The terminal velocity of a copper ball of radius $5 ext{ mm}$ falling through a tank of oil at room temperat

Question

The terminal velocity of a copper ball of radius $5 	ext{ mm}$ falling through a tank of oil at room temperature is $10 	ext{ cm s}^{-1}$. If the viscosity of oil at room temperature is $0.9 	ext{ kg m}^{-1} 	ext{ s}^{-1}$, the viscous drag force is:

Accepted Answer

According to Stokes' Law, the viscous drag force ($F$) acting on a spherical body of radius $r$ moving with terminal velocity $v$ in a fluid of viscosity $\eta$ is given by the formula:
$$F = 6\pi \eta r v$$

**Given values:**
*   Viscosity, $\eta = 0.9 	ext{ kg m}^{-1} 	ext{ s}^{-1}$
*   Radius, $r = 5 	ext{ mm} = 5 	imes 10^{-3} 	ext{ m}$
*   Velocity, $v = 10 	ext{ cm s}^{-1} = 0.1 	ext{ m s}^{-1}$

**Calculation:**
Substitute the values into the formula:
$$F = 6 	imes 3.14159 	imes 0.9 	imes (5 	imes 10^{-3}) 	imes 0.1$$
$$F = 6 	imes 0.9 	imes 0.5 	imes 10^{-3} 	imes 3.14159$$
$$F = 5.4 	imes 5 	imes 10^{-4} 	imes 3.14159$$
$$F = 2.7 	imes 10^{-3} 	imes 3.14159$$
$$F \approx 8.48 	imes 10^{-3} 	ext{ N}$$

Step-by-Step Solution

Practice All PHYSICS Questions