30 NEET previous-year questions on Mechanical Properties of Fluids, each with the correct answer and a step-by-step solution. Filter by topic and expand any question to see how to solve it.
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Solution
Gauge pressure (above atmospheric) at depth is .
500 N
5000 N
50000 N
50 N
Solution
.
0.4
0.6
0.8
0.2
Solution
Fraction submerged = . Fraction above = 1 − 0.6 = 0.4.
10 N
40 N
50 N
90 N
Solution
Apparent loss in weight = buoyant force = .
0.2 m/s
2 m/s
5 m/s
10 m/s
Solution
.
Solution
.
Stays zero
Increases continuously
Increases until it reaches terminal velocity, then stays constant
Oscillates
Solution
Initially gravity dominates, so the ball accelerates downward. As speed grows, viscous drag (proportional to speed) grows. When drag + buoyancy = gravity, the net force is zero — the ball continues at constant terminal velocity.
Solution
.
Solution
A soap bubble has two surfaces (inner and outer), so the excess pressure is . (For a single droplet with one surface, it would be .)
Solution
. Doubling halves .
0.78
0.89
0.92
1.0
Solution
Fraction submerged = .
Higher pressure
Lower pressure
Same pressure
Zero pressure
Solution
is constant. Where is larger, must be smaller (at the same height).
Bernoulli's equation
Hydraulic press
Buoyancy
Capillary rise
Solution
Pascal's law (pressure transmitted equally in a confined fluid) is the principle behind hydraulic press and hydraulic lift.
Conservation of energy
Conservation of momentum
Conservation of mass
Bernoulli's principle
Solution
comes from the fact that the same mass of fluid that enters one section must exit another in steady, incompressible flow.
N/m²
N/m
N·m
J
Solution
Surface tension = force per unit length, units N/m. Equivalently J/m² (energy per unit area) — same numerical value.
~5 m
~10 m
~20 m
~50 m
Solution
Need .
.
Solution
. Memorise the factor of 6π.
Solution
Bernoulli: .
. Positive (so ) since in the constriction.
Solution
Soap bubble has 2 surfaces. Initial area . Final area .
.
.
Solution
Half submerged → buoyant force = weight of fluid displaced by half-volume = .
Reading = real weight − buoyant force = .
Solution
1 atm . Roughly Pa.
Solution
. Ratio .
Solution
Continuity: . Bernoulli (horizontal): .
Water is incompressible
Surface tension minimises surface area
Air pressure is uniform
Density is constant
Solution
Surface tension acts to minimise the surface area for a given volume. The geometric shape with minimum surface area for a fixed volume is a sphere.
0.4
0.5
0.6
0.7
Solution
Apparent loss in water = 30 − 20 = 10 N → buoyant force in water.
Apparent loss in liquid = 30 − 24 = 6 N → buoyant force in liquid.
Relative density = = ratio of buoyant forces = 6/10 = 0.6.
Solution
. So entry : exit = 1 : 4 (entry is narrower!).
Wait — re-checking. Higher speed means smaller area. at entry, so entry has smaller area. → entry area is 1/4 of exit area. So ratio is 1 : 4.
Decreasing with distance
Equally and undiminished
Only along the direction of the applied force
Increases with depth
Solution
Pascal's law: pressure transmitted equally in all directions, undiminished.
Solution
Fraction submerged = .
.
Closest option: 925 kg/m³.
Solution
Surface tension = force/length. Force is , so .
Pascal's law
Archimedes' principle
Bernoulli's principle
Stokes' law
Solution
Bernoulli's principle: faster air over the curved upper surface of the wing means lower pressure, slower air below means higher pressure → net upward lift.
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