Newton's Law of Viscosity Questions in English

(D) The viscous force $F$ is given by Newton's law of viscosity: $F = \eta A \frac{dv}{dx}$.
Here, the area of the square plate $A = (0.1 \; m)^2 = 0.01 \; m^2$.
The coefficient of viscosity $\eta = 0.01 \; \text{poise} = 0.001 \; \text{Pa} \cdot s$ (since $1 \; \text{poise} = 0.1 \; \text{Pa} \cdot s$).
The velocity gradient is $\frac{dv}{dx} = \frac{v}{dx}$, where $v = 0.1 \; m/s$ and $dx$ is the distance between the plates.
Substituting the values into the formula: $0.002 = 0.001 \times 0.01 \times \frac{0.1}{dx}$.
Rearranging for $dx$: $dx = \frac{0.001 \times 0.01 \times 0.1}{0.002} = \frac{0.000001}{0.002} = 0.0005 \; m$.

(A) In a laminar flow of a viscous liquid through a pipe,the fluid adheres to the walls of the pipe due to the no-slip condition.
As a result,the velocity of the liquid layer in direct contact with the stationary walls is $0$.
The velocity profile is parabolic,increasing from $0$ at the walls to a maximum value along the central axis of the tube.
Therefore,the velocity at the walls is $0$.

(B) Water flowing in a river experiences viscous drag due to the interaction with the river banks and the riverbed.
Near the banks,the water molecules experience significant friction with the stationary surface,which significantly reduces their velocity.
In the middle of the river,the water molecules are surrounded by other water molecules moving at similar speeds,resulting in minimal viscous resistance.
Therefore,the velocity of water is maximum in the middle and minimum near the banks.

(B) The viscosity of a liquid is a measure of its resistance to flow,which arises due to the cohesive forces between its molecules.
In liquids,as the temperature increases,the kinetic energy of the molecules increases.
This increased kinetic energy allows the molecules to overcome the cohesive forces more easily,thereby reducing the internal friction between the layers of the liquid.
Consequently,the viscosity of water (and most liquids) decreases as the temperature increases.
Therefore,the correct option is $B$.

(A) The viscosity of gases is primarily due to the transfer of momentum between layers of gas molecules moving at different velocities.
According to the kinetic theory of gases,the coefficient of viscosity $\eta$ is given by $\eta = \frac{1}{3} \rho v_{avg} \lambda$,where $\rho$ is the density,$v_{avg}$ is the average speed of molecules,and $\lambda$ is the mean free path.
As temperature increases,the average speed of gas molecules $(v_{avg} \propto \sqrt{T})$ increases.
Although the mean free path $\lambda$ may change,the increase in the average speed of molecules dominates,leading to an increase in the coefficient of viscosity with temperature.
Therefore,the coefficient of viscosity for hot air is greater than that for cold air.

(A) good lubricant must have high viscosity so that machine parts can operate efficiently.
Higher viscosity of the lubricant increases its ability to provide boundary lubrication,which creates a protective film between moving parts to reduce friction and wear.

(D) Given: Side of the square plate $L = 0.1 \, m$,Velocity $dv = 0.1 \, m/s$,Viscosity $\eta = 0.01 \, poise = 0.001 \, N \cdot s/m^2$ ($SI$ unit),Viscous force $F = 0.002 \, N$.
Area of the plate $A = L^2 = (0.1)^2 = 0.01 \, m^2$.
According to Newton's law of viscosity,the viscous force is given by $F = \eta A \frac{dv}{dx}$,where $dx$ is the distance between the plates.
Rearranging the formula for $dx$: $dx = \frac{\eta A dv}{F}$.
Substituting the values: $dx = \frac{0.001 \times 0.01 \times 0.1}{0.002} = \frac{0.000001}{0.002} = 0.0005 \, m$.

(B) Given: Area $A = 0.2\; m^{2}$,mass $m = 0.02\; kg$,thickness $l = 0.6\; mm = 0.6 \times 10^{-3}\; m$,velocity $v = 0.17\; m/s$,$g = 10\; m/s^{2}$.
The block moves with a constant speed,so the net force on it is zero. The tension $T$ in the string is equal to the weight of the hanging mass:
$T = m \cdot g = 0.02\; kg \times 10\; m/s^{2} = 0.2\; N$.
This tension is balanced by the viscous force $F$ acting on the block:
$F = \eta A \frac{v}{l} \implies T = \eta A \frac{v}{l}$.
Rearranging for the coefficient of viscosity $\eta$:
$\eta = \frac{T \cdot l}{A \cdot v} = \frac{0.2 \times 0.6 \times 10^{-3}}{0.2 \times 0.17}$.
$\eta = \frac{0.6 \times 10^{-3}}{0.17} \approx 3.53 \times 10^{-3} \; Pa \cdot s$.
Comparing with the given options,the closest value is $3.45 \times 10^{-3} \; Pa \cdot s$.

(B) The velocity gradient is defined as the rate of change of velocity with respect to the perpendicular distance between the layers,given by the formula: $\frac{dv}{dx}$.
Given:
Relative velocity $(dv)$ = $8 \ cm/s$
Perpendicular distance $(dx)$ = $0.1 \ cm$
Velocity gradient = $\frac{dv}{dx} = \frac{8 \ cm/s}{0.1 \ cm} = 80 \ s^{-1}$.
Therefore,the correct option is $B$.

(C) According to Newton's law of viscosity,the shear stress $\tau$ is proportional to the velocity gradient $\frac{du}{dy}$.
$\tau = \mu \frac{du}{dy}$
Since $\tau = \frac{F}{A}$,where $F$ is the applied force and $A$ is the surface area,we have $\frac{F}{A} = \mu \frac{u}{y}$.
This implies $F = \left( \frac{A \mu}{y} \right) u$,which means $F \propto u$.
Therefore,$\frac{F_1}{u_1} = \frac{F_2}{u_2}$.
Given $F_1 = 800$ $N$,$u_1 = 1.5$ $cm/s$,and $F_2 = 2.4$ $kN = 2400$ $N$.
Substituting these values: $\frac{800}{1.5} = \frac{2400}{u_2}$.
Solving for $u_2$: $u_2 = \frac{2400 \times 1.5}{800} = 3 \times 1.5 = 4.5$ $cm/s$.

(A) The block moves with a constant velocity $v$,which implies that the net force acting on the block along the inclined plane is zero.
The component of the gravitational force acting down the incline is $F_g = mg \sin \theta$.
The viscous drag force acting up the incline is given by Newton's law of viscosity: $F_v = \eta A \frac{v}{t}$,where $A = a^2$ is the contact area.
Equating the forces: $F_v = F_g$
$\eta (a^2) \frac{v}{t} = mg \sin 37^{\circ}$
Given that the mass $m = \text{density} \times \text{volume} = \rho a^3$ and $\sin 37^{\circ} = \frac{3}{5}$:
$\eta (a^2) \frac{v}{t} = (\rho a^3) g \left(\frac{3}{5}\right)$
Solving for $\eta$:
$\eta = \frac{\rho a^3 g \cdot 3}{5 \cdot a^2 \cdot v / t} = \frac{3 \rho a g t}{5 v}$

(B) The force required to move the plate is given by Newton's law of viscosity: $F = \eta A \frac{dv}{dy}$.
Given values:
Area $A = 100 \text{ cm}^2 = 100 \times 10^{-4} \text{ m}^2 = 10^{-2} \text{ m}^2$.
Gap $dy = 1 \text{ mm} = 10^{-3} \text{ m}$.
Velocity $v = 7 \text{ cm/s} = 0.07 \text{ m/s}$.
Coefficient of viscosity $\eta = 1.0 \text{ kg/(m} \cdot \text{s)}$.
Substituting these values into the formula:
$F = 1.0 \times 10^{-2} \times \frac{0.07}{10^{-3}}$
$F = 1.0 \times 10^{-2} \times 70$
$F = 0.7 \text{ N}$.
Therefore,the required force is $0.7 \text{ N}$.

(C) Let $\eta$ be the coefficient of viscosity and $x$ be the thickness of the oil film.
Consider an elemental ring of radius $r$ and width $dr$ on the disc.
The velocity of the disc at radius $r$ is $v = \omega r$.
The velocity gradient across the film is $\frac{dv}{dx} = \frac{\omega r - 0}{x} = \frac{\omega r}{x}$.
The viscous force $dF$ acting on this elemental ring is given by $dF = \eta A \frac{dv}{dx}$,where $A = 2\pi r dr$.
$dF = \eta (2\pi r dr) \frac{\omega r}{x} = \frac{2\pi \eta \omega}{x} r^2 dr$.
The torque $d\tau$ required to overcome this viscous force is $d\tau = (dF) r$.
$d\tau = \left( \frac{2\pi \eta \omega}{x} r^2 dr \right) r = \frac{2\pi \eta \omega}{x} r^3 dr$.
Integrating from $r = 0$ to $r = R$ to find the total torque $\tau$:
$\tau = \int_{0}^{R} \frac{2\pi \eta \omega}{x} r^3 dr = \frac{2\pi \eta \omega}{x} \left[ \frac{r^4}{4} \right]_{0}^{R} = \frac{\pi \eta \omega}{2x} R^4$.
Thus,the torque $\tau$ is proportional to $R^4$.

(D) According to Newton's law of viscosity,the shear stress $\tau$ is given by $\tau = \eta \frac{du}{dy}$.
Given the velocity profile $u = \alpha \left[ \frac{y}{h} - 2\left( \frac{y}{h} \right)^2 \right]$.
Differentiating $u$ with respect to $y$:
$\frac{du}{dy} = \alpha \left[ \frac{1}{h} - 2 \cdot 2 \left( \frac{y}{h} \right) \cdot \frac{1}{h} \right] = \alpha \left[ \frac{1}{h} - \frac{4y}{h^2} \right]$.
At the base of the plate,$y = 0$.
Therefore,the velocity gradient at the base is $\left( \frac{du}{dy} \right)_{y=0} = \alpha \left[ \frac{1}{h} - 0 \right] = \frac{\alpha}{h}$.
The magnitude of the shear stress at the base is $\tau = \eta \left| \frac{du}{dy} \right|_{y=0} = \eta \left( \frac{\alpha}{h} \right) = \frac{\alpha \eta}{h}$.

(B) Let the velocity of the thin plate be $v$ and its area be $A$. The force required to pull the plate is the sum of the viscous forces from both liquids:
$F = F_1 + F_2 = \eta A \frac{v}{d_1} + 4\eta A \frac{v}{d_2}$
Given that $d_1 + d_2 = d$ (constant total distance),we can write $d_2 = d - d_1$. Substituting this:
$F = \eta A v \left( \frac{1}{d_1} + \frac{4}{d - d_1} \right)$
To find the minimum force,we differentiate $F$ with respect to $d_1$ and set it to zero:
$\frac{dF}{dd_1} = \eta A v \left( -\frac{1}{d_1^2} + \frac{4}{(d - d_1)^2} \right) = 0$
$\frac{1}{d_1^2} = \frac{4}{(d - d_1)^2}$
Taking the square root on both sides:
$\frac{1}{d_1} = \frac{2}{d - d_1}$
$d - d_1 = 2d_1$
$d = 3d_1 \Rightarrow d_1 = \frac{d}{3}$
Then $d_2 = d - \frac{d}{3} = \frac{2d}{3}$.
Therefore,the ratio $\frac{d_2}{d_1} = \frac{2d/3}{d/3} = 2$.

(B) Given: Distance between surfaces $d = 2.5\ cm = 2.5 \times 10^{-2}\ m$. The plate is at the midway,so the distance from each surface is $dz = d/2 = 1.25 \times 10^{-2}\ m$.
Area of the plate $A = 0.5\ m^2$.
Velocity of the plate $v = 0.5\ m/s$.
Force applied $F = 1\ N$.
According to Newton's law of viscosity,the viscous force on one side of the plate is $F_{viscous} = \eta A \frac{dv}{dz}$.
Since the plate is dragged between two surfaces,the total viscous force is the sum of the forces from both sides: $F = 2 \times \eta A \frac{v}{dz}$.
Substituting the values: $1 = 2 \times \eta \times 0.5 \times \frac{0.5}{1.25 \times 10^{-2}}$.
$1 = \eta \times \frac{0.5}{1.25 \times 10^{-2}}$.
$\eta = \frac{1.25 \times 10^{-2}}{0.5} = 2.5 \times 10^{-2}\ kg/(m \cdot s)$.

(B) The viscous force $F$ required to move the plate is the sum of the viscous drag forces from both liquids. The formula for viscous force is $F = \eta A \frac{dv}{dx}$.
Given: side of square plate $L = 2\ m$, so area $A = L^2 = 4\ m^2$. Velocity $v = 10\ m/s$. Viscosities in $SI$ units: $\eta_1 = 1 \text{ poise} = 0.1 \text{ Pa} \cdot \text{s}$ and $\eta_2 = 4 \text{ poise} = 0.4 \text{ Pa} \cdot \text{s}$.
Total force $F = F_1 + F_2 = \eta_1 A \frac{v}{h_1} + \eta_2 A \frac{v}{h_2}$.
Substituting the values: $F = 0.1 \times 4 \times \frac{10}{h_1} + 0.4 \times 4 \times \frac{10}{h_2} = \frac{4}{h_1} + \frac{16}{h_2}$.
Since $h_1 + h_2 = 3$, we have $h_1 = 3 - h_2$. Thus, $F(h_2) = \frac{4}{3 - h_2} + \frac{16}{h_2}$.
To find the minimum force, set $\frac{dF}{dh_2} = 0$:
$\frac{d}{dh_2} [4(3 - h_2)^{-1} + 16(h_2)^{-1}] = 4(3 - h_2)^{-2} - 16(h_2)^{-2} = 0$.
$\frac{4}{(3 - h_2)^2} = \frac{16}{h_2^2} \implies \frac{1}{(3 - h_2)^2} = \frac{4}{h_2^2}$.
Taking the square root: $\frac{1}{3 - h_2} = \frac{2}{h_2} \implies h_2 = 6 - 2h_2 \implies 3h_2 = 6 \implies h_2 = 2\ m$.
Then $h_1 = 3 - 2 = 1\ m$.
Substituting back: $F_{\min} = \frac{4}{1} + \frac{16}{2} = 4 + 8 = 12\ N$.

(D) According to Newton's law of viscosity, the viscous force $F$ is given by $F = \eta A \frac{dv}{dy}$.
Here, area $A = 10 \, cm^2$, velocity gradient $\frac{dv}{dy} = \frac{v}{y} = \frac{1 \, cm/s}{1 \, mm} = \frac{1 \, cm/s}{0.1 \, cm} = 10 \, s^{-1}$.
The coefficient of viscosity $\eta = 20 \, \text{poise} = 20 \, dyne \cdot s/cm^2$.
Substituting the values: $F = 20 \times 10 \times 10 = 2000 \, dyne$.

(C) Given:
Velocity $v = 18\, km/h = 18 \times \frac{5}{18} = 5\, m/s$.
Depth $l = 5\, m$.
Coefficient of viscosity $\eta = 10^{-2}\, \text{poise} = 10^{-2} \times 0.1\, N\cdot s/m^2 = 10^{-3}\, N\cdot s/m^2$.
The velocity gradient (strain rate) is given by $\frac{dv}{dx} = \frac{v}{l} = \frac{5\, m/s}{5\, m} = 1\, s^{-1}$.
According to Newton's law of viscosity,the shearing stress $\tau$ is given by:
$\tau = \eta \times \frac{dv}{dx}$.
Substituting the values:
$\tau = 10^{-3}\, N\cdot s/m^2 \times 1\, s^{-1} = 10^{-3}\, N/m^2$.
Therefore,the correct option is $C$.

(C) The shearing stress $\tau$ is given by the formula: $\tau = \eta \frac{dv}{dy}$.
Here,$\eta = 10^{-3}\,SI\,unit$ is the coefficient of viscosity.
The velocity gradient $\frac{dv}{dy}$ can be approximated as $\frac{v}{h}$,where $v = 5\,m/s$ is the velocity and $h = 10\,m$ is the depth of the river.
Substituting the values:
$\tau = 10^{-3} \times \frac{5}{10}$
$\tau = 10^{-3} \times 0.5$
$\tau = 0.5 \times 10^{-3}\,N/m^2$.

(A) To move the plate with a constant velocity,the applied force must be equal to the viscous drag force.
The formula for viscous force is $F = \eta A \frac{\Delta v}{\Delta z}$.
Given values:
Coefficient of viscosity $\eta = 1.0\, kg/(m\cdot s)$.
Area $A = 100\, cm^2 = 100 \times 10^{-4}\, m^2 = 10^{-2}\, m^2$.
Velocity gradient $\frac{\Delta v}{\Delta z} = \frac{7\, cm/s}{1\, mm} = \frac{7 \times 10^{-2}\, m/s}{10^{-3}\, m} = 70\, s^{-1}$.
Substituting these values into the formula:
$F = 1.0 \times 10^{-2} \times 70 = 0.7\, N$.
Therefore,the force required is $0.7\, N$.

(A) Since the block moves with constant velocity,the net force acting on it is zero.
The component of the gravitational force acting down the incline is $F = mg \sin \theta$.
Here,$m = \rho V = \rho a^3$,so $F = \rho a^3 g \sin \theta$.
According to Newton's law of viscosity,the viscous force is $F = \eta A \left(\frac{dv}{dx}\right)$,where $A = a^2$ is the contact area and $\frac{dv}{dx} = \frac{v}{t}$ is the velocity gradient.
Equating the forces: $\rho a^3 g \sin \theta = \eta a^2 \left(\frac{v}{t}\right)$.
Solving for the coefficient of viscosity $\eta$: $\eta = \frac{\rho a^3 g \sin \theta \cdot t}{a^2 v} = \frac{\rho agt \sin \theta}{v}$.

(A) The viscosity of a liquid is inversely proportional to its temperature. As the temperature decreases,the viscosity of the lubricant increases.
According to Newton's law of viscosity,the viscous drag force is given by $F = -\eta A \frac{dv}{dx}$,where $\eta$ is the coefficient of viscosity.
Since the viscosity $\eta$ increases at low temperatures (in winter),the viscous drag force increases significantly.
This increased resistance makes it difficult for the machine parts to move,causing them to jam.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) The metal block moves to the right due to the tension in the string. Since the block moves at a constant speed,the net force on it is zero. The tension $T$ in the string is equal to the weight of the suspended mass $m$.
$T = m g = 0.010 \; kg \times 9.8 \; m s^{-2} = 0.098 \; N$
This tension is balanced by the viscous drag force $F$ acting on the block.
$F = \eta A \frac{dv}{dx}$
Where $\eta$ is the coefficient of viscosity,$A = 0.10 \; m^2$ is the area,$v = 0.085 \; m s^{-1}$ is the velocity,and $dx = 0.30 \; mm = 0.30 \times 10^{-3} \; m$ is the film thickness.
Rearranging for $\eta$:
$\eta = \frac{F \cdot dx}{A \cdot v}$
$\eta = \frac{(0.098 \; N) \times (0.30 \times 10^{-3} \; m)}{(0.10 \; m^2) \times (0.085 \; m s^{-1})}$
$\eta = \frac{0.0294 \times 10^{-3}}{0.0085} \; Pa \; s$
$\eta \approx 3.46 \times 10^{-3} \; Pa \; s$

(A) solid can sustain a shearing stress by undergoing a small,finite deformation (shear strain).
However,a fluid at rest cannot sustain any shearing stress.
If a fluid is subjected to a shearing stress,it will continue to deform (flow) as long as the stress is applied.
Therefore,the shearing stress of a fluid at rest is $0$.

(B) The shear modulus (also known as the modulus of rigidity) is defined only for solids because it measures the resistance of a material to shear deformation. Fluids,such as liquids (like sulfuric acid) and gases,cannot sustain a shear stress because they flow when subjected to it. Therefore,the shear modulus for any fluid is $0$.

(N/A) Viscosity is the property of a fluid by virtue of which it opposes the relative motion between its adjacent layers.
Cause of Viscosity:
In a laminar flow,any two consecutive layers of a fluid have a relative velocity between them. Due to this relative motion,a resistive force is produced tangentially at the surface of the layers in contact. This internal frictional force is known as the viscous force.
Consider a fluid (like oil) enclosed between two parallel glass plates. The bottom plate is fixed,while the top plate is moved with a constant velocity $v$ relative to the fixed plate.
The fluid layer in contact with a surface has the same velocity as that surface. Thus,the layer of liquid in contact with the top plate moves with velocity $v$,and the layer in contact with the bottom (stationary) plate has zero velocity.
The velocities of different layers increase uniformly from the bottom to the top (from $0$ to $v$).
For any layer of liquid,the upper layer pulls it forward,while the lower layer pulls it backward. This interaction results in a resistive force between the layers,which is the cause of viscosity.

(N/A) $1$. Velocity Gradient: When a fluid flows over a fixed surface,the velocity of the fluid layers increases with distance from the surface. The rate of change of velocity $(v)$ with respect to the perpendicular distance $(z)$ from the fixed surface is called the velocity gradient. It is given by $\frac{dv}{dz}$. Its $SI$ unit is $s^{-1}$.
$2$. Coefficient of Viscosity: According to Newton's law of viscosity,the viscous force $(F)$ acting between two layers of a fluid is directly proportional to the area of contact $(A)$ and the velocity gradient $(\frac{dv}{dz})$. Thus,$F = \eta A \frac{dv}{dz}$,where $\eta$ is the coefficient of viscosity. It is defined as the tangential force per unit area required to maintain a unit velocity gradient between two parallel layers of fluid. Its $SI$ unit is $Pa \cdot s$ or $N \cdot s/m^2$ (also known as Poiseuille).

(N/A) Consider a laminar flow over a horizontal surface as shown in the figure.
Suppose two layers $P$ and $Q$ are at distances $x$ and $x+dx$ from the stationary surface.
The velocity difference between these two layers separated by a distance $dx$ is $dv$.
The ratio $\frac{dv}{dx}$ is known as the velocity gradient.
Velocity Gradient: The rate of change of velocity with respect to distance perpendicular to the direction of flow is called the velocity gradient. Its $SI$ unit is $s^{-1}$.
The viscous force $F$ between two layers depends on the following factors:
$(1)$ It is directly proportional to the area $A$ of the contact surface: $F \propto A$.
$(2)$ It is directly proportional to the velocity gradient: $F \propto \frac{dv}{dx}$.
Combining these,we get $F \propto A \frac{dv}{dx}$,which leads to $F = -\eta A \frac{dv}{dx}$.
Here,$\eta$ is the coefficient of viscosity. The negative sign indicates that the viscous force acts in the direction opposite to the relative motion of the layers.
The $SI$ unit of $\eta$ is $N \cdot s \cdot m^{-2}$ or $Pa \cdot s$ (Pascal-second).

(N/A) The figure shows the laminar flow of a fluid between two parallel plates.
The fluid is placed between two glass plates. The bottom plate is stationary,so the fluid layer in contact with it is also at rest.
The top plate moves with a velocity $v$,causing the fluid layer in contact with it to move with the same velocity $v$.
Due to this motion,the fluid initially in the shape $ABCD$ takes the shape $AEFD$ after a small time interval $\Delta t$.
During this process,the shear strain produced is $\frac{\Delta x}{l}$. As the top plate continues to move,this strain increases continuously with time.
In this case,the stress depends not on the strain itself,but on the rate of change of strain,which is $\frac{(\Delta x / l)}{\Delta t} = \frac{\Delta x}{l \Delta t} = \frac{v}{l}$ (where $\frac{\Delta x}{\Delta t} = v$ is the velocity).
Here,the shear stress is $\frac{F}{A}$,where $A$ is the area of the contact surface and $F$ is the tangential viscous force.
For a fluid,the coefficient of viscosity $\eta$ is defined as:
$\eta = \frac{\text{Shear Stress}}{\text{Rate of Shear Strain}}$
$\therefore \eta = \frac{(F / A)}{(v / l)} = \frac{Fl}{vA}$
Or,$F = \eta A \left(\frac{v}{l}\right)$.

(N/A) Viscous force is the internal frictional force that acts between the layers of a fluid in motion,opposing the relative motion between them.
According to Newton's law of viscosity,the viscous force $F$ between two layers of a fluid is given by $F = -\eta A \frac{dv}{dx}$,where:
$1$. $\eta$ is the coefficient of viscosity of the fluid.
$2$. $A$ is the area of contact between the layers.
$3$. $\frac{dv}{dx}$ is the velocity gradient,which represents the change in velocity with respect to the distance perpendicular to the flow.
Thus,the viscous force depends on the nature of the fluid (viscosity),the surface area of the layers,and the velocity gradient.

(B) When milk is stirred,the different layers of the liquid move with different velocities,creating a velocity gradient between them.
According to the property of viscosity,a fluid exerts an internal frictional force that opposes the relative motion between adjacent layers.
This viscous drag force acts in the direction opposite to the flow,gradually dissipating the kinetic energy of the liquid into heat.
Consequently,the milk eventually comes to rest.

(N/A) The velocity gradient is defined as the rate of change of velocity with respect to the distance perpendicular to the direction of flow.
Mathematically,it is expressed as $\frac{dv}{dx}$,where $dv$ is the change in velocity and $dx$ is the change in distance.
The $SI$ unit of velocity gradient is $\text{per second}$ $(s^{-1})$.
The dimensional formula is calculated as: $\frac{[LT^{-1}]}{[L]} = [M^0L^0T^{-1}]$.

(N/A) The coefficient of viscosity of a liquid decreases,while that of a gas increases,with an increase in temperature.
This difference arises because the mechanism of viscosity in gases is based on molecular momentum exchange,whereas in liquids,it is based on cohesive forces.
$\Rightarrow$ In gases,an increase in temperature leads to higher molecular kinetic energy and increased molecular momentum exchange,which results in an increase in viscosity.
$\Rightarrow$ In liquids,an increase in temperature weakens the intermolecular cohesive forces,which leads to a decrease in viscosity.

(N/A) The coefficient of viscosity,denoted by the Greek letter $\eta$ (eta),is a measure of a fluid's resistance to flow or deformation under shear stress.
According to Newton's law of viscosity,the viscous force $F$ acting between two layers of a fluid is given by $F = \eta A \frac{dv}{dx}$,where $A$ is the area of contact,and $\frac{dv}{dx}$ is the velocity gradient.
Rearranging the formula,we get $\eta = \frac{F}{A(dv/dx)}$.
Thus,the coefficient of viscosity is defined as the tangential force per unit area required to maintain a unit velocity gradient between two parallel layers of a fluid.
The $SI$ unit of the coefficient of viscosity is $\text{Pa} \cdot \text{s}$ or $\text{N} \cdot \text{s/m}^2$ (also known as Poiseuille,$\text{Pl}$).

(N/A) The coefficient of viscosity $\eta$ is defined by the relation $F = \eta A \frac{dv}{dx}$,where $F$ is the viscous force,$A$ is the area,and $\frac{dv}{dx}$ is the velocity gradient.
Rearranging for $\eta$,we get $\eta = \frac{F}{A (dv/dx)}$.
In $SI$ units,the unit of force is $N$,area is $m^2$,and velocity gradient is $s^{-1}$. Thus,the $SI$ unit is $\frac{N}{m^2 \cdot s^{-1}} = N \cdot s \cdot m^{-2} = Pa \cdot s$ (Pascal-second) or $kg \cdot m^{-1} \cdot s^{-1}$.
In $CGS$ units,the unit of force is $dyne$,area is $cm^2$,and velocity gradient is $s^{-1}$. Thus,the $CGS$ unit is $\frac{dyne}{cm^2 \cdot s^{-1}} = dyne \cdot s \cdot cm^{-2}$,which is also known as $Poise$ $(P)$.
Conversion: $1 \text{ } Pa \cdot s = 10 \text{ } Poise$.

(N/A) One practical use of viscosity is in the lubrication of machinery. Lubricating oils with appropriate viscosity are used in engines and machines to reduce friction between moving parts,thereby preventing wear and tear and improving efficiency.

(N/A) The viscosity of a liquid is defined by its coefficient of viscosity,denoted by $\eta$.
As the temperature of a liquid increases,the intermolecular forces of attraction decrease.
Consequently,the coefficient of viscosity $\eta$ decreases for hot liquids.
Since the viscous drag force between consecutive layers is given by $F = -\eta A \frac{dv}{dx}$,a lower value of $\eta$ results in a smaller viscous force.
Therefore,hot liquids experience less internal resistance and flow more easily than cold liquids.

(N/A) In winter,the ambient temperature decreases. As the temperature of the lubricating oil drops,its viscosity increases significantly. This increase in viscosity makes the oil thicker and more resistant to flow,which creates greater internal friction between the moving surfaces of the machine. Consequently,the machine parts do not move easily and appear to become sticky.

(N/A) The flow of water in a river is governed by the property of viscosity.
$1$. Near the banks,the water molecules experience adhesive forces between the water and the solid surface of the river bank.
$2$. These adhesive forces create a resistive force (viscous drag) that opposes the motion of the water,causing the velocity to decrease significantly near the banks.
$3$. In the middle of the river,the water molecules are far from the banks,so the effect of adhesive forces is minimal.
$4$. Consequently,the velocity of the water flow is higher in the middle compared to the banks.

(B) Viscosity is a physical property of a fluid that characterizes its resistance to flow.
Since it does not have a specific direction associated with it,it is a scalar quantity.

(D) According to Newton's law of viscosity,the shearing stress $\tau$ is given by $\tau = \eta \frac{dv}{dy}$.
Here,$\tau = 10^{-3} \, N/m^2$,$\eta = 10^{-2} \, Pa \cdot s$,and the velocity gradient $\frac{dv}{dy} = \frac{v}{h}$,where $v$ is the velocity of the upper layer and $h$ is the depth of the river.
First,convert the velocity $v$ into $SI$ units: $v = 36 \, km/h = 36 \times \frac{5}{18} \, m/s = 10 \, m/s$.
Substituting the values into the formula: $10^{-3} = 10^{-2} \times \frac{10}{h}$.
Rearranging for $h$: $h = \frac{10^{-2} \times 10}{10^{-3}} = \frac{10^{-1}}{10^{-3}} = 10^2 = 100 \, m$.
Thus,the depth of the river is $100 \, m$.

(D) The correct answer is $D$.
The property of a fluid (liquid or gas) by which it opposes the relative motion between its adjacent layers is known as viscosity.
This internal frictional force acts to resist the flow of the fluid.

(B) Given:
Viscous drag force, $F = 1 \, N$
Coefficient of viscosity, $\eta = 0.02 \, \text{decapoise} = 0.02 \, N \cdot s/m^2$
Cross-sectional area, $A = 20 \, m^2$
According to Newton's law of viscosity, the viscous drag force is given by:
$F = \eta A \frac{dv}{dx}$
Rearranging the formula to find the velocity gradient $\frac{dv}{dx}$:
$\frac{dv}{dx} = \frac{F}{\eta A}$
Substituting the given values:
$\frac{dv}{dx} = \frac{1}{0.02 \times 20}$
$\frac{dv}{dx} = \frac{1}{0.4}$
$\frac{dv}{dx} = 2.5 \, s^{-1}$
Thus, the velocity gradient is $2.5 \, s^{-1}$.

(B) According to Newton's law of viscosity,the viscous force $F$ is given by the formula: $F = \eta A \frac{dv}{dx}$.
Here,$\eta = 1.5 \,N \cdot s \cdot m^{-2}$ is the coefficient of viscosity,$A = 0.1 \,m^2$ is the area of the plate,$v = 1 \,mm \cdot s^{-1} = 10^{-3} \,m \cdot s^{-1}$ is the velocity,and $dx = 10^{-5} \,m$ is the thickness of the oil film.
Substituting these values into the formula:
$F = 1.5 \times 0.1 \times \frac{10^{-3}}{10^{-5}}$
$F = 0.15 \times 10^2$
$F = 0.15 \times 100 = 15 \,N$.
Therefore,the force required is $15 \,N$.

(D) The viscous drag force $F$ acting on a body moving through a fluid is given by Newton's law of viscosity,which can be expressed as $F = \eta A \frac{dv}{dx}$.
Here:
$1$. $\eta$ is the coefficient of viscosity of the fluid.
$2$. $A$ is the surface area of the body,which relates to the size of the body.
$3$. $v$ is the velocity with which the body moves through the fluid.
Since the force $F$ depends on the viscosity $(\eta)$,the area $(A)$,and the velocity $(v)$,it depends on all of the given factors.
Therefore,the correct option is $(d)$.

(A) According to Newton's law of viscosity,the shear stress $\tau$ is given by $\tau = \eta \frac{dv}{dy}$.
Given the velocity profile $v = k \left( \frac{2y^2}{a^2} - \frac{y^3}{a^3} \right)$.
Differentiating with respect to $y$,we get $\frac{dv}{dy} = k \left( \frac{4y}{a^2} - \frac{3y^2}{a^3} \right)$.
Substituting this into the stress formula,$\tau = \eta k \left( \frac{4y}{a^2} - \frac{3y^2}{a^3} \right)$.
At $y = a$,the shear stress is $\tau = \eta k \left( \frac{4a}{a^2} - \frac{3a^2}{a^3} \right) = \eta k \left( \frac{4}{a} - \frac{3}{a} \right) = \frac{\eta k}{a}$.

(B) Viscosity is defined as the internal friction of a fluid. It is the property of a fluid (liquid or gas) by virtue of which it opposes the relative motion between its adjacent layers. When a fluid flows,the layers moving at different velocities exert a tangential force on each other,which acts to retard the flow. This internal resistance is known as viscosity.

(B) According to Newton's law of viscosity,the viscous force $F$ is given by:
$F = \eta A \frac{dv}{dx}$
Where:
$F = 0.1 \,N$ (applied force)
$\eta = 5.0 \times 10^{-3} \,Pa-s$ (viscosity)
$A = 0.20 \,m^2$ (base area)
$dx = 0.25 \,mm = 0.25 \times 10^{-3} \,m$ (thickness of the film)
Since the block moves with a constant speed,the applied force equals the viscous force:
$0.1 = (5.0 \times 10^{-3}) \times 0.20 \times \frac{v}{0.25 \times 10^{-3}}$
$0.1 = \frac{1.0 \times 10^{-3} \times v}{0.25 \times 10^{-3}}$
$0.1 = 4v$
$v = \frac{0.1}{4} = 0.025 \,m/s$
$v = 25 \times 10^{-3} \,m/s$
Thus,the speed of the block is $25 \times 10^{-3} \,m/s$.

(C) According to Newton's law of viscosity,the viscous force $F$ acting on a plate of area $A$ moving with velocity $u_0$ over a liquid layer of thickness $h$ is given by:
$F = \eta A \frac{dv}{dx} = \eta A \frac{u_0}{h}$
From this expression:
$1$. The resistive force $F$ is inversely proportional to $h$ $(F \propto 1/h)$. Thus,statement $(A)$ is true.
$2$. The resistive force $F$ is directly proportional to the area $A$ $(F \propto A)$. Thus,statement $(B)$ is false.
$3$. The tangential (shear) stress $\tau$ is defined as $\tau = F/A = \eta \frac{u_0}{h}$.
$4$. Since $\tau = \eta \frac{u_0}{h}$,the shear stress on the floor (or plate) is directly proportional to $u_0$. Thus,statement $(C)$ is true.
$5$. Since $\tau = \eta \frac{u_0}{h}$,the shear stress is directly proportional to the viscosity $\eta$. Thus,statement $(D)$ is true.
Therefore,statements $(A), (C),$ and $(D)$ are true.