Q.(a) Derive the expression for the terminal velocity of a sphere moving in a high viscous fluid using Stoke's law. OR

Stokes' law: F = 6πηrv — the viscous drag force on a small sphere moving through a fluid.

When a small spherical body moves slowly through a viscous fluid, the fluid exerts a retarding (drag) force on it that opposes its motion. Sir George Stokes showed that, for a sphere of radius r moving with speed v through a fluid of viscosity η (at low Reynolds number, i.e. streamline/laminar flow), this drag force is F = 6πηrv. This law explains why small objects (like raindrops, or particles settling in a liquid) …

Terminal velocity is proportional to r^2, so doubling r gives 4x. Answer (B).

For a small sphere falling through a viscous medium, the terminal velocity is:

v_t = 2 r^2 (rho - sigma) g /(9 eta),

which is proportional to r^2 (other quantities fixed).

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If terminal velocities are in ratio 1:4, the masses of the raindrops are in ratio 1:8.

By Stokes' law, terminal velocity of a spherical drop is:

$v_t = \dfrac{2r^2(\rho-\sigma)g}{9\eta} \propto r^2$

Given $v_1 : v_2 = 1 : 4$:

$r_1^2 : r_2^2 = 1 : 4 \Rightarrow r_1 : r_2 = 1 : 2$

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Terminal velocity depends on radius squared, but mass depends on radius cubed — so working from the mass ratio (8:1) back to the radius ratio (2:1) gives a terminal-velocity ratio of 4:1.

By Stokes' law, the terminal velocity of a sphere falling through a viscous fluid is:

$v_t = \left[2r^2(\rho - \sigma)g\right]/(9\eta)$

For spheres of the same material (same density $\rho$) falling in the same fluid (same $\sigma, \eta$), this reduces to $v_t \propto r^2$.

Mass of a sphere of the same material: $m \propto r^3$ (density constant).

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Acceleration is zero once terminal velocity is reached.

As a body falls through a viscous fluid, the upward viscous drag force ($6\pi\eta rv$ by Stokes' law) increases with speed until it, together with buoyancy, exactly balances the downward weight. At that point the net force is zero, so by …

Terminal velocity is, by definition, the constant velocity reached when net force = 0, so acceleration at and after that point is zero.

As a body falls through a viscous fluid, it experiences: weight (mg, downward), buoyant force (upward), and viscous drag (upward, increasing with speed, given by Stokes' law $F = 6\pi\eta r v$ for a sphere). As speed increases, drag increases until:

$mg = F_{buoyant} + F_{viscous}$

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Terminal velocity is, by definition, the constant speed reached once the net force (and hence acceleration) on the falling body becomes zero.

As a body falls through a viscous fluid, it experiences three forces: weight (mg, downward), buoyant force (upward), and viscous drag (upward, increasing with speed per Stokes' law, F = 6 pi eta r v). Initially, weight exceeds the upward forces, so the body accelerates. As its speed increases, the drag force grows until, at some speed, the upward forces (drag + buoyancy) exactly balance the downward weight: …

A rain drop's speed increases only up to a point; once the upward viscous drag force equals the downward weight, it reaches a constant 'terminal velocity' and stops accelerating.

As a rain drop falls through air, two forces act on it: its weight (mg, downward, constant) and a viscous drag force (upward, given by Stokes' law as F = 6pietarv, which increases as the drop's speed v increases). Initially, weight exceeds drag, so the drop accelerates and speeds up. But as v grows, drag grows too, until eventually drag balances weight:

6pietarv_terminal = mg

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