The viscous drag acting on a metal sphere of diameter $1 ext{ mm}$, falling through a fluid of viscosity $0.

Question

The viscous drag acting on a metal sphere of diameter $1 	ext{ mm}$, falling through a fluid of viscosity $0.8 	ext{ Pa-s}$ with a velocity of $2 	ext{ m s}^{-1}$ is nearly equal to:

Accepted Answer

According to Stokes' Law, the viscous drag force ($F$) acting on a spherical body moving through a viscous fluid is given by:
$$F = 6\pi \eta r v$$

**Given:**
*   Viscosity, $\eta = 0.8 	ext{ Pa-s}$ (or $	ext{kg m}^{-1} 	ext{s}^{-1}$)
*   Velocity, $v = 2 	ext{ m s}^{-1}$
*   Diameter, $d = 1 	ext{ mm} \Rightarrow$ Radius, $r = 0.5 	ext{ mm} = 0.5 	imes 10^{-3} 	ext{ m}$

**Calculation:**
$$F = 6 	imes 3.14 	imes 0.8 	imes (0.5 	imes 10^{-3}) 	imes 2$$
$$F = 6 	imes 3.14 	imes 0.8 	imes 10^{-3}$$
$$F = 15.072 	imes 10^{-3} 	ext{ N}$$

Rounding to the nearest integer, the force is approximately $15 	imes 10^{-3} 	ext{ N}$.

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