Quick Summary
Concept Tested: Viscous drag on a small sphere (Stokes’ law)
Chapter – Subtopic: Physics – Solid State – Motion of Small Particles
Difficulty: ★★☆☆☆ (2/5)
Estimated Time: 2 minutes
Key Formula: $$v=\frac{2r^{2}(\rho-\rho_a)g}{9\eta},\qquad F=6\pi\eta r v$$
Answer: Option B
One‑line Reason: Substituting the given data yields a viscous force of $3.9\times10^{-10}\,\text{N}$, which matches option B.
The Question
An oil drop of radius $r = 2.0\times10^{-5}\,\text{m}$ and density $\rho = 1.2\times10^{3}\,\text{kg/m}^3$ falls through air of viscosity $\eta = 1.8\times10^{-5}\,\text{N·s/m}^2$. The density of air is to be neglected ($\rho_a = 0$). Find the magnitude of the viscous force acting on the drop at terminal velocity.
(A) $3.8\times10^{-10}\,\text{N}$
(B) $3.9\times10^{-10}\,\text{N}$
(C) $1.8\times10^{-10}\,\text{N}$
(D) $5.8\times10^{-10}\,\text{N}$
Quick Answer
The correct choice is Option B. Using Stokes’ law, the terminal velocity is $5.81\times10^{-2}\,\text{m/s}$ and the corresponding viscous drag $F = 6\pi\eta r v$ evaluates to $3.9\times10^{-10}\,\text{N}$.
Why Other Options Are Incorrect
Option A (3.8×10⁻¹⁰ N): This value arises if the radius is not squared correctly in the velocity formula, leading to a slightly lower terminal speed and hence a lower force.
Option C (1.8×10⁻¹⁰ N): This result is obtained when the viscosity of the oil (instead of air) is mistakenly used, drastically reducing the drag force.
Option D (5.8×10⁻¹⁰ N): This corresponds to neglecting the factor $2/9$ in Stokes’ law or applying an incorrect buoyancy correction, which inflates the calculated force.
Video Solution
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Understanding the Concept
When a small sphere moves slowly through a viscous fluid, the drag force is given by Stokes’ law: $F_{\text{drag}} = 6\pi\eta r v$. At terminal velocity the net force on the sphere is zero, so the drag balances the effective weight $(\rho-\rho_a)g\,\frac{4}{3}\pi r^3$. Solving for $v$ gives the terminal‑velocity expression shown above.
Detailed Step-by-Step Solution
Step 1: List the Given Quantities
Radius of oil drop: $r = 2.0\times10^{-5}\,\text{m}$
Density of oil: $\rho = 1.2\times10^{3}\,\text{kg/m}^3$
Viscosity of air: $\eta = 1.8\times10^{-5}\,\text{N·s/m}^2$
Density of air (neglected): $\rho_a = 0$
Acceleration due to gravity: $g = 9.8\,\text{m/s}^2$
Step 2: Compute the Terminal Velocity
Using Stokes’ law for terminal velocity:
$$v=\frac{2r^{2}(\rho-\rho_a)g}{9\eta}$$
Substituting the values:
$$v=\frac{2\,(2.0\times10^{-5})^{2}\,(1.2\times10^{3})\,(9.8)}{9\,(1.8\times10^{-5})}$$
Calculate the numerator:
$$2\,(2.0\times10^{-5})^{2}=2\,(4.0\times10^{-10})=8.0\times10^{-10}$$
$$8.0\times10^{-10}\times1.2\times10^{3}=9.6\times10^{-7}$$
$$9.6\times10^{-7}\times9.8=9.408\times10^{-6}$$
Denominator:
$$9\,(1.8\times10^{-5})=1.62\times10^{-4}$$
Thus
$$v=\frac{9.408\times10^{-6}}{1.62\times10^{-4}}=5.807\times10^{-2}\,\text{m/s}$$
Step 3: Compute the Viscous Force
Viscous drag at terminal velocity is
$$F=6\pi\eta r v$$
Substituting:
$$F=6\pi\,(1.8\times10^{-5})\,(2.0\times10^{-5})\,(5.807\times10^{-2})$$
First multiply the constants:
$$6\pi = 18.8496$$
$$18.8496\times1.8\times10^{-5}=3.3929\times10^{-4}$$
$$3.3929\times10^{-4}\times2.0\times10^{-5}=6.7858\times10^{-9}$$
$$6.7858\times10^{-9}\times5.807\times10^{-2}=3.94\times10^{-10}\,\text{N}$$
Rounded to two significant figures, $F \approx 3.9\times10^{-10}\,\text{N}$.
Final Answer
✓ The viscous force acting on the oil drop is $3.9\times10^{-10}\,\text{N}$ (Option B).
Essential Formulas for This Topic
$$v=\frac{2r^{2}(\rho-\rho_a)g}{9\eta}\qquad\text{(Terminal velocity of a sphere)}$$
$$F_{\text{drag}}=6\pi\eta r v\qquad\text{(Stokes’ drag force)}$$
Note: $\rho$ and $\rho_a$ must be in the same units (kg m⁻³), $r$ in meters, $\eta$ in N·s m⁻², and $g$ in m s⁻² for consistency.
Common Mistakes to Avoid
Mistake 1: Forgetting to Square the Radius
In the velocity formula $v$, the radius appears as $r^{2}$. Omitting the square reduces $v$ by a factor of $r$, leading to an incorrect force.
Mistake 2: Using the Oil’s Viscosity Instead of Air’s
The drag is produced by the surrounding fluid (air). Inserting the oil’s viscosity $\eta_{\text{oil}}$ dramatically changes the magnitude of $F$.
Mistake 3: Applying a Buoyancy Correction When It Is Neglected
The problem explicitly states “neglect buoyancy”, i.e., $\rho_a = 0$. Including a buoyancy term $(\rho-\rho_a)$ when $\rho_a=0$ double‑counts the effect and yields a larger force.
Key Concept Summary
- Stokes’ law relates drag force to viscosity, radius, and velocity of a sphere moving slowly in a fluid.
- At terminal velocity, the drag force equals the effective weight of the sphere.
- Accurate substitution of numerical values (including proper powers) is essential for correct results.
- Neglecting buoyancy simplifies the effective density to the sphere’s own density.
Golden Rule: Always verify the units and the presence of squared terms before performing the final calculation.
Frequently Asked Questions
Q: Why is the density of air taken as zero?
A: The problem explicitly instructs to “neglect buoyancy”. Setting $\rho_a = 0$ removes the buoyancy term, simplifying the calculation.
Q: Can Stokes’ law be applied to larger drops?
A: Stokes’ law is valid only for low Reynolds numbers (laminar flow). For larger drops, inertial effects become significant and the law no longer holds.
Q: What would happen if the viscosity value were off by a factor of 10?
A: Since $F = 6\pi\eta r v$, the drag force scales linearly with $\eta$. A ten‑fold error in viscosity would produce a ten‑fold error in the computed force.
Q: How does the terminal velocity change if the drop’s density doubles?
A: From $v = \frac{2r^{2}(\rho-\rho_a)g}{9\eta}$, $v$ is directly proportional to $(\rho-\rho_a)$. Doubling $\rho$ (with $\rho_a=0$) would double the terminal velocity.
Prerequisites to Solve This Question
- Understanding of Stokes’ law and its derivation.
- Ability to identify and apply the terminal‑velocity condition (net force = 0).
- Proficiency with unit conversion and scientific notation.
- Basic algebraic manipulation of equations.
After Solving This, You Can:
- ✔ Accurately compute viscous forces on microscopic particles.
- ✔ Distinguish when Stokes’ law is applicable versus when other drag models are needed.
- ✔ Perform quick estimations of terminal velocities in low‑Reynolds‑number regimes.
- ✔ Interpret experimental data from Millikan‑oil‑drop‑type setups.
Study Tips for This Topic
- Memorize the two core Stokes’ formulas and the conditions under which they hold.
- Practice converting all quantities to SI units before plugging them into equations.
- Work through numerical examples emphasizing the correct placement of squares and powers.
- Review past JEE problems on viscous drag to recognize common traps (e.g., buoyancy, wrong viscosity).
Difficulty Rating & Exam Frequency
Difficulty: ★★☆☆☆ (2/5)
JEE Main Frequency: Occasionally (≈ 5 % of physics questions)
JEE Advanced Frequency: Rare (≈ 2 % of physics questions)
Importance: High for understanding experimental physics and fluid dynamics concepts.
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Last Updated: August 2026
Question Source: JEE Main 2021 PYQ
Topic: Solid State