Q.A tall cylinder is filled with viscous oil. A round pebble is dropped from the top of the cylinder with zero initial velocity, so it falls downward through the oil. Consider four possible plots of the pebble's speed $v$ (vertical axis) against time $t$ (horizontal axis):

The Intuition: Why a Falling Object Stops Accelerating

Drop a steel ball bearing into a jar of honey. You expect it to fall, but it doesn't crash down like it would in air. Instead, it sinks slowly and steadily, at a constant speed, almost from the moment you let go. That constant speed is the terminal velocity.

Why does it happen? Two forces are at work. Gravity pulls the ball down. But the honey resists its motion — that's viscous drag. At the start, the ball is moving slowly, so the drag is small. Gravity wins, and the ball accelerates. But as speed increases, drag grows. Eventually, drag becomes exactly as strong as the net downward pull (weight minus buoyancy). At that point, the net force is zero. No net force means no acceleration — the ball continues at whatever speed it has reached. That speed is terminal velocity.

The Precise Statement

For a small, smooth sphere falling through a viscous fluid at low speeds (so the flow is laminar, not turbulent), the drag force is given by Stokes' law:

where $\eta$ is the fluid's viscosity, $r$ is the sphere's radius, and $v$ is its speed.

The other forces are:

• Weight downward: $W = \frac{4}{3}\pi r^3 \rho g$ (where $\rho$ is the sphere's density)
• Buoyant force upward: $F_b = \frac{4}{3}\pi r^3 \sigma g$ (where $\sigma$ is the fluid's density)

The net downward force (weight minus buoyancy) is:

At terminal velocity, drag equals this net downward force:

Solve for $v_t$:

What the Formula Tells You

• Radius squared — a larger sphere falls much faster. Double the radius, quadruple the terminal speed.
• Density difference — if the sphere is only slightly denser than the fluid, it falls slowly. If it's lighter than the fluid ($\rho < \sigma$), $v_t$ becomes negative — it rises.
• Viscosity — thicker fluids (higher $\eta$) give a lower terminal speed. Honey stops a ball far more than water does.

A Concrete Example

A steel ball bearing ($\rho = 7800\ \text{kg/m}^3$, radius $1\ \text{mm} = 0.001\ \text{m}$) falls through glycerine ($\sigma = 1260\ \text{kg/m}^3$, $\eta = 1.5\ \text{Pa·s}$). Take $g = 9.8\ \text{m/s}^2$.

First, $\rho - \sigma = 6540\ \text{kg/m}^3$. Then $r^2 = 10^{-6}\ \text{m}^2$. So numerator: $2 \times 10^{-6} \times 6540 \times 9.8 \approx 0.128$. Denominator: $13.5$. So $v_t \approx 0.0095\ \text{m/s}$, or about 1 cm/s. That matches the slow, steady sinking you'd see.

The Big Picture

Terminal velocity is a universal idea, not just for spheres in honey. A skydiver reaches terminal velocity when air drag balances weight — though the drag law is different (proportional to $v^2$, not $v$). A raindrop falls at terminal speed; that's why small drops drift and large ones hurt when they hit you. The core concept is always the same: when the resisting force grows with speed, eventually it catches up to the driving force, and motion becomes uniform.

Terminal velocity, derived from Stokes' law, is part of the NCERT/CBSE Class 11 Physics fluid mechanics curriculum, and "terminal velocity formula and derivation class 11 physics" is a heavily searched revision topic. It's also a classic JEE Main and NEET important-question type combining viscosity, buoyancy, and weight in one numerical.