Q.Write down Stokes' law. Using this law find the expression for the terminal velocity of a body falling freely in an infinitely extended homogeneous liquid. Also draw the variation graph of terminal velocity with time. OR Define angle of contact and capillary action. Find the relation between surface tension and the angle of contact for a liquid rising in a column. Also give any two examples of capillarity in daily life.

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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