Goyal Brothers Solutions for ICSE Class 9 Physics Chapter 4 Pressure In Fluids

Practice Problems 1

Question 1. Calculate pressure exerted by 0.8 m vertical length of alcohol of density \( 0.80 \text{ g cm}^{-3} \) in SI units. [Take \( g = 10 \text{ ms}^{-2} \)].
Answer: Vertical length of alcohol column = \( h = 0.8 \text{ m} \)
Density of alcohol = \( \rho = 0.80 \text{ g cm}^{-3} \)

\( \implies \rho = \frac{0.80 \times 10^6}{10^3} \text{ kgm}^{-3} \)

\( \implies \rho = 0.80 \times 1000 \text{ kgm}^{-3} \)

\( \implies \rho = 800 \text{ kg m}^{-3} \)
Pressure = \( P = ? \)
\( P = h\rho g \)

\( \implies P = 0.8 \times 800 \times 10 \)

\( \implies P = 6400 \text{ Pa} \)
In simple words: To find the pressure, we multiply the height of the liquid, its density, and gravity together. It is like calculating the weight of a column of liquid pressing down on a small area.

📝 Teacher's Note: Always remind students to convert density from \( \text{g cm}^{-3} \) to \( \text{kg m}^{-3} \) by multiplying by 1000 before using the S.I. formula.

🎯 Exam Tip: Writing the formula \( P = h\rho g \) carries 1 mark even if the final calculation is slightly off. Don't skip it!

Question 2. What is the pressure exerted by 75 cm vertical column of mercury of density \( 13600 \text{ kgm}^{-3} \) in SI units. [Take \( g = 9.8 \text{ ms}^{-2} \)].
Answer: Vertical length of mercury column = \( h = 75 \text{ cm} = 0.75 \text{ m} \)
Density of mercury = \( \rho = 13600 \text{ kg m}^{-3} \)
Acceleration due to gravity = \( g = 9.8 \text{ ms}^{-2} \)
Pressure = \( P = ? \)
\( P = h\rho g \)

\( \implies P = 0.75 \times 13600 \times 9.8 \)

\( \implies P = 99960 \text{ Pa} \)
In simple words: Mercury is very heavy, so even a relatively short column of it creates a lot of pressure. We use the same height-density-gravity formula here.

📝 Teacher's Note: Make sure students convert cm to m. This is a common trap where students use 75 instead of 0.75.

🎯 Exam Tip: Pay attention to the value of \( g \) provided in the question; sometimes it is 10 and sometimes 9.8.

Practice Problems 2

Question 1. 66640 Pa pressure is exerted by 0.50 m vertical column of a liquid. If \( g = 9.8 \text{ Nkg}^{-1} \), calculate density of the liquid.
Answer: Pressure = \( P = 66640 \text{ Pa} \)
Vertical length of liquid column = \( h = 0.50 \text{ m} \)
Acceleration due to gravity \( g = 9.8 \text{ ms}^{-2} \)
\( P = h\rho g \)

\( \implies \rho = \frac{P}{h \times g} \)

\( \implies \rho = \frac{66640}{0.50 \times 9.8} \)

\( \implies \rho = 13600 \text{ kg m}^{-3} \)
In simple words: Here we know the pressure and height, and we need to figure out what the liquid is by finding its density. We rearrange the pressure formula to solve for density.

📝 Teacher's Note: Explain that \( \text{Nkg}^{-1} \) is the same unit as \( \text{ms}^{-2} \). It represents the gravitational field strength.

🎯 Exam Tip: When rearranging formulas, perform the denominator multiplication first (\( 0.5 \times 9.8 \)) to make the final division easier.

Question 2. What vertical height of water will exert pressure of 333200 Pa? Density of water is \( 1000 \text{ kgm}^{-3} \) and \( g = 9.8 \text{ ms}^{-2} \).
Answer: Vertical height of water = \( h = ? \)
Pressure due to water column = \( P = 333200 \text{ Pa} \)
Acceleration due to gravity = \( g = 9.8 \text{ ms}^{-2} \)
\( P = h\rho g \)

\( \implies h = \frac{P}{\rho g} \)

\( \implies h = \frac{333200}{1000 \times 9.8} \)

\( \implies h = 34 \text{ m} \)
In simple words: To get this amount of pressure from water, you would need a pipe that is 34 metres tall—that's roughly the height of a 10-story building!

📝 Teacher's Note: This problem demonstrates why water isn't used in standard barometers; the tube would need to be over 10 metres tall.

🎯 Exam Tip: Ensure the units of the final answer are correct. Since we used S.I. units for everything, height will be in metres.

Question 3. Pressure at bottom of sea at some particular place is 8968960 Pa. If density of sea water is \( 1040 \text{ kgm}^{-3} \) calculate the depth of sea. Take \( g = 9.8 \text{ ms}^{-2} \). Neglect the pressure of the atmosphere.
Answer: Pressure at the bottom of the sea = \( P = 8968960 \text{ Pa} \)
Density of sea water = \( \rho = 1040 \text{ kg m}^{-3} \)
Acceleration due to gravity \( g = 9.8 \text{ ms}^{-2} \)
\( P = h\rho g \)

\( \implies h = \frac{P}{\rho g} \)

\( \implies h = \frac{8968960}{1040 \times 9.8} \)

\( \implies h = 880 \text{ m} \)
In simple words: The pressure at the bottom of the ocean is enormous because of all the water piled on top. At a depth of 880 metres, the pressure reaches nearly 9 million Pascals.

📝 Teacher's Note: Point out that sea water is denser than fresh water (\( 1040 \) vs \( 1000 \text{ kg/m}^3 \)) due to dissolved salts.

🎯 Exam Tip: "Neglect atmospheric pressure" means you only use the pressure from the liquid itself (\( h\rho g \)). If you weren't told this, you'd have to add about \( 10^5 \text{ Pa} \).

Practice Problems 3

Question 1. Atmospheric pressure at sea level is 76 cm of mercury. Calculate the vertical height of air column exerting the above pressure. Assume the density of air \( 1.29 \text{ kgm}^{-3} \) and that of mercury is \( 13600 \text{ kgm}^{-3} \). Why the height calculated by you is far less than actual height of atmosphere?
Answer: Height of mercury column = \( h_{Hg} = 76 \text{ cm} = 0.76 \text{ m} \)
Vertical height of air column = \( h_{air} = ? \)
Density of mercury = \( \rho_{Hg} = 13600 \text{ kgm}^{-3} \)
Density of air = \( \rho_{air} = 1.29 \text{ kgm}^{-3} \)
According to question:
Pressure due to air column = Pressure due to Hg column
\( \rho_{air} \times h_{air} \times g = \rho_{Hg} \times h_{Hg} \times g \)

\( \implies h_{air} = \frac{\rho_{Hg} \times h_{Hg}}{\rho_{air}} \)

\( \implies h_{air} = \frac{13600 \times 0.76}{1.29} = 8012.4 \text{ m} \)
The calculated height is far less than the actual height because the density of air decreases as we go higher up in the atmosphere, whereas we assumed it to be constant.
In simple words: We calculated that an 8 km tall column of air weighs the same as 76 cm of mercury. In reality, the atmosphere is much taller because the air gets "thinner" and lighter as you go up.

📝 Teacher's Note: Use the analogy of a pile of blankets: the bottom ones are squashed and dense, while the top ones are fluffy and light.

🎯 Exam Tip: The reasoning for the height difference (varying density of air) is as important as the numerical calculation for full marks.

Question 2. Calculate the equivalent height of mercury, which will exert as much pressure as 960 m of sea water of density \( 1040 \text{ kgm}^{-3} \). Density of mercury is \( 13600 \text{ kgm}^{-3} \).
Answer: Density of mercury = \( \rho_{Hg} = 13600 \text{ kgm}^{-3} \)
Density of sea water = \( \rho_{water} = 1040 \text{ kgm}^{-3} \)
Height of sea water column = \( h_{water} = 960 \text{ m} \)
Height of mercury column = \( h_{Hg} = ? \)
According to question,
Pressure due to mercury column = Pressure due to sea water column
\( \rho_{Hg} \times h_{Hg} \times g = \rho_{water} \times h_{water} \times g \)

\( \implies h_{Hg} = \frac{\rho_{water} \times h_{water}}{\rho_{Hg}} \)

\( \implies h_{Hg} = \frac{1040 \times 960}{13600} = 73.41 \text{ m} \)
In simple words: Because mercury is about 13 times denser than water, you only need 73 metres of mercury to create the same crushing pressure as nearly a kilometre of sea water.

📝 Teacher's Note: Notice how 'g' cancels out from both sides of the equation. This makes the calculation simpler.

🎯 Exam Tip: When equating pressures of two different liquids, always use the formula \( h_1\rho_1 = h_2\rho_2 \).

Practice Problems 4

Question 1. The pressure of water on the ground floor, in a water pipe is 150000 Pa, whereas pressure on the fourth floor is 30000 Pa. Calculate height of fourth floor. Take \( g = 10 \text{ ms}^{-2} \).
Answer: Pressure of water at fourth floor = \( P_2 = 30000 \text{ Pa} \)
Difference in pressure of water at ground floor and fourth floor
\( = P_1 - P_2 = 150000 - 30000 = 120000 \text{ Pa} \)
Pressure of water due to height \( (h) = h\rho g \)

\( \implies P_1 - P_2 = h \rho g \)

\( \implies 120000 = h \times 1000 \times 10 \)
[ \( \rho = 1000 \text{ kgm}^{-3} = \) Density of water ]

\( \implies h = \frac{120000}{10000} = 12 \text{ m} \)
In simple words: As you go higher up in a building, the water pressure drops because there is less "weight" of water above you in the supply system. We find the height by seeing how much the pressure changed.

📝 Teacher's Note: This is a practical example of why high-rise buildings need pumps; gravity isn't enough to maintain pressure at the top.

🎯 Exam Tip: The density of water (\( 1000 \text{ kg/m}^3 \)) is often not given in the question. Students are expected to memorize it.

Question 2. The pressure of water on ground floor is 160000 Pa. Calculate the pressure at the fifth floor, at a height of 15 m.
Answer: Pressure of water at ground floor = \( P_1 = 160000 \text{ Pa} \)
Pressure of water at fifth floor = \( P_2 = ? \)
Height of fifth floor = \( h = 15 \text{ m} \)
Density of water = \( \rho = 1000 \text{ kgm}^{-3} \)
Difference in pressure of water at ground and fifth floor = \( P_1 - P_2 \)
Pressure of water due to height \( (h) = h\rho g \)

\( \implies P_1 - P_2 = h\rho g \)

\( \implies 160000 - P_2 = 15 \times 1000 \times 10 \)

\( \implies P_2 = 160000 - 150000 \)

\( \implies P_2 = 10000 \text{ Pa} \)
In simple words: The pressure at the top floor is much lower (only 10,000 Pa) because the water has lost 150,000 Pa of pressure just by climbing 15 metres against gravity.

📝 Teacher's Note: Emphasize that pressure *decreases* with height. A common mistake is adding the height-pressure instead of subtracting it.

🎯 Exam Tip: When moving upward, use \( P_{top} = P_{bottom} - h\rho g \). When moving downward, use \( P_{bottom} = P_{top} + h\rho g \).

Practice Problems 5

Question 1. (a) The area of cross-sections of the pump plunger and press plunger of a hydraulic press are \( 0.02 \text{ m}^2 \) and \( 8 \text{ m}^2 \) respectively. If the hydraulic press overcomes a load of 800 kgf, calculate the force acting on pump plunger. (b) If the mechanical advantage of the handle of pump plunger is 8, calculate the force applied at the end of the handle of pump plunger.
Answer:
(a) Load on the press plunger = \( L = 800 \text{ kgf} \)
Let the effort acting on the pump plunger = \( E \)
Area of cross-section of pump plunger = \( A_1 = 0.02 \text{ m}^2 \)
Area of cross-section of press plunger = \( A_2 = 8 \text{ m}^2 \)
Now, \( \frac{L}{E} = \frac{A_2}{A_1} \)

\( \implies \frac{800}{E} = \frac{8}{0.02} = \frac{800}{2} = 400 \)

\( \implies E = \frac{800}{400} = 2 \text{ kgf} \)

\( \implies \) Force acting on the pump plunger = \( 2 \text{ kgf} \)

(b) Mechanical advantage = 8
But mechanical advantage = \( \frac{L}{E} \)
Here, \( L = 2 \text{ kgf} \)
Force applied at the of handle of pump plunger = \( E = ? \)

\( \implies \text{M.A.} = \frac{L}{E} \)

\( \implies 8 = \frac{2}{E} \)

\( \implies E = \frac{2}{8} = \frac{1}{4} = 0.25 \text{ kgf} \).
In simple words: A hydraulic press is a "force multiplier." By pushing with just 2 kg of force on a small pipe, we can lift 800 kg. If we use a lever handle too, we only need to push with 0.25 kg—that's about the weight of two tomatoes to lift a small car!

📝 Teacher's Note: This is an application of Pascal's Law. Pressure is the same everywhere in the liquid, so \( F_1/A_1 = F_2/A_2 \).

🎯 Exam Tip: Note that for part (b), the "Load" is the force we just calculated in part (a). This is a common multi-step exam problem.

Question 2. The radii of the press plunger and pump plunger are in ratio of 50 : 4. If an effort of 20 kgf acts on the pump plunger, calculate the maximum effort which the press plunger can over come.
Answer: Effort acting on the pump plunger \( E = 20 \text{ kgf} \)
Load acting on the press plunger = \( L = ? \)
So let radius of pump plunger = \( r = 4x \) and radius of press plunger = \( R = 50x \)
Now, \( \frac{L}{E} = \frac{\pi R^2}{\pi r^2} \)

\( \implies \frac{L}{20} = \frac{(50x)^2}{(4x)^2} = \frac{2500}{16} = \frac{625}{4} \)

\( \implies L = \frac{625}{4} \times 20 \)

\( \implies L = 625 \times 5 = 3125 \text{ kgf} \).
In simple words: When we know the radius instead of the area, we must square the numbers because the area of a circle depends on the radius squared. This is why a small increase in radius leads to a huge increase in lifting power.

📝 Teacher's Note: Remind students that Area \( A = \pi r^2 \). Since \( \pi \) appears in both top and bottom, it cancels out, leaving only the ratio of radii squared.

🎯 Exam Tip: Don't forget to square the ratio! Many students mistakenly write \( 50/4 \) instead of \( 2500/16 \).

QUESTIONS BASED ON ICSE EXAMINATION

(A) Objective Questions

Question 1. Unit of thrust in SI system is
(a) dynes
(b) joule
(c) \( \text{N/m}^2 \)
(d) newton
Answer: (d) newton
In simple words: Thrust is just another name for the total force acting perpendicularly on a surface. Since it is a force, its unit is the Newton.

📝 Teacher's Note: Distinguish between Thrust (total force, unit N) and Pressure (force per area, unit Pa).

🎯 Exam Tip: Examiners often put \( \text{N/m}^2 \) as a trap. Remember: Thrust = Newton, Pressure = \( \text{N/m}^2 \).

Question 2. The unit \( \text{Nm}^{-2} \) is the unit of
(a) force
(b) pressure
(c) thrust
(d) momentum
Answer: (b) pressure
In simple words: This unit tells you how many Newtons of force are spread across one square metre. That is exactly what pressure is.

📝 Teacher's Note: \( 1 \text{ Nm}^{-2} = 1 \text{ Pascal} \). These units are interchangeable.

🎯 Exam Tip: Always look for "per square metre" (\( \text{m}^{-2} \)) to identify pressure units.

Question 3. One Pascal is equal to:
(a) \( \text{Nm}^2 \)
(b) \( \text{Nm}^{-2} \)
(c) \( \text{Nm}^2 \)
(d) \( \text{Nm}^{-1} \)
Answer: (b) \( \text{Nm}^{-2} \)
In simple words: Pascal is just the scientific name given to the unit of pressure (\( \text{Newtons divided by square metres} \)).

📝 Teacher's Note: This honors Blaise Pascal, who did pioneering work in fluid mechanics.

🎯 Exam Tip: The options (a) and (c) are identical in the source text; usually, this is a typo for \( \text{N m} \). Focus on the correct one: \( \text{N m}^{-2} \).

Question 4. Thrust acting perpendicularly on the unit surface area is called :
(a) pressure
(b) moment of force
(c) down thrust
(d) none of the options
Answer: (a) pressure
In simple words: If you push on a wall, the "force" is your total effort, but the "pressure" is how much of that effort hits each tiny square inch of the wall.

📝 Teacher's Note: Emphasize the word "perpendicularly." Pressure only accounts for the component of force at 90 degrees to the surface.

🎯 Exam Tip: This is a standard definition. Use the words "normal force" or "perpendicular thrust" when defining pressure in subjective answers.

Question 5. Pressure applied in liquids is transmitted with undiminished force:
(a) in downward direction
(b) upward direction only
(c) sides of containing vessel
(d) in all directions
Answer: (d) in all directions
In simple words: If you squeeze a balloon filled with water, the water doesn't just push down; it pushes out in every single direction equally.

📝 Teacher's Note: This is Pascal's Law. It is the reason hydraulic brakes and jacks work.

🎯 Exam Tip: The keyword is "undiminished" and "all directions." Make sure both are in your head.

Question 6. As we move upwards, the atmospheric pressure :
(a) increases
(b) decreases
(c) remains same
(d) cannot be said
Answer: (b) decreases
In simple words: Think of the atmosphere as a stack of heavy blankets. On the ground, you have the whole stack on you. If you go up a mountain, there are fewer blankets above you, so there is less pressure.

📝 Teacher's Note: This is why ears "pop" in airplanes—the air pressure outside your body is changing rapidly.

🎯 Exam Tip: Pressure and Altitude have an inverse relationship—as one goes up, the other goes down.

Question 7. A dam for water reservoir is built thicker at the bottom than at the top because :
(a) pressure of water is very large at the bottom due to its large depth
(b) water is likely to have more density at the bottom due to its large depth
(c) quantity of water at the bottom is large
(d) variation in value of ‘g’
Answer: (a) pressure of water is very large at the bottom due to its large depth
In simple words: Water pressure gets stronger as you go deeper. A dam has to be extra thick at the bottom so it doesn't crack under the massive weight of the deep water.

📝 Teacher's Note: Pressure is \( h\rho g \). Since 'h' (depth) is maximum at the bottom, pressure is also maximum there.

🎯 Exam Tip: Explain that the wall must withstand the horizontal force (thrust) exerted by this high pressure.

Question 8. The pressure exerted by 50 kg (\( g = 10 \text{ m/s}^2 \)) on an area of cross section of \( 2 \text{ m}^2 \) is :
(a) 50 Pa
(b) 200 Pa
(c) 250 Pa
(d) 1000 Pa
Answer: (c) 250 Pa
Explanation : \( m = 50 \text{ kg} \); \( g = 10 \text{ ms}^{-2} \)
Area of cross-section = \( A = 2 \text{ m}^2 \)
\( \text{Pressure } P = \frac{F}{A} = \frac{mg}{A} = \frac{50 \times 10}{2} = 250 \text{ Pa} \)
In simple words: First we find the weight (force) by multiplying mass by gravity (\( 50 \times 10 = 500 \text{ N} \)). Then we divide that force by the area (\( 500 / 2 = 250 \)).

📝 Teacher's Note: Students often forget to multiply the mass by \( g \) to get the force. Remind them that kg is mass, not force.

🎯 Exam Tip: Always check if units are in S.I. (kg, m, s) before calculating. Here they are, so no conversion is needed.

Question 9. Pressure at a point inside a liquid does not depend on :
(a) The depth of the point below the surface of the liquid
(b) The nature of the liquid
(c) The acceleration due to gravity at that point
(d) The shape of the containing vessel
Answer: (d) The shape of the containing vessel
In simple words: Whether water is in a tall skinny glass or a wide bowl, if the depth is the same, the pressure at the bottom is the same. The shape of the container doesn't matter!

📝 Teacher's Note: This is called the "Hydrostatic Paradox." Pressure only depends on the vertical height of the liquid column.

🎯 Exam Tip: Memorize the three things pressure *does* depend on (\( h, \rho, g \)) so you can easily spot the "odd one out."

Question 10. The atmospheric pressure on earth’s surface is approximately
(a) \( 10^5 \text{ Pa} \)
(b) \( 10^4 \text{ Pa} \)
(c) \( 9.6 \times 10^4 \text{ N/m}^2 \)
(d) \( 10^{-4} \text{ Pa} \)
Answer: (a) \( 10^5 \text{ Pa} \)
In simple words: Air might seem light, but there is so much of it above us that it pushes on every square metre of our skin with a force equal to the weight of a large truck!

📝 Teacher's Note: Standard atmospheric pressure is \( 101,325 \text{ Pa} \), which scientists round to \( 10^5 \text{ Pa} \) for simplicity.

🎯 Exam Tip: Remember that \( 1 \text{ atmosphere (atm)} \approx 10^5 \text{ Pa} \approx 76 \text{ cm of Hg} \).

(B) Subjective Questions

Question 1. State three factors on which the pressure at a point in a liquid depends.
Answer: Factors on which the pressure at a point in a liquid depends are:
1. Pressure in a liquid is directly proportional to its height or depth.
2. Pressure in a liquid is directly proportional to its density.
3. Pressure in a liquid is directly proportional to the acceleration due to gravity.
4. Pressure in a liquid is independent of the area of cross-section.
In simple words: Pressure depends on how deep you are, how heavy the liquid is, and how strong gravity is. It does NOT matter how wide the container is.

📝 Teacher's Note: Use the formula \( P = h\rho g \) as a mnemonic for these three points.

🎯 Exam Tip: The question asks for three factors, but adding the fourth point (what it *doesn't* depend on) shows you have a complete understanding.

Question 2. The normal pressure of air is 76 cm of mercury. Calculate the pressure in SI units. [Density of mercury = \( 13600 \text{ kg/m}^3 \) and \( g = 10 \text{ m/s}^2 \)]
Answer: Height of mercury column = \( h = 76 \text{ cm} = 0.76 \text{ m} \)
Density of mercury = \( \rho = 13600 \text{ kg/m}^3 \)
Acceleration due to gravity = \( g = 10 \text{ m/s}^2 \)
Pressure = \( P = ? \)
\( P = h\rho g \)

\( \implies P = 0.76 \times 13600 \times 10 = 103360 \text{ N/m}^2 \)
In simple words: This is how scientists turn a length measurement (76 cm) into a pressure measurement (Pascals). We just need to know the density of the liquid in the tube.

📝 Teacher's Note: Explain that "76 cm of Hg" is a shortcut for saying "the pressure exerted by a 76 cm column of mercury."

🎯 Exam Tip: SI unit for pressure is Pascal (Pa), but \( \text{N/m}^2 \) is also correct. Most examiners prefer 'Pa'.

Question 3. At a given place, a barometer records 70 cm of Hg. If the mercury in barometer is replaced by water, what would be resulting reading? (Density of Hg = \( 13600 \text{ kg/m}^3 \); Density of water = \( 1000 \text{ kg/m}^3 \))
Answer: Height of mercury column = \( h = 70 \text{ cm} = 0.70 \text{ m} \)
Density of mercury = \( \rho = 13600 \text{ kg/m}^3 \)
Acceleration due to gravity = \( g = 10 \text{ m/s}^2 \)
Pressure = \( h\rho g \)
For water:
Height of water column = \( h' = ? \)
Pressure due to water column = \( P' \)
Density of water = \( \rho_w = 1000 \text{ kg/m}^3 \)
\( P' = h'\rho_w g \)
But, \( P' = P \)

\( \implies h'\rho_w g = h\rho g \)

\( \implies h'\rho_w = h\rho \)

\( \implies h' = \frac{h\rho}{\rho_w} = \frac{0.7 \times 13600}{1000} = 9.52 \text{ m} \)
In simple words: Since water is much lighter than mercury, the atmosphere can push it up much higher. If you used water instead of mercury, your barometer would have to be nearly 10 metres tall!

📝 Teacher's Note: This problem is a classic comparison using the principle that the same atmospheric pressure supports different heights of different liquids.

🎯 Exam Tip: Note that 'g' cancels out. Don't waste time multiplying by 9.8 or 10 on both sides.

Question 4. The base of cylindrical vessel measures \( 300 \text{ cm}^2 \). Water is poured into it upto a depth of 6 cm. Calculate the pressure of water on the base in vessel.
Answer: Area of base of cylinder = \( A = 300 \text{ cm}^2 \)
\( A = 300 \times 10^{-4} \text{ m}^2 \)
\( A = 3 \times 10^{-2} \text{ m}^2 \)
Height (or depth) of water column = \( h = 6 \text{ cm} = 0.06 \text{ m} \)
Density of water = \( \rho = 1000 \text{ kg/m}^3 \)
Acceleration due to gravity = \( g = 10 \text{ m/s}^2 \)
Pressure at the base in vessel = \( P = h\rho g \)

\( \implies P = \frac{6}{100} \times 1000 \times 10 \)

\( \implies P = 600 \text{ Pa} \)
In simple words: To find the pressure at the bottom, we only need to know how deep the water is (6 cm). The width of the container (300 cm²) actually doesn't affect the pressure!

📝 Teacher's Note: This is a logic test. The area is "extra info" meant to confuse students. Liquid pressure at a point depends only on depth, not the base area.

🎯 Exam Tip: If the question asks for "Thrust," then you use the Area (\( \text{Thrust} = \text{Pressure} \times \text{Area} \)). For "Pressure," just use \( h\rho g \).

Question 5. The pressure in water pipe on the ground floor of a building is 40000 pascals, whereas on the first floor it’s 10000 pascals. Find the height of first floor. (Acceleration due to gravity \( g = 10 \text{ ms}^{-2} \))
Answer: Pressure on the ground floor of a building = \( P_1 = 40000 \text{ Pa} \)
Pressure on the first floor of a building = \( P_2 = 10000 \text{ Pa} \)
\( \rho = 1000 \text{ kg/m}^3 = \) density of water
Acceleration due to gravity = \( g = 10 \text{ m/s}^2 \)
Difference in pressure = \( P_1 – P_2 = 40000 – 10000 = 30000 \text{ Pa} \)
Let \( h = \) height of first floor
Pressure of water due to height \( (h) = h\rho g \)

\( \implies 30000 = h \times 1000 \times 10 \)

\( \implies h = \frac{30000}{10000} = 3 \text{ m} \)
In simple words: We find the "missing" pressure between the floors and calculate how much water height that pressure represents. 30,000 Pascals of pressure is equal to a 3-metre column of water.

📝 Teacher's Note: This is a standard architectural physics problem. Every floor usually adds about 3 metres of height.

🎯 Exam Tip: Clearly show the subtraction \( P_1 - P_2 \) to get the pressure difference before solving for \( h \).

Question 6. (a) Define SI unit of pressure. (b) The atmospheric pressure at a place is 650 mm of Hg. Calculate this pressure in Pascals (Pa).
Answer: (a) SI unit of pressure is pascal (Pa) or \( \text{Nm}^{-2} \)
One Pascal: When a force of one newton acts normally on an area of one square metre (\( 1 \text{ m}^2 \)) then pressure acting on the surface is called one Pascal.
(b) Height of mercury column = \( h = 650 \text{ mm} = 0.65 \text{ m} \)
Density of mercury = \( \rho = 13600 \text{ kg/m}^3 \)
Acceleration due to gravity = \( g = 10 \text{ ms}^{-2} \)
Pressure \( (P) = h\rho g \)

\( \implies P = 0.65 \times 13600 \times 10 \)

\( \implies P = 88400 \text{ Pa} \)
In simple words: A Pascal is the pressure of a very light touch (1 Newton) spread over a whole square metre. 650 mm of mercury is less than standard sea-level pressure, coming out to about 88,400 Pa.

📝 Teacher's Note: To convert mm to m, divide by 1000. Many students divide by 100 by mistake.

🎯 Exam Tip: For the definition of 1 Pascal, the words "normally" (meaning perpendicularly) and "one square metre" are the key terms that examiners look for.

Question 7. Pressure in a water pipe on the ground floor of a building is 100,000 Pa. Calculate the pressure in water pipe on first floor at a height of 3 m. [Density of water = \( 1000 \text{ kgm}^{-3} \); \( g = 10 \text{ ms}^{-2} \)]
Answer: Pressure of water at ground floor = \( P_1 = 1,00,000 \text{ Pa} \)
Pressure of water at first floor = \( P_2 = ? \)
Height of first floor = \( h = 3 \text{ m} \)
Density of water = \( \rho = 1000 \text{ kgm}^{-3} \)
Acceleration due to gravity = \( g = 10 \text{ ms}^{-2} \)
Difference in pressure between ground floor and first floor = \( P_1 - P_2 = h\rho g \)

\( \implies 1,00,000 – P_2 = 3 \times 1000 \times 10 \)

\( \implies P_2 = 1,00,000 - 30,000 \)

\( \implies P_2 = 70000 \text{ Pa} \)
\( \therefore \) Pressure of water in pipe at first floor of a building is 70000 Pa.
In simple words: The water loses some of its pressure as it travels uphill. For every 3 metres of height, it loses about 30,000 Pa of pressure.

📝 Teacher's Note: This is the reverse of Question 5. Here we know the height and need the pressure.

🎯 Exam Tip: Be sure to write the final statement with units (Pa) to conclude your calculation.

Question 8. P is the pressure at some point in a liquid. State whether pressure P is a scalar or vector quantity.
Answer: Pressure exerted on an enclosed fluid gets transmitted equally and undiminishingly in all directions. So no particular direction is associated with pressure. That is why Pressure is a scalar quantity.
In simple words: Even though force has a direction, pressure in a fluid acts in all directions at once. Because it doesn't point in one specific way, it is a scalar.

📝 Teacher's Note: This is a common point of confusion. Force is a vector, but Pressure (\( \text{Force/Area} \)) is a scalar because the force at a point in a fluid acts in every direction equally.

🎯 Exam Tip: Always justify the "scalar" label by mentioning that pressure is transmitted in "all directions."

Question 9. A beaker contains a liquid of density ‘\( \rho \)’ upto height ‘\( h \)’ such that ‘\( P_A \)’ is atmospheric pressure and ‘\( g \)’ is acceleration due to gravity. Answer the following questions :
(a) What is the pressure on the free surface of liquid?
(b) What is the pressure on the base of beaker?
(c) What is the lateral pressure at the base on the inner walls of beaker?
Answer:
(a) Pressure on the free surface of liquid is equal to the atmospheric pressure (\( P_a \)).
(b) Consider a liquid contained in a beaker, such that ‘\( \rho \)’ is the density of liquid.
Consider a point B at the base of liquid and the liquid column of area of cross-section ‘a’ around it, such that ‘h’ is the height of the liquid column.

\( \therefore \text{Volume of imaginary column of liquid} = \text{area of cross-section} \times \text{length} = ah \)

\( \therefore \text{Mass of liquid column} = \text{Volume} \times \text{density} = ah\rho \)

\( \therefore \text{Weight of liquid column} = \text{mass} \times g = ah\rho g \)

\( \therefore \text{Thrust exerted by liquid column on the base of the beaker} = ah\rho g \)
Pressure due to liquid column \( P = \frac{\text{Force}}{\text{Area}} = \frac{F}{a} = \frac{ah\rho g}{a} = h\rho g \)
So, pressure on the base of beaker = \( h\rho g \)
Total pressure at base of beaker = \( \text{Atmospheric pressure} + h\rho g = P_a + h\rho g \)
(c) Also lateral pressure at the base on the inner walls of beaker = \( P_a + h\rho g \)
In simple words: The surface only feels the air pushing down. The bottom feels the air push PLUS the weight of all the liquid. The sides feel this same pressure pushing outwards.

📝 Teacher's Note: "Lateral pressure" is the pressure on the walls. One of the unique properties of fluids is that at any depth, the pressure pushing down is the same as the pressure pushing sideways.

🎯 Exam Tip: For the total pressure at a depth, never forget to add the atmospheric pressure \( P_a \) unless the question specifically asks for "gauge pressure" or "liquid pressure."

Question 10. State the law of transmission of pressure in liquids.
Answer: Pascal’s law : “The pressure applied on the surface of a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.”
In simple words: If you apply pressure to any part of a contained liquid, that exact same pressure travels to every other part of the liquid instantly.

📝 Teacher's Note: This only works for "confined" (trapped) liquids. If the liquid can flow out, the pressure won't transmit perfectly.

🎯 Exam Tip: This is a very common 2-mark theory question. Memorize the statement word-for-word.

Question 11. Calculate the hydrostatic pressure exerted by water at the bottom of a beaker. Take the depth of water as 40 cm, the density of water \( 1000 \text{ kgm}^{-3} \) and \( g = 9.8 \text{ ms}^{-2} \).
Answer: Pressure at the bottom of beaker = \( P = ? \)
Height (or depth) of water in beaker = \( h = 40 \text{ cm} = 0.4 \text{ m} \)
Density of water = \( \rho = 1000 \text{ kgm}^{-3} \)
Acceleration due to gravity = \( g = 9.8 \text{ ms}^{-2} \)
\( P = h\rho g \)

\( \implies P = 0.4 \times 1000 \times 9.8 \)

\( \implies P = 3920 \text{ Pa} \)
In simple words: We find the pressure by multiplying the depth, density, and gravity. 40 cm of water creates about 4,000 Pascals of pressure at the bottom.

📝 Teacher's Note: "Hydrostatic" just means the pressure of a liquid that isn't moving.

🎯 Exam Tip: In numericals, always show your given data with converted units first (\( 40 \text{ cm} \rightarrow 0.4 \text{ m} \)) to get partial credit.

Question 12. State Pascal’s law of transmission of pressure in a liquid.
Answer: Pascal’s law : “The pressure applied on the surface of a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.”
In simple words: If you poke a hole in a water-filled balloon, the water squirts out because the pressure you apply at the top travels everywhere inside.

📝 Teacher's Note: This is a repeat of Question 10, highlighting how important this law is for the syllabus.

🎯 Exam Tip: If asked for an application, mention the hydraulic jack or hydraulic brakes.

Question 13. State briefly, how and why the atmospheric pressure of a place varies with the altitude. Draw an approximate graph to illustrate this variation.
Answer:
1. We know atmospheric pressure = height of air column \( \times \) density of air \( \times \) acceleration due to gravity; \( P = h\rho g \). So, as we go up i.e. at higher altitudes, height of air column and hence atmospheric pressure decreases.
2. Also with the increase in altitude, density of air decreases and hence atmospheric pressure decreases.
If we take average density of air as \( 1.29 \text{ kgm}^{-3} \) and the density of mercury as \( 13600 \text{ kgm}^{-3} \), we can find the height column which will exert as much pressure as is exerted by 1 cm or (0.01 m) column of mercury as:
Height of air column \( \times \) density of air = height of mercury column \( \times \) density of mercury

\( \implies \text{height of air column} \times 1.29 \text{ kg/m}^3 = 0.01 \text{ m} \times 13600 \text{ kg/m}^3 \)

\( \implies \text{Height of air column} = \frac{136}{1.29} = 105 \text{ m} \text{ (approx).} \)
Thus, 105 m of air column, on the average, will exert as much pressure as 1 cm column of mercury. Further, 1 cm of mercury column exerts pressure = 105 m of air column. 76 cm of mercury column exerts pressure = \( 105 \times 76 \text{ m} = 7980 \text{ m} = 8 \text{ km} \text{ (approx).} \)
Thus, 8 km of air column will exert as much pressure as 76 cm of mercury column. However, it does not mean that atmosphere extends to only 8 km. As it is pointed out earlier, the density of atmosphere also changes with height. On higher altitudes the vertical height of air is far in excess of 105 m, because of low density of air.
In simple words: Air pressure drops as you go up because there is less air above you and the air that is there is thinner (less dense). The graph shows that pressure falls very quickly at first and then levels off.

📝 Teacher's Note: The relationship is non-linear. This is why it's much harder to breathe at the top of Mt. Everest than at base camp; the pressure and density are extremely low.

🎯 Exam Tip: The graph should be a curve sloping downwards from left to right, representing that pressure decreases as altitude increases.

Question 14. The blood pressure reading of a patient is recorded 160/ 100. Express the lower pressure in SI units. [Take density of mercury as \( 13.6 \times 10^3 \text{ kgm}^{-3} \) and the value of ‘g’ as \( 10 \text{ ms}^{-2} \)]
Answer: Lower pressure of the patient = 100 mm of Hg column
Height of mercury column = \( h = 100 \text{ mm} = 10 \text{ cm} = 0.1 \text{ m} \)
Density of mercury = \( \rho = 13.6 \times 10^3 \text{ kgm}^{-3} \)
Acceleration due to gravity = \( g = 10 \text{ ms}^{-2} \)
So lower pressure of patient = \( P = h\rho g \)

\( \implies P = 0.1 \times 13.6 \times 10^3 \times 10 \)

\( \implies P = 13600 \text{ Pa} \)
In simple words: Doctors measure blood pressure in millimetres of mercury. In physics units, a "low" reading of 100 means your blood is pushing with a pressure of 13,600 Pascals.

📝 Teacher's Note: The "lower" number in blood pressure is the diastolic pressure. Help students see the connection between biology and physics here.

🎯 Exam Tip: \( 13.6 \times 10^3 \) is just scientific notation for \( 13600 \). Don't let the powers of 10 confuse your calculation.

Question 15. State two advantages of aneroid barometer.
Answer: Advantages of aneroid barometer :
1. It is compact, portable and hence can be carried anywhere.
2. It does not contain any liquid and there is no chance of spilling over of liquid as in mercury barometer.
In simple words: An aneroid barometer is like a watch—it's small, has no messy liquids, and you can take it hiking or on a plane easily.

📝 Teacher's Note: The word "aneroid" means "without liquid." It uses a flexible metal box instead of mercury.

🎯 Exam Tip: Focus on "portability" and "lack of liquid" as the two distinct advantages.

Question 16. Explain, why a gas bubble released at the bottom of a lake grows in size as it rises to the surface of the lake.
Answer: Bubble released at the bottom of a lake grows in size as it rises to the surface of the lake because the pressure exerted on it by water of the lake DECREASES hence by BOYLE’S LAW \( PV = \text{constant} \) the VOLUME of bubble INCREASES and the bubble grows in size.
In simple words: At the bottom, the water squeezes the bubble very hard. As it floats up, there is less water on top, so the squeezing force (pressure) drops, letting the air inside the bubble expand.

📝 Teacher's Note: This is a perfect qualitative application of Boyle's Law (\( P \propto 1/V \)). As \( P \) goes down, \( V \) must go up.

🎯 Exam Tip: You must mention "Boyle's Law" or the relationship between "Pressure and Volume" to get full marks for the explanation.