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1. Which of the following physical quantities is described by “Mechanical Properties of Fluids”?
ⓐ. Elasticity of solids
ⓑ. Behavior of gases at low pressure
ⓒ. Properties of liquids and gases under external forces
ⓓ. Heat conduction in metals
Correct Answer: Properties of liquids and gases under external forces
Explanation: The chapter “Mechanical Properties of Fluids” deals with how fluids (liquids and gases) behave when external forces like pressure, gravity, and shear are applied. Elasticity belongs to solids, gases at low pressure are studied in kinetic theory, and heat conduction is a thermal property, not a fluid property.
2. Fluids are defined as substances that:
ⓐ. Resist shear stress indefinitely
ⓑ. Cannot flow under any condition
ⓒ. Offer no resistance to compression
ⓓ. Flow when a shear stress is applied
Correct Answer: Flow when a shear stress is applied
Explanation: A fluid is a substance that deforms continuously (flows) under the influence of a shear stress, no matter how small. Solids resist shear stress, while fluids like liquids and gases cannot resist shear stress permanently.
3. Which of the following is NOT a mechanical property of fluids?
ⓐ. Pressure
ⓑ. Viscosity
ⓒ. Surface tension
ⓓ. Specific heat
Correct Answer: Specific heat
Explanation: Mechanical properties include pressure, viscosity, buoyancy, and surface tension as they are related to force and motion. Specific heat is a thermal property, relating to heat capacity rather than mechanical behavior.
4. The SI unit of pressure is:
ⓐ. Dyne/cm$^2$
ⓑ. Pascal (Pa)
ⓒ. Torr
ⓓ. Bar
Correct Answer: Pascal (Pa)
Explanation: Pressure is defined as force per unit area, $P = \frac{F}{A}$. In SI units, force is in newtons and area in m$^2$, so the SI unit of pressure is N/m$^2$, called Pascal (Pa). Other units like bar and torr are used in specific applications but are not SI units.
5. Which of the following laws explains why atmospheric pressure decreases with height?
ⓐ. Newton’s law of gravitation
ⓑ. Archimedes’ principle
ⓒ. Pascal’s law
ⓓ. Hydrostatic law
Correct Answer: Hydrostatic law
Explanation: Hydrostatic law states $\frac{dP}{dh} = -\rho g$, meaning pressure decreases with height in a fluid under gravity. This explains why atmospheric pressure reduces as altitude increases. Archimedes’ principle is about buoyancy, Pascal’s law relates pressure transmission, and Newton’s law explains gravity.
6. Density ($\rho$) of a fluid is defined as:
ⓐ. Mass per unit volume
ⓑ. Weight per unit volume
ⓒ. Mass per unit area
ⓓ. Volume per unit mass
Correct Answer: Mass per unit volume
Explanation: Density is given by $\rho = \frac{m}{V}$, where $m$ is the mass of the fluid and $V$ is its volume. Option B is weight density (specific weight), option C is incorrect, and option D is the reciprocal of density.
7. The pressure at a depth $h$ in a liquid of density $\rho$ is given by:
ⓐ. $P = \rho gh$
ⓑ. $P = \frac{gh}{\rho}$
ⓒ. $P = \frac{\rho}{gh}$
ⓓ. $P = gh$
Correct Answer: $P = \rho gh$
Explanation: Hydrostatic pressure at depth $h$ is derived from the hydrostatic law: $P = \rho g h$. This shows pressure increases linearly with depth. Other options are incorrect rearrangements and do not satisfy dimensional analysis.
8. Who formulated Pascal’s Law for pressure transmission in fluids?
ⓐ. Isaac Newton
ⓑ. Blaise Pascal
ⓒ. Robert Boyle
ⓓ. Evangelista Torricelli
Correct Answer: Blaise Pascal
Explanation: Blaise Pascal discovered that pressure applied to a confined fluid is transmitted equally in all directions. This is Pascal’s law, which underlies hydraulic systems. Boyle studied gas laws, Newton formulated mechanics and gravitation, and Torricelli invented the barometer.
9. Which instrument is used to measure atmospheric pressure?
ⓐ. Hydrometer
ⓑ. Barometer
ⓒ. Manometer
ⓓ. Thermometer
Correct Answer: Barometer
Explanation: A barometer measures atmospheric pressure, usually using a column of mercury. A manometer measures pressure differences, a hydrometer measures density of liquids, and a thermometer measures temperature.
10. If the density of mercury is $13.6 \times 10^3 \, \text{kg/m}^3$, what is the pressure at a depth of 0.5 m in mercury? (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. $6.67 \times 10^3 \, \text{Pa}$
ⓑ. $6.67 \times 10^4 \, \text{Pa}$
ⓒ. $3.34 \times 10^3 \, \text{Pa}$
ⓓ. $3.34 \times 10^4 \, \text{Pa}$
Correct Answer: $6.67 \times 10^4 \, \text{Pa}$
Explanation: Hydrostatic pressure is $P = \rho g h = (13.6 \times 10^3)(9.8)(0.5) = 6.67 \times 10^4 \, \text{Pa}$. Other options are incorrect due to missing factors of 10.
11. Why is the study of fluid mechanics important in physics and engineering?
ⓐ. It helps understand only solid mechanics
ⓑ. It explains the behavior of gases and liquids under forces
ⓒ. It deals only with thermal properties of matter
ⓓ. It studies chemical reactions of fluids
Correct Answer: It explains the behavior of gases and liquids under forces
Explanation: Fluid mechanics studies how fluids (liquids and gases) respond to forces like pressure, gravity, and shear stress. This knowledge is essential in designing pumps, dams, aircrafts, and medical devices. It is not restricted to solids or chemical properties, but focuses on mechanical behavior of fluids.
12. Which of the following is a direct application of fluid mechanics in the human body?
ⓐ. Growth of bones
ⓑ. Circulation of blood through arteries and veins
ⓒ. Transmission of nerve signals
ⓓ. Formation of DNA strands
Correct Answer: Circulation of blood through arteries and veins
Explanation: Blood is a fluid and its flow in the circulatory system obeys fluid mechanics principles, including pressure, viscosity, and laminar/turbulent flow. Other options like bones, DNA, and nerves are not fluid-mechanical processes.
13. Why do civil engineers need to study fluid mechanics?
ⓐ. To design circuits for electronics
ⓑ. To understand soil chemistry
ⓒ. To construct dams, bridges, and water supply systems
ⓓ. To analyze chemical bonds in structures
Correct Answer: To construct dams, bridges, and water supply systems
Explanation: Civil engineers must analyze water flow, pressure, and resistance to design hydraulic structures like dams, canals, and pipelines. Circuits and chemical bonds belong to other fields, not fluid mechanics.
14. Which branch of engineering applies fluid mechanics extensively in the design of aircrafts?
ⓐ. Civil engineering
ⓑ. Electrical engineering
ⓒ. Aerospace engineering
ⓓ. Chemical engineering
Correct Answer: Aerospace engineering
Explanation: Aerospace engineers use fluid mechanics to study airflow (aerodynamics) around wings, turbines, and jet engines. This ensures flight stability and efficiency. Civil engineers focus on structures, and chemical engineers on chemical processes.
15. In which of the following medical technologies is fluid mechanics most relevant?
ⓐ. MRI scanning
ⓑ. Blood pressure monitoring and IV drips
ⓒ. X-ray imaging
ⓓ. Genetic sequencing
Correct Answer: Blood pressure monitoring and IV drips
Explanation: Medical applications like blood pressure measurement, flow of IV fluids, and design of artificial heart valves are based on fluid mechanics. MRI and X-ray are based on electromagnetism, and genetic sequencing on molecular biology.
16. Which field of study uses both fluid mechanics and heat transfer together?
ⓐ. Thermofluids
ⓑ. Crystallography
ⓒ. Astronomy
ⓓ. Solid mechanics
Correct Answer: Thermofluids
Explanation: Thermofluids combines thermodynamics and fluid mechanics to study how fluids transfer both mass and heat. It is widely applied in engines, turbines, and refrigeration systems. Astronomy and crystallography do not directly focus on fluid mechanics in this sense.
17. How does fluid mechanics help in environmental studies?
ⓐ. Predicting motion of tectonic plates
ⓑ. Understanding rainfall patterns, ocean currents, and pollution spread
ⓒ. Determining hardness of minerals
ⓓ. Studying atomic structure
Correct Answer: Understanding rainfall patterns, ocean currents, and pollution spread
Explanation: Environmental scientists use fluid mechanics to analyze wind flow, ocean currents, groundwater flow, and pollutant transport in air and water. These models help predict climate and environmental impacts.
18. Which of the following is an example of fluid mechanics in everyday life?
ⓐ. Operation of a hydraulic lift in car workshops
ⓑ. Expansion of metals on heating
ⓒ. Magnetism in electric motors
ⓓ. Formation of shadows
Correct Answer: Operation of a hydraulic lift in car workshops
Explanation: Hydraulic lifts apply Pascal’s law, a fundamental principle of fluid mechanics, to lift heavy vehicles. Thermal expansion, magnetism, and optics are unrelated to fluid mechanics.
19. Why is fluid mechanics essential in chemical engineering?
ⓐ. To calculate boiling point of chemicals only
ⓑ. To understand and design fluid transport in reactors and pipelines
ⓒ. To make alloys of metals
ⓓ. To study atomic bonding of compounds
Correct Answer: To understand and design fluid transport in reactors and pipelines
Explanation: Chemical engineers deal with liquid and gas flows in reactors, separators, and pipelines. Fluid mechanics helps in calculating pressure drops, flow rates, and energy losses. Alloy formation and atomic bonding are outside its scope.
20. Which of the following summarizes the overall importance of studying fluid mechanics?
ⓐ. It is only theoretical and has no applications
ⓑ. It explains the properties of solids in stress conditions
ⓒ. It provides a foundation for applications in engineering, environment, and medicine
ⓓ. It is only used for research in pure physics
Correct Answer: It provides a foundation for applications in engineering, environment, and medicine
Explanation: Fluid mechanics is a multidisciplinary subject applied in engineering (dams, aircraft, pipelines), medicine (blood flow, IV systems), and environment (weather, pollution). It is not limited to theory or solids.
21. Which principle of fluid mechanics is applied in hydraulic brakes of automobiles?
ⓐ. Archimedes’ principle
ⓑ. Pascal’s law
ⓒ. Bernoulli’s principle
ⓓ. Stokes’ law
Correct Answer: Pascal’s law
Explanation: Hydraulic brakes work on Pascal’s law, which states that pressure applied at one point of an enclosed fluid is transmitted equally in all directions. This allows a small force applied on the brake pedal to generate a large force on the brake shoes.
22. The lift of an airplane wing is explained by:
ⓐ. Newton’s first law
ⓑ. Bernoulli’s principle
ⓒ. Pascal’s principle
ⓓ. Stokes’ law
Correct Answer: Bernoulli’s principle
Explanation: Airflow over the curved wing surface increases velocity, reducing pressure (by Bernoulli’s principle). The pressure difference between the upper and lower wing surfaces generates lift. Pascal’s law deals with static fluids, not moving airflow.
23. The working of a perfume spray is based on:
ⓐ. Archimedes’ principle
ⓑ. Bernoulli’s theorem
ⓒ. Newton’s law of cooling
ⓓ. Boyle’s law
Correct Answer: Bernoulli’s theorem
Explanation: In a perfume atomizer, fast air flow creates a low-pressure region, causing liquid perfume to rise and spray out. This is a direct application of Bernoulli’s theorem.
24. Why do ships made of steel float on water despite being denser than water?
ⓐ. Due to surface tension
ⓑ. Because steel resists gravity
ⓒ. Due to Archimedes’ principle of buoyancy
ⓓ. Because of Pascal’s law
Correct Answer: Due to Archimedes’ principle of buoyancy
Explanation: A ship floats because its hollow structure displaces a large volume of water. According to Archimedes’ principle, the buoyant force equals the weight of displaced fluid. This upward force balances the ship’s weight, enabling floating.
25. Which of the following fluid mechanics concepts is applied in the working of a carburetor in automobiles?
ⓐ. Boyle’s law
ⓑ. Pascal’s law
ⓒ. Bernoulli’s principle
ⓓ. Archimedes’ principle
Correct Answer: Bernoulli’s principle
Explanation: A carburetor mixes fuel with air. High-speed air in a narrow region (Venturi) creates low pressure, drawing fuel into the airflow. This relies on Bernoulli’s principle of pressure variation with velocity.
26. The flow of blood through arteries can be modeled using:
ⓐ. Ohm’s law analogy
ⓑ. Bernoulli’s equation and Poiseuille’s law
ⓒ. Archimedes’ principle
ⓓ. Hooke’s law
Correct Answer: Bernoulli’s equation and Poiseuille’s law
Explanation: Blood flow in arteries involves viscous laminar flow. Poiseuille’s law gives the volume flow rate:
$$
Q = \frac{\pi r^4 \Delta P}{8 \eta L}
$$
where $r$ is artery radius, $\eta$ viscosity, and $L$ length. Bernoulli’s equation also applies for pressure-velocity relations.
27. Which of the following explains why tall buildings use water tanks at the top for water supply?
ⓐ. Bernoulli’s principle
ⓑ. Hydrostatic pressure $P = \rho g h$
ⓒ. Stokes’ law
ⓓ. Surface tension law
Correct Answer: Hydrostatic pressure $P = \rho g h$
Explanation: The pressure in water pipes depends on height $h$. A water tank placed high provides sufficient pressure $P = \rho g h$ to ensure water reaches every floor without pumps.
28. The design of submarines is mainly based on:
ⓐ. Pascal’s principle
ⓑ. Archimedes’ principle and buoyancy control
ⓒ. Bernoulli’s equation
ⓓ. Newton’s law of motion
Correct Answer: Archimedes’ principle and buoyancy control
Explanation: Submarines adjust their buoyancy by changing water levels in ballast tanks. Archimedes’ principle explains how buoyant force changes depending on water displaced, enabling sinking or floating.
29. Which of the following applications is explained by surface tension in fluids?
ⓐ. Rising of kerosene in a wick lamp
ⓑ. Buoyancy of a ship
ⓒ. Hydraulic lifts
ⓓ. Bernoulli’s airflow in wings
Correct Answer: Rising of kerosene in a wick lamp
Explanation: Capillary action due to surface tension allows kerosene to rise in the fine pores of the wick against gravity. Buoyancy and hydraulic lifts are due to pressure principles, not surface tension.
30. The Venturi meter measures:
ⓐ. Density of liquids
ⓑ. Viscosity of fluids
ⓒ. Pressure difference and flow rate of fluid
ⓓ. Surface tension of liquids
Correct Answer: Pressure difference and flow rate of fluid
Explanation: The Venturi meter uses Bernoulli’s equation to measure fluid velocity and flow rate. Pressure difference is recorded between a wide and narrow section. Neither density nor surface tension are measured directly by it.
31. Which of the following best distinguishes liquids from gases?
ⓐ. Liquids have no mass, gases do
ⓑ. Liquids have definite volume, gases do not
ⓒ. Gases cannot be compressed, liquids can
ⓓ. Gases have definite shape, liquids do not
Correct Answer: Liquids have definite volume, gases do not
Explanation: Liquids have a definite volume but no definite shape, whereas gases have neither definite volume nor definite shape. Gases can be compressed easily due to large intermolecular spaces, unlike liquids.
32. Which property explains why gases are highly compressible compared to liquids?
ⓐ. Strong intermolecular forces in gases
ⓑ. Large intermolecular spaces in gases
ⓒ. Weak gravitational effect in liquids
ⓓ. Higher density of gases
Correct Answer: Large intermolecular spaces in gases
Explanation: Gases have very weak intermolecular forces and large intermolecular distances, making them compressible. Liquids, being denser with smaller intermolecular spaces, are nearly incompressible.
33. Liquids transmit pressure equally in all directions because:
ⓐ. Liquids can expand indefinitely
ⓑ. Liquids resist shear stress
ⓒ. Liquids have fixed volume and weak intermolecular spaces
ⓓ. Liquids can be compressed like gases
Correct Answer: Liquids have fixed volume and weak intermolecular spaces
Explanation: Liquids are nearly incompressible but allow molecules to move, so they transmit pressure equally, as explained in Pascal’s law. Gases also transmit pressure but can expand freely.
34. Which of the following is true about gases but NOT about liquids?
ⓐ. Gases exert pressure on container walls
ⓑ. Gases expand to fill the entire container
ⓒ. Gases have viscosity
ⓓ. Gases have density
Correct Answer: Gases expand to fill the entire container
Explanation: A liquid takes the shape of its container but does not fill it entirely. A gas expands freely to occupy the whole container, a major distinction. Both exert pressure and have viscosity and density.
35. Which equation describes the behavior of gases but not liquids?
ⓐ. $P = \rho g h$
ⓑ. $F = ma$
ⓒ. $PV = nRT$
ⓓ. $Q = \frac{\pi r^4 \Delta P}{8 \eta L}$
Correct Answer: $PV = nRT$
Explanation: The ideal gas equation applies to gases, relating pressure, volume, and temperature. Liquids are not described by this equation. Hydrostatic pressure and Poiseuille’s law apply to liquids.
36. Liquids differ from gases in their response to shear stress because:
ⓐ. Liquids resist shear stress temporarily, gases do not resist at all
ⓑ. Gases are stronger than liquids
ⓒ. Liquids flow faster than gases
ⓓ. Liquids have lower density than gases
Correct Answer: Liquids resist shear stress temporarily, gases do not resist at all
Explanation: Liquids offer resistance to shear stress for a short time before flowing, while gases immediately deform under shear stress. This difference affects their viscosity and flow behavior.
37. Which of the following is almost incompressible?
ⓐ. Oxygen gas
ⓑ. Steam
ⓒ. Water
ⓓ. Hydrogen gas
Correct Answer: Water
Explanation: Liquids like water have very little compressibility due to tightly packed molecules. Gases such as oxygen and hydrogen are highly compressible, and steam (gaseous water) behaves like gas.
38. The density of gases is generally much lower than liquids because:
ⓐ. Gases have negligible mass
ⓑ. Gas molecules have larger intermolecular distances
ⓒ. Liquids have no intermolecular forces
ⓓ. Liquids contain dissolved gases
Correct Answer: Gas molecules have larger intermolecular distances
Explanation: In gases, molecules are far apart, leading to lower density. Liquids have closer molecular packing, making them denser. Gases do have mass, so option A is incorrect.
39. Which statement is correct about both liquids and gases?
ⓐ. Both have definite shape
ⓑ. Both resist compression strongly
ⓒ. Both can flow and are called fluids
ⓓ. Both follow the ideal gas law
Correct Answer: Both can flow and are called fluids
Explanation: Liquids and gases are collectively called fluids because they flow under shear stress. Liquids have definite volume, gases do not. Gases are compressible, while liquids are nearly incompressible.
40. Why does a gas exert uniform pressure in all directions inside a container, unlike a liquid?
ⓐ. Gas molecules are lighter
ⓑ. Gas molecules move randomly in all directions
ⓒ. Gases have definite volume
ⓓ. Gases are less viscous
Correct Answer: Gas molecules move randomly in all directions
Explanation: Due to random motion, gas molecules collide with container walls equally in all directions, creating uniform pressure. Liquids exert hydrostatic pressure, which increases with depth but is not uniform everywhere.
41. Pressure is defined as:
ⓐ. Force × Area
ⓑ. Force ÷ Area
ⓒ. Mass ÷ Area
ⓓ. Weight × Area
Correct Answer: Force ÷ Area
Explanation: Pressure is defined as the force applied per unit area, $P = \frac{F}{A}$. If a force is distributed over a larger area, the pressure decreases. The other options are dimensionally incorrect.
42. The SI unit of pressure is:
ⓐ. Dyne/cm$^2$
ⓑ. Pascal (Pa)
ⓒ. Atmosphere (atm)
ⓓ. Torr
Correct Answer: Pascal (Pa)
Explanation: In SI units, pressure is measured in pascals (Pa), where $1 \, \text{Pa} = 1 \, \text{N/m}^2$. Atmosphere, torr, and dyne/cm$^2$ are other units but not SI units.
43. The dimensional formula of pressure is:
ⓐ. $[M^1 L^1 T^{-2}]$
ⓑ. $[M^0 L^1 T^{-2}]$
ⓒ. $[M^1 L^{-1} T^{-2}]$
ⓓ. $[M^1 L^2 T^{-2}]$
Correct Answer: $[M^1 L^{-1} T^{-2}]$
Explanation: Since $P = \frac{F}{A}$ and force $F = ma = [M L T^{-2}]$, dividing by area $[L^2]$, we get $[M^1 L^{-1} T^{-2}]$.
44. Which of the following correctly represents 1 atmosphere in pascals?
ⓐ. $1.013 \times 10^3 \, \text{Pa}$
ⓑ. $1.013 \times 10^4 \, \text{Pa}$
ⓒ. $1.013 \times 10^5 \, \text{Pa}$
ⓓ. $1.013 \times 10^6 \, \text{Pa}$
Correct Answer: $1.013 \times 10^5 \, \text{Pa}$
Explanation: Standard atmospheric pressure is $1 \, \text{atm} = 760 \, \text{mm Hg} = 1.013 \times 10^5 \, \text{Pa}$.
45. A force of 50 N acts on a surface of area 0.25 m$^2$. What is the pressure exerted?
ⓐ. 100 Pa
ⓑ. 150 Pa
ⓒ. 200 Pa
ⓓ. 250 Pa
Correct Answer: 200 Pa
Explanation: $P = \frac{F}{A} = \frac{50}{0.25} = 200 \, \text{Pa}$. The calculation shows how pressure increases when force is concentrated over smaller area.
46. Which of the following is NOT a unit of pressure?
ⓐ. Pascal
ⓑ. Bar
ⓒ. Newton
ⓓ. Torr
Correct Answer: Newton
Explanation: Newton is a unit of force, not pressure. Pressure units include Pascal (N/m$^2$), bar ($10^5 \, \text{Pa}$), and torr ($\approx 133.3 \, \text{Pa}$).
47. Pressure can also be expressed in terms of energy density because:
ⓐ. Pressure is energy per unit mass
ⓑ. Pressure is energy per unit length
ⓒ. Pressure is energy per unit volume
ⓓ. Pressure is energy per unit time
Correct Answer: Pressure is energy per unit volume
Explanation: Work done by pressure is $W = P \Delta V$. Thus, pressure is equivalent to energy stored per unit volume, linking mechanics with thermodynamics.
48. Which conversion is correct?
ⓐ. $1 \, \text{bar} = 10^6 \, \text{Pa}$
ⓑ. $1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa}$
ⓒ. $1 \, \text{torr} = 100 \, \text{Pa}$
ⓓ. $1 \, \text{Pa} = 1 \, \text{dyne/cm}^2$
Correct Answer: $1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa}$
Explanation: $1 \, \text{bar} = 10^5 \, \text{Pa}$, $1 \, \text{torr} \approx 133.3 \, \text{Pa}$, and $1 \, \text{Pa} = 10 \, \text{dyne/cm}^2$.
49. Which instrument is based on the principle of pressure measurement?
ⓐ. Thermometer
ⓑ. Barometer
ⓒ. Hygrometer
ⓓ. Calorimeter
Correct Answer: Barometer
Explanation: Barometer measures atmospheric pressure using the height of a mercury column. Thermometer measures temperature, hygrometer measures humidity, and calorimeter measures heat.
50. Pressure at a point in a liquid depends on:
ⓐ. Area of the liquid surface only
ⓑ. Depth, density of the liquid, and gravitational acceleration
ⓒ. Shape of the container
ⓓ. Volume of liquid only
Correct Answer: Depth, density of the liquid, and gravitational acceleration
Explanation: Hydrostatic pressure is given by $P = \rho g h$. It depends on liquid density ($\rho$), depth ($h$), and gravity ($g$), but is independent of container shape or liquid volume.
51. Atmospheric pressure at sea level is approximately equal to:
ⓐ. $1.013 \times 10^3 \, \text{Pa}$
ⓑ. $1.013 \times 10^4 \, \text{Pa}$
ⓒ. $1.013 \times 10^5 \, \text{Pa}$
ⓓ. $1.013 \times 10^6 \, \text{Pa}$
Correct Answer: $1.013 \times 10^5 \, \text{Pa}$
Explanation: Standard atmospheric pressure at sea level is defined as 1 atmosphere (atm), which equals $1.013 \times 10^5 \, \text{Pa}$ or 760 mm Hg. Other values are either too small or too large.
52. Who invented the mercury barometer to measure atmospheric pressure?
ⓐ. Isaac Newton
ⓑ. Blaise Pascal
ⓒ. Evangelista Torricelli
ⓓ. Robert Boyle
Correct Answer: Evangelista Torricelli
Explanation: Evangelista Torricelli (1643) invented the mercury barometer. He showed that atmospheric pressure supports a mercury column about 76 cm high, leading to the unit “torr” (1 mm Hg).
53. The height of a mercury column in a barometer at sea level is about:
ⓐ. 10 cm
ⓑ. 76 cm
ⓒ. 1 m
ⓓ. 5 m
Correct Answer: 76 cm
Explanation: At standard atmospheric pressure, the mercury column rises to 76 cm. This corresponds to $1.013 \times 10^5 \, \text{Pa}$. Water would need a column about 10.3 m for the same pressure due to its lower density.
54. Atmospheric pressure decreases with altitude because:
ⓐ. Gravity becomes zero at higher altitude
ⓑ. Density of air decreases with height
ⓒ. Sunlight reduces air pressure
ⓓ. Winds balance pressure at high altitude
Correct Answer: Density of air decreases with height
Explanation: Atmospheric pressure results from the weight of the air column above. As height increases, the density and weight of air above decrease, leading to lower pressure. Gravity decreases only slightly, not enough to explain this effect.
55. Which unit is commonly used in meteorology to express atmospheric pressure?
ⓐ. Pascal (Pa)
ⓑ. Bar
ⓒ. Torr
ⓓ. Millibar (mb)
Correct Answer: Millibar (mb)
Explanation: Meteorologists commonly use millibars: $1 \, \text{mb} = 100 \, \text{Pa}$. Standard atmospheric pressure is 1013 mb. Although Pa is SI, mb is convenient for weather systems.
56. If the atmospheric pressure is 760 mm Hg, what is the corresponding pressure in pascals? (Take $ \rho_{Hg} = 13.6 \times 10^3 \, \text{kg/m}^3$, $ g = 9.8 \, \text{m/s}^2$)
ⓐ. $1.01 \times 10^4 \, \text{Pa}$
ⓑ. $1.01 \times 10^5 \, \text{Pa}$
ⓒ. $1.01 \times 10^6 \, \text{Pa}$
ⓓ. $1.01 \times 10^7 \, \text{Pa}$
Correct Answer: $1.01 \times 10^5 \, \text{Pa}$
Explanation: Pressure $P = \rho g h = (13.6 \times 10^3)(9.8)(0.76) \approx 1.01 \times 10^5 \, \text{Pa}$. This matches 1 atm.
57. Why can’t water be used in barometers instead of mercury?
ⓐ. Water evaporates quickly
ⓑ. Water is too dense
ⓒ. Water is colorless
ⓓ. A very tall column would be required
Correct Answer: A very tall column would be required
Explanation: Pressure $P = \rho g h$. Since water has much lower density than mercury, the height $h$ needed would be around 10.3 m, which is impractical. Mercury’s high density allows a shorter 76 cm column.
58. What will happen to the height of the mercury column in a barometer if taken to the top of a mountain?
ⓐ. It will rise
ⓑ. It will fall
ⓒ. It will remain constant
ⓓ. It will oscillate continuously
Correct Answer: It will fall
Explanation: At higher altitudes, atmospheric pressure decreases due to less air above. Therefore, the mercury column supported by atmospheric pressure is lower than 76 cm.
59. The instrument used in aircrafts to measure altitude based on atmospheric pressure is:
ⓐ. Hydrometer
ⓑ. Altimeter
ⓒ. Thermometer
ⓓ. Manometer
Correct Answer: Altimeter
Explanation: An altimeter is a modified barometer calibrated to display altitude instead of pressure. Since pressure decreases with height, altimeters give altitude readings.
60. A barometer reads 74 cm of Hg in a city. What is the atmospheric pressure in pascals? (Take $ \rho_{Hg} = 13.6 \times 10^3 \, \text{kg/m}^3$, $ g = 9.8 \, \text{m/s}^2$)
ⓐ. $9.85 \times 10^4 \, \text{Pa}$
ⓑ. $1.01 \times 10^5 \, \text{Pa}$
ⓒ. $7.40 \times 10^3 \, \text{Pa}$
ⓓ. $1.36 \times 10^6 \, \text{Pa}$
Correct Answer: $9.85 \times 10^4 \, \text{Pa}$
Explanation: Pressure $P = \rho g h = (13.6 \times 10^3)(9.8)(0.74) \approx 9.85 \times 10^4 \, \text{Pa}$. This is slightly less than 1 atm, consistent with a lower reading than 76 cm Hg.
61. Which of the following devices is used to measure the pressure of a gas in a closed container?
ⓐ. Barometer
ⓑ. Manometer
ⓒ. Altimeter
ⓓ. Thermometer
Correct Answer: Manometer
Explanation: A manometer measures the pressure of a gas inside a container by comparing it with atmospheric pressure. A barometer measures only atmospheric pressure, an altimeter measures altitude, and a thermometer measures temperature.
62. A simple U-tube manometer contains mercury and is connected to a gas container. If the mercury level difference between two arms is 20 cm and the atmospheric pressure is 76 cm Hg, then the gas pressure is:
ⓐ. 56 cm Hg
ⓑ. 76 cm Hg
ⓒ. 96 cm Hg
ⓓ. 120 cm Hg
Correct Answer: 96 cm Hg
Explanation: Gas pressure = atmospheric pressure + manometer height difference = $76 + 20 = 96 \, \text{cm Hg}$.
63. Why is mercury commonly used in barometers and manometers instead of water?
ⓐ. Mercury has low density
ⓑ. Mercury is opaque and visible
ⓒ. Mercury has high density and low vapor pressure
ⓓ. Mercury evaporates quickly
Correct Answer: Mercury has high density and low vapor pressure
Explanation: Due to its high density ($13.6 \times 10^3 \, \text{kg/m}^3$), mercury requires only 76 cm to balance atmospheric pressure, unlike water (10.3 m). Its low vapor pressure prevents evaporation errors, making it ideal.
64. Which type of manometer is used for measuring very small pressure differences?
ⓐ. Inclined manometer
ⓑ. Digital manometer
ⓒ. Bourdon gauge
ⓓ. Aneroid barometer
Correct Answer: Inclined manometer
Explanation: An inclined manometer increases sensitivity by spreading small height differences over a longer scale, making it suitable for detecting small pressure differences.
65. The aneroid barometer differs from a mercury barometer because:
ⓐ. It uses liquid mercury
ⓑ. It uses a flexible sealed metal box
ⓒ. It requires water instead of mercury
ⓓ. It cannot measure pressure
Correct Answer: It uses a flexible sealed metal box
Explanation: An aneroid barometer contains a sealed metallic box that expands or contracts with pressure changes. It avoids mercury use and is portable compared to mercury barometers.
66. Which device is used in weather stations to continuously record atmospheric pressure over time?
ⓐ. Simple barometer
ⓑ. Aneroid barometer
ⓒ. Barograph
ⓓ. Manometer
Correct Answer: Barograph
Explanation: A barograph uses an aneroid capsule and a rotating drum with a pen to record pressure variations continuously. Simple barometers and manometers cannot record continuously.
67. In a U-tube manometer, if both ends are open to the atmosphere, the mercury levels in both arms:
ⓐ. Rise continuously
ⓑ. Fall continuously
ⓒ. Stay at the same level
ⓓ. Show random oscillations
Correct Answer: Stay at the same level
Explanation: When both arms are open to the atmosphere, the pressures on both sides are equal. Hence, the mercury levels remain the same.
68. Which instrument is used in aircraft cabins to regulate pressure by comparing internal and external air pressure?
ⓐ. Mercury barometer
ⓑ. Manometer
ⓒ. Pressure gauge
ⓓ. Altimeter
Correct Answer: Pressure gauge
Explanation: A pressure gauge monitors and regulates cabin pressure by comparing with external atmospheric pressure. Altimeters are used for altitude, not cabin regulation.
69. A U-tube manometer connected to a gas container shows the mercury level on the container side higher by 15 cm compared to the open side. The atmospheric pressure is 76 cm Hg. What is the gas pressure?
ⓐ. 61 cm Hg
ⓑ. 76 cm Hg
ⓒ. 91 cm Hg
ⓓ. 101 cm Hg
Correct Answer: 61 cm Hg
Explanation: Gas pressure = atmospheric pressure – difference = $76 – 15 = 61 \, \text{cm Hg}$. Since the mercury is higher on the container side, gas pressure is less than atmospheric pressure.
70. Which of the following is NOT a pressure-measuring device?
ⓐ. Manometer
ⓑ. Barometer
ⓒ. Thermocouple
ⓓ. Bourdon gauge
Correct Answer: Thermocouple
Explanation: A thermocouple measures temperature, not pressure. Pressure devices include manometers, barometers, and Bourdon gauges (used for high pressures in engineering systems).
71. The pressure at a depth $h$ in a liquid of density $\rho$ is given by:
ⓐ. $P = \frac{\rho}{gh}$
ⓑ. $P = \rho g h$
ⓒ. $P = \frac{gh}{\rho}$
ⓓ. $P = \frac{1}{\rho g h}$
Correct Answer: $P = \rho g h$
Explanation: Hydrostatic pressure at depth $h$ is derived from the hydrostatic law: $P = \rho g h$. It increases linearly with both depth and density. The other options are incorrect rearrangements and dimensionally inconsistent.
72. Which statement correctly explains why pressure increases with depth in a fluid?
ⓐ. Because gravitational force increases with depth
ⓑ. Because more liquid is above the point, adding more weight
ⓒ. Because density decreases with depth
ⓓ. Because viscosity increases with depth
Correct Answer: Because more liquid is above the point, adding more weight
Explanation: At greater depths, a larger column of liquid is above, exerting more weight per unit area. This increases pressure. Gravity is nearly constant, and density does not decrease in incompressible liquids.
73. At the same depth in a liquid, pressure is:
ⓐ. Higher at the sides than at the bottom
ⓑ. The same in all directions
ⓒ. Zero on vertical surfaces
ⓓ. Independent of density
Correct Answer: The same in all directions
Explanation: Pressure in a fluid at rest is isotropic—it acts equally in all directions at a given depth. This is why fluids exert pressure sideways on container walls as well as downward.
74. If a diver is 20 m below the surface of freshwater ($\rho = 1000 \, \text{kg/m}^3$), what is the pressure due to water at that depth? (Take $ g = 9.8 \, \text{m/s}^2$)
ⓐ. $1.96 \times 10^3 \, \text{Pa}$
ⓑ. $1.96 \times 10^4 \, \text{Pa}$
ⓒ. $1.96 \times 10^5 \, \text{Pa}$
ⓓ. $1.96 \times 10^6 \, \text{Pa}$
Correct Answer: $1.96 \times 10^5 \, \text{Pa}$
Explanation: $P = \rho g h = (1000)(9.8)(20) = 1.96 \times 10^5 \, \text{Pa}$. This is in addition to atmospheric pressure at the surface.
75. Which of the following is independent of the shape of the container?
ⓐ. Hydrostatic pressure at a given depth
ⓑ. Height of fluid
ⓒ. Volume of fluid
ⓓ. Mass of fluid
Correct Answer: Hydrostatic pressure at a given depth
Explanation: Hydrostatic pressure depends only on depth, density, and gravity, not on container shape. This is known as the hydrostatic paradox.
76. If pressure at 5 m depth in a liquid is $P_1$, then the pressure at 10 m depth is:
ⓐ. $2 P_1$
ⓑ. $\frac{1}{2} P_1$
ⓒ. $P_1 + \rho g$
ⓓ. Independent of depth
Correct Answer: $2 P_1$
Explanation: Since pressure varies linearly with depth, doubling the depth doubles the hydrostatic pressure. Thus, pressure at 10 m is twice that at 5 m, ignoring atmospheric contribution.
77. The pressure difference between two points in a fluid at depths $h_1$ and $h_2$ is:
ⓐ. $\Delta P = \rho g (h_1 + h_2)$
ⓑ. $\Delta P = \frac{\rho g}{h_1 – h_2}$
ⓒ. $\Delta P = \rho g (h_2 – h_1)$
ⓓ. $\Delta P = \rho g \frac{h_1}{h_2}$
Correct Answer: $\Delta P = \rho g (h_2 – h_1)$
Explanation: Pressure difference depends only on the vertical depth difference. Hence, $\Delta P = \rho g \Delta h$.
78. A swimmer experiences greater pressure at the bottom of a pool than near the surface because:
ⓐ. Temperature increases with depth
ⓑ. Gravity increases with depth
ⓒ. The water column above is greater at the bottom
ⓓ. Density decreases with depth
Correct Answer: The water column above is greater at the bottom
Explanation: The pressure is due to the weight of the water column above. At the bottom, the column height is larger, so pressure is greater.
79. A tank contains oil of density $800 \, \text{kg/m}^3$. If the depth of oil is 3 m, what pressure does it exert at the bottom? (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. $2.35 \times 10^3 \, \text{Pa}$
ⓑ. $2.35 \times 10^4 \, \text{Pa}$
ⓒ. $2.35 \times 10^5 \, \text{Pa}$
ⓓ. $2.35 \times 10^6 \, \text{Pa}$
Correct Answer: $2.35 \times 10^4 \, \text{Pa}$
Explanation: $P = \rho g h = (800)(9.8)(3) = 2.35 \times 10^4 \, \text{Pa}$.
80. Which law governs the pressure variation with depth in a static fluid?
ⓐ. Boyle’s law
ⓑ. Hydrostatic law
ⓒ. Pascal’s law
ⓓ. Bernoulli’s law
Correct Answer: Hydrostatic law
Explanation: Hydrostatic law states that in a static fluid under gravity, the pressure increases linearly with depth: $\frac{dP}{dh} = \rho g$. This forms the basis of hydrostatics.
81. A container holds water of density $1000 \, \text{kg/m}^3$ up to a height of 4 m. What is the pressure at the bottom of the container due to water? (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. $2.92 \times 10^3 \, \text{Pa}$
ⓑ. $3.92 \times 10^4 \, \text{Pa}$
ⓒ. $4.92 \times 10^4 \, \text{Pa}$
ⓓ. $5.92 \times 10^4 \, \text{Pa}$
Correct Answer: $3.92 \times 10^4 \, \text{Pa}$
Explanation: Using hydrostatic formula $P = \rho g h = (1000)(9.8)(4) = 3.92 \times 10^4 \, \text{Pa}$.
82. A tank has oil of density $800 \, \text{kg/m}^3$ up to a height of 5 m. Find the pressure at the bottom due to oil. (Take $g = 10 \, \text{m/s}^2$)
ⓐ. $2.0 \times 10^4 \, \text{Pa}$
ⓑ. $3.0 \times 10^4 \, \text{Pa}$
ⓒ. $4.0 \times 10^4 \, \text{Pa}$
ⓓ. $5.0 \times 10^4 \, \text{Pa}$
Correct Answer: $4.0 \times 10^4 \, \text{Pa}$
Explanation: $P = \rho g h = (800)(10)(5) = 4.0 \times 10^4 \, \text{Pa}$.
83. A diver is 25 m deep in a lake of water. Calculate the absolute pressure he experiences if atmospheric pressure is $1.01 \times 10^5 \, \text{Pa}$ and water density is $1000 \, \text{kg/m}^3$. (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. $1.21 \times 10^5 \, \text{Pa}$
ⓑ. $2.46 \times 10^5 \, \text{Pa}$
ⓒ. $4.46 \times 10^5 \, \text{Pa}$
ⓓ. $3.46 \times 10^5 \, \text{Pa}$
Correct Answer: $3.46 \times 10^5 \, \text{Pa}$
Explanation: Gauge pressure = $\rho g h = (1000)(9.8)(25) = 2.45 \times 10^5 \, \text{Pa}$. Absolute pressure = $1.01 \times 10^5 + 2.45 \times 10^5 = 3.46 \times 10^5 \, \text{Pa}$.
84. A cube of side 0.2 m is immersed fully in water. If water density is $1000 \, \text{kg/m}^3$, calculate the force acting on the top face at depth 2 m. (Take $g = 10 \, \text{m/s}^2$)
ⓐ. 200 N
ⓑ. 400 N
ⓒ. 800 N
ⓓ. 1000 N
Correct Answer: 800 N
Explanation: Pressure at depth $h = 2$ m, $P = \rho g h = (1000)(10)(2) = 2 \times 10^4 \, \text{Pa}$. Area of face = $(0.2)^2 = 0.04 \, \text{m}^2$. Force = $P \times A = 2 \times 10^4 \times 0.04 = 800 \, \text{N}$.
85. The pressure at the bottom of a swimming pool 3 m deep is: (Take $ \rho = 1000 \, \text{kg/m}^3, g = 9.8 \, \text{m/s}^2$)
ⓐ. $2.94 \times 10^3 \, \text{Pa}$
ⓑ. $2.94 \times 10^4 \, \text{Pa}$
ⓒ. $2.94 \times 10^5 \, \text{Pa}$
ⓓ. $2.94 \times 10^6 \, \text{Pa}$
Correct Answer: $2.94 \times 10^4 \, \text{Pa}$
Explanation: $P = \rho g h = (1000)(9.8)(3) = 2.94 \times 10^4 \, \text{Pa}$.
86. The gauge pressure at 50 m depth in the ocean ($\rho = 1025 \, \text{kg/m}^3$) is: (Take $ g = 9.8 \, \text{m/s}^2$)
ⓐ. $5.02 \times 10^4 \, \text{Pa}$
ⓑ. $2.51 \times 10^5 \, \text{Pa}$
ⓒ. $5.02 \times 10^5 \, \text{Pa}$
ⓓ. $1.01 \times 10^6 \, \text{Pa}$
Correct Answer: $5.02 \times 10^5 \, \text{Pa}$
Explanation: $P = \rho g h = (1025)(9.8)(50) \approx 5.02 \times 10^5 \, \text{Pa}$.
87. A tank contains two immiscible liquids, oil (density $ 800 \, \text{kg/m}^3$) of height 2 m over water (density $ 1000 \, \text{kg/m}^3$) of height 3 m. Find the pressure at the bottom. (Take $ g = 10 \, \text{m/s}^2$)
ⓐ. $2.6 \times 10^4 \, \text{Pa}$
ⓑ. $3.6 \times 10^4 \, \text{Pa}$
ⓒ. $4.6 \times 10^4 \, \text{Pa}$
ⓓ. $5.6 \times 10^4 \, \text{Pa}$
Correct Answer: $4.6 \times 10^4 \, \text{Pa}$
Explanation: Pressure due to oil = $(800)(10)(2) = 1.6 \times 10^4 \, \text{Pa}$. Pressure due to water = $(1000)(10)(3) = 3.0 \times 10^4 \, \text{Pa}$. Total = $4.6 \times 10^4 \, \text{Pa}$.
88. A pressure of 2 atm is equivalent to:
ⓐ. $1.01 \times 10^5 \, \text{Pa}$
ⓑ. $2.02 \times 10^5 \, \text{Pa}$
ⓒ. $3.03 \times 10^5 \, \text{Pa}$
ⓓ. $4.04 \times 10^5 \, \text{Pa}$
Correct Answer: $2.02 \times 10^5 \, \text{Pa}$
Explanation: $1 \, \text{atm} = 1.01 \times 10^5 \, \text{Pa}$. Therefore, $2 \, \text{atm} = 2.02 \times 10^5 \, \text{Pa}$.
89. A dam holds water of depth 50 m. Find the pressure at the base of the dam. (Take $ \rho = 1000 \, \text{kg/m}^3, g = 10 \, \text{m/s}^2$)
ⓐ. $2.5 \times 10^5 \, \text{Pa}$
ⓑ. $5.0 \times 10^5 \, \text{Pa}$
ⓒ. $7.5 \times 10^5 \, \text{Pa}$
ⓓ. $1.0 \times 10^6 \, \text{Pa}$
Correct Answer: $5.0 \times 10^5 \, \text{Pa}$
Explanation: Pressure at base = $\rho g h = (1000)(10)(50) = 5.0 \times 10^5 \, \text{Pa}$.
90. The absolute pressure at 10 m depth in water is: (Take $\rho = 1000 \, \text{kg/m}^3, g = 9.8 \, \text{m/s}^2, P_{atm} = 1.01 \times 10^5 \, \text{Pa}$)
ⓐ. $1.01 \times 10^5 \, \text{Pa}$
ⓑ. $1.99 \times 10^5 \, \text{Pa}$
ⓒ. $2.99 \times 10^5 \, \text{Pa}$
ⓓ. $3.99 \times 10^5 \, \text{Pa}$
Correct Answer: $1.99 \times 10^5 \, \text{Pa}$
Explanation: Gauge pressure = $\rho g h = (1000)(9.8)(10) = 9.8 \times 10^4 \, \text{Pa}$. Absolute pressure = $1.01 \times 10^5 + 9.8 \times 10^4 = 1.99 \times 10^5 \, \text{Pa}$.
91. A cylindrical tank of height 6 m is filled with water ($\rho = 1000 \, \text{kg/m}^3$). Find the pressure at the bottom of the tank. (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. $4.9 \times 10^4 \, \text{Pa}$
ⓑ. $5.9 \times 10^4 \, \text{Pa}$
ⓒ. $6.9 \times 10^4 \, \text{Pa}$
ⓓ. $7.9 \times 10^4 \, \text{Pa}$
Correct Answer: $5.9 \times 10^4 \, \text{Pa}$
Explanation: Hydrostatic pressure $P = \rho g h = (1000)(9.8)(6) = 5.88 \times 10^4 \, \text{Pa}$. Rounded value is $5.9 \times 10^4 \, \text{Pa}$.
92. A gas exerts a pressure of $2.5 \times 10^5 \, \text{Pa}$. If the force exerted is 500 N, calculate the area of the surface.
ⓐ. $1.0 \times 10^{-2} \, \text{m}^2$
ⓑ. $2.0 \times 10^{-2} \, \text{m}^2$
ⓒ. $3.0 \times 10^{-2} \, \text{m}^2$
ⓓ. $5.0 \times 10^{-2} \, \text{m}^2$
Correct Answer: $2.0 \times 10^{-2} \, \text{m}^2$
Explanation: $P = \frac{F}{A} \Rightarrow A = \frac{F}{P} = \frac{500}{2.5 \times 10^5} = 2.0 \times 10^{-2} \, \text{m}^2$.
93. A container has a liquid of density $1200 \, \text{kg/m}^3$ up to 2.5 m height. Find the pressure at the bottom due to the liquid. (Take $g = 10 \, \text{m/s}^2$)
ⓐ. $2.5 \times 10^4 \, \text{Pa}$
ⓑ. $3.0 \times 10^4 \, \text{Pa}$
ⓒ. $3.5 \times 10^4 \, \text{Pa}$
ⓓ. $4.0 \times 10^4 \, \text{Pa}$
Correct Answer: $3.0 \times 10^4 \, \text{Pa}$
Explanation: $P = \rho g h = (1200)(10)(2.5) = 3.0 \times 10^4 \, \text{Pa}$.
94. The pressure difference between two depths in water at 8 m and 12 m is: (Take $\rho = 1000 \, \text{kg/m}^3, g = 9.8 \, \text{m/s}^2$)
ⓐ. $1.96 \times 10^4 \, \text{Pa}$
ⓑ. $2.94 \times 10^4 \, \text{Pa}$
ⓒ. $3.92 \times 10^4 \, \text{Pa}$
ⓓ. $4.90 \times 10^4 \, \text{Pa}$
Correct Answer: $3.92 \times 10^4 \, \text{Pa}$
Explanation: Pressure difference = $\rho g \Delta h = (1000)(9.8)(4) = 3.92 \times 10^4 \, \text{Pa}$.
95. A solid block of area $0.2 \, \text{m}^2$ rests at the bottom of a tank 10 m deep filled with water. Find the total force acting on the top surface of the block. (Take $\rho = 1000 \, \text{kg/m}^3, g = 10 \, \text{m/s}^2$)
ⓐ. $2.0 \times 10^3 \, \text{N}$
ⓑ. $2.0 \times 10^4 \, \text{N}$
ⓒ. $2.0 \times 10^5 \, \text{N}$
ⓓ. $2.0 \times 10^6 \, \text{N}$
Correct Answer: $2.0 \times 10^4 \, \text{N}$
Explanation: Pressure at depth = $P = \rho g h = (1000)(10)(10) = 1.0 \times 10^5 \, \text{Pa}$. Force = $P \times A = (1.0 \times 10^5)(0.2) = 2.0 \times 10^4 \, \text{N}$.
96. An iron ball of radius 0.1 m is immersed at 15 m depth in water. Find the pressure acting on its surface. (Take $\rho = 1000 \, \text{kg/m}^3, g = 9.8 \, \text{m/s}^2$)
ⓐ. $9.8 \times 10^4 \, \text{Pa}$
ⓑ. $1.47 \times 10^5 \, \text{Pa}$
ⓒ. $1.96 \times 10^5 \, \text{Pa}$
ⓓ. $2.94 \times 10^5 \, \text{Pa}$
Correct Answer: $1.96 \times 10^5 \, \text{Pa}$
Explanation: Pressure = $\rho g h = (1000)(9.8)(20) = 1.96 \times 10^5 \, \text{Pa}$.
97. If atmospheric pressure is $1.01 \times 10^5 \, \text{Pa}$, what is the total pressure at 30 m depth in water? (Take $\rho = 1000 \, \text{kg/m}^3, g = 10 \, \text{m/s}^2$)
ⓐ. $2.01 \times 10^5 \, \text{Pa}$
ⓑ. $3.01 \times 10^5 \, \text{Pa}$
ⓒ. $4.01 \times 10^5 \, \text{Pa}$
ⓓ. $5.01 \times 10^5 \, \text{Pa}$
Correct Answer: $4.01 \times 10^5 \, \text{Pa}$
Explanation: Hydrostatic pressure = $\rho g h = (1000)(10)(30) = 3.0 \times 10^5 \, \text{Pa}$. Absolute pressure = $1.01 \times 10^5 + 3.0 \times 10^5 = 4.01 \times 10^5 \, \text{Pa}$.
98. A tank is filled with two liquids: kerosene ($ \rho = 800 \, \text{kg/m}^3$) up to 2 m and water ($ \rho = 1000 \, \text{kg/m}^3$) up to 3 m above kerosene. Calculate the pressure at the bottom. (Take $g = 10 \, \text{m/s}^2$)
ⓐ. $4.1 \times 10^4 \, \text{Pa}$
ⓑ. $5.2 \times 10^4 \, \text{Pa}$
ⓒ. $4.6 \times 10^4 \, \text{Pa}$
ⓓ. $7.6 \times 10^4 \, \text{Pa}$
Correct Answer: $4.6 \times 10^4 \, \text{Pa}$
Explanation: Pressure due to kerosene = $(800)(10)(2) = 1.6 \times 10^4 \, \text{Pa}$. Pressure due to water = $(1000)(10)(3) = 3.0 \times 10^4 \, \text{Pa}$. Total = $4.6 \times 10^4 \, \text{Pa}$.
99. A column of mercury of height 76 cm balances atmospheric pressure. If density of mercury = $13.6 \times 10^3 \, \text{kg/m}^3$ and $g = 9.8 \, \text{m/s}^2$, calculate the atmospheric pressure.
ⓐ. $0.76 \times 10^5 \, \text{Pa}$
ⓑ. $1.01 \times 10^5 \, \text{Pa}$
ⓒ. $1.36 \times 10^5 \, \text{Pa}$
ⓓ. $1.50 \times 10^5 \, \text{Pa}$
Correct Answer: $1.01 \times 10^5 \, \text{Pa}$
Explanation: $P = \rho g h = (13.6 \times 10^3)(9.8)(0.76) \approx 1.01 \times 10^5 \, \text{Pa}$.
100. A liquid exerts a pressure of $4 \times 10^4 \, \text{Pa}$ at 5 m depth. Find the density of the liquid. (Take $g = 10 \, \text{m/s}^2$)
ⓐ. $600 \, \text{kg/m}^3$
ⓑ. $700 \, \text{kg/m}^3$
ⓒ. $800 \, \text{kg/m}^3$
ⓓ. $900 \, \text{kg/m}^3$
Correct Answer: $800 \, \text{kg/m}^3$
Explanation: $P = \rho g h \Rightarrow \rho = \frac{P}{gh} = \frac{4 \times 10^4}{(10)(5)} = 800 \, \text{kg/m}^3$.
Welcome to Class 11 Physics MCQs – Chapter 10: Mechanical Properties of Fluids (Part 1). This page is a chapter-wise question bank for the NCERT/CBSE Class 11 Physics syllabus—built for quick revision and exam speed. Practice MCQs / objective questions / Physics quiz items with solutions and explanations, ideal for CBSE Boards, JEE Main, NEET, competitive exams, and Board exams.
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What you will learn & practice
- Introduction to Mechanical Properties of Fluids
- Pressure in fluids, hydrostatics, and Pascal’s law
- Effect of gravity on fluid pressure (depth dependence, hydrostatic pressure)
- Viscosity, Stokes’ law, and terminal velocity
- Streamline vs turbulent flow; Reynolds number and flow criteria
- Bernoulli’s principle and applications (flow speed, pressure, height)
- Surface tension, surface energy, and angle of contact
- Drops and bubbles (excess pressure) and capillary rise
- Detergents and surface tension: wetting & cleaning effects
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FAQs on Mechanical Properties of Fluids ▼
▸ What are Mechanical Properties of Fluids MCQs in Class 11 Physics?
These are multiple-choice questions from Chapter 10 of NCERT Class 11 Physics – Mechanical Properties of Fluids. They cover concepts like pressure, Pascal’s law, Archimedes’ principle, viscosity, surface tension, Bernoulli’s theorem, and streamline flow.
▸ How many MCQs are available in this chapter?
There are a total of 700 MCQs from Mechanical Properties of Fluids. They are divided into 7 sets of 100 questions each for easy and structured practice.
▸ Are these MCQs useful for NCERT/CBSE and state board exams?
Yes, these MCQs are designed from the NCERT/CBSE Class 11 syllabus and are equally useful for state board exams. They improve clarity of fundamental concepts and exam readiness.
▸ Are Mechanical Properties of Fluids MCQs important for JEE and NEET?
Yes, this chapter is highly important for JEE and NEET. Questions from Pascal’s law, Bernoulli’s principle, viscosity, and surface tension are frequently asked in competitive exams.
▸ Do these MCQs include correct answers and explanations?
Yes, every MCQ comes with the correct answer along with explanations wherever required. This ensures that students not only practice but also understand the logic behind each question.
▸ Which subtopics are covered in Mechanical Properties of Fluids MCQs?
These MCQs cover all key subtopics: pressure in fluids, Pascal’s law, Archimedes’ principle, viscosity and Poiseuille’s law, surface tension, capillarity, Bernoulli’s theorem, and applications in real-life systems.
▸ Who should practice Mechanical Properties of Fluids MCQs?
These MCQs are ideal for Class 11 students, CBSE and state board exam aspirants, and students preparing for JEE, NEET, NDA, UPSC, and other entrance tests involving Physics.
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Yes, solving these MCQs helps in quick revision, boosts memory retention, and improves exam performance by enhancing speed and accuracy under timed conditions.
▸ Do these MCQs cover both basic and advanced concepts?
Yes, the MCQs cover basics like pressure and density as well as advanced topics like Bernoulli’s theorem, Poiseuille’s law of viscosity, and applications of surface tension and capillarity.
▸ Why are the 700 MCQs divided into 7 parts?
The MCQs are divided into 7 sets of 100 questions each to make practice organized and less overwhelming, helping students progress step by step with better focus.
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