car being lifted.

Question 32S. Describe the working of hydraulic brakes in a car.
Answer: Working: To apply brakes, the foot pedal is pressed due to which pressure is exerted on the liquid in the master cylinder P, so liquid runs out from the master cylinder P to the wheel cylinder Q. As a result, the pressure is transmitted equally and undiminished through the liquid to the pistons \( B_1 \) and \( B_2 \) of the wheel cylinder. Therefore, the pistons \( B_1 \) and \( B_2 \) get pushed outwards and brake shoes get pressed against the rim of the wheel due to which the motion of the vehicle retards. Due to transmission of pressure through the liquid, equal pressure is exerted on all the wheels of the vehicle connected to the pipe line R.
On releasing the pressure on the pedal, the liquid runs back from the wheel cylinder Q to the master cylinder P and the spring pulls the break shoes to their original position and forces the pistons \( B_1 \) and \( B_2 \) to return back into the wheel cylinder Q. Thus, the brakes get released.
In simple words: When you step on the brake pedal, you push liquid through pipes to every wheel. The liquid pushes pads against the wheels to stop them. When you let go, springs pull the pads back so you can drive again.

📝 Teacher's Note: Highlight the safety benefit: because pressure is transmitted equally, all wheels brake with the same force, preventing the car from skidding sideways.

🎯 Exam Tip: Mention the "Master Cylinder" and "Wheel Cylinder" as the two main components of the system.

Question 33S. Fill in the blanks:
Answer:
(a) \( h \rho g \)
(b) same
(c) the same
(d) directly proportional
(e) directly proportional.
In simple words: These terms summarize how liquid pressure works—it's consistent and follows strict mathematical rules.

📝 Teacher's Note: These often cover the "Laws of Liquid Pressure" in a summarized format.

🎯 Exam Tip: Review the basic laws to ensure you can fill these in even if the wording changes slightly.

Question 1M. The S.I. unit of pressure is:
(a) N
(b) \( \text{N/m} \)
(c) \( \text{Pa} \)
(d) \( \text{kg} \)
Answer: (c) Pa
In simple words: Pascal (Pa) is the name given to the pressure of 1 Newton per square meter.

📝 Teacher's Note: "N" is for force, "kg" is for mass. Students must distinguish between units of force and units of pressure.

🎯 Exam Tip: If "Pa" isn't an option, look for \( \text{N/m}^2 \).

Question 2M. Pressure inside a liquid is given by the formula:
(a) \( h/ \rho g \)
(b) \( h \rho g \)
(c) \( h \rho / g \)
(d) \( \rho g / h \)
Answer: (b) h \(\rho\) g
In simple words: Just multiply depth by the liquid's density and gravity to find the pressure.

📝 Teacher's Note: This formula is derived from the weight of a column of liquid.

🎯 Exam Tip: Remember the order doesn't matter, but all three terms must be multiplied.

Question 3M. If two points in a liquid are at different depths \( h_1 \) and \( h_2 \) (\( h_1 < h_2 \)), then:
(a) \( P_1 < P_2 \)
(b) \( P_1 > P_2 \)
(c) \( P_1 = P_2 \)
(d) None of the options
Answer: (a) \( P_1 < P_2 \)
In simple words: The shallower point (h1) has less pressure than the deeper point (h2).

📝 Teacher's Note: This reinforces the concept that pressure increases linearly with depth.

🎯 Exam Tip: More depth always equals more pressure.

Question 4M. The difference in pressure between two points at a vertical distance h in a liquid is:
(a) \( h / \rho g \)
(b) \( \rho g \)
(c) \( h \rho g \)
(d) \( h \rho \)
Answer: (c) h \(\rho\) g
In simple words: The "extra" pressure you get by going down a distance \( h \) is just the weight of that extra slice of water.

📝 Teacher's Note: This is a common way to ask for the "relative" or "gauge" pressure between two points.

🎯 Exam Tip: This formula \( \Delta P = \Delta h \cdot \rho \cdot g \) is very useful for solving differential pressure problems.

Question 1N. Two nails A and B have cross-sectional areas of \( 2\text{mm}^2 \) and \( 6\text{mm}^2 \) respectively. If a force of 1.5 N is applied to each, calculate the pressure on each in pascal.
Answer:
1. Force exerted, \( F = 1.5 \text{ N} \)
Area of cross-section of tip of nail A, \( a_1 = 2\text{mm}^2 = 2 \times 10^{-6} \text{ m}^2 \)
Area of cross-section of tip of nail B, \( a_2 = 6\text{mm}^2 = 6 \times 10^{-6} \text{ m}^2 \)
Pressure on nail A = \( \frac{F}{a_1} = \frac{1.5}{2 \times 10^{-6}} = 7.5 \times 10^5 \text{ pascal} \)
Pressure on nail B = \( \frac{F}{a_2} = \frac{1.5}{6 \times 10^{-6}} = 2.5 \times 10^5 \text{ pascal} \)
In simple words: Even with the same push, the thinner nail creates three times more pressure because its tip is three times smaller.

📝 Teacher's Note: Note the unit conversion: \( 1\text{mm}^2 = 10^{-6} \text{m}^2 \). This is where most students make errors.

🎯 Exam Tip: Always convert area to square meters before calculating pressure in Pascals.

Question 2N. A block of mass 7.5 kg has base dimensions \( 12\text{cm} \times 8\text{cm} \). Calculate the thrust and pressure exerted on the table. (Take \( g = 10 \text{ m s}^{-2} \))
Answer:
\( 1 \text{ kgf} = 10 \text{ N} \)
\( \therefore g = 10 \text{ m s}^{-2} \)

a) Thrust is the weight of the block:
\( F = mg = 7.5 \text{ kg} \times 10 = 75 \text{ N} \)

b) Pressure is force per area:
\( \therefore P = \frac{F}{A_{\text{base}}} = \frac{75}{12 \times 10^{-2} \times 8 \times 10^{-2}} \)
\( \therefore P = 7812.5 \text{ Pa} \)
In simple words: The total downward push is 75 Newtons. Because that push is spread over a small rectangle on the table, it creates a pressure of about 7812 units.

📝 Teacher's Note: Convert cm to m before calculating the area. \( 12\text{cm} = 0.12\text{m} \) and \( 8\text{cm} = 0.08\text{m} \).

🎯 Exam Tip: "Thrust" is just the weight (\( mg \)) when an object is resting on a horizontal surface.

Question 3N. A vessel contains water up to a height of 1.5 m. Find (a) pressure and (b) thrust at the base. (Density = \( 10^3 \text{ kg m}^{-3} \), Area = \( 100\text{cm}^2 \), \( g = 9.8 \text{ m/s}^2 \))
Answer:
Given height, \( h = 1.5\text{m} \)
Density of water, \( \rho = 10^3 \text{ kg m}^{-3} \)
Acceleration due to gravity, \( g = 9.8 \text{ m/s}^2 \)
Area of base of the vessel, \( a = 100 \text{ cm}^2 = 100 \times 10^{-4} \text{ m}^2 \)

(a) Pressure, \( P = h\rho g \)
or, \( P = 1.5 \times 10^3 \times 9.8 \)
or, \( P = 1.47 \times 10^4 \text{ Nm}^{-2} \)

(b) Thrust = Pressure \( \times \) area
or, \( \text{thrust} = 1.47 \times 10^4 \times 100 \times 10^{-4} \text{ N} \)
or, \( \text{thrust} = 147 \text{ N} \)
In simple words: The water at the bottom of this deep tank is pressing down with a huge pressure. Because the tank bottom has a specific size, we can multiply that pressure to find the total weight pushing down.

📝 Teacher's Note: Emphasize that pressure depends only on height, but thrust depends on both pressure and base area.

🎯 Exam Tip: Watch your exponents! \( 10^4 \times 10^{-4} \) equals 1, which simplifies the final calculation.

Question 4N. A cylinder has base area \( 300\text{cm}^2 \) and contains water up to \( 6\text{cm} \). Find (a) pressure and (b) thrust. (Density = \( 1000 \text{ kg m}^{-3} \), \( g = 10 \text{ ms}^{-2} \))
Answer:
Given Area of base of vessel, \( a = 300 \text{ cm}^2 = 300 \times 10^{-4} \text{ m}^2 \)
Density of water, \( \rho = 1000 \text{ kg m}^{-3} \)
Depth, \( h = 6 \text{ cm} = 0.06 \text{ m} \)
Acceleration due to gravity, \( g = 10 \text{ ms}^{-2} \)

(a) Pressure \( = h\rho g = 0.06 \times 1000 \times 10 = 600 \text{ pascal} \)
(b) Thrust, \( T = \text{pressure} \times \text{area} = 600 \times 300 \times 10^{-4} = 18\text{N} \)
In simple words: This shallow container has a relatively low pressure. The total weight of the water inside is only 18 Newtons.

📝 Teacher's Note: Remind students to convert the height from cm to m (\( 6\text{cm} = 0.06\text{m} \)) to maintain SI unit consistency.

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

Question 5N. Density of mercury is \( 13.6 \text{ gcm}^{-3} \). If mercury height is 70 cm, what is the equivalent height of a water column?
Answer:
(a) Given, density of mercury \( \rho' = 13.6 \text{ gcm}^{-3} \)
Height of mercury column, \( h' = 70 \text{ cm} \)
Acceleration due to gravity, \( g = 9.8 \text{ ms}^{-2} \)

Let \( h \) be the height of the water column.
Density of water \( \rho = 1 \text{ gcm}^{-3} \)

Given, pressure exerted by mercury column = pressure exerted by water column
or, \( h' \rho' g = h \rho g \)

\( \implies h = \frac{h' \rho'}{\rho} = \frac{70 \times 13.6}{1} = 952 \text{ cm} \) or \( 9.52 \text{ m} \)

(b) No, the height of the water column shall not change.
In simple words: Mercury is very heavy. It takes a huge 9.5-meter tall column of water to push down as hard as just a 70-centimeter column of mercury.

📝 Teacher's Note: This principle is used to show why mercury is used in barometers instead of water—a water barometer would be too tall to fit in a house!

🎯 Exam Tip: Since 'g' is on both sides of the equation, you can cancel it out before starting the calculation.

Question 6N. Pressure of water on ground floor = 40,000 pascal. Pressure of water on first floor = 10,000 pascal. Density of water, \( \rho = 1000 \text{ kg m}^{-3} \). Acceleration due to gravity, \( g = 10 \text{ ms}^{-2} \). Let h be the height of the first floor. Find the height of the first floor.
Answer: 5. Pressure of water on ground floor = 40,000 pascal
Pressure of water on first floor = 10,000 pascal
Density of water, \( \rho = 1000 \text{ kg m}^{-3} \)
Acceleration due to gravity, \( g = 10 \text{ ms}^{-2} \)
Let h be the height of the first floor.
Difference in water pressure between ground and first floor = \( h \rho g \)
or, \( (40,000 - 10,000) = h (1000) (10) \)
or, \( 30,000 = 10,000h \)

\( \implies h = 3 \text{ m} \)
In simple words: The water pressure drops as you go higher up in a building. By measuring how much the pressure dropped between floors, we can calculate that the first floor is 3 meters above the ground.

📝 Teacher's Note: This problem illustrates the relationship \( P = h\rho g \). Remind students that the pressure at the bottom must be higher to "push" the water up against gravity.

🎯 Exam Tip: Always calculate the "Difference in Pressure" (\( \Delta P \)) first before substituting into the height formula.

Question 7N. Length of mercury columns in two arms is equal. Height to which water is poured in one arm, h = 13.6 cm. Let h' be the rise in the mercury level in the other arm. Given, density of mercury = \( 13.6 \times 10^{3} \text{ kg m}^{-3} \). Density of water = \( 10^{3} \text{ kg m}^{-3} \). Find h'.
Answer: 6. Length of mercury columns in two arms is equal. 1 cm
Height to which water is poured in one arm, \( h = 13.6 \text{ cm} \)
Let h' be the rise in the mercury level in the other arm.
Given, density of mercury = \( 13.6 \times 10^{3} \text{ kg m}^{-3} \)
Density of water = \( 10^{3} \text{ kg m}^{-3} \)
Pressure due to water on one arm = Pressure on mercury column in the other arm
or, \( 13.6 \times 10^{3} \times g = h' \times 13.6 \times 10^{3} \times g \)

\( \implies h' = 1 \text{ cm} \)
In simple words: Mercury is much heavier than water. To balance out a tall 13.6 cm column of water, the heavy mercury only needs to rise by 1 cm.

📝 Teacher's Note: This is a classic U-tube manometer problem. The principle is that at the same horizontal level in a continuous stationary fluid, the pressure is the same.

🎯 Exam Tip: Since \( g \) appears on both sides of the equation, you can cancel it out to simplify your calculations.

Question 8N. Force on narrow piston, \( F_{1} = 2 \text{N} \). Area of cross-section of narrow piston, \( A_{1} = 10 \text{ cm}^{2} \). Let Force on wider piston be \( F_{2} \). Area of cross-section of wider piston, \( A_{2} = 100 \text{ cm}^{2} \). By the principle of hydraulic machine, find \( F_{2} \).
Answer: 7. Force on narrow piston, \( F_{1} = 2 \text{N} \)
Area of cross-section of narrow piston, \( A_{1} = 10 \text{ cm}^{2} \)
Let Force on wider piston be \( F_{2} \)
Area of cross-section of wider piston, \( A_{2} = 100 \text{ cm}^{2} \)
By the principle of hydraulic machine,
Pressure on narrow piston = pressure on wider piston
or, \( \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \)
or, \( \frac{2}{10} = \frac{F_{2}}{100} \)

\( \implies F_{2} = 20 \text{N} \)
In simple words: A hydraulic machine acts like a strength multiplier. Because the second pipe is 10 times wider, the small 2N push you give on the small side turns into a 20N push on the big side.

📝 Teacher's Note: This demonstrates Pascal's Law. Ensure students understand that while force increases, the distance the larger piston moves will be less than the smaller one.

🎯 Exam Tip: Check that the units for area (\( \text{cm}^{2} \)) are the same on both sides; if they are, you don't need to convert them to \( \text{m}^{2} \).

Question 9N. Let the ratio of area of cross-section of the master cylinder and wheel cylinder be \( A_{1} : A_{2} \). Force on pedal, \( F_{1} = 0.5 \text{N} \). Force on break shoe, \( F_{2} = 15 \text{N} \). By the principle of hydraulic machine, find the ratio.
Answer: 8. Let the ratio of area of cross-section of the master cylinder and wheel cylinder be \( A_{1} : A_{2} \)
Force on pedal, \( F_{1} = 0.5 \text{N} \)
Force on break shoe, \( F_{2} = 15 \text{N} \)
By the principle of hydraulic machine,
Pressure on narrow piston = pressure on wider piston
or, \( \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \)
or, \( \frac{F_{1}}{F_{2}} = \frac{A_{1}}{A_{2}} \)
or, \( \frac{A_{1}}{A_{2}} = \frac{0.5}{15} \)

\( \implies \frac{A_{1}}{A_{2}} = \frac{1}{30} \)
Thus, the required ratio is 1:30
In simple words: Car brakes use this trick. To turn a tiny 0.5N tap of your foot into a strong 15N grip on the wheels, the wheel cylinder needs to be 30 times bigger than the one at your pedal.

📝 Teacher's Note: Explain that "break shoe" refers to the output of the hydraulic system in a car's braking mechanism.

🎯 Exam Tip: When asked for a ratio, always express the final answer in the \( x:y \) format.

Question 10N. Area of small piston \( A_{1} = 5 \text{ cm}^{2} \). Area of wider piston, \( A_{2} = 625 \text{ cm}^{2} \). Force on small piston be \( F_{1} \). Force on wider piston or load, \( F_{2} = 1250 \text{N} \). Find \( F_{1} \).
Answer: 9. Area of small piston \( A_{1} = 5 \text{ cm}^{2} \)
Area of wider piston, \( A_{2} = 625 \text{ cm}^{2} \)
Force on small piston be \( F_{1} \)
Force on wider piston or load, \( F_{2} = 1250 \text{N} \)
By the principle of hydraulic machine,
Pressure on narrow piston = pressure on wider piston
or, \( \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \)
or, \( \frac{F_{1}}{5} = \frac{1250}{625} \)
or, \( F_{1} = \frac{1250}{625} \times 5 \)

\( \implies F_{1} = 10 \text{ N} \)
Assumption: No friction or leakage of liquid happens.
In simple words: To lift a heavy 1250N load on a big platform, you only need to push down with 10N of force on the small side. It's like lifting a car with one finger!

📝 Teacher's Note: Emphasize the "Assumption" part. In reality, some force is lost to friction, so the actual force needed would be slightly higher.

🎯 Exam Tip: The phrase "Force on wider piston or load" tells you that \( F_{2} \) is the weight being lifted.

Question 11N. (i) Diameter of the neck of the bottle, \( d_{1} = 2 \text{ cm} \). Diameter of the bottom of the bottle, \( d_{2} = 10 \text{ cm} \). Force on the cork in the neck, \( F_{1} = 1.2 \text{ kgf} \). Force on the bottom be \( F_{2} \). Find \( F_{2} \). (ii) Which law has been used?
Answer: 10. (i) Diameter of the neck of the bottle, \( d_{1} = 2 \text{ cm} \)
Diameter of the bottom of the bottle, \( d_{2} = 10 \text{ cm} \)
Force on the cork in the neck, \( F_{1} = 1.2 \text{ kgf} \)
Force on the bottom be \( F_{2} \)
By the principle of hydraulic machine,
Pressure on neck = pressure on bottom
or, \( \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \)
or, \( \frac{1.2}{\pi(d_{1}/2)^{2}} = \frac{F_{2}}{\pi(d_{2}/2)^{2}} \)
or, \( F_{2} = \frac{1.2}{(2)^{2}} \times (10)^{2} = 30 \text{ kgf} \)

(ii) Pascal's law has been used to find the force.
In simple words: If you push a cork into a bottle, that pressure spreads to the whole bottom. Since the bottom is much wider than the neck, the total force pushing on the bottom is much stronger—30 kgf instead of just 1.2 kgf.

📝 Teacher's Note: Remind students that Area is proportional to the square of the diameter (\( A \propto d^{2} \)). This is why \( \pi \) and the \( /2 \) cancel out in the ratio.

🎯 Exam Tip: Be careful with units like "kgf" (kilogram-force). You don't always need to convert to Newtons if the question doesn't ask for it.

Question 12N. Ratio of diameter of smaller piston to bigger piston = 5 : 25. Force applied on smaller piston, \( F_{1} = 50 \text{ kgf} \). Let \( F_{2} \) be the force on the bigger piston. Find \( F_{2} \).
Answer: 11. Ratio of diameter of smaller piston to bigger piston = 5 : 25
\( \therefore \) Ratio of area of smaller piston to bigger piston = \( 25 : 625 \)
Force applied on smaller piston, \( F_{1} = 50 \text{ kgf} \)
Let \( F_{2} \) be the force on the bigger piston.
By the principle of hydraulic machine,
Pressure on narrow piston = pressure on wider piston
or, \( \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \)
or, \( \frac{F_{1}}{F_{2}} = \frac{A_{1}}{A_{2}} \)
or, \( \frac{50}{F_{2}} = \frac{25}{625} \)
or, \( F_{2} = 50 \times \frac{625}{25} \)

\( \implies F_{2} = 1250 \text{ kgf} \)
In simple words: When the diameters are in a 5:25 ratio (which is 1:5), the areas are in a 25:625 ratio (which is 1:25). This means the force is multiplied by 25 times!

📝 Teacher's Note: This is a common mental math shortcut: if diameter increases by \( k \), area (and force) increases by \( k^{2} \).

🎯 Exam Tip: When diameters are given as a ratio, square them immediately to get the ratio of areas.

Question 13N. 
Answer: Data is incomplete
In simple words: Sometimes a problem doesn't give enough clues (like missing numbers) to find a solution.

📝 Teacher's Note: In such cases, identify what is missing (e.g., area or force) to show you understand the required variables.

🎯 Exam Tip: In an actual exam, if you suspect data is missing, clearly state what information is required to solve the problem.

Question 14N. In a hydraulic machine, pressure on narrow piston = Pressure on wider piston. \( F_{1} = ? \), \( A_{1} = 2 \times 10^{-4} \), \( F_{2} = 150 \), \( A_{2} = 12 \times 10^{-4} \). Find \( F_{1} \).
Answer: In a hydraulic machine
Pressure on narrow piston = Pressure on wider piston
\( \therefore P_{1} = P_{2} \)
\( \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \)
\( \frac{F_{1}}{2 \times 10^{-4}} = \frac{150}{12 \times 10^{-4}} \)
\( F_{1} = \frac{150}{12 \times 10^{-4}} \times 2 \times 10^{-4} \)
\( F_{1} = \frac{150}{12} \times 2 = 25 \text{ N} \)

\( \implies F_{1} = 25 \text{ N} \)
In simple words: To balance a 150N weight on a large pipe, you only need to push with 25N on the smaller pipe.

📝 Teacher's Note: Help students see that \( 10^{-4} \) cancels out on both sides, making the calculation much simpler than it looks.

🎯 Exam Tip: When dealing with scientific notation (\( 10^{-4} \)), look for opportunities to cancel them out across the equals sign.

Exercise 4(B)

Question 1S. Define atmospheric pressure.
Answer: The thrust exerted per unit area of the earth surface due to column of air, is called the atmospheric pressure on the earth surface.
In simple words: We live at the bottom of a giant ocean of air. The weight of all that air pushing down on us is called atmospheric pressure.

📝 Teacher's Note: Explain that even though we don't feel it, air has mass and weight just like everything else.

🎯 Exam Tip: Mention that atmospheric pressure acts in all directions, not just downward.

Question 2S. State the value of standard atmospheric pressure in pascal.
Answer: \( 1.013 \times 10^5 \text{ pascal} \)
In simple words: This is a very large number—about 100,000 Newtons of force on every square meter of your body!

📝 Teacher's Note: Compare this to the weight of an elephant spread over a table to help students visualize the magnitude.

🎯 Exam Tip: Memorize this numerical value as it is standard for all physics problems involving STP.

Question 3S. What is torr? How is it related to mm of Hg?
Answer: Atmospheric pressure is measured in ‘torr’.
\( 1 \text{ torr} = 1 \text{ mm of Hg} \).
In simple words: Torr is just a fancy name for measuring pressure by looking at how high mercury rises in a tube. 1 torr is exactly 1 millimeter of mercury.

📝 Teacher's Note: The unit is named after Evangelista Torricelli, the inventor of the barometer.

🎯 Exam Tip: 760 torr is equal to 1 standard atmosphere.

Question 4S. Relate atmosphere to cm of mercury and pascal.
Answer: At normal temperature and pressure, the barometric height is 0.76 m of Hg at sea level which is taken as one atmosphere.
1 atmosphere = 0.76 m of Hg = \( 1.013 \times 10^5 \text{ pascal} \)
In simple words: 1 "Atmosphere" is the normal air pressure at the beach. It's strong enough to hold up a 76-centimeter-tall tube of mercury.

📝 Teacher's Note: Show the math: \( P = h\rho g = 0.76 \times 13600 \times 9.8 \approx 10^5 \text{ Pa} \).

🎯 Exam Tip: Use 76 cm or 0.76 m depending on the units required in your numerical problem.

Question 5S. Why do we not feel uneasy under enormous atmospheric pressure?
Answer: We do not feel uneasy under enormous pressure of the atmosphere above as well as around us because of the pressure of our blood, known as blood pressure, is slightly more than the atmospheric pressure. Thus, our blood pressure balances the atmospheric pressure.
In simple words: Your body is pushing out just as hard as the air is pushing in. It's like a balanced tug-of-war, so you don't feel the squeeze.

📝 Teacher's Note: This is a great way to explain why astronauts need pressurized suits—there is no air in space to push back against their blood pressure.

🎯 Exam Tip: Use the word "balances" to explain the equilibrium between internal and external pressure.

Question 6S. Describe an experiment to demonstrate that air exerts pressure.
Answer: Take a thin can fitted with an airtight stopper. The stopper is removed and a small quantity of water is boiled in the can. Gradually the steam occupies the entire space of can by expelling the air from it [Fig (a)]. Then stopper is then tightly replaced and simultaneously the flame beneath the can is removed. Cold water is then poured over the can. It is observed that the can collapses inwards as shown in fig (b).
The reason is that the pressure due to steam inside the can is same as the air pressure outside the can [Fig (a)]. However, on pouring cold water over the can, fitted with a stopper [fig (b)], the steam inside the can condenses producing water and water vapour at very low pressure. Thus, the air pressure outside the can becomes more than the vapour pressure inside the closed can. Consequently, the excess atmospheric pressure outside the can causes it to collapse inwards.
In simple words: When you cool the steam inside a sealed can, the steam turns back to water and leaves an empty vacuum. The outside air is now much stronger than the inside and crushes the can like a soda can in your hand.

📝 Teacher's Note: This "crushing can" experiment is very dramatic and proves that air pressure is a real, powerful force.

🎯 Exam Tip: Mention "condensation of steam" and "creation of vacuum/low pressure" as the key scientific steps.

Question 7S. Explain the following: (i) Balloon collapses when air is removed. (ii) Liquid stays in a dropper. (iii) Why two holes are better than one to pour oil from a can?
Answer:
(i) When air is removed from the balloon, the pressure inside the balloon (which was due to air in it) is much less than the atmospheric pressure outside and hence the balloon collapses.
(ii) Water is held inside the dropper against the atmospheric pressure because the pressure due to height column of liquid inside the dropper is less than the atmospheric pressure. By pressing the dropper we increase the pressure inside the dropper and when it becomes greater than the atmospheric pressure the liquid comes out of the dropper.
(iii) There is no air inside a completely filled and sealed can. When a single hole is made to drain out the oil from the can, some of the oil will come out and due to that the volume of air above the oil will increase and hence the pressure of air will decrease. But if two holes are made on the top cover of the can, air outside the can will enter it through one hole and exert atmospheric pressure on the oil from inside along with the pressure due to oil column, and it will come out of the can from the other hole.
In simple words: (i) Outside air wins the push-war. (ii) Outside air is strong enough to hold the liquid up in the tube. (iii) One hole creates a vacuum that pulls the oil back; a second hole lets air in to push the oil out.

📝 Teacher's Note: These are "Everyday Physics" examples. Use a simple drinking straw experiment to demonstrate part (ii).

🎯 Exam Tip: For the oil can, focus on how the second hole "equalizes the internal and external pressure."

Question 8S. How does a syringe work?
Answer: When syringe is kept with its opening just inside a liquid and its plunger is pulled up in the barrel, the pressure of air inside the barrel below the plunger becomes much less than the atmospheric pressure acting on the liquid. As a result, the atmospheric pressure forces the liquid to rise up in the syringe.
In simple words: By pulling the plunger, you make an empty space (low pressure). The air outside pushes the liquid into that empty space.

📝 Teacher's Note: Clarify that the syringe doesn't "suck" the liquid; the atmospheric pressure "pushes" it in.

🎯 Exam Tip: The key phrase is "atmospheric pressure forces the liquid to rise."

Question 9S. How is water pushed up in a water pump?
Answer: In a water pump, on pulling the piston up, the pressure of air inside the siphon decreases and the atmospheric pressure on the water outside increases. As a result, the atmospheric pressure pushes the water up in pump.

In simple words: Pulling the pump handle creates a low-pressure zone inside the pipe. The regular air pressure outside then pushes the water up through the pipe and out of the spout.

📝 Teacher's Note: Explain that pumps work on the same principle of creating a partial vacuum as the syringe or a drinking straw.

🎯 Exam Tip: Use keywords like "decreased pressure" and "atmospheric pressure pushes" to explain the movement of water accurately.

Question 10S. What happens to the pressure inside a bell jar and a balloon during an experiment?
Answer: (a) Pressure increases inside the bell jar.
(b) Pressure decreases inside the balloon.

In simple words: If you pump air into a jar, the pressure on the outside of the balloon goes up. If the balloon expands, the air pressure inside it gets stretched out and decreases.

📝 Teacher's Note: This demonstrates the inverse relationship between volume and pressure (Boyle's Law); as volume increases, pressure decreases.

🎯 Exam Tip: Remember that if the air is being removed from a jar, the balloon inside will expand because the external pressure is dropping.

Question 11S. Name the instrument used to measure atmospheric pressure.
Answer: A barometer is used to measure atmospheric pressure.

In simple words: A barometer is like a scale, but instead of weighing your body, it weighs the air in the sky.

📝 Teacher's Note: Distinguish between a manometer, which measures gas pressure in a container, and a barometer, which specifically measures the atmosphere.

🎯 Exam Tip: This is a common one-word answer question; ensure the spelling is correct as it is a technical term.

Question 12S. Describe the construction of a simple barometer.
Answer: A barometer is an instrument which is used to measure the atmospheric pressure.
Construction of a simple barometer:
A simple mercury barometer can be made with a clear, dry, thick-walled glass tube about 1 metre ling. The glass tube is sealed at one end and is filled with mercury completely. While filling the tube with mercury care has to be taken so that there are no air bubbles present in the mercury column. Close the open end with thumb and turn the tube upside down carefully over a trough containing mercury. Dip the open end under the mercury level in the trough and remove the thumb. The mercury level in the tube falls until it is about 76 cm (\( h = 760 \text{ mm} \)) vertically above the mercury level. It is the atmospheric pressure acting on the surface of the mercury in the trough that supports the vertical mercury column. The empty space above the mercury column is called the ‘Torricellian vacuum’.
In simple words: You fill a long glass tube with mercury and flip it into a bowl of mercury. The air pushes down on the bowl, which holds up a column of mercury about 76 centimeters high.

📝 Teacher's Note: Point out that mercury is used because its high density allows for a reasonably sized instrument. A water barometer would need to be over 10 meters tall!

🎯 Exam Tip: Mention the specific length of the tube (1 metre) and the fact that it must be free of air bubbles to ensure full marks for the construction description.

Question 13S. Explain the theory of how a simple barometer works and define barometric height.
Answer: In given figure, at all points such as C on the surface of mercury in trough, only the atmospheric pressure acts. When the mercury level in the tube becomes stationary, the pressure inside tube at the point A, which is at the level of point C, must be same as that at the point C. The pressure at point A is due to the weight (or thrust) of the mercury column AB above it. Thus, the vertical height of the mercury column from the mercury surface in trough to the level in tube is a measure of the atmospheric pressure. The vertical of the mercury column in it (i.e., AB = h) is called the barometric height. Had the pressure at points A and C be not equal, the level of mercury in the tube would not have been stationary.

In simple words: The pressure of the air pushing down on the liquid in the bowl must be equal to the pressure of the liquid pushing down inside the tube. This balance tells us how strong the air pressure is.

📝 Teacher's Note: Use the analogy of a balance scale where the weight of the mercury column is balanced against the weight of the air column above the trough.

🎯 Exam Tip: "Barometric height" refers specifically to the vertical height. Remember that tilting the tube will change the length of mercury in the glass but not the vertical height.

Question 15S. Why is barometric height used as a unit to express atmospheric pressure?
Answer: It is the atmospheric pressure acting on the surface of the mercury in the trough that supports the vertical mercury column. Hence, barometric height is used as unit to express the atmospheric pressure.

In simple words: Because the height of the mercury is directly controlled by the air pressure, saying the air is "76 cm high" is an easy way to measure how hard it is pushing.

📝 Teacher's Note: Help students understand that the height of the column is independent of the tube's shape, which is why it serves as a reliable standard.

🎯 Exam Tip: Clearly state that the vertical column is supported by atmospheric pressure to link the two concepts.

Question 16S. What does it mean when we say atmospheric pressure is 76 cm of Hg? Express this in Pascal.
Answer: The atmospheric pressure at a place is 76 cm of Hg means at normal temperature and pressure, the height of the mercury column supported by the atmospheric pressure is 76 cm.
\( 76 \text{ cm of Hg} = 1.013 \times 10^5 \text{ pascal} \)

In simple words: This means the air is pushing down just hard enough to hold up a column of heavy mercury that is 76 centimeters tall.

📝 Teacher's Note: \( 76 \text{ cm} \) is the average pressure at sea level. Use this value to help students practice converting pressure between different units like Torr and Pascal.

🎯 Exam Tip: Memorize the Pascal value \( 1.013 \times 10^5 \text{ Pa} \) as it is frequently required for numerical problems in this chapter.

Question 17S. What is Torricellian vacuum? What happens if air enters this space?
Answer: The space above mercury is a vacuum. This empty space is called ‘Torricellian vacuum’. This can be shown by tilting the tube till the mercury fills the tube completely. Again when the mercury column becomes stationary, the empty space is created above the mercury column. If somehow air enters into the empty space or a drop of water gets into the tube, it will immediately vaporize and the air will exert pressure on mercury column due to which the barometric height will decrease.

In simple words: The empty space at the top of the tube is a vacuum. If a tiny bit of air or water gets in there, it will push the mercury down, giving you a wrong, lower reading.

📝 Teacher's Note: Explain that the Torricellian vacuum is nearly perfect but actually contains a very small amount of mercury vapor.

🎯 Exam Tip: If air enters the vacuum, it is called a "faulty barometer." The presence of air always causes the mercury level to drop.

Question 18S. How is the barometric height affected when: (a) tube is tilted, (b) tube is pushed down, (c) a small amount of air is introduced?
Answer: (a) The barometric height remains unaffected.
(b) The barometric height remains unaffected.
(c) The barometric height decreases.

In simple words: Moving or tilting the glass doesn't change the air pressure outside, so the height stays the same. But adding air inside creates a new push from the top that makes the mercury fall.

📝 Teacher's Note: This is a crucial concept. Demonstrate with a straw in water that tilting doesn't change the vertical level to reinforce that only vertical height matters.

🎯 Exam Tip: Remember that "vertical height" is the key term. Only air leakage or moisture causes the height to drop in these scenarios.

Question 19S. State two uses of a barometer.
Answer: Two uses of barometer:
1. To measure the atmospheric pressure.
2. For weather forecasting

In simple words: We use it to see how heavy the air is and to predict if it's going to be a sunny day or a stormy one.

📝 Teacher's Note: Barometers are vital for predicting storms, as a rapid drop in pressure usually indicates incoming low-pressure weather systems.

🎯 Exam Tip: List these as two distinct points to score full marks in descriptive sections.

Question 20S. State two advantages of using mercury as a barometric liquid.
Answer: Two advantages of using mercury as barometric liquid:
1. The density of mercury is greater than that of all the liquids, so only 0.76m height of mercury column is needed to balance the normal atmospheric pressure.
2. The mercury neither wets nor sticks to the glass tube therefore it gives the correct reading.

In simple words: Mercury is very heavy, so the tube doesn't have to be too tall. Also, it doesn't leave a messy coating on the glass, so it's easy to read.

📝 Teacher's Note: Mercury also has very low vapor pressure, which prevents the vacuum at the top from being filled with gas, unlike water.

🎯 Exam Tip: "High density" and "non-wetting property" are the two primary technical keywords examiners look for.

Question 21S. Why is water not suitable as a barometric liquid?
Answer: Water is not a suitable barometric liquid because:
1. The vapour pressure of water is high, so its vapours in the vacuum space will make the reading inaccurate.
2. Water sticks with the glass tube and wets it, so the reading becomes inaccurate.

In simple words: Water evaporates easily, and its steam would push down on the column. Also, it makes the glass blurry and wet, which makes it hard to see the exact level.

📝 Teacher's Note: Mention that a water barometer would need to be 10.4 meters high (about 3 stories tall!), which is the most practical reason it isn't used.

🎯 Exam Tip: Focus on "high vapour pressure" as it is the most scientific reason why water fails to maintain a proper vacuum.

Question 22S. Mention two advantages of Fortin’s barometer over a simple barometer.
Answer: In a simple barometer, there is no protection for the glass tube but in Fortin’s barometer, this defect has been removed by enclosing the glass tube in a brass case.
In a simple barometer, a scale cannot be fixed with the tube (or it cannot be marked on the tube) to measure the atmospheric pressure but Fortin’s barometer is provided with a vernier calipers to measure the accurate reading.

In simple words: Fortin's barometer has a metal case to keep it from breaking and a special sliding ruler to get much more exact measurements.

📝 Teacher's Note: Fortin's barometer is essentially a high-precision version of the simple one used in scientific laboratories.

🎯 Exam Tip: Highlight "protection" and "precision/accuracy" as the two main advantages.

Question 23S. How is a Fortin’s barometer used to measure atmospheric pressure?
Answer: To measure the atmospheric pressure, first the leather cup is raised up or lowered down with the help of the screw S so that the ivory pointer I just touches the mercury level in the glass vessel. The position of the mercury level in the barometer tube is noted with the help of main scale and the vernier scale. The sum of the vernier scale reading to the main scale reading gives the barometric height.
In simple words: You turn a screw to line up the mercury with a little pointer, then you read the numbers on the side just like using a ruler with a sliding magnifying glass for extra detail.

📝 Teacher's Note: Explain that the ivory pointer ensures that the mercury level in the reservoir is always at the "zero" point of the scale.

🎯 Exam Tip: The total reading is calculated as: \( \text{Total Reading} = \text{Main Scale Reading} + \text{Vernier Scale Reading} \). Mentioning both scales is essential.

Question 24S. What is an aneroid barometer? Describe its construction and working.
Answer: A barometer calibrated to read directly the atmospheric pressure is called an aneroid barometer. It has no liquid, it is light and portable.
Construction: Figure above shows the main parts of an aneroid barometer. It consists of a metallic box B which is partially evacuated. The top D of the box is springy and corrugated in form of a diaphragm as shown. At the middle of diaphragm, there is a thin rod L toothed at its upper end. The teeth of rod fit well into the teeth of a wheel S attached with a pointer P which can slide over a circular scale. The circular scale is initially calibrated with a standard barometer so as to read the atmospheric pressure directly in terms of the barometric height.
Working: When atmospheric pressure increases, it presses the diaphragm D and the rod L gets depressed. The wheel S rotates clockwise and pointer P moves to the right on the circular scale. When atmospheric pressure decreases, the diaphragm D bulges out due to which the rod L moves up and the wheel S rotates anti-clockwise. Consequently, the pointer moves to the left.
In simple words: An aneroid barometer is like a clock that measures air pressure. It uses a squishy metal box that gets crushed by thick air or expands in thin air, moving a needle to show the pressure level.

📝 Teacher's Note: Emphasize that "aneroid" means "without liquid," making it safer and more portable than mercury barometers. Use the analogy of a bag of chips expanding at high altitudes to explain the corrugated box.

🎯 Exam Tip: When describing the working, clearly link the increase in pressure to the "depression" of the diaphragm and the clockwise movement of the pointer.

Question 25S. State the main features of an aneroid barometer.
Answer: Aneroid barometer has no liquid and it is portable. It is calibrated to read directly the atmospheric pressure.
In simple words: It doesn't use messy mercury and is small enough to carry around, showing the air pressure on a simple dial.

📝 Teacher's Note: Contrast this with the simple barometer which is bulky and fragile due to the glass tube and mercury.

🎯 Exam Tip: The keywords are "no liquid," "portable," and "calibrated to read directly."

Question 26S. How does the reading of a barometer change with location?
Answer:
(i) In a mine, reading of a barometer increases.
(ii) On hills, reading of barometer decreases.
In simple words: The further down you go (like in a mine), the more air is above you, so the pressure goes up. The higher you climb (like on a hill), the less air is above you, so the pressure drops.

📝 Teacher's Note: Use the analogy of a stack of blankets—the person at the bottom feels more weight (pressure) than the person near the top.

🎯 Exam Tip: Remember that pressure is directly related to the depth of the fluid column above you.

Question 28S. List the factors that affect the atmospheric pressure.
Answer: Factors that affect the atmospheric pressure are:

• Height of air column
• Density of air
  In simple words: Air pressure depends on how much air is above you and how "packed together" that air is.

📝 Teacher's Note: Explain that as altitude increases, both the height of the air column and the density of air decrease, leading to a non-linear drop in pressure.

🎯 Exam Tip: Atmospheric pressure decreases exponentially with height, as shown in the graph.

Question 29S. Why does a fountain pen leak at high altitudes?
Answer: A fountain pen filled with ink contains some air at a pressure equal to atmospheric pressure on earth’s surface. When pen is taken to an altitude, atmospheric pressure is low so the excess pressure inside the rubber tube forces the ink to leak out.
In simple words: The air inside the pen is pushing out hard. High up on a mountain, the outside air is weak and can't push back, so the inside air wins and pushes the ink out

📝 Teacher's Note: This is a classic "pressure difference" example. The pen is like a tiny pressurized cabin.

🎯 Exam Tip: Use the phrase "excess pressure inside" to explain the cause of the leakage

Question 30S. Why does nose bleeding often occur at high altitudes?
Answer: On mountains, the atmospheric pressure is quite low. As such, nose bleeding may occur due to excess pressure of blood over the atmospheric pressure.
In simple words: Your blood is pushing against your veins from the inside. At high altitudes, there isn't enough air pressure on the outside to balance it, so tiny veins in the nose can pop.

📝 Teacher's Note: This explains why aircraft cabins must be pressurized for passenger safety.

🎯 Exam Tip: Focus on the imbalance between internal blood pressure and low external atmospheric pressure.

Question 31S. What is an altimeter? State its principle and mention its use.
Answer: An altimeter is a device used in aircraft to measure its altitude.
Principle: Atmospheric pressure decreases with the increase in height above the sea level; therefore, a barometer measuring the atmospheric pressure can be used to determine the altitude of a place above the sea level.
The scale of altimeter is graduated with height increasing towards left because the atmospheric pressure decreases with increase of height above the sea level.
In simple words: It is a special barometer that tells pilots how high they are by checking how thin the air is outside the plane.

📝 Teacher's Note: Clarify that an altimeter is essentially an aneroid barometer with a scale marked in meters or feet instead of pressure units.

🎯 Exam Tip: The key principle is that atmospheric pressure decreases with altitude.

Question 32S. What weather changes are indicated by a barometer?
Answer:
(a) It indicates that the moisture is increasing i.e., there is a possibility of rain.
(b) It indicates the coming of a storm or cyclone.
(c) It indicates that the moisture is decreasing i.e., it indicates dry weather.
In simple words: A dropping barometer means wet or stormy weather is coming. A rising barometer means the sky will clear up and be dry.

📝 Teacher's Note: Explain that moist air is lighter than dry air, which is why a fall in pressure indicates moisture/rain.

🎯 Exam Tip: Differentiate between a "slow fall" (rain) and a "sudden fall" (storm) for precise marks.

Question 1M. 1 torr is equal to
(a) 1 mm of Hg
(b) 1 cm of Hg
(c) 1 Pa
(d) 1 bar
Answer: (a) 1 mm of Hg
In simple words: Torr is just another name for measuring how many millimeters high the mercury rises in a tube.

📝 Teacher's Note: The unit is named after Torricelli, the inventor of the barometer.

🎯 Exam Tip: Remember \( 1 \text{ torr} = 1 \text{ mm of Hg} \) as a basic conversion.

Question 2M. Standard atmospheric pressure is
(a) 76 cm of Hg
(b) 760 cm of Hg
(c) 7.6 cm of Hg
(d) 0.76 cm of Hg
Answer: (a) 76 cm of Hg
In simple words: This is the average air pressure at sea level on a normal day.

📝 Teacher's Note: Remind students that this is also equal to 760 mm of Hg or 101,325 Pascals.

🎯 Exam Tip: Ensure the units (cm vs mm) are checked carefully in the options.

Question 1N. Calculate the pressure in pascal for 1 mm of Hg. (Density of Hg = \( 13.6 \times 10^3 \text{ kgm}^{-3} \), \( g = 9.8 \text{ ms}^{-2} \))
Answer: 1. Density of \( \text{Hg} = 13.6 \times 10^3 \text{ kgm}^{-3} \)
Acceleration due to gravity, \( g = 9.8 \text{ ms}^{-2} \)
Height of mercury column \( = 1 \text{ mm} = 0.001 \text{ m} \)
\( \therefore \text{Pressure}, P = h\rho g \)
or, \( P = ( 0.001 )( 13.6 \times 10^3 )( 9.8 ) \text{ pascal} \)
or, \( P = 133.28 \text{ Pa} \)
In simple words: We find the pressure by multiplying the height of the mercury by its density and gravity. A tiny 1mm of mercury exerts about 133 units of pressure.

📝 Teacher's Note: This is the fundamental calculation to convert Torr to Pascals. Ensure units are converted to SI (meters) first.

🎯 Exam Tip: Use the formula \( P = h\rho g \) and always show the conversion of mm to m.

Question 2N. At a place, the barometric height is 0.70 m. Calculate (i) the pressure in pascal and (ii) the equivalent height of a water column.
Answer: 2. Relative Density of \( \text{Hg} = 13.6 \implies \text{Density} = 13.6 \times 10^3 \text{ kgm}^{-3} \)
Acceleration due to gravity, \( g = 9.8 \text{ ms}^{-2} \)
Height of mercury column \( = 0.70 \text{ m} \)
\( \therefore \text{Pressure}, P = h\rho g \)
or, \( P = ( 0.7 )( 13.6 \times 10^3 )( 9.8 ) \text{ pascal} \)
or, \( P = 93.3 \times 10^3 \text{ Pa} \)
Let h be the height of water column
Then, \( P = h \times (\text{density of water}) \times g \)
or, \( 93.3 \times 10^3 = h \times 10^3 \times 9.8 \)
or, \( h = 9.52 \text{ m} \)
In simple words: The mercury is 0.7m high. To push back as hard as that heavy mercury, a column of light water would need to be almost 10 meters tall!

📝 Teacher's Note: This illustrates why we use mercury instead of water in barometers—the water version would be too tall for most buildings.

🎯 Exam Tip: For the second part, equate the pressures: \( h_1\rho_1 = h_2\rho_2 \) to find the height of the second liquid easily.

Question 3N. Calculate the height of a hill if the barometric height at the bottom is 76 cm Hg and at the top is 70 cm Hg. Assume pressure falls 1 cm Hg for every 120 m ascent.
Answer: 3. Atmospheric pressure, \( P = 76 \text{ cm Hg} \)
Rate at which pressure falls \( = 10 \text{ mm of Hg per } 120 \text{ m of ascent} = 1 \text{ cm of Hg per } 120 \text{ m of ascent} \)
Pressure at hill, \( P' = 70 \text{ cm Hg} \)
Total fall in pressure \( = P - P' = (76-70) \text{ cm Hg} = 6 \text{ cm Hg} \)
Now, fall in pressure is \( 1 \text{ cm Hg} \) for every \( 120 \text{ m} \) increase in height.
Thus, if the fall in pressure is \( 6 \text{ cm Hg} \), increase in height shall be \( (6 \times 120) \text{ m} = 720 \text{ m} \)
\( \therefore \text{Height of the hill} = 720 \text{ m} \)
Assumption: Atmospheric pressure falls linearly with ascent.
In simple words: Since we lost 6 cm of pressure, and we know each cm means climbing 120 meters, we just multiply 6 times 120 to find the hill's height.

📝 Teacher's Note: This is a ratio-based approach. It assumes the air density is constant, which is only approximately true for small heights.

🎯 Exam Tip: Clearly state the total fall in pressure (\( 76 - 70 = 6 \)) before doing the calculation.

Question 4N. Calculate the height of the atmosphere if the atmospheric pressure is \( 1.04 \times 10^5 \text{ Pa} \). (Density of air = \( 1.3 \text{ kgm}^{-3} \), \( g = 10 \text{ ms}^{-2} \))
Answer: 4. Atmospheric pressure, \( P = 1.04 \times 10^5 \text{ Pa} \)
Acceleration due to gravity, \( g = 10 \text{ ms}^{-2} \)
Density, \( \rho = 1.3 \text{ kgm}^{-3} \)
Let h be the height of the atmosphere.
\( P = h\rho g \)

\( \implies h = \frac{P}{\rho g} = \frac{1.04 \times 10^5}{1.3 \times 10} = 8000 \text{ m} \)
In simple words: If the air stayed equally thick all the way up, our atmosphere would be exactly 8 kilometers high.

📝 Teacher's Note: This is the "equivalent height" of a uniform atmosphere. In reality, the atmosphere is much taller because it gets thinner (less dense) as you go up.

🎯 Exam Tip: Re-arrange the formula to \( h = P / (\rho g) \) to solve for height.

Question 5N. At a certain place, the height above sea level is 107 m. Find the fall in barometric height. (Density of air = \( 1.295 \text{ kgm}^{-3} \), Density of Hg = \( 13.6 \times 10^3 \text{ kgm}^{-3} \))
Answer: Let \( h = 107 \text{ m} \) be the height above sea level
\( \therefore P_h - P_{sea} = \rho_{air} g h \)
\( \therefore \rho_m g h_f - \rho_m g h_i = \rho_{air} g h \)
\( \therefore \rho_m g \Delta h = \rho_{air} g h \)

\( \implies \Delta h = \frac{\rho_{air} h}{\rho_m} = \frac{1.295 \times 107}{13.6 \times 10^3} \)
\( \therefore \Delta h = 0.010 \text{ m of Hg} \)
\( \therefore \Delta h = 10 \text{ mm of Hg} \)
In simple words: When you climb up 107 meters, the mercury in your barometer will drop by exactly 10 millimeters.

📝 Teacher's Note: This problem shows how a barometer can act as a measuring tool for height. The ratio of densities determines the scale.

🎯 Exam Tip: Final answers are usually requested in mm of Hg for small changes in altitude. Be sure to convert from meters to millimeters.