Class 11 Physics (Physics Part-II) Chapter 9 Mechanical Properties Of Fluids

Key Concept: Basic Formula

c) $\frac{4Sla}{r}$
[Solution Description] The excess pressure inside a bubble with two interfaces is given by the formula $(Pi – Po) = \frac{4 Sla}{r}$. This represents the pressure difference between the inside and outside of the bubble, accounting for both liquid surfaces.

Key Concept: Basic Definition

d) Pascal

[Solution Description]

The SI unit of pressure is derived from its formula, which is force per unit area ($P = \frac{F}{A}$). The unit of force is Newton (N), and area is measured in square meters (m²). Therefore, the unit for pressure is Newton per square meter, which is called Pascal.

Hence, the SI unit of pressure is Pascal (Pa).

Key Concept: Terminal Velocity Concept

c) It moves with constant velocity

[Solution Description] When an object reaches terminal velocity, the net force on it becomes zero because the sum of the upward forces (viscous force and buoyant force) equals the downward gravitational force. As a result, the object falls with constant velocity.

Key Concept: Advanced Conceptual, Theoretical Analysis

c) $h_A = 2 h_B$

[Solution Description]
Using the formula for capillary rise $h = \frac{2S \cos \theta}{\rho g a}$, it’s clear that $h$ is directly proportional to $S$. Therefore, if Liquid A has twice the surface tension of Liquid B, its capillary rise ($h_A$) will be twice as high as that of Liquid B ($h_B$), given all other factors remain constant.

Key Concept: Basic Concept

c) Obtuse angle

[Solution Description] The angle of contact ($\theta$) for mercury on glass is high, which means it is an obtuse angle. This occurs because the solid-liquid interfacial tension ($S_{sl}$) is greater than the liquid-air interfacial tension ($S_{la}$), leading to droplet formation.

Key Concept: Surface Energy Interface, Molecular Interaction

c) The liquid will spread over the surface due to adhesion.

[Solution Description]

The strong adhesive forces between the liquid and the solid surface compared to the cohesive forces within the liquid molecules suggest that the liquid will spread on the surface. This occurs because the energy required to form new surfaces (surface energy) is counteracted by the gain in adhesion energy.

Key Concept: Pressure Calculation

d) 29.2 Pa

[Solution Description] Using the relation for pressure difference across a curved liquid-air interface:
$\Delta P = \frac{2S}{a}$
where $S = 0.073 \, \text{N/m}$ and $a = 0.005 \, \text{m}$. Substituting values:
$\Delta P = \frac{2 \times 0.073}{0.005} = \frac{0.146}{0.005} = 29.2 \, \text{Pa}$

Key Concept: Pressure Difference

a) $$2 S_{la}/r$$

[Solution Description]

The pressure difference $$(P_i – P_o)$$ for a spherical drop in equilibrium is given by the formula $$\frac{2 S_{la}}{r}$$, where $$S_{la}$$ is the surface tension and $$r$$ is the radius of the drop.

Key Concept: Equation of Continuity Application

d) The velocity triples.

[Solution Description] According to the equation of continuity for incompressible fluids, $Av = \text{constant}$, where $A$ is the cross-sectional area and $v$ is the fluid velocity. If the cross-sectional area $A$ is reduced to one-third, the velocity $v$ must increase to maintain the constant value. Let’s denote the initial area as $A_1$, the final area as $A_2 = \frac{1}{3}A_1$, the initial velocity as $v_1$, and the final velocity as $v_2$.
Using the continuity equation:
$A_1 v_1 = A_2 v_2$
Substituting the values:
$A_1 v_1 = \left(\frac{1}{3}A_1\right) v_2$
Dividing both sides by $A_1$:
$v_1 = \frac{1}{3} v_2$
This implies:
$v_2 = 3 v_1$
Therefore, the velocity increases three times.

Key Concept: Torricelli’s Law

c) It calculates the speed of efflux of fluid from an open tank

[Solution Description] Torricelli’s law states that the speed of efflux of a fluid from an open tank is equal to the square root of 2 times the acceleration due to gravity times the height of the fluid column. This can be related to the flow of air over an aerofoil.

Key Concept: Basic Definition

c) Pascal

[Solution Description]

The SI unit for any type of pressure, including gauge pressure, is the pascal.

Key Concept: Comparative Analysis

c) Both have the same speed of efflux

[Solution Description] According to Torricelli’s Law, the speed of efflux $v$ is given by $v = \sqrt{2gh}$ and depends only on the height $h$ of the liquid column above the opening, not on the cross-sectional area $A$. Therefore, both tanks will have the same speed of efflux.

Key Concept: Basic Concept

b) Force per unit length

[Solution Description] Surface tension is defined as the force per unit length or surface energy per unit area acting in the plane of the interface.

Key Concept: Theoretical Analysis, Multi-Concept Integration

c) $S_{sl} = 0.055 \, N/m$

[Solution Description] For an obtuse angle of contact, the condition is $S_{sl} > S_{la}$.
Using the relation:
$S_{la} \cos \theta + S_{sl} = S_{sa}$
Since the angle is obtuse, $\cos \theta < 0$, meaning: $S_{sl} > S_{la}$
This implies $S_{sl}$ must satisfy:
$S_{sl} > 0.05$
To maintain balance in the equation:
$S_{sl} > 0.035$
But since it must also be greater than $S_{la}$, options show possible ranges:

Key Concept: Basic Definition

b) Ability to flow

[Solution Description]

Solids have a fixed shape and do not flow like fluids. A fluid’s ability to flow is what distinguishes it from a solid. The fundamental characteristic of fluids is their ability to adapt to the shape of their container by flowing.

Key Concept: Continuity Equation

b) 8 m/s

[Solution Description] According to the equation of continuity for incompressible fluid flow, $A_1v_1 = A_2v_2$, where $A_1$ and $v_1$ are the area and velocity at point A, and $A_2$ and $v_2$ are the area and velocity at point B. Substituting the given values:
$3 \times 4 = 1.5 \times v_2$
Simplifying,
$v_2 = \frac{12}{1.5} = 8 \text{ m/s}$
Therefore, the velocity at point B is 8 m/s.

Key Concept: Streamline Concept

d) A curve whose tangent at any point is in the direction of the fluid velocity at that point

[Solution Description]

A streamline in fluid dynamics is defined as a curve whose tangent at any point is in the direction of the fluid velocity at that point. This means it represents the path that a massless particle will follow if placed in the flow.

Key Concept: Real-World Application

c) 8 m/s

[Solution Description]
To solve this, we first calculate the cross-sectional areas for both sections of the hose. The area $A$ is given by:
$A = \pi r^2$
The radius $r$ is half of the diameter.
For the initial section, diameter = 3 cm or $0.03$ m, so $r_1 = 0.015$ m:
$A_1 = \pi (0.015)^2 = \pi \times 0.000225 \approx 0.00070686 \text{ m}^2$
For the narrowed section, diameter = 1.5 cm or $0.015$ m, so $r_2 = 0.0075$ m:
$A_2 = \pi (0.0075)^2 = \pi \times 0.00005625 \approx 0.00017672 \text{ m}^2$
Using the equation of continuity:
$A_1v_1 = A_2v_2$
Substitute known values:
$0.00070686 \times 2 = 0.00017672 \times v_2$
Solve for $v_2$:
$v_2 = \frac{0.00070686 \times 2}{0.00017672} \approx 8 \text{ m/s}$

Key Concept: Cause and Effect

b) Increases ink absorption

[Solution Description]

Wetting agents reduce the surface tension of liquids, allowing them to spread more easily over surfaces. When added to ink, the reduced surface tension enhances its ability to penetrate the fibers of the blotting paper, leading to increased absorption. This is because the decreased surface tension diminishes the resistance against the liquid spreading, increasing its affinity for the surface.

Key Concept: Basic Concept

c) They reduce the surface tension of water

[Solution Description] Detergents primarily work by reducing surface tension, which allows liquids to spread more easily and penetrate surfaces for effective cleaning.

Key Concept: Critical Thinking, Real-world Problem

c) 0.5 m$^2$

[Solution Description]
Using the formula for force transmission in hydraulic systems:
$F_2 = \frac{A_2}{A_1} \times F_1$
Rearrange to solve for $A_2$:
$A_2 = \frac{F_2 \cdot A_1}{F_1}$
Substitute the known values:
$A_2 = \frac{5000 \, \text{N} \times 0.02 \, \text{m}^2}{200 \, \text{N}} = 0.5 \, \text{m}^2$
Therefore, to transmit sufficient force for effective braking, the area of the larger piston must be approximately 0.5 m$^2$.

Key Concept: Application of Continuity Equation

b) $8 \, \text{m/s}$

[Solution Description]
Using the equation of continuity, $A_1v_1 = A_2v_2$, where $A_1$ and $A_2$ are the cross-sectional areas of the wider and narrower sections respectively, and $v_1$ and $v_2$ are the velocities at these sections.
The area of a circle is given by $A = \pi r^2$.
For the wider section,
$A_1 = \pi \left(\frac{0.1}{2}\right)^2 = \frac{\pi \times 0.01}{4} = 0.00785 \, \text{m}^2$
For the narrower section,
$A_2 = \pi \left(\frac{0.05}{2}\right)^2 = \frac{\pi \times 0.0025}{4} = 0.001963 \, \text{m}^2$
Using $A_1v_1 = A_2v_2$:
$0.00785 \times 2 = 0.001963 \times v_2$
$v_2 = \frac{0.0157}{0.001963} = 8 \, \text{m/s}$
Therefore, the velocity in the narrower section is $8 \, \text{m/s}$.

Key Concept: Equation Application

b) 8 m/s

[Solution Description]
According to the equation of continuity, $A_1v_1 = A_2v_2$, where $A_1$ and $A_2$ are the cross-sectional areas at sections A and B, and $v_1$ and $v_2$ are the velocities at these sections.
The area $A_1$ can be calculated as:
$A_1 = \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4} \, \text{m}^2$
Similarly, $A_2$ is:
$A_2 = \pi \left(\frac{0.5}{2}\right)^2 = \frac{\pi}{16} \, \text{m}^2$
Using the continuity equation:
$\frac{\pi}{4} \cdot 2 = \frac{\pi}{16} \cdot v_2$
Solving for $v_2$:
$v_2 = \frac{8\pi}{\pi} = 8 \, \text{m/s}$
Hence, the velocity at section B is 8 m/s.

Key Concept: Pressure Units

d) $1.01 \times 10^5$ Pa
[Solution Description] The atmospheric pressure 1 atm is equivalent to $1.01 \times 10^5$ pascals (Pa).

Key Concept: Basic Concept

b) Spherical

[Solution Description] Liquid drops and bubbles tend to minimize their surface area due to surface tension, which results in a spherical shape. A sphere has the smallest surface area for a given volume, thus minimizing the surface energy.

Key Concept: Definition Recall

b) A substance with the ability to flow and no definite shape

[Solution Description] A fluid is a substance that can flow and takes the shape of its container, which includes both liquids and gases.

Key Concept: Real-world Application

a) Oil has weaker intermolecular forces compared to honey.

[Solution Description] Viscosity is influenced by intermolecular forces. Honey is more viscous than oil because it contains larger molecules with stronger intermolecular forces, leading to greater resistance to flow. Temperature also affects viscosity; at room temperature, honey’s viscosity remains higher due to these molecular interactions.

Key Concept: Basic Definition

d) Force per unit area

[Solution Description] Pressure is defined as the normal force acting per unit area. Mathematically, it can be expressed as $P = \frac{F}{A}$, where $F$ is the force and $A$ is the area.

Key Concept: Pressure Difference

b) 7.28 Pa

[Solution Description] The pressure difference $\Delta P$ across a spherical drop is given by the formula:
$\Delta P = \frac{2T}{r},$
where $T$ is the surface tension and $r$ is the radius of the droplet.
Given $T = 0.0728 \, \text{N/m}$ and $r = 0.02 \, \text{m}$,
$\Delta P = \frac{2 \times 0.0728}{0.02} = 7.28 \, \text{Pa}.$
Therefore, the pressure difference is $7.28 \, \text{Pa}$.

Key Concept: Basic Concept

b) Surface Tension

[Solution Description] Capillary rise is caused by surface tension and adhesion between the liquid and solid surfaces of the tube.

Key Concept: Fluid Properties

d) Ability to flow

[Solution Description] A key property that distinguishes fluids from solids is the ability to flow. Fluids do not have a definite shape and can take the shape of their container, whereas solids have a definite shape and high resistance to deformation.

Key Concept: Stokes’ Law Application

c) Halves

[Solution Description]
According to Stokes’ Law, the viscous force $F = 6\pi \eta av$, where $\eta$ is the viscosity, $a$ is the radius of the raindrop, and $v$ is the velocity.
If the radius $a$ is halved, then the new viscous force will be:
$F_{\text{new}} = 6\pi \eta \left(\frac{a}{2}\right) v = \frac{F}{2}$
So, the viscous force will halve.

Key Concept: Stokes’ Law

a) $F = 6\pi \eta av$

[Solution Description]

Stokes’ Law states that the force $F$ experienced by a sphere moving through a viscous fluid is given by $F = 6\pi \eta av$. Here, $\eta$ is the viscosity of the fluid, $a$ is the radius of the sphere, and $v$ is the velocity of the sphere.

Key Concept: Pascal’s Law

b) Same

[Solution Description]

Pascal’s Law explains that the pressure at any given point in a fluid is the same when measured horizontally, meaning that within an enclosed fluid at rest, a change in pressure applied at any point in the fluid is transmitted undiminished throughout the fluid. Thus, for two locations at the same height in a homogeneous fluid at rest, the pressure remains constant.

Key Concept: Cause and Effect

c) Pressure fluctuates

[Solution Description] As fluid velocity increases beyond the critical speed, the flow transitions into a turbulent state. This results in irregularities and fluctuations in both velocity and pressure. Therefore, the pressure does not remain constant but fluctuates due to chaotic eddies and vortices.

Key Concept: Conservation of Mass

c) 8 m/s

[Solution Description] Applying the equation of continuity:

$A_1 v_1 = A_2 v_2$

where $A_1 = 0.5$ m², $v_1 = 4$ m/s, and $A_2 = 0.25$ m². Solving for $v_2$:

$0.5 \times 4 = 0.25 \times v_2$

$2 = 0.25 \times v_2$

$v_2 = \frac{2}{0.25}$

$v_2 = 8 \text{ m/s}$

Key Concept: Bernoulli’s Principle

c) Equal pressure on all sides

[Solution Description] For a non-spinning ball, the streamlines are symmetric and the fluid velocity above and below the ball are equal. Hence, by Bernoulli’s principle, which states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure, the pressure will be equal at corresponding points because the velocities are the same.

Key Concept: Bernoulli’s Principle

d) By relating pressure and velocity

[Solution Description] Bernoulli’s principle states that an increase in the velocity of a fluid occurs simultaneously with a decrease in pressure. On an aircraft wing, air moves faster over the top surface than underneath, leading to reduced pressure above the wing. This pressure difference provides dynamic lift.

Key Concept: Volume Comparison

c) Gas

[Solution Description] Gases expand to fill any container they are placed in because their particles move freely and spread apart from each other. This characteristic allows them to occupy the entire volume available.

Key Concept: Application of Stokes’ Law

b) 0.00028 N

[Solution Description] Using Stokes’ Law, $F = 6\pi \eta av$, where $\eta = 0.05$ Pa s, $a = 0.002$ m, and $v = 0.15$ m/s:
$F = 6\pi \times 0.05 \times 0.002 \times 0.15$
$F = 6 \times 3.14159 \times 0.05 \times 0.002 \times 0.15$
$F = 0.0002827 \text{ N} \approx 0.00028 \text{ N}$
The correct option is therefore approximately 0.00028 N.

Key Concept: Capillary Action

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion correctly states that capillary rise is caused by surface tension, which acts along the liquid-air interface to pull the liquid up the tube. The reason also accurately explains this phenomenon by stating that surface tension results in a pressure difference across the curved interface, facilitating the upward movement against gravity. Therefore, both statements are true, and the reason provided is indeed the correct explanation for the assertion.

Key Concept: Viscosity and Flow

d) Laminar

[Solution Description]

High viscosity fluids tend to exhibit laminar flow. Laminar flow is characterized by smooth, orderly layers of fluid moving parallel to one another without mixing, which occurs when a fluid’s internal friction resists chaotic or turbulent motion.

Key Concept: Formula Application

c) $(Pi – Po) = \frac{4 S_{la}}{r}$
[Solution Description] The pressure difference inside a bubble with two interfaces is given by the formula: $(Pi – Po) = \frac{4 S_{la}}{r}$. Here, $S_{la}$ is the surface tension and $r$ is the radius of the bubble.

Key Concept: Equation Application

c) 7.5 m/s

[Solution Description]
According to the equation of continuity for incompressible fluids, the product of cross-sectional area and velocity remains constant:
$A_1 v_1 = A_2 v_2.$
Given that $A_1 = 5 \, \text{m}^2$, $v_1 = 3 \, \text{m/s}$, and $A_2 = 2 \, \text{m}^2$, we substitute these values into the equation:
$5 \times 3 = 2 \times v_2.$
Simplifying gives us:
$15 = 2v_2.$
Solving for $v_2$:
$v_2 = \frac{15}{2} = 7.5 \, \text{m/s}.$
Thus, the new velocity $v_2$ is $7.5 \, \text{m/s}$.

Key Concept: Interface Dynamics, Real-World Application

d) The liquid will form droplets on the solid surface.

[Solution Description] For spreading to occur, the surface energy at the liquid-solid interface should be less than the sum of the surface energies at the solid-air and liquid-air interfaces. Here, $0.08 \, \text{J/m}^2$ is greater than $0.04 \, \text{J/m}^2 + 0.03 \, \text{J/m}^2 = 0.07 \, \text{J/m}^2$, the liquid will not spread; instead, it will form droplets as the cohesive forces are stronger than adhesive forces.