---
title: "Terminal velocity of steel sphere in glycerine"
subject: "Fluid Mechanics"
chapter: 9
key: "semester_3/fluid_mechanics:sphere-terminal-velocity-glycerine"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/sphere-terminal-velocity-glycerine
site: BCT Engineering
---

# Terminal velocity of steel sphere in glycerine

Fluid Mechanics · Chapter 9: Flow past through submerged bodies

- Asked in: 2075 Chaitra, 4 marks
- Video: [Coefficient of Viscosity of glycerine](https://www.youtube.com/watch?v=vFNTHarVPG0) — NCERT OFFICIAL
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/sphere-terminal-velocity-glycerine
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-9

## Question

A 3 mm diameter sphere made of steel (sp. wt. 75 KN/m³) falls in glycerine (sp. wt. 12.5 KN/m³) of viscosity 0.893 NS/m² at a terminal velocity. Determine the terminal velocity and drag force on the sphere.

## Short answer

Terminal velocity under Stokes' flow: V = (γ_s-γ_f)d²/(18μ) = (75000-12500)×0.003²/(18×0.893) = 0.035 m/s. Drag force F_D = 3πμVd = 8.84×10^-4 N (Re=0.15).

## Answer

At terminal velocity, the weight minus buoyancy equals the drag. For a small sphere in a viscous liquid, assume Stokes' law, and check $Re$ afterwards.

#### Given

$d = 3$ mm $= 0.003$ m, $\gamma_s = 75\ \text{kN/m}^3$, $\gamma_f = 12.5\ \text{kN/m}^3$, $\mu = 0.893\ \text{N s/m}^2$.

#### Terminal velocity

Net downward force: $(\gamma_s - \gamma_f)\dfrac{\pi d^3}{6}$. Stokes' drag: $F_D = 3\pi\mu V d$.

$$
\begin{aligned}
3\pi\mu V d &= (\gamma_s - \gamma_f)\frac{\pi d^3}{6} \\
V &= \frac{(\gamma_s - \gamma_f)\,d^2}{18\mu} = \frac{(75\,000 - 12\,500)(0.003)^2}{18(0.893)} = 0.0350\ \text{m/s}
\end{aligned}
$$

#### Check of Reynolds number

$\rho_f = 12\,500/9.81 = 1274\ \text{kg/m}^3$

$$
Re = \frac{\rho_f V d}{\mu} = \frac{1274(0.0350)(0.003)}{0.893} = 0.15 < 1
$$

So Stokes' law is valid.

#### Drag force

$$
F_D = 3\pi\mu V d = 3\pi(0.893)(0.0350)(0.003) = 8.84\times10^{-4}\ \text{N}
$$

Check: $(\gamma_s - \gamma_f)\pi d^3/6 = 62\,500 \times 1.414\times10^{-8} = 8.84\times10^{-4}$ N.

**Answer: terminal velocity $\approx 0.035$ m/s (3.5 cm/s); drag force $\approx 0.88$ mN ($8.84\times10^{-4}$ N).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What balance defines terminal velocity?
   A. Weight equals buoyancy
   B. Drag equals buoyancy only
   C. Weight equals drag only
   D. Weight minus buoyancy equals drag
   Answer: D. Net downward force equals drag. (Key point: Terminal condition)

2. Using Stokes' law, V =
   A. (gamma_s + gamma_f) d^2/(18 mu)
   B. (gamma_s - gamma_f) d/(3 mu)
   C. (gamma_s - gamma_f) d^2/(18 mu) = 0.035 m/s
   D. 3.5 m/s
   Answer: C. Equating 3 pi mu V d with (gs - gf) pi d^3/6. (Key point: Terminal velocity)

3. Why check Re afterwards?
   A. Because Re must be above 2000
   B. Stokes' law is valid only if Re < 1; here Re = 0.15
   C. To find the density
   D. To find buoyancy
   Answer: B. The Stokes assumption must be verified. (Key point: Check)

4. What is the drag?
   A. About 8.84 x 10^-4 N, equal to the net weight
   B. About 0.88 N
   C. About 3.5 N
   D. Zero
   Answer: A. F_D = 3 pi mu V d equals the net force. (Key point: Drag force)

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