Mechanical Properties of Fluids - Practice Questions with Answers
68 free MCQs on Mechanical Properties of Fluids with worked answers and explanations. Pressure, buoyancy, Bernoulli equation, viscosity, and surface tension.
Below are 68 practice questions on Mechanical Properties of Fluids, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Mechanical Properties of Fluids notes.
Since the same volume of fluid must pass every cross-section per second (continuity), the fluid speeds up where the pipe narrows; Bernoulli's equation then says this faster-moving fluid has lower pressure - the principle behind a venturi meter and an aircraft wing's lift.
Easy - 20 questions
Q1.
Pressure at depth h in a fluid of density ρ is:
A P<sub>0</sub> + ρgh
B ρgh
C P<sub>0</sub> - ρgh
D ρg/h
Show answer & explanation
Answer: A. P<sub>0</sub> + ρgh
Why: Total pressure = atmospheric pressure P<sub>0</sub> plus the pressure due to the fluid column: P = P<sub>0</sub> + ρgh.
Q2.
Pascal's law states:
A Pressure applied to an enclosed fluid is transmitted equally in all directions
B Pressure in a fluid increases linearly with depth below the surface as frequently described
C The buoyant force equals the weight of fluid displaced by the body in most textbook accounts
D Fluid usually flows from a region of high pressure to low pressure during normal conditions
Show answer & explanation
Answer: A. Pressure applied to an enclosed fluid is transmitted equally in all directions
Why: Pascal's law: a change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of its container.
Q3.
According to Archimedes' principle, the buoyant force on an object equals:
A Weight of fluid displaced by the object
B Weight of the object
C Volume of fluid displaced times g
D Density of fluid times volume of object
Show answer & explanation
Answer: A. Weight of fluid displaced by the object
Why: Buoyant force = weight of fluid displaced = ρ_fluid × V<sub>submerged</sub> × g. This is Archimedes' principle.
Q4.
An object floats when:
A The buoyant force equals its weight
B Its density equals the fluid density
C Its volume exceeds the fluid volume
D The fluid pressure equals atmospheric pressure
Show answer & explanation
Answer: A. The buoyant force equals its weight
Why: Floating condition: F<sub>buoyancy</sub> = Weight. This means ρ_fluid × V<sub>submerged</sub> × g = m × g. For complete submersion: ρ_object = ρ_fluid.
Q5.
The equation of continuity A<sub>1</sub> v<sub>1</sub> = A<sub>2</sub> v<sub>2</sub> represents conservation of:
A Mass (for incompressible fluid)
B Kinetic plus potential energy along the streamline
C Linear momentum of the fluid element
D Static pressure across the pipe cross-section
Show answer & explanation
Answer: A. Mass (for incompressible fluid)
Why: The continuity equation follows from conservation of mass for incompressible (constant density) fluids. The volume flow rate Q = Av must be the same everywhere in a pipe.
Q6.
In a narrower section of a pipe, the fluid flows:
A Faster (by continuity equation)
B Slower
C At the same speed
D Depends on fluid density
Show answer & explanation
Answer: A. Faster (by continuity equation)
Why: A<sub>1</sub> v<sub>1</sub> = A<sub>2</sub> v<sub>2</sub>. Smaller cross-section A<sub>2</sub> requires larger velocity v<sub>2</sub> for the same volume flow rate. This is seen in nozzles and hoses.
Q7.
Bernoulli's equation states that for steady flow of an ideal fluid:
A P + ½ρv² + ρgh = constant along a streamline
B P = ρgh holds at every point regardless of flow speed
C P + ρv stays constant along the streamline
D P equals ½ρv² with no dependence on height
Show answer & explanation
Answer: A. P + ½ρv² + ρgh = constant along a streamline
Why: Bernoulli's equation: P + ½ρv² + ρgh = constant. It is an expression of energy conservation (work-energy theorem) for fluid flow.
Q8.
Torricelli's theorem gives the velocity of fluid efflux from a hole in a tank at depth h as:
A v = √(2gh)
B v = 2gh
C v = √(gh)
D v = gh
Show answer & explanation
Answer: A. v = √(2gh)
Why: Applying Bernoulli's equation between the top surface and the hole: v = √(2gh). This is identical to the speed of a freely falling object from height h.
Q9.
Surface tension is defined as:
A Force per unit length along the surface
B Force per unit area
C Energy per unit volume
D Pressure difference across a surface
Show answer & explanation
Answer: A. Force per unit length along the surface
Why: Surface tension T = F/L (force per unit length, or equivalently, energy per unit area). It acts along the surface of a liquid, minimizing surface area.
Q10.
In a hydraulic press with areas A<sub>1</sub> = 10 cm² and A<sub>2</sub> = 1000 cm², a force of 100 N on the small piston produces a force on the large piston of:
A 10,000 N
B 1 N
C 100 N
D 1000 N
Show answer & explanation
Answer: A. 10,000 N
Why: Pascal's law: F<sub>1</sub>/A<sub>1</sub> = F<sub>2</sub>/A<sub>2</sub> (pressure transmitted). F<sub>2</sub> = F<sub>1</sub> × A<sub>2</sub>/A<sub>1</sub> = 100 × 1000/10 = 10,000 N. This mechanical advantage is the basis of hydraulic machines.
Q11.
The excess pressure inside a soap bubble is:
A 4T/r (two surfaces: inner and outer)
B 2T/r, as for a single liquid drop with one surface
C T/r, treating the film as having only one surface
D 8T/r, double-counting both surfaces twice over
Show answer & explanation
Answer: A. 4T/r (two surfaces: inner and outer)
Why: A soap bubble has two surfaces (inside and outside). Each surface contributes 2T/r, giving total excess pressure = 4T/r. For a single liquid droplet, it is 2T/r.
Q12.
Viscosity of a liquid generally:
A Decreases with increasing temperature
B Increases with increasing temperature
C Is independent of temperature
D Equals surface tension
Show answer & explanation
Answer: A. Decreases with increasing temperature
Why: As temperature increases, liquid viscosity decreases (molecular bonds weaken). Gas viscosity, however, increases with temperature (more molecular collisions).
Q13.
The buoyant force on a submerged object depends on:
A The volume of the object (not its weight or density)
B The weight of the object in air before submersion
C The chemical composition or material of the object
D Only the geometric shape of the object, regardless of size
Show answer & explanation
Answer: A. The volume of the object (not its weight or density)
Why: Buoyant force = ρ_fluid × V<sub>submerged</sub> × g. It depends on the density of the fluid and the volume submerged, NOT on the object weight, material, or shape.
Q14.
Capillary rise of water in a glass tube is due to:
A Surface tension (adhesive forces between water and glass)
B Atmospheric pressure acting only inside the narrow tube
C Viscous drag dragging water along the tube walls
D Gravitational attraction pulling water upward into the tube
Show answer & explanation
Answer: A. Surface tension (adhesive forces between water and glass)
Why: Water wets glass (adhesive force > cohesive force). Surface tension causes the meniscus to be concave. The pressure difference across this curved surface pushes water up.
Q15.
The density of water is 1000 kg/m³. The buoyant force on a 1 litre submerged object is:
A 9.8 N
B 1 N
C 98 N
D 10 N
Show answer & explanation
Answer: A. 9.8 N
Why: F<sub>b</sub> = ρ V g = 1000 × 10⁻³ × 9.8 = 9.8 N. This is also the weight of 1 litre of water.
Q16.
Relative density (specific gravity) of a material is:
A Density of material / density of water
B Density of water / density of material
C Mass of material / mass of water
D Volume of material / volume of water
Show answer & explanation
Answer: A. Density of material / density of water
Why: Relative density = density of substance / density of water at 4°C (approximately 1000 kg/m³). It is dimensionless. A relative density greater than 1 means the material sinks in water.
Q17.
In streamline (laminar) flow, the fluid layers:
A Flow smoothly in parallel without mixing
B Mix chaotically as generally observed
C Flow turbulently in typical laboratory settings
D Create eddies and vortices under usual circumstances
Show answer & explanation
Answer: A. Flow smoothly in parallel without mixing
Why: In laminar flow, fluid flows in smooth, parallel layers. This occurs at low velocities (low Reynolds number). At high velocities, flow becomes turbulent.
Q18.
An airplane wing generates lift because:
A Air flows faster above the wing (lower pressure) than below it
B The wing's trailing edge pushes a downward jet of air with no pressure change
C The wing is tilted so gravity components add to lift
D Engine thrust is redirected vertically beneath the wing
Show answer & explanation
Answer: A. Air flows faster above the wing (lower pressure) than below it
Why: By Bernoulli's principle, faster-moving air above the wing has lower pressure than slower-moving air below. The pressure difference creates an upward lift force.
Q19.
The angle of contact for water on a glass surface is:
A Less than 90° (acute, water wets glass)
B Greater than 90°, as for mercury on glass
C Exactly 90°, where the meniscus is perfectly flat
D Exactly 0°, where the surface is completely non-wetting
Show answer & explanation
Answer: A. Less than 90° (acute, water wets glass)
Why: For water on glass, the contact angle is about 0° to 20° (acute). Water wets glass, so the meniscus is concave and capillary rise occurs. For mercury on glass, contact angle > 90° (capillary depression).
Q20.
Reynolds number determines:
A Whether flow is laminar or turbulent
B The absolute viscosity coefficient of the fluid
C The static pressure head inside a pipe
D The surface tension coefficient of the liquid
Show answer & explanation
Answer: A. Whether flow is laminar or turbulent
Why: Reynolds number Re = ρvd/η. Re < 2000: laminar flow. Re > 4000: turbulent flow. 2000-4000: transition region.
Medium - 20 questions
Q21.
Water flows in a pipe that narrows from diameter 4 cm to 2 cm. If water enters at 1 m/s, what is the speed in the narrow section?
An object of mass 10 kg and density 2000 kg/m³ is submerged in water. The net downward force is:
A 49 N
B 98 N
C 0 N
D 24.5 N
Show answer & explanation
Answer: A. 49 N
Why: V = m/ρ = 10/2000 = 5×10⁻³ m³. F<sub>b</sub> = 1000 × 5×10⁻³ × 10 = 50 N. Weight = 10 × 10 = 100 N. Net downward = 100 - 50 = 50 N... Hmm let me recalculate: Weight = mg = 10 × 9.8 = 98 N. F<sub>b</sub> = ρ_w V g = 1000 × (10/2000) × 9.8 = 49 N. Net = 98 - 49 = 49 N.
Q25.
A capillary tube of radius 0.1 mm is dipped in water (T = 0.07 N/m, contact angle = 0, ρ = 1000 kg/m³). Height of capillary rise is:
A 14.3 cm
B 1.43 cm
C 143 cm
D 0.143 cm
Show answer & explanation
Answer: A. 14.3 cm
Why: h = 2T cosθ/(ρgr) = 2 × 0.07 × 1/(1000 × 10 × 0.1×10⁻³) = 0.14/(1) = 0.14 m = 14 cm ≈ 14.3 cm.
Q26.
The velocity of efflux from a tank depends only on the depth of the hole below the water surface (not tank dimensions). This is called:
A Torricelli's theorem
B Bernoulli's principle
C Pascal's law
D Archimedes' principle
Show answer & explanation
Answer: A. Torricelli's theorem
Why: Torricelli's theorem: v = √(2gh). The efflux velocity is determined solely by the depth h of the hole below the free surface, regardless of tank shape or size.
Q27.
A steel ball (density 7800 kg/m³) falls in water. As it reaches terminal velocity, the viscous drag force equals:
A Weight - Buoyant force (net weight in fluid)
B The full weight of the ball measured in air
C Zero, since the ball is still accelerating at terminal velocity
D The buoyant force alone, ignoring the ball's actual weight
Show answer & explanation
Answer: A. Weight - Buoyant force (net weight in fluid)
Why: At terminal velocity, acceleration = 0. Net upward force = 0. Drag + Buoyancy = Weight. So Drag = Weight - Buoyancy (apparent weight in fluid).
Q28.
Two identical bubbles of soap merge. The radius of the resulting bubble (same temperature and pressure) is:
A r<sub>new</sub> = r × 2<sup>1/3</sup> ≈ 1.26r
B r<sub>new</sub> = 2r, as if the radii generally added together
C r<sub>new</sub> = r√2, as if the surface areas generally added
D r<sub>new</sub> = r/√2, shrinking instead of growing on merging
Show answer & explanation
Answer: A. r<sub>new</sub> = r × 2<sup>1/3</sup> ≈ 1.26r
Why: Volume is conserved: (4/3)π r<sub>new</sub>³ = 2 × (4/3)π r³. r<sub>new</sub> = r × 2<sup>1/3</sup> ≈ 1.26r.
Q29.
The Venturi meter principle uses Bernoulli's equation. If a fluid of density ρ passes through a constriction of area A<sub>2</sub> (< A<sub>1</sub>), the pressure drop ΔP is:
A ΔP = ½ρ v<sub>2</sub>²(1 - (A<sub>2</sub>/A<sub>1</sub>)²)
B ΔP = ρ v<sub>2</sub>² as frequently described
C ΔP = ρg(h<sub>1</sub>-h<sub>2</sub>) in most textbook accounts
D ΔP = ½ρ(v<sub>2</sub>-v<sub>1</sub>)² during normal conditions
Show answer & explanation
Answer: A. ΔP = ½ρ v<sub>2</sub>²(1 - (A<sub>2</sub>/A<sub>1</sub>)²)
Why: By continuity and Bernoulli: ΔP = ½ρ(v<sub>2</sub>² - v<sub>1</sub>²). Using continuity A<sub>1v</sub>_1 = A<sub>2v</sub>_2: ΔP = ½ρv<sub>2</sub>²(1-(A<sub>2</sub>/A<sub>1</sub>)²). This pressure drop can be measured to find flow rate.
Q30.
Terminal velocity of a sphere of radius r falling in a fluid of viscosity η (Stokes law) is:
In a dam, the force on the wall increases with depth because:
A Pressure increases with depth, so average pressure × area gives total force
B Water at every depth pushes on the wall with exactly equal pressure
C Temperature variation with depth is what raises the hydrostatic pressure
D The flow velocity of standing water increases with depth
Show answer & explanation
Answer: A. Pressure increases with depth, so average pressure × area gives total force
Why: Pressure P = ρgh increases linearly with depth h. Total force = ∫ P dA. For a wall of width w and height H: F = ρg(H/2) × wH = ρgwH²/2.
Q32.
A boat carrying iron cubes floats in a pond. If the cubes are thrown overboard and sink, the water level:
A Falls (iron displaced less water when submerged than when floating)
B Rises, since the total mass of boat plus cubes hasn't changed
C Stays exactly the same regardless of how the cubes are distributed
D Depends only on the density of iron relative to the boat's hull
Show answer & explanation
Answer: A. Falls (iron displaced less water when submerged than when floating)
Why: When iron is on the boat, it displaces water equal to its weight (dense iron displaces a lot). When submerged, it displaces only its own volume (much less). So water level falls.
Q33.
The dynamic pressure of a fluid flowing at velocity v is:
A ½ρv²
B ρv²
C ρv
D 2ρv²
Show answer & explanation
Answer: A. ½ρv²
Why: Dynamic pressure = ½ρv². In Bernoulli's equation: static pressure P + dynamic pressure + hydrostatic pressure = constant. The dynamic pressure represents kinetic energy per unit volume.
Q34.
Poiseuille's law for viscous flow through a tube states that flow rate Q is proportional to:
A r⁴/L (fourth power of radius divided by length)
B r²/L, as if flow rate scaled only with cross-sectional area
C r/L, a linear dependence on radius alone
D r³/L, an intermediate power between area and the true relation
Show answer & explanation
Answer: A. r⁴/L (fourth power of radius divided by length)
Why: Poiseuille's law: Q = πr⁴ΔP/(8ηL). Doubling the tube radius increases flow rate 16 times. This explains why even small arterial narrowing severely reduces blood flow.
Q35.
An ice cube floats in a glass of water. When it melts completely, the water level:
A Stays the same
B Rises
C Falls
D Depends on ice density
Show answer & explanation
Answer: A. Stays the same
Why: Floating ice displaces water equal to its weight. When ice melts, it produces exactly that same weight of water. The water level remains unchanged.
Q36.
The pressure difference between the inside and outside of a liquid droplet of radius r is:
A 2T/r
B 4T/r
C T/r
D T/2r
Show answer & explanation
Answer: A. 2T/r
Why: A liquid droplet has one surface (unlike a soap bubble with two). Excess pressure inside = 2T/r, where T is surface tension and r is the radius of the droplet.
Q37.
A swimmer underwater feels a greater pressure than at the surface. At 10 m depth in sea water (ρ = 1025 kg/m³), the pressure is approximately:
A 2 atm (1 atm atmospheric + 1 atm from water column)
B 10 atm, mistakenly equating 10 m of depth to 10 atm directly
C 1 atm, ignoring the added pressure of the overlying water column
D 5 atm, from incorrectly halving the depth-to-pressure conversion
Show answer & explanation
Answer: A. 2 atm (1 atm atmospheric + 1 atm from water column)
A gas bubble rises from the bottom of a lake (depth 10 m). When it reaches the surface, its volume:
A Doubles (pressure halves)
B Stays the same as generally observed
C Quadruples in typical laboratory settings
D Halves under usual circumstances
Show answer & explanation
Answer: A. Doubles (pressure halves)
Why: At 10 m depth, pressure ≈ 2 atm. At surface, 1 atm. By Boyle's law: PV = constant. V<sub>surface</sub> = 2 × V<sub>depth</sub>. Volume doubles.
Q39.
The work done to blow a soap bubble of radius R from zero is:
A 8πR² T
B 2πR² T
C 4πR² T
D 16πR² T
Show answer & explanation
Answer: A. 8πR² T
Why: Work = surface tension × increase in surface area = T × 2 × 4πR² = 8πR² T. The factor 2 is because a soap bubble has two surfaces.
Q40.
The Magnus effect (a spinning ball curves in flight) is due to:
A Pressure difference created by Bernoulli effect on two sides of spinning ball
B Gravity alone curving the ball's path sideways during flight according to most researchers
C Plain air resistance acting uniformly on a non-spinning ball in the majority of cases studied
D A centrifugal force acting outward on the spinning ball's surface as widely reported
Show answer & explanation
Answer: A. Pressure difference created by Bernoulli effect on two sides of spinning ball
Why: A spinning ball drags air faster on one side and slower on the other. By Bernoulli, faster air has lower pressure. The resulting pressure difference causes the ball to curve toward the lower-pressure side.
Hard - 28 questions
Q41.
A tank of large cross-section has a hole of area a at depth h. The range of the horizontal jet of water on the ground (if the hole is at height H from ground) is:
A x = 2√(h(H-h))
B x = √(2gH)
C x = √(gh)
D x = 2H
Show answer & explanation
Answer: A. x = 2√(h(H-h))
Why: Efflux speed v = √(2gh). Time to fall height (H-h): (H-h) = ½gt², t = √(2(H-h)/g). Range x = vt = √(2gh)×√(2(H-h)/g) = 2√(h(H-h)).
Q42.
Two holes are at depths h<sub>1</sub> and h<sub>2</sub> from the water surface of a tank. Their ranges on the ground are equal. The relationship is:
A h<sub>1</sub> + h<sub>2</sub> = H (total tank height)
B h<sub>1</sub> = h<sub>2</sub>, meaning the two holes are at the same depth
C h<sub>1</sub> × h<sub>2</sub> = H², an incorrect product relation with the tank height
D h<sub>1</sub> = 2h<sub>2</sub>, an arbitrary fixed ratio between the two depths
Show answer & explanation
Answer: A. h<sub>1</sub> + h<sub>2</sub> = H (total tank height)
Why: Range = 2√(h(H-h)). For equal ranges: h<sub>1</sub>(H-h<sub>1</sub>) = h<sub>2</sub>(H-h<sub>2</sub>). This holds when h<sub>1</sub> and h<sub>2</sub> are symmetric about H/2, i.e., h<sub>1</sub> + h<sub>2</sub> = H.
Q43.
A sphere of density ρ_s, radius r, falls in a liquid of density ρ_l and viscosity η. At terminal velocity, the coefficient of viscosity is:
A η = 2r²(ρ_s - ρ_l)g/9v<sub>t</sub>
B η = 9v<sub>t</sub>/(2r² g) in standard practice
C η = 6πrv_t under most conditions encountered
D η = 2r² v<sub>t</sub> as frequently observed in practice
Show answer & explanation
Answer: A. η = 2r²(ρ_s - ρ_l)g/9v<sub>t</sub>
Why: Stokes: v<sub>t</sub> = 2r²(ρ_s - ρ_l)g/(9η). Rearranging: η = 2r²(ρ_s - ρ_l)g/(9v<sub>t</sub>). This is how viscometry works experimentally.
Q44.
Two soap bubbles of radii r<sub>1</sub> and r<sub>2</sub> are connected by a tube. If r<sub>1</sub> > r<sub>2</sub>, what happens?
A Smaller bubble (r<sub>2</sub>) shrinks and larger (r<sub>1</sub>) grows
B The larger bubble shrinks while the smaller one grows instead
C Both bubbles settle to exactly the same final equilibrium radius
D Neither bubble's size changes once they are connected by the tube
Show answer & explanation
Answer: A. Smaller bubble (r<sub>2</sub>) shrinks and larger (r<sub>1</sub>) grows
Why: Excess pressure in a soap bubble = 4T/r. The smaller bubble (r<sub>2</sub>) has higher pressure than the larger one (r<sub>1</sub>). Air flows from smaller to larger. The smaller bubble shrinks.
Q45.
A siphon transfers liquid from a container to a lower level. The maximum height h the siphon can reach above the container level is:
A h = P<sub>atm</sub>/(ρg) ≈ 10.3 m for water
B h = ∞, since a siphon can theoretically rise to any height
C h = 5 m, a fixed value independent of atmospheric pressure
D h = 1 m, a fixed value independent of fluid density or pressure
Show answer & explanation
Answer: A. h = P<sub>atm</sub>/(ρg) ≈ 10.3 m for water
Why: The siphon works as long as pressure at the top does not go below zero (vacuum). Maximum height: P<sub>atm</sub> - ρgh = 0 → h = P<sub>atm</sub>/(ρg) = 10⁵/(1000×10) = 10 m for water.
Q46.
Blood flows in an artery of diameter 4 mm with speed 0.4 m/s. If the artery narrows to 2 mm diameter, the speed becomes:
The angle of contact θ of a liquid in a capillary tube determines the direction of capillary flow. For mercury on glass (θ ≈ 135°):
A Capillary depression occurs (mercury is pushed down)
B Capillary rise occurs, exactly as it does for water in a glass tube
C No capillary effect occurs for an obtuse angle of contact
D Mercury overflows out of the top of the capillary tube largely
Show answer & explanation
Answer: A. Capillary depression occurs (mercury is pushed down)
Why: When θ > 90°, cosθ < 0. Capillary height h = 2T cosθ/(ρgr) becomes negative. This means capillary depression - mercury is pushed down. Mercury meniscus is convex.
Q48.
A fluid undergoes rotational flow where velocity v = ωr at all points (rigid rotation). The pressure increases radially as:
Two pistons in a hydraulic system have areas 1 cm² and 50 cm². For a load of 500 N on the large piston, the required force on the small piston is 10 N. If the small piston moves down by 50 cm, the large piston moves up by:
A 1 cm
B 50 cm
C 100 cm
D 0.02 cm
Show answer & explanation
Answer: A. 1 cm
Why: Volume conservation: A<sub>1</sub> × d<sub>1</sub> = A<sub>2</sub> × d<sub>2</sub>. 1 × 50 = 50 × d<sub>2</sub>. d<sub>2</sub> = 1 cm. Energy is conserved: work in = work out (ideal case).
Q50.
The rate of flow through a capillary tube is measured by Poiseuille: Q = πr⁴ΔP/8ηL. If the tube radius is halved, the flow rate:
A Decreases by factor 16
B Decreases by factor 2
C Increases by factor 16
D Decreases by factor 4
Show answer & explanation
Answer: A. Decreases by factor 16
Why: Q ∝ r⁴. If r halves (r/2), Q = (r/2)⁴ × (original Q/r⁴) = Q/16. The flow rate decreases by factor 16.
Q51.
In Stokes' law, the drag force on a sphere of radius r moving at velocity v is F = 6πηrv. This gives terminal velocity ∝ r². If two spheres have radii in ratio 2:1, their terminal velocities are in ratio:
A 4:1
B 2:1
C 1:4
D 8:1
Show answer & explanation
Answer: A. 4:1
Why: v<sub>t</sub> = 2r²(ρ-σ)g/9η ∝ r². If r<sub>1</sub>:r<sub>2</sub> = 2:1, then v<sub>1</sub>:v<sub>2</sub> = 4:1.
Q52.
A hollow sphere and a solid sphere of the same external dimensions are placed in water. The hollow sphere floats while the solid sphere sinks. This is because:
A The hollow sphere has lower average density than water while solid sphere has higher density
B The hollow sphere is generally lighter in absolute weight than the solid one in many documented cases
C Water seeps into the hollow sphere's interior cavity through tiny pores according to conventional understanding
D The hollow sphere has a larger exposed surface area than the solid one in routine practice
Show answer & explanation
Answer: A. The hollow sphere has lower average density than water while solid sphere has higher density
Why: Floating depends on average density. Hollow sphere: total mass / total volume < ρ_water. Solid sphere: density of material > ρ_water. Ships float because their average density (steel hull + hollow interior with air) is less than water.
Q53.
The work done per unit area to create a new surface of liquid is the surface energy. For a soap film of area A, the total surface energy is:
A 2TA (factor 2 for two surfaces)
B TA, counting only a single surface of the soap film
C 4TA, double-counting both surfaces of the film twice over
D T/A, dividing the surface tension by the area instead of multiplying
Show answer & explanation
Answer: A. 2TA (factor 2 for two surfaces)
Why: A soap film has two surfaces (front and back). Surface energy = T × total area = T × 2A = 2TA.
Q54.
Bernoulli's equation breaks down for:
A Turbulent, viscous, or compressible flows
B Slow, steady laminar flows of an ideal incompressible fluid
C Flows occurring at a uniformly low absolute pressure
D Flows confined to perfectly horizontal pipes of constant area
Show answer & explanation
Answer: A. Turbulent, viscous, or compressible flows
Why: Bernoulli's equation applies to ideal fluid flow: inviscid, incompressible, steady, and irrotational (streamline). It fails for turbulent flow, viscous fluids, or compressible gases at high speeds.
Q55.
An object of density ρ = 0.6 g/cm³ floats in water. The fraction of its volume above water is:
In a horizontal Venturi meter, the throat (narrow section) has area A<sub>t</sub> = A/2 where A is pipe area. Fluid density ρ, velocity at pipe entrance v. The height difference Δh in the manometer (density ρ_m) is:
Dimensional analysis shows that surface tension has dimensions of:
A [MT⁻²] (force per length = energy per area)
B [MLT⁻²], the dimensional formula for force itself
C [ML⁻¹T⁻²], the dimensional formula for pressure or stress
D [ML²T⁻²], the dimensional formula for energy or work
Show answer & explanation
Answer: A. [MT⁻²] (force per length = energy per area)
Why: T = F/L = [MLT⁻²/L] = [MT⁻²]. Equivalently, energy/area = [ML²T⁻²/L²] = [MT⁻²]. Both confirm the same dimensions.
Q58.
The critical velocity above which flow becomes turbulent in a tube of diameter d is approximately (Reynolds number ≈ 2000):
A v<sub>c</sub> = 2000η/(ρd)
B v<sub>c</sub> = 2000ρd
C v<sub>c</sub> = η/(ρd)
D v<sub>c</sub> = ρd/η
Show answer & explanation
Answer: A. v<sub>c</sub> = 2000η/(ρd)
Why: Re = ρvd/η. For Re = 2000 (laminar-turbulent transition): v<sub>c</sub> = 2000η/(ρd). For water in a 1 cm pipe: v<sub>c</sub> ≈ 0.2 m/s.
Q59.
A barometer uses mercury (ρ = 13600 kg/m³). Atmospheric pressure 10⁵ Pa corresponds to a mercury column height of:
A 0.735 m (73.5 cm)
B 10 m according to standard textbooks
C 13.6 m in general practice
D 1 m as frequently described
Show answer & explanation
Answer: A. 0.735 m (73.5 cm)
Why: P = ρgh → h = P/(ρg) = 10⁵/(13600 × 10) = 10⁵/1.36×10⁵ ≈ 0.735 m = 73.5 cm ≈ 760 mmHg.
Q60.
A steel needle can float on water even though steel is denser than water. This is due to:
A Surface tension creating a curved water surface that supports the needle
B Ordinary buoyant force alone, as with any floating dense object in most textbook accounts
C The needle being hollow and filled with trapped air inside during normal conditions
D A pocket of air trapped directly underneath the solid needle as generally observed
Show answer & explanation
Answer: A. Surface tension creating a curved water surface that supports the needle
Why: When gently placed, the needle rests on the water surface. Surface tension creates a dent in the surface, and the vertical component of the surface tension forces supports the needle weight. If the surface tension is broken, the needle sinks.
Q61.
Water flows out of a small hole in a tank at a depth of 5 m below the surface (g = 10 m/s²). The efflux speed is:
A 5 m/s
B 10 m/s
C 20 m/s
D 7 m/s
Show answer & explanation
Answer: B. 10 m/s
Why: By Torricelli theorem, v = √(2gh) = √(2·10·5) = 10 m/s.
Q62.
Water flows through a pipe whose cross-sectional area changes from 4A to A. The ratio of speeds in the wide to the narrow section is:
A 4:1
B 1:4
C 2:1
D 1:2
Show answer & explanation
Answer: B. 1:4
Why: By continuity A₁v₁ = A₂v₂, so v<sub>wide</sub>/v<sub>narrow</sub> = A<sub>narrow</sub>/A<sub>wide</sub> = 1/4.
Q63.
A spherical drop falls at terminal velocity through a viscous fluid. If its radius is doubled, its terminal velocity becomes:
A 2×
B 4×
C 8×
D unchanged
Show answer & explanation
Answer: B. 4×
Why: Terminal velocity ∝ r², so doubling the radius multiplies it by 4.
Q64.
A soap bubble of radius 2 mm has surface tension 0.03 N/m. The excess pressure inside it is:
A 15 Pa
B 30 Pa
C 60 Pa
D 120 Pa
Show answer & explanation
Answer: C. 60 Pa
Why: For a soap bubble (two surfaces), excess pressure = 4T/r = 4·0.03/0.002 = 60 Pa.
Q65.
In a hydraulic press, the small piston has area 0.01 m² and the large piston 0.1 m². A force of 100 N on the small piston produces on the large piston a force of:
A 100 N
B 1000 N
C 10 N
D 10000 N
Show answer & explanation
Answer: B. 1000 N
Why: F<sub>large</sub> = F<sub>small</sub>·(A<sub>large</sub>/A<sub>small</sub>) = 100·(0.1/0.01) = 1000 N.
Q66.
The rate of laminar flow through a tube is proportional to the fourth power of its radius. If the radius is halved, the flow rate becomes:
A 1/2
B 1/4
C 1/8
D 1/16
Show answer & explanation
Answer: D. 1/16
Why: By Poiseuille law Q ∝ r⁴, so (1/2)⁴ = 1/16 of the original.
Q67.
An iceberg floats in sea water with density 900 kg/m³ for ice and 1000 kg/m³ for water. The fraction of its volume below the surface is: