Welcome to Lecture 82 of our comprehensive JEE Physics Master Series. In this lecture, we explore the dynamics of internal fluid friction: the physical concept of viscosity, velocity gradients, Newton’s law of viscous shear, units of dynamic and kinematic viscosity, temperature dependence in liquids vs gases, and the mathematical formulation of Stoke’s drag law for spherical bodies moving through viscous media.
1. Detailed Concept Card
1.1 Physical Nature & Molecular Origin of Viscosity
Viscosity represents the internal friction of a fluid that opposes relative motion between adjacent fluid layers moving with different velocities. Just as dry kinetic friction opposes sliding between solid surfaces, viscosity opposes the continuous shearing deformation of a fluid.
Molecular Mechanisms:
- In Liquids: Intermolecular cohesive forces dominate. Molecules in a faster-moving layer experience backward attractive pulls from neighboring molecules in slower-moving layers. As temperature rises, thermal agitation increases average intermolecular separation, weakening cohesive intermolecular bonds and drastically decreasing liquid viscosity.
- In Gases: Cohesion is negligible. Viscosity arises purely from the random thermal diffusion and transport of molecules across adjacent layers. Faster layers transfer forward momentum to slower layers, and slower molecules migrate into faster layers. As temperature rises, mean molecular thermal speed $\bar{v} \propto \sqrt{T}$ increases, intensifying molecular momentum exchange and increasing gas viscosity ($\eta_{\text{gas}} \propto \sqrt{T}$).
1.2 Velocity Gradient & Newton’s Law of Viscosity
Consider steady laminar (streamline) flow of a fluid over a fixed, flat horizontal solid boundary. Due to the no-slip condition, fluid particles in direct contact with the solid boundary have zero velocity ($v = 0$). Successive layers above the surface move with progressively higher velocities.
Let two parallel fluid layers separated by a perpendicular distance $dy$ move with velocities $v$ and $v + dv$. The rate of change of fluid velocity with respect to normal distance is defined as the Velocity Gradient:
$$\text{Velocity Gradient} = \frac{dv}{dy}$$
According to Newton’s Law of Viscous Flow, the tangential viscous drag force $F$ acting between two parallel fluid layers of contact area $A$ is directly proportional to the area $A$ and to the velocity gradient $\frac{dv}{dy}$:
$$F = -\eta A \frac{dv}{dy}$$
where the negative sign indicates that the viscous force opposes the relative motion between layers. The corresponding Viscous Shear Stress $\tau$ is:
$$\tau = \frac{F}{A} = -\eta \frac{dv}{dy}$$
1.3 Coefficient of Dynamic Viscosity ($\eta$) & Units
The proportionality constant $\eta$ is called the Coefficient of Dynamic Viscosity (or simply viscosity):
$$\eta = \frac{F/A}{dv/dy} = \frac{\tau}{dv/dy}$$
Physical Definition: The coefficient of viscosity is defined as the tangential viscous force per unit contact area required to maintain a unit velocity gradient between two parallel fluid layers.
Units of Dynamic Viscosity:
- SI Unit: $\text{N}\cdot\text{s/m}^2 = \text{Pa}\cdot\text{s} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}$ (also called Decapoise or Poiseuille, $\text{Pl}$).
- CGS Unit: $\text{dyne}\cdot\text{s/cm}^2 = \text{Poise}$ (symbol: $\text{P}$).
- Unit Conversion: $$1\text{ Pa}\cdot\text{s} = 10\text{ Poise} = 1\text{ Decapoise} = 1\text{ Poiseuille}$$ $$1\text{ Poise} = 0.1\text{ Pa}\cdot\text{s} = 0.1\text{ N}\cdot\text{s/m}^2$$
- Centipoise ($\text{cP}$): $1\text{ cP} = 10^{-2}\text{ Poise} = 10^{-3}\text{ Pa}\cdot\text{s}$. (Water at $20^\circ\text{C}$ has $\eta \approx 1.0\text{ cP} = 10^{-3}\text{ Pa}\cdot\text{s}$).
- Dimensional Formula: $$[\eta] = \frac{[F]}{[A][dv/dy]} = \frac{[M L T^{-2}]}{[L^2][T^{-1}]} = [M^1 L^{-1} T^{-1}]$$
Kinematic Viscosity ($\nu$): The ratio of dynamic viscosity to fluid density:
$$\nu = \frac{\eta}{\rho}$$
- SI Unit: $\text{m}^2/\text{s}$.
- CGS Unit: $\text{Stokes}$ ($\text{St}$), where $1\text{ Stokes} = 1\text{ cm}^2/\text{s} = 10^{-4}\text{ m}^2/\text{s}$.
- Dimension: $[L^2 T^{-1}]$.
1.4 Stoke’s Law of Viscous Resistance
Sir George Gabriel Stokes showed that when a small, smooth, rigid sphere of radius $r$ moves with velocity $v$ through an infinite, continuous, and homogeneous fluid of viscosity $\eta$, the retarding viscous drag force experienced by the sphere is given by:
$$F_v = 6\pi\, \eta\, r\, v$$
Dimensional Derivation:
Assume $F_v \propto \eta^a r^b v^c$, where $k = 6\pi$ is a dimensionless proportionality constant determined hydrodynamically.
$$[M L T^{-2}] = [M L^{-1} T^{-1}]^a [L]^b [L T^{-1}]^c = [M^a L^{-a + b + c} T^{-a – c}]$$
- Equating powers of $M$: $a = 1$
- Equating powers of $T$: $-a – c = -2 \implies c = 2 – 1 = 1$
- Equating powers of $L$: $-a + b + c = 1 \implies -1 + b + 1 = 1 \implies b = 1$
Thus, $F_v = 6\pi \eta r v$.
Critical Assumptions for Validity of Stoke’s Law:
- 1. Low Reynolds Number (Laminar Creeping Flow): Flow around the sphere must be strictly non-turbulent without wake separation or vortex shedding ($\text{Re} = \frac{\rho v (2r)}{\eta} < 1$). At higher speeds, inertial drag ($F_d \propto v^2$) supersedes Stoke's linear drag ($F_v \propto v$).
- 2. Infinite Medium: The fluid boundaries (container walls) must be situated at distances very large compared to the sphere’s radius $r$.
- 3. Rigid & Smooth Geometry: The body must be perfectly spherical and undeformable.
- 4. Continuum Condition: The sphere radius $r$ must be much larger than the mean free path of the fluid molecules (no slip at the boundary).
1.5 Key Insights & Critical Pitfalls for JEE
- Pitfall 1 (Temperature Trend Inversion): Never assume all fluids become less viscous with heat! Liquids become less viscous as temperature increases, whereas gases become more viscous with temperature ($d\eta/dT > 0$ for gases).
- Pitfall 2 (Non-Uniform Velocity Gradient): The expression $dv/dy = v/h$ holds strictly when the velocity profile is linear (e.g., small gap between parallel flat plates). For flow in pipes or around rotating cylinders, $dv/dr$ varies with radial position and must be integrated.
- Pitfall 3 (Stokes Law Breakdown at High Speeds): Stoke’s linear law $F_v \propto v$ is invalid for large macroscopic objects falling through air (like skydivers, cricket balls, or heavy rocks) where the drag force scales quadratically with velocity ($F_{\text{drag}} = \frac{1}{2} C_d \rho A v^2$). Stoke’s law applies to micron-sized droplets, dust, colloidal particles, or bodies falling through highly viscous glycerine/oil.
2. Four Graded Solved Examples
Example 1: Flat Plate Pulled Over an Oil Film (JEE Main Level)
Problem Statement: A rectangular metal plate of dimensions $40\text{ cm} \times 50\text{ cm}$ rests on a horizontal layer of oil of thickness $t = 2.0\text{ mm}$ spreading over a large horizontal flat floor. The dynamic viscosity of the oil is $\eta = 0.8\text{ Pa}\cdot\text{s}$.
(a) Determine the horizontal force $F$ required to drag the plate at a constant velocity of $v = 1.5\text{ m/s}$.
(b) Calculate the rate of mechanical energy dissipated as heat in the oil layer.
Solution:
Part (a): Pulling Force
Contact area of the plate in contact with the oil layer is:
$$A = 0.40\text{ m} \times 0.50\text{ m} = 0.20\text{ m}^2$$
The thickness of the liquid film is $t = 2.0\text{ mm} = 2.0 \times 10^{-3}\text{ m}$. Assuming a linear velocity profile between the stationary floor ($v = 0$) and the moving plate ($v = 1.5\text{ m/s}$), the velocity gradient is uniform:
$$\frac{dv}{dy} = \frac{v}{t} = \frac{1.5}{2.0 \times 10^{-3}} = 750\text{ s}^{-1}$$
The viscous retarding force opposing the motion is given by Newton’s law of viscosity:
$$F_{\text{viscous}} = \eta A \frac{dv}{dy} = 0.8 \times 0.20 \times 750 = 120\text{ N}$$
To pull the plate at constant velocity ($a = 0$), the applied force must balance the viscous drag force:
$$F = F_{\text{viscous}} = 120\text{ N}$$
Part (b): Rate of Energy Dissipation (Power)
The mechanical power supplied by the applied force is converted entirely into heat through viscous dissipation across the fluid layers:
$$P = F \cdot v = 120\text{ N} \times 1.5\text{ m/s} = 180\text{ W (Joules/sec)}$$
Key Takeaway: In planar laminar shear, viscous shear stress is $\tau = \eta \frac{v}{t}$, and power dissipation per unit volume is $\frac{P}{\text{Volume}} = \eta \left(\frac{dv}{dy}\right)^2$.
Example 2: Stoke’s Drag on an Airborne Water Micro-Droplet (JEE Main Level)
Problem Statement: A spherical cloud droplet of water of radius $r = 20\, \mu\text{m}$ falls through air of viscosity $\eta = 1.8 \times 10^{-5}\text{ Pa}\cdot\text{s}$ and density $\rho_{\text{air}} = 1.2\text{ kg/m}^3$. At an instant when the droplet is falling at a speed of $v = 4.0\text{ cm/s}$:
(a) Calculate the viscous drag force acting on the droplet.
(b) Compare the viscous drag force with the gravitational weight of the droplet. (Take $\rho_{\text{water}} = 1000\text{ kg/m}^3$, $g = 9.8\text{ m/s}^2$).
(c) Verify whether the flow satisfies the low Reynolds number criterion ($\text{Re} < 1$).
Solution:
Part (a): Viscous Drag Force
Convert parameters into standard SI units:
$r = 20 \times 10^{-6}\text{ m}$, $v = 4.0 \times 10^{-2}\text{ m/s}$, $\eta = 1.8 \times 10^{-5}\text{ Pa}\cdot\text{s}$.
According to Stoke’s Law:
$$F_v = 6\pi \eta r v = 6 \times 3.1416 \times (1.8 \times 10^{-5}) \times (20 \times 10^{-6}) \times (4.0 \times 10^{-2})$$
$$F_v = 18.8496 \times 1.8 \times 20 \times 4.0 \times 10^{-13} = 2.713 \times 10^{-10}\text{ N}$$
Part (b): Comparison with Weight
The volume of the spherical water droplet is $V = \frac{4}{3}\pi r^3$. The true gravitational weight is:
$$W = m g = \rho_{\text{water}} \left(\frac{4}{3}\pi r^3\right) g = 1000 \times \frac{4}{3}\pi \times (20 \times 10^{-6})^3 \times 9.8$$
$$W = 1000 \times 4.1888 \times (8.0 \times 10^{-15}) \times 9.8 \approx 3.284 \times 10^{-10}\text{ N}$$
The ratio of viscous force to weight is:
$$\frac{F_v}{W} = \frac{2.713 \times 10^{-10}}{3.284 \times 10^{-10}} \approx 0.826$$
The viscous drag force already counteracts $\approx 83\%$ of the droplet’s weight, indicating that the droplet is rapidly accelerating toward its terminal velocity.
Part (c): Reynolds Number Check
Diameter $D = 2r = 40 \times 10^{-6}\text{ m}$.
$$\text{Re} = \frac{\rho_{\text{air}}\, v\, D}{\eta} = \frac{1.2 \times (0.04) \times (40 \times 10^{-6})}{1.8 \times 10^{-5}} = \frac{1.92 \times 10^{-6}}{1.8 \times 10^{-5}} \approx 0.107$$
Since $\text{Re} = 0.107 \ll 1$, the flow is strictly in the laminar creeping regime where Stoke’s law is valid.
Example 3: Viscous Resistance Torque on a Rotating Coaxial Cylinder (JEE Advanced Level)
Problem Statement: A cylindrical rotor of radius $R_1 = 5.0\text{ cm}$ and length $L = 20\text{ cm}$ is mounted coaxially inside a stationary hollow cylinder of radius $R_2 = 5.2\text{ cm}$. The clearance gap between the cylinders is filled with a viscous lubricant of dynamic viscosity $\eta = 0.5\text{ Pa}\cdot\text{s}$. The rotor rotates at a constant speed of $N = 600\text{ rpm}$.
(a) Derive an expression for the viscous torque opposing the rotation.
(b) Calculate the numerical magnitude of the resisting torque.
(c) Find the mechanical power needed to maintain this steady rotation.
Solution:
Part (a): Derivation of Viscous Torque
The gap thickness between the inner rotor and outer sleeve is $d = R_2 – R_1 = 5.2 – 5.0 = 0.2\text{ cm} = 2.0 \times 10^{-3}\text{ m}$. Since $d \ll R_1$ ($d/R_1 = 0.2/5.0 = 0.04 \ll 1$), we can approximate the fluid flow in the annulus as planar Couette flow with a uniform velocity gradient.
The tangential speed of the rotor surface is:
$$v = \omega R_1, \quad \text{where } \omega = \frac{2\pi N}{60} = \frac{2\pi \times 600}{60} = 20\pi\text{ rad/s} \approx 62.83\text{ rad/s}$$
The velocity gradient across the gap is:
$$\frac{dv}{dr} = \frac{\omega R_1}{d}$$
The viscous shear stress acting along the cylindrical surface of the rotor is:
$$\tau = \eta \frac{dv}{dr} = \eta \frac{\omega R_1}{d}$$
The surface area of the rotor cylinder is $A = 2\pi R_1 L$. The total tangential viscous drag force acting on the perimeter is:
$$F = \tau A = \left(\eta \frac{\omega R_1}{d}\right)(2\pi R_1 L) = \frac{2\pi \eta \omega L R_1^2}{d}$$
The resisting torque is the product of the tangential force and the lever arm $R_1$:
$$\tau_{\text{torque}} = F \cdot R_1 = \frac{2\pi \eta \omega L R_1^3}{d}$$
Part (b): Numerical Calculation of Torque
Substitute $\eta = 0.5\text{ Pa}\cdot\text{s}$, $\omega = 20\pi\text{ rad/s}$, $L = 0.20\text{ m}$, $R_1 = 0.05\text{ m}$, $d = 2.0 \times 10^{-3}\text{ m}$:
$$\tau_{\text{torque}} = \frac{2\pi \times 0.5 \times (20\pi) \times 0.20 \times (0.05)^3}{2.0 \times 10^{-3}}$$
$$\tau_{\text{torque}} = \frac{20\pi^2 \times 0.20 \times (1.25 \times 10^{-4})}{2.0 \times 10^{-3}} = \frac{4\pi^2 \times 1.25 \times 10^{-4}}{2.0 \times 10^{-3}} = \frac{5\pi^2 \times 10^{-4}}{2.0 \times 10^{-3}} = 2.5\pi^2 \times 10^{-1} = 0.25 \pi^2\text{ N}\cdot\text{m}$$
Taking $\pi^2 \approx 9.87$:
$$\tau_{\text{torque}} \approx 0.25 \times 9.8696 \approx 2.467\text{ N}\cdot\text{m}$$
Part (c): Power Required
Power required to overcome viscous torque is:
$$P = \tau_{\text{torque}} \cdot \omega = 2.4674 \times (20\pi) = 49.348 \times 3.1416 \approx 155.0\text{ W}$$
Key Takeaway: In concentric cylindrical viscometers (Couette flow), torque scales with $R_1^3 / d$. This principle is used in laboratory viscometers to measure $\eta$ precisely.
Example 4: Spinning Flat Disc Resting on a Viscous Layer (JEE Advanced Edge Case)
Problem Statement: A circular disc of mass $M$ and radius $R$ is placed horizontally on a thin layer of oil of thickness $h$ and dynamic viscosity $\eta$. Initially, the disc is given an angular velocity $\omega_0$ about its central vertical axis and then left to spin freely.
(a) Derive an expression for the viscous torque $d\tau$ acting on an elemental circular ring of radius $r$ and width $dr$ on the bottom face of the disc.
(b) Find the total viscous torque $\tau_{\text{total}}$ acting on the disc at any instantaneous angular velocity $\omega$.
(c) Determine the time $t$ required for the angular velocity of the disc to decrease to half of its initial value ($\omega_0 / 2$).
Solution:
Part (a) & (b): Elemental Torque and Total Torque
Consider an elemental circular ring on the bottom face of the disc of radius $r$ and radial thickness $dr$.
Area of this ring: $dA = 2\pi r\, dr$.
At radius $r$, the linear speed of the disc surface is $v(r) = \omega r$. Assuming linear velocity profile across the thin oil layer $h$:
$$\tau_{\text{shear}}(r) = \eta \frac{v(r)}{h} = \frac{\eta \omega r}{h}$$
The tangential viscous drag force on the annular element is:
$$dF = \tau_{\text{shear}}\, dA = \left(\frac{\eta \omega r}{h}\right)(2\pi r\, dr) = \frac{2\pi \eta \omega}{h} r^2\, dr$$
The retarding torque about the central axis contributed by this ring is:
$$d\tau = dF \cdot r = \frac{2\pi \eta \omega}{h} r^3\, dr$$
Integrating over the entire disc from $r = 0$ to $r = R$:
$$\tau_{\text{total}} = \int_0^R \frac{2\pi \eta \omega}{h} r^3\, dr = \frac{2\pi \eta \omega}{h} \left[ \frac{r^4}{4} \right]_0^R = \frac{\pi \eta \omega R^4}{2h}$$
Part (c): Time to Halve Angular Velocity
The rotational equation of motion for the disc (with moment of inertia $I = \frac{1}{2} M R^2$) is:
$$I \frac{d\omega}{dt} = -\tau_{\text{total}} \implies \left(\frac{1}{2} M R^2\right) \frac{d\omega}{dt} = -\frac{\pi \eta \omega R^4}{2h}$$
Canceling $\frac{1}{2} R^2$ from both sides:
$$M \frac{d\omega}{dt} = -\frac{\pi \eta R^2}{h} \omega \implies \frac{d\omega}{\omega} = -\left( \frac{\pi \eta R^2}{M h} \right) dt$$
Integrating from $t = 0$ ($\omega = \omega_0$) to time $t$ ($\omega = \omega_0 / 2$):
$$\int_{\omega_0}^{\omega_0 / 2} \frac{d\omega}{\omega} = -\frac{\pi \eta R^2}{M h} \int_0^t dt$$
$$\ln\left(\frac{1}{2}\right) = -\frac{\pi \eta R^2}{M h} t \implies -\ln 2 = -\frac{\pi \eta R^2}{M h} t$$
$$t = \frac{M h \ln 2}{\pi \eta R^2}$$
Key Takeaway: Because viscous torque is linearly proportional to angular speed ($\tau \propto \omega$), the spinning disc undergoes exponential decay $\omega(t) = \omega_0 e^{-t/\tau_c}$, where the characteristic decay time constant is $\tau_c = \frac{M h}{\pi \eta R^2}$.
3. Practice Worksheet (10 Questions)
Questions 1 to 5: Single Correct (JEE Main level). Questions 6 to 8: One or More than One Correct (JEE Advanced level). Questions 9 and 10: Numerical Value (JEE Advanced level).
Question 1 (JEE Main Single Correct)
The dimensional formula of dynamic viscosity $\eta$ is:
(A) $[M^1 L^{-1} T^{-1}]$
(B) $[M^1 L^1 T^{-1}]$
(C) $[M^1 L^{-2} T^{-2}]$
(D) $[M^1 L^{-1} T^{-2}]$
Question 2 (JEE Main Single Correct)
With an increase in temperature, the viscosity of liquids and the viscosity of gases respectively:
(A) Increases, increases
(B) Decreases, increases
(C) Increases, decreases
(D) Decreases, decreases
Question 3 (JEE Main Single Correct)
A metal plate of area $100\text{ cm}^2$ rests on a layer of castor oil $2.0\text{ mm}$ thick. The coefficient of viscosity of castor oil is $1.5\text{ Pa}\cdot\text{s}$. The horizontal force required to move the plate with a velocity of $3.0\text{ cm/s}$ is:
(A) $0.225\text{ N}$
(B) $0.0225\text{ N}$
(C) $2.25\text{ N}$
(D) $0.450\text{ N}$
Question 4 (JEE Main Single Correct)
According to Stoke’s law, the viscous drag force $F$ acting on a spherical raindrop of radius $r$ moving with velocity $v$ in air of dynamic viscosity $\eta$ satisfies:
(A) $F \propto r^2 v$
(B) $F \propto r v^2$
(C) $F \propto r v$
(D) $F \propto r^2 v^2$
Question 5 (JEE Main Single Correct)
One Poise is equal to how many SI units ($\text{Pa}\cdot\text{s}$ or $\text{N}\cdot\text{s/m}^2$):
(A) $10\text{ Pa}\cdot\text{s}$
(B) $1\text{ Pa}\cdot\text{s}$
(C) $0.1\text{ Pa}\cdot\text{s}$
(D) $0.01\text{ Pa}\cdot\text{s}$
Question 6 (JEE Advanced One or More Correct)
Which of the following conditions is/are strictly necessary for Stoke’s law $F = 6\pi \eta r v$ to remain valid?
(A) The moving body must be rigid and perfectly spherical.
(B) The flow around the sphere must be laminar with very small Reynolds number ($\text{Re} \ll 1$).
(C) The fluid medium must be infinite in extent with no boundary walls nearby.
(D) The fluid must be highly compressible so that shockwaves cushion the front surface.
Question 7 (JEE Advanced One or More Correct)
Two liquid layers are sheared relative to each other under steady state. Which of the following statements is/are correct regarding the viscosity and shear stress?
(A) Viscous shear stress has dimensions of pressure ($[M^1 L^{-1} T^{-2}]$).
(B) The rate of mechanical energy dissipated per unit volume in a fluid sheared at velocity gradient $\frac{dv}{dy}$ is $\eta \left(\frac{dv}{dy}\right)^2$.
(C) Kinematic viscosity has the dimensional formula $[L^2 T^{-1}]$.
(D) Water at room temperature has dynamic viscosity greater than that of honey.
Question 8 (JEE Advanced One or More Correct)
A spherical steel ball of radius $r$ is falling through a tall column of glycerine of dynamic viscosity $\eta$. Which of the following forces act on the ball?
(A) Downward gravitational force of magnitude $\frac{4}{3}\pi r^3 \rho_{\text{steel}} g$.
(B) Upward buoyant force of magnitude $\frac{4}{3}\pi r^3 \rho_{\text{gly}} g$.
(C) Viscous drag force directed opposite to velocity of magnitude $6\pi \eta r v$.
(D) The viscous drag force depends directly on the acceleration $\frac{dv}{dt}$ rather than the velocity $v$.
Question 9 (JEE Advanced Numerical Value)
A square plate of side length $L = 20\text{ cm}$ is pulled horizontally at a constant speed of $v = 0.5\text{ m/s}$ across an oil film of thickness $h = 1.0\text{ mm}$ resting on a flat floor. The dynamic viscosity of the oil is $\eta = 2.5\text{ Pa}\cdot\text{s}$. Find the horizontal force $F$ (in Newtons) required to pull the plate.
Question 10 (JEE Advanced Numerical Value)
Two spherical droplets of water with radii $r_1 = 1.0\text{ mm}$ and $r_2 = 2.0\text{ mm}$ move through air of dynamic viscosity $\eta = 2.0 \times 10^{-5}\text{ Pa}\cdot\text{s}$ at speeds of $v_1 = 3.0\text{ m/s}$ and $v_2 = 1.5\text{ m/s}$ respectively. Find the ratio $F_1 / F_2$ of the viscous drag forces acting on the two droplets.
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Type | Correct Answer |
|---|---|---|
| 1 | Single Correct | (A) |
| 2 | Single Correct | (B) |
| 3 | Single Correct | (A) |
| 4 | Single Correct | (C) |
| 5 | Single Correct | (C) |
| 6 | One or More Correct | (A, B, C) |
| 7 | One or More Correct | (A, B, C) |
| 8 | One or More Correct | (A, B, C) |
| 9 | Numerical | 50 |
| 10 | Numerical | 1 |
Step-by-Step Solutions
Solution 1:
From Newton’s law of viscosity, $\tau = \frac{F}{A} = \eta \frac{dv}{dy}$:
$$[\eta] = \frac{[F/A]}{[dv/dy]} = \frac{[M L^{-1} T^{-2}]}{[L T^{-1} / L]} = \frac{[M L^{-1} T^{-2}]}{[T^{-1}]} = [M^1 L^{-1} T^{-1}]$$
Correct Answer: (A)
Solution 2:
In liquids, viscosity arises from cohesive intermolecular forces, which weaken as temperature rises, so liquid viscosity decreases.
In gases, viscosity arises from the thermal molecular diffusion and momentum exchange across adjacent layers. Since mean molecular speed increases with temperature ($\bar{v} \propto \sqrt{T}$), gas viscosity increases.
Correct Answer: (B)
Solution 3:
Area $A = 100\text{ cm}^2 = 100 \times 10^{-4}\text{ m}^2 = 10^{-2}\text{ m}^2$.
Thickness $t = 2.0\text{ mm} = 2.0 \times 10^{-3}\text{ m}$.
Velocity $v = 3.0\text{ cm/s} = 3.0 \times 10^{-2}\text{ m/s}$.
Velocity gradient $\frac{dv}{dy} = \frac{v}{t} = \frac{3.0 \times 10^{-2}}{2.0 \times 10^{-3}} = 15\text{ s}^{-1}$.
Viscous force $F = \eta A \frac{dv}{dy} = 1.5 \times 10^{-2} \times 15 = 0.225\text{ N}$.
Correct Answer: (A)
Solution 4:
According to Stoke’s law, $F_v = 6\pi \eta r v$. Therefore, $F \propto r v$.
Correct Answer: (C)
Solution 5:
$1\text{ Poise} = 1\text{ dyne}\cdot\text{s/cm}^2 = \frac{10^{-5}\text{ N}\cdot\text{s}}{10^{-4}\text{ m}^2} = 0.1\text{ N}\cdot\text{s/m}^2 = 0.1\text{ Pa}\cdot\text{s}$.
Correct Answer: (C)
Solution 6:
Stoke’s law requires:
– A rigid and smooth spherical body.
– Low Reynolds number laminar creeping flow ($\text{Re} \ll 1$) with no turbulence.
– An infinite, unbounded continuous medium.
Compressibility of the fluid is not an assumption (in fact, Stoke’s law assumes incompressible fluid flow).
Correct Answer: (A, B, C)
Solution 7:
(A) Shear stress $\tau = F/A$ has units of $\text{N/m}^2$, which has the dimensions of pressure: $[M^1 L^{-1} T^{-2}]$. Statement (A) is correct.
(B) Power dissipated per unit volume is $\tau \frac{dv}{dy} = \left(\eta \frac{dv}{dy}\right)\frac{dv}{dy} = \eta \left(\frac{dv}{dy}\right)^2$. Statement (B) is correct.
(C) Kinematic viscosity $\nu = \eta / \rho$ has dimension $[M L^{-1} T^{-1}] / [M L^{-3}] = [L^2 T^{-1}]$. Statement (C) is correct.
(D) Honey is significantly more viscous than water ($\eta_{\text{honey}} \gg \eta_{\text{water}}$). Statement (D) is incorrect.
Correct Answer: (A, B, C)
Solution 8:
For a sphere falling through a viscous liquid column:
– Downward gravity: $W = V \rho_{\text{steel}} g = \frac{4}{3}\pi r^3 \rho_{\text{steel}} g$.
– Upward buoyant force: $F_B = V \rho_{\text{gly}} g = \frac{4}{3}\pi r^3 \rho_{\text{gly}} g$.
– Upward viscous drag force: $F_v = 6\pi \eta r v$, which depends directly on velocity $v$, not acceleration $\frac{dv}{dt}$.
Correct Answer: (A, B, C)
Solution 9:
Side length $L = 20\text{ cm} = 0.2\text{ m}$. Area $A = L^2 = (0.2)^2 = 0.04\text{ m}^2$.
Oil film thickness $h = 1.0\text{ mm} = 1.0 \times 10^{-3}\text{ m}$.
Velocity $v = 0.5\text{ m/s}$.
Viscosity $\eta = 2.5\text{ Pa}\cdot\text{s}$.
Applied force $F = \eta A \frac{v}{h} = 2.5 \times 0.04 \times \frac{0.5}{1.0 \times 10^{-3}} = 2.5 \times 0.04 \times 500 = 2.5 \times 20 = 50\text{ N}$.
Correct Answer: 50
Solution 10:
By Stoke’s law, $F = 6\pi \eta r v$.
For droplet 1: $F_1 = 6\pi \eta r_1 v_1$.
For droplet 2: $F_2 = 6\pi \eta r_2 v_2$.
Ratio:
$$\frac{F_1}{F_2} = \frac{r_1 v_1}{r_2 v_2} = \frac{(1.0\text{ mm}) \times (3.0\text{ m/s})}{(2.0\text{ mm}) \times (1.5\text{ m/s})} = \frac{3.0}{3.0} = 1$$
Correct Answer: 1