31 Year NEET Previous Questions: Mechanical Properties of Fluids Free MCQ

Curvature = 1 / ROC = |d²y/dx²| / (1 + (dy/dx)²)^3/2 ≈ |d²y/dx²| / (1 + 0)^3/2 = d²y/dx²
(dy/dx) ≈ tan θ ≈ 0 (small angle approximation)
Change in pressure, ΔP = S × curvature
ΔP = S × d²y/dx²
Also, ΔP = ρgy
⇒ ρgy = S × d²y/dx²
⇒ d²y/dx² = (ρg / S) y

Answer: (D) √(2T/(πρg))

For a long thin wire resting on the liquid surface, the maximum upward force provided by surface tension occurs from the two contact edges (one on each side). The total upward force due to surface tension is F_surface = 2T L.

The weight of the wire is its mass times g. Mass = density × volume = ρ × π r² L, so the weight is W = ρ π r² L g.

At the limiting (maximum) radius the upward surface-tension force just balances the weight:

2T L = ρ π r² L g

Cancel L and solve for r:

r² = 2T / (π ρ g)

r = √(2T/(π ρ g)), which corresponds to option (D).

To calculate the pressure experienced by the swimmer at a depth of 20 m below the water surface, we need to use the formula for the pressure at a depth in a fluid:
P = P₀ + ρgh
Where:

• P is the total pressure at depth,
• P₀ is the atmospheric pressure at the surface (1 atm),
• ρ is the density of the water,
• g is the acceleration due to gravity,
• h is the depth below the surface.

Given:

• ρ = 10³ kg/m³ (density of water),
• g = 10 m/s² (acceleration due to gravity),
• h = 20 m (depth),
• P₀ = 1 atm = 10⁵ Pa (atmospheric pressure).

Now, we calculate the pressure at the depth:
First, calculate the hydrostatic pressure due to the water column:
ρgh = (10³ kg/m³) × (10 m/s²) × (20 m) = 2 × 10⁵ Pa
Total pressure at the depth is the sum of atmospheric pressure and the hydrostatic pressure:
P = 1 atm + 2 × 10⁵ Pa
Since 1 atm = 10⁵ Pa, we have:
P = 10⁵ Pa + 2 × 10⁵ Pa = 3 × 10⁵ Pa
Now, convert the pressure to atmospheres:
P = 3 × 10⁵ Pa / 10⁵ Pa = 3 atm
Thus, the correct answer is: (c) 3 atm.

The excess force required to remove a thin flat circular disc from the surface of water can be calculated using the formula for the force due to surface tension:
Force (F) = 2πr × T
Where:

• r is the radius of the disc,
• T is the surface tension of the water.

Given:
r = 4.5 cm = 0.045 m,
T = 0.07 N/m.
Now, substituting the values into the formula:
F = 2π × 0.045 × 0.07
F ≈ 2 × 3.1416 × 0.045 × 0.07
F ≈ 0.0198 N
Converting to milli Newtons (mN):
F ≈ 19.8 mN
So, the correct answer is:
(d) 19.8 mN.

To solve this problem, we need to apply the principle of conservation of mechanical energy (also known as Bernoulli's theorem) and understand the relationship between velocity and kinetic energy in a fluid flow.

Bernoulli's Principle:

Bernoulli's equation states that for an ideal, incompressible fluid flowing through a non-uniform cross-sectional tube, the total mechanical energy (sum of pressure energy, kinetic energy, and potential energy) remains constant along a streamline.

The equation is given by:

P + ½ ρv² + ρgh = constant

Where:

• P is the pressure of the fluid,
• ρ is the density of the fluid,
• v is the velocity of the fluid,
• g is the acceleration due to gravity,
• h is the height of the fluid at a point (potential energy term).

In the given problem, K₁ and K₂ represent the kinetic energy per unit volume at points X and Y. The kinetic energy per unit volume is:
K = ½ ρv²

Now, we are given that the cross-section of the tube is non-uniform, which means the velocity of the fluid will change as it moves from one section to another.

Relationship Between Kinetic Energy and Velocity:
Since the tube is non-uniform, the velocity of the fluid will change based on the cross-sectional area. According to the continuity equation:
A₁v₁ = A₂v₂
Where:

• A₁ and A₂ are the cross-sectional areas at points X and Y,
• v₁ and v₂ are the velocities of the fluid at points X and Y.

If the cross-sectional area at point Y is smaller than at point X, the velocity of the fluid will increase at point Y (since the fluid must maintain the same flow rate). This implies that the kinetic energy at point Y (K2) will be higher than at point X (K1).

Therefore, based on the continuity equation and Bernoulli's principle, the correct relationship between the kinetic energy per unit volume at points X and Y is:

K₁ > K₂

This is because, at point X, the velocity is lower (due to a larger cross-sectional area), and at point Y, the velocity is higher (due to a smaller cross-sectional area), which leads to a greater kinetic energy at X.
Thus, the correct answer is: (c) K₁ > K₂.

B. Bernoulli's principle is correct.

Bernoulli's equation: P + 1/2 ρv² + ρgh = constant along a streamline.

For flow in a horizontal tube the ρgh terms at two points are equal and cancel, so differences in pressure are due to differences in the kinetic energy per unit volume (that is, the 1/2 ρv² term).

Continuity equation: A₁v₁ = A₂v₂. A reduction in area causes an increase in velocity at the throat.

Therefore the throat of the meter has a higher speed and a lower pressure than the wider section; the resulting pressure difference (measured between the two sections) is related to the flow rate. This combination of the continuity equation and Bernoulli's principle is the basis of the instrument.

Final answer: B.

To calculate the energy required to form a soap bubble, we need to consider the work done to increase the surface area of the bubble.
When a soap bubble is formed, it has two surfaces (inner and outer) exposed to the air. The energy required is related to the surface area of the bubble and the surface tension of the soap solution.

Formula for Energy:
The energy required to form a soap bubble is given by the formula:
Energy (E) = 4πr² × T
Where:

• r is the radius of the bubble,
• T is the surface tension of the soap solution.

Since a soap bubble has two surfaces, the total surface area exposed is 2 * 4πr². The formula for the energy required to form a soap bubble is:
E = 8πr² × T
Given:

• r = 2 cm = 0.02 m,
• T = 0.03 N/m (surface tension).

Now, substituting the given values into the formula:
E = 8π × (0.02)² × 0.03
E = 8π × 0.0004 × 0.03
E = 8 × 3.1416 × 0.000012
E ≈ 3.01 × 10^-4 J
Thus, the correct answer is: (d) 3.01 × 10^-4 J.

(a) The coefficient of viscosity is a scalar quantity - TRUE
The coefficient of viscosity (η) represents the internal friction of a fluid and has only magnitude, no direction.

(b) Surface tension is a scalar quantity - TRUE
Surface tension is measured as energy per unit area or force per unit length, and it has only magnitude, no direction.

(c) Pressure is a vector quantity - FALSE
Pressure is defined as force per unit area. While force is a vector, when we calculate pressure, we consider only the magnitude of the force acting perpendicular to the surface area. Pressure at a point acts equally in all directions (Pascal's principle) and is characterized only by its magnitude, making it a scalar quantity.

(d) Relative density is a scalar quantity - TRUE
Relative density (or specific gravity) is the ratio of the density of a substance to a reference substance (usually water). Being a ratio of two scalar quantities, it is also a scalar quantity.
Therefore, option (c) is the incorrect statement, making it the correct answer to this question.

To calculate the viscous drag on a metal sphere falling through a fluid, we use Stokes' Law, which is given by the formula:
F = 6πηrv
Where:

• F is the viscous drag force (in N),
• η is the dynamic viscosity of the fluid (in Pa-s),
• r is the radius of the sphere (in m),
• v is the velocity of the sphere (in m/s).

Given:

• Diameter of the sphere = 1 mm = 1 × 10^-3 m
• Radius, r = (1 × 10^-3 m) / 2 = 0.5 × 10^-3 m = 5 × 10^-4 m
• Viscosity, η = 0.8 Pa-s
• Velocity, v = 2 m/s

Calculation:
Substitute the given values into the formula:
F = 6π × 0.8 × (5 × 10^-4) × 2
F = 6 × 3.14 × 0.8 × 5 × 10^-4 × 2
F = 15.072 × 10^-3 N
Thus, the viscous drag is approximately 15 × 10^-3 N.
Final Answer: (a) 15 × 10⁻³ N

Excess pressure inside the bubble =
P = P₀ + 4T/R
as ‘R’ increases ‘P’ decreases

Initial speed of ball is zero and it finally attains terminal speed

Mass = M

Density of ball = d
Density of glycerine = d/2

Density of liquid, ρl = 760 kg /m³
Density of mercury, ρm = 13600 kg /m³
Height of liquid column in mercury barometer,
hm = 76 cm = 0.76 m
If height of liquid in liquid column be hl , then

A liquid does not wet the solid surface, if the angle of contact is obtuse i.e., θ > 90º.

Excess pressure = 4T/R, Gauge pressure

Rate of liquid flow

According to Bernoulli’s theorem,

= constant and Av = constant

If A is minimum, v is maximum, P is minimum.

Wetability of a surface by a liquid primarily depends on angle of contact between the surface and liquid.
If angle of contact is acute liquids wet the solid and vice-versa.