This MCQ module is based on: Pressure Pascal
TOPIC 5 OF 33
Pressure Pascal
🎓 Class 11
Physics
CBSE
Theory
Ch 9 – Mechanical Properties of Fluids
⏱ ~14 min
🌐 Language: [gtranslate]
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Pressure Pascal
9.1 Introduction: What Are Fluids?
Chapter 8 dealt with the mechanical response of solids, which possess a definite shape and resist shear forces. We now turn to fluids — the collective name for liquids and gases. Unlike solids, fluids cannot support a permanent shear stress; they flow to take the shape of their container.
Key distinctions:
- Solid: fixed shape and fixed volume.
- Liquid: no fixed shape (adopts container's shape), but has a fixed (nearly incompressible) volume.
- Gas: no fixed shape and no fixed volume — expands to fill the entire container available.
Because fluids flow, the way they transmit forces is very different from solids. Instead of stress at a point acting in one direction, a fluid at rest transmits force as pressure acting equally in all directions at that point. This chapter develops the laws that govern fluids at rest (fluid statics) and in motion (fluid dynamics): Pascal's law, Bernoulli's principle, viscosity, surface tension and more.
9.2 Pressure
When you stand on soft snow in ordinary shoes, your feet sink; but when you wear snow-shoes of large area, you do not sink. Your weight is the same in both cases — what differs is the pressure on the snow.
Pressure is defined as the normal force acting per unit area:
\[P = \frac{F_{\perp}}{A}\]
Pressure is a scalar quantity. SI unit: pascal (Pa) where \(1\text{ Pa} = 1\text{ N/m}^2\). Dimensions: \([ML^{-1}T^{-2}]\).
Other commonly used pressure units:
| Unit | Conversion | Where used |
|---|---|---|
| 1 atmosphere (atm) | \(1.013\times 10^5\) Pa | Standard atmospheric pressure at sea level |
| 1 bar | \(10^5\) Pa | Meteorology (weather charts) |
| 1 torr = 1 mm Hg | 133 Pa | Blood pressure, vacuum systems |
| 1 psi (lb/in²) | 6895 Pa | Tyre pressure (engineering, USA) |
9.2.1 Density and Specific Gravity
The density of a fluid is mass per unit volume: \[\rho = \frac{m}{V}\] SI unit: kg/m³. Water at 4 °C has \(\rho_w = 1000\) kg/m³. The specific gravity (or relative density) is the ratio of the density of a substance to that of water: \[\text{Specific gravity} = \frac{\rho}{\rho_w}\] being a ratio, it is dimensionless. Mercury has specific gravity 13.6; olive oil about 0.92.
9.2.2 Pressure Variation with Depth
Consider a fluid of density \(\rho\) at rest. Take an imaginary vertical cylinder of cross-section \(A\) and height \(h\) with its top at the free surface (Fig 9.3). For equilibrium of the cylinder, the upward force on the bottom face must balance the downward atmospheric push plus the weight of the fluid column:
P(bottom) · A = P₀ · A + (ρ · A · h) · g
Dividing by \(A\) we obtain the fundamental relation for hydrostatic pressure:
\[\boxed{\,P = P_0 + \rho g h\,}\]
The pressure at depth \(h\) below the free surface of a liquid of density \(\rho\) exceeds atmospheric pressure \(P_0\) by the amount \(\rho g h\). The quantity \(\rho g h\) is called the gauge pressure, while \(P\) itself is the absolute pressure.
Pascal's principle (same height, same pressure): All points of a connected fluid at rest which are at the same horizontal level have the same pressure, regardless of the shape of the container. This is why water in a U-tube stands at the same height in both arms.
9.2.3 Atmospheric Pressure and the Mercury Barometer
The column of air above us exerts a pressure at sea-level of about 1.013 × 10⁵ Pa. Torricelli (1643) measured it by inverting a tube (closed at one end, about a metre long) filled with mercury into a trough of mercury (Fig 9.4). The mercury in the tube fell until its weight balanced atmospheric pressure at the level of the trough. At this height, the vacuum above the mercury column exerts essentially no pressure.
Balancing pressure at the mercury surface in the trough: \(P_0 = \rho_{Hg} g h\). With \(\rho_{Hg}=13.6\times 10^3\) kg/m³, \(g = 9.8\) m/s², \(h = 0.76\) m, we get \(P_0 = 1.013 \times 10^5\) Pa.
9.2.4 Open-Tube Manometer
The pressure of a gas in a container can be measured with an open U-tube manometer (Fig 9.5). One arm is connected to the gas, the other is open to the atmosphere. The difference in heights of the liquid in the two arms gives the gauge pressure \(P - P_0 = \rho g h\).
9.2.5 Pascal's Law for Transmission of Fluid Pressure
Pascal's Law: A change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every point of the fluid and to the walls of the containing vessel.
In plain terms: if you squeeze one end of a sealed bottle of water, the pressure rises by the same amount everywhere inside. This principle is the basis of every hydraulic device — lifts, brakes, presses, jackhammers.
9.2.6 Hydraulic Machines
Imagine two connected cylinders of different cross-sectional areas \(A_1\) (small) and \(A_2\) (large) filled with an incompressible liquid (Fig 9.8). A small force \(F_1\) applied on piston 1 produces a pressure \(P = F_1/A_1\). By Pascal's law this pressure is the same throughout the liquid, so the upward force on piston 2 is:
\[\frac{F_1}{A_1} = \frac{F_2}{A_2} \quad\Longrightarrow\quad F_2 = F_1 \cdot \frac{A_2}{A_1}\]
The factor \(A_2/A_1\) is the mechanical advantage of the hydraulic lift.
Energy is conserved: the volumes swept by the two pistons are equal, \(A_1 d_1 = A_2 d_2\), so the small piston must travel a larger distance \(d_1 = d_2(A_2/A_1)\). You pay in distance what you gain in force — exactly as with a lever.
Hydraulic brakes in vehicles use the same idea. When the driver presses the brake pedal, the force is transmitted to the master cylinder, which pushes brake fluid through narrow pipes to wheel cylinders at each wheel. The larger area at the wheel cylinder multiplies the force and pushes the brake shoes against the drum/disc. Because the fluid is incompressible and Pascal's law is instantaneous, all four wheels feel the braking force almost simultaneously.
Worked Examples
Example 9.1: Pressure at the bottom of a tank
A rectangular tank contains water to a depth of 10.0 m. Take \(\rho_w = 1000\) kg/m³ and \(g = 9.8\) m/s². Find (a) the gauge pressure and (b) the absolute pressure at the bottom. Atmospheric pressure at the surface = \(1.01\times 10^5\) Pa.
Given: \(h = 10.0\) m, \(\rho = 1000\) kg/m³, \(g = 9.8\) m/s², \(P_0 = 1.01\times 10^5\) Pa.
(a) Gauge pressure: \[P_g = \rho g h = 1000 \times 9.8 \times 10 = 9.8\times 10^4 \text{ Pa}\] (b) Absolute pressure: \[P = P_0 + \rho g h = 1.01\times 10^5 + 0.98\times 10^5 = 1.99\times 10^5 \text{ Pa}\] That is about 2 atm — the pressure nearly doubles at a depth of only 10 m of water!
Answer: \(P_g = 0.98\times 10^5\) Pa; \(P = 1.99\times 10^5\) Pa.
(a) Gauge pressure: \[P_g = \rho g h = 1000 \times 9.8 \times 10 = 9.8\times 10^4 \text{ Pa}\] (b) Absolute pressure: \[P = P_0 + \rho g h = 1.01\times 10^5 + 0.98\times 10^5 = 1.99\times 10^5 \text{ Pa}\] That is about 2 atm — the pressure nearly doubles at a depth of only 10 m of water!
Answer: \(P_g = 0.98\times 10^5\) Pa; \(P = 1.99\times 10^5\) Pa.
Example 9.2: Hydraulic lift — force and distance (NCERT Example 9.2)
Two syringes of different cross-sections (without needles) are filled with water and connected by a rubber tube. The radii of the pistons are \(r_1 = 1.0\) cm and \(r_2 = 3.0\) cm. If a force of 10 N is applied to the smaller piston, what force is needed on the larger piston to keep the system in equilibrium? If the smaller piston is pushed in by 6.0 cm, how much does the larger piston rise?
Step 1 — Areas:
\[A_1 = \pi r_1^2 = \pi(1.0)^2 = \pi\text{ cm}^2,\quad A_2 = \pi(3.0)^2 = 9\pi\text{ cm}^2\]
Step 2 — Force using Pascal's law:
\[F_2 = F_1 \cdot \frac{A_2}{A_1} = 10 \times \frac{9\pi}{\pi} = 10\times 9 = 90\text{ N}\]
Step 3 — Distance (volume conservation):
\[A_1 d_1 = A_2 d_2 \;\Rightarrow\; d_2 = d_1 \cdot \frac{A_1}{A_2} = 6.0 \times \frac{1}{9} = 0.67\text{ cm}\]
Answer: Force required on the larger piston \(= \boxed{90\text{ N}}\); it rises by \(\boxed{0.67\text{ cm}}\).
Note: Force is multiplied by 9×, but distance is reduced by the same factor — total work done \(F_1 d_1 = F_2 d_2\) is conserved.
Note: Force is multiplied by 9×, but distance is reduced by the same factor — total work done \(F_1 d_1 = F_2 d_2\) is conserved.
Example 9.3: Barometer reading in a different liquid
A barometer is constructed using water (\(\rho = 10^3\) kg/m³) instead of mercury. Atmospheric pressure is \(1.013\times 10^5\) Pa. What height of water column would balance the atmosphere? Why is mercury preferred?
From \(P_0 = \rho g h\):
\[h = \frac{P_0}{\rho g} = \frac{1.013\times 10^5}{10^3 \times 9.8} = 10.34\text{ m}\]
That is over 10 metres of water — a completely impractical tube! Mercury, being 13.6 times denser, gives a column of only 76 cm. Mercury also has extremely low vapour pressure, so the "vacuum" at the top is a genuine vacuum. Water would evaporate at the top, reducing accuracy. Hence mercury is preferred.
Answer: \(h_{\text{water}} = \boxed{10.34\text{ m}}\).
Answer: \(h_{\text{water}} = \boxed{10.34\text{ m}}\).
Example 9.4: Hydraulic car lift
A hydraulic lift in a service station is used to raise a car of mass 1800 kg. The piston that supports the car has radius 20 cm. What is the maximum gauge pressure (in pascal and atm) in the hydraulic fluid when the car is being lifted? Take \(g = 9.8\) m/s².
Weight of the car \(W = mg = 1800\times 9.8 = 17640\) N.
Area of piston \(A = \pi r^2 = \pi(0.20)^2 = 0.1257\) m².
\[P = \frac{W}{A} = \frac{17640}{0.1257} = 1.40\times 10^5\text{ Pa} \approx 1.39\text{ atm}\] Answer: Gauge pressure \(= \boxed{1.40\times 10^5\text{ Pa} \approx 1.39\text{ atm}}\).
Area of piston \(A = \pi r^2 = \pi(0.20)^2 = 0.1257\) m².
\[P = \frac{W}{A} = \frac{17640}{0.1257} = 1.40\times 10^5\text{ Pa} \approx 1.39\text{ atm}\] Answer: Gauge pressure \(= \boxed{1.40\times 10^5\text{ Pa} \approx 1.39\text{ atm}}\).
Example 9.5: Pressure difference in the oceans
Find the pressure at a depth of 1000 m below the ocean surface. Density of sea-water = 1030 kg/m³, \(P_0 = 1.01\times 10^5\) Pa, \(g = 9.8\) m/s². Would a window in a submarine at this depth need to withstand this pressure?
\[P = P_0 + \rho g h = 1.01\times 10^5 + 1030\times 9.8\times 1000\]
\[= 1.01\times 10^5 + 1.0094\times 10^7 = 1.02\times 10^7\text{ Pa} \approx 101\text{ atm}\]
So the window must withstand roughly 100 times atmospheric pressure. Submarine viewports are built from very thick acrylic or titanium frames for this reason.
Answer: \(P \approx \boxed{1.02\times 10^7\text{ Pa} \approx 101\text{ atm}}\).
Answer: \(P \approx \boxed{1.02\times 10^7\text{ Pa} \approx 101\text{ atm}}\).
Interactive: Hydraulic Lift Simulator L3 Apply
Change the piston radii and the input force — watch the output force and the mechanical advantage update in real time.
Output force F₂ = ... N
Activity — Prove Pressure Depends on Depth
L3 Apply
Predict: If you make two small holes in the side of a tall plastic bottle — one near the top, one near the bottom — which hole will shoot water further?
- Take a 2-litre plastic bottle. With a pin, make a small hole 5 cm below the rim and another 5 cm above the base.
- Cover both holes with tape, fill the bottle with water, and place on the edge of a sink.
- Remove both tapes simultaneously.
- Observe which hole produces the stronger, longer-range jet.
Observation: The jet from the lower hole travels much further and has a much stronger push.
Explanation: The pressure at the lower hole is \(P_0 + \rho g h\) where \(h\) is the depth of water above it. Greater depth means greater pressure, which (via Torricelli's theorem that we will meet in Part 2) translates into greater efflux speed \(v = \sqrt{2gh}\).
Explanation: The pressure at the lower hole is \(P_0 + \rho g h\) where \(h\) is the depth of water above it. Greater depth means greater pressure, which (via Torricelli's theorem that we will meet in Part 2) translates into greater efflux speed \(v = \sqrt{2gh}\).
Competency-Based Questions
A hydraulic press in a forge workshop has a small piston of area 5 cm² and a large piston of area 200 cm². The large piston lifts a steel block of mass 400 kg. The worker pushes the small piston down by 40 cm in a single stroke. (Use \(g = 10\) m/s² for quick estimates.)
Q1. L1 Remember State Pascal's law in one sentence.
Pressure applied to an enclosed incompressible fluid is transmitted equally and undiminished to every point of the fluid and the walls of the container.
Q2. L3 Apply What minimum force must the worker apply on the small piston to just lift the steel block?
Weight \(= 400\times 10 = 4000\) N. By \(F_1/A_1 = F_2/A_2\): \(F_1 = F_2(A_1/A_2) = 4000\times (5/200) = \boxed{100\text{ N}}\).
Q3. L3 Apply By how much does the steel block rise in one stroke?
Volume conservation: \(A_1 d_1 = A_2 d_2\). \(d_2 = d_1(A_1/A_2) = 40\times (5/200) = \boxed{1.0\text{ cm}}\).
Q4. L4 Analyse Show that the work done by the worker equals the work done against gravity on the block.
Work by worker \(= F_1 d_1 = 100\times 0.40 = 40\) J.
Work against gravity \(= F_2 d_2 = 4000\times 0.01 = 40\) J.
The two are equal — the hydraulic press does not create energy; it trades distance for force, just like a mechanical lever.
Work against gravity \(= F_2 d_2 = 4000\times 0.01 = 40\) J.
The two are equal — the hydraulic press does not create energy; it trades distance for force, just like a mechanical lever.
Q5. L5 Evaluate Engineers are replacing water with mineral oil as the hydraulic fluid. Two advantages?
(i) Mineral oil does not rust metal components, whereas water corrodes steel; (ii) oil lubricates pistons and seals, reducing wear. Additionally, oil has a lower freezing point, reducing the risk of winter seizure. Both are virtually incompressible, so Pascal's law applies equally well.
Assertion-Reason Questions
Assertion (A): The pressure at the bottom of a container full of water depends only on the depth, not on the shape of the vessel.
Reason (R): In a fluid at rest, all points at the same horizontal level have the same pressure.
Answer: A. This is the hydrostatic paradox. R directly implies A.
Assertion (A): A hydraulic lift allows a small input force to lift a heavy load.
Reason (R): Hydraulic lifts multiply energy in violation of energy conservation.
Answer: C. A is true. R is false — hydraulic lifts multiply force, not energy. The small piston travels a proportionally larger distance so work is conserved.
Assertion (A): A mercury barometer column is about 760 mm, while a water barometer would require about 10 m.
Reason (R): Mercury has much higher density than water; for the same atmospheric pressure \(P_0 = \rho g h\), greater \(\rho\) gives smaller \(h\).
Answer: A. \(h = P_0/(\rho g)\). With \(\rho_{Hg}=13.6\times 10^3\) and \(\rho_w=10^3\), the ratio of heights is exactly the inverse ratio of densities, i.e. 13.6.
Frequently Asked Questions - Pressure Pascal
What is the main concept covered in Pressure Pascal?
In NCERT Class 11 Physics Chapter 9 (Mechanical Properties of Fluids), "Pressure Pascal" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Pressure Pascal useful in real-life applications?
Real-life applications of Pressure Pascal from NCERT Class 11 Physics Chapter 9 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Pressure Pascal?
Key formulas in Pressure Pascal (NCERT Class 11 Physics Chapter 9 Mechanical Properties of Fluids) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 9?
NCERT Class 11 Physics Chapter 9 (Mechanical Properties of Fluids) is structured so each part builds on the previous one. Pressure Pascal connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Pressure Pascal?
CBSE board questions from Pressure Pascal typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Pressure Pascal lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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