Newton's Law of Viscosity Questions in English

(C) In a steady flow of a river,the velocity $v$ of a layer at a height $y$ from the bottom is given by the relation $v \propto y$,assuming a linear velocity profile (laminar flow).
Thus,$\frac{v_A}{y_A} = \frac{v_B}{y_B}$.
Given: $v_A = 12 \ cm/s$,$y_A = 40 \ cm$,and $y_B = 90 \ cm$.
Substituting the values: $\frac{12}{40} = \frac{v_B}{90}$.
$v_B = \frac{12 \times 90}{40} = \frac{1080}{40} = 27 \ cm/s$.
Therefore,the velocity of layer $B$ is $27 \ cm/s$.

(C) According to Newton's law of viscosity,the viscous force $F$ acting on an area $A$ is given by $\tau = \frac{F}{A} = -\eta \frac{dv}{dz}$.
Viscous force acts like internal friction between adjacent layers of a fluid moving with different velocities.
This force opposes the relative motion between the layers,causing a shearing action.
$A$ shear force always acts tangential to the plane of the fluid layers.
Therefore,the viscous force is tangential to the liquid surface.

(B) The viscosity of a fluid is a measure of its resistance to flow.
For liquids,the viscosity is primarily due to cohesive forces between molecules. As temperature increases,the kinetic energy of the molecules increases,which overcomes these cohesive forces,causing the viscosity of liquids to decrease.
For gases,the viscosity is primarily due to the transfer of momentum between layers of gas molecules due to random thermal motion. As temperature increases,the random motion of gas molecules increases,leading to more frequent collisions and a higher rate of momentum transfer,which causes the viscosity of gases to increase.
Therefore,when the temperature increases,the viscosity of gases increases and the viscosity of liquids decreases.

(B) The viscous drag force is given by $F = \eta A \frac{dv}{dx}$.
Since the box moves with a constant speed,the applied force $F$ must be equal to the viscous drag force.
Here,$F = \frac{1}{3} \ N$,$A = 0.01 \ m^2$,$dv = 0.09 \ m/s$,and $dx = 0.3 \ mm = 0.3 \times 10^{-3} \ m$.
Substituting these values into the formula:
$\eta = \frac{F \cdot dx}{A \cdot dv}$
$\eta = \frac{(1/3) \times (0.3 \times 10^{-3})}{0.01 \times 0.09}$
$\eta = \frac{0.1 \times 10^{-3}}{0.0009} = \frac{10^{-4}}{9 \times 10^{-4}} = \frac{1}{9} \approx 0.11 \ Pa \cdot s$.
Thus,$\eta \approx 1.1 \times 10^{-1} \ Pa \cdot s$.

(B) An ideal fluid is defined as a fluid that is incompressible and non-viscous.
Since an ideal fluid is non-viscous,it does not offer any resistance to the relative motion between its layers.
Therefore,the coefficient of viscosity for an ideal fluid is $0$.

$(A)$ Given:
Side of the cube, $L = 10 \,cm = 0.1 \,m$.
Area of the base, $A = L^2 = (0.1 \,m)^2 = 0.01 \,m^2$.
Thickness of the liquid film, $dx = 0.2 \,mm = 0.2 \times 10^{-3} \,m$.
Applied force, $F = 0.1 \,N$.
Constant velocity, $v = 0.08 \,m/s$.
Since the cube moves with a constant speed, the net force is zero, meaning the applied force equals the viscous drag force: $F = F_{drag}$.
According to Newton's law of viscosity, $F = \eta A \frac{dv}{dx}$.
Substituting the values: $0.1 = \eta \times 0.01 \times \frac{0.08}{0.2 \times 10^{-3}}$.
$0.1 = \eta \times 0.01 \times 400$.
$0.1 = \eta \times 4$.
$\eta = \frac{0.1}{4} = 0.025 \,Ns/m^2 = 2.5 \times 10^{-2} \,Ns/m^2$.

(D) The shearing stress $\tau$ is given by the formula: $\tau = \eta \left( \frac{dv}{dx} \right)$.
Here,$\eta = 10^{-3} \ SI \ units$ is the coefficient of viscosity.
The velocity gradient $\frac{dv}{dx}$ is calculated as the velocity $v$ divided by the depth $x$.
Given $v = 10 \ ms^{-1}$ and $x = 20 \ m$,we have $\frac{dv}{dx} = \frac{10}{20} = 0.5 \ s^{-1}$.
Substituting these values into the formula:
$\tau = 10^{-3} \times 0.5 = 0.5 \times 10^{-3} \ Nm^{-2}$.

(B) The shearing stress is defined as the viscous force per unit area,given by the formula: $\text{Shearing Stress} = \frac{F}{A} = \eta \frac{dv}{dx}$.
In a river,the velocity $v$ changes from $V$ at the surface to $0$ at the bottom (depth $H$).
Thus,the velocity gradient is $\frac{dv}{dx} = \frac{V}{H}$.
Substituting this into the stress formula,we get: $\text{Shearing Stress} = \eta \frac{V}{H}$.

(B) The viscous force $F$ acting on the plate is given by Newton's law of viscosity: $F = \eta A \frac{dv}{dx}$.
Here,$\eta = 1.55 \,Ns\, m^{-2}$ is the coefficient of viscosity,$A = 10^{-2} \,m^2$ is the area of the plate,$v = 3 \times 10^{-2} \,ms^{-1}$ is the velocity,and $h = dx = 2 \times 10^{-3} \,m$ is the thickness of the oil layer.
Substituting the values: $F = 1.55 \times 10^{-2} \times \frac{3 \times 10^{-2}}{2 \times 10^{-3}}$.
$F = 1.55 \times 10^{-2} \times 1.5 \times 10^1$.
$F = 1.55 \times 1.5 \times 10^{-1} = 2.325 \times 10^{-1} = 0.2325 \,N$.