Question

If the terminal speed of spere of gold density = 19.5 kg /m cube is 0.2 m/s in a viscous liquid density = 1.5 kg / m cbe , find the terminal speed of a spere of silver (density = 10.5 kg /m cube ) of the same size in same liquid .

Answer 1

If the terminal speed of sphere of gold density = 19.5 kg/m³ is 0.2 m/s in a viscous liquid density = 1.5 kg/m³, find the terminal speed of a sphere of silver (density = 10.5 kg/m³) of the same size in the same liquid.

Terminal speed can be calculated from:

$$V _ {t} = \sqrt {\frac {2 m g}{\rho_ {f} A C _ {d}}}$$

where, $m$ - is the mass of the falling object, $g$ - the acceleration due to gravity,

$C_d$ is the drag coefficient, $\rho_f$ is the density of the fluid through which the object is falling, and $A$ is the projected area of the object.

Assuming $m = V\rho$ we can write

$$v _ {t} = \alpha \sqrt {\rho}$$

where $\alpha = \sqrt{\frac{2Vg}{\rho AC_d}}$ - the same for all spheres equal size in the same liquid.

Therefore, terminal speed of sphere of gold equals:

$$v _ {t} (g) = \alpha \sqrt {\rho_ {g}} \quad \alpha = \frac {v _ {t} (g)}{\sqrt {\rho_ {g}}}$$

where $\rho_{g}$ - density of gold

And terminal speed of sphere of silver equals:

$$v _ {t} (s) = \alpha \sqrt {\rho_ {s}} = \frac {v _ {t} (g)}{\sqrt {\rho_ {g}}} \sqrt {\rho_ {s}} = 0.2 \frac {m}{s} \sqrt {\frac {10.5}{19.5}} = 0.15 \frac {m}{s}$$

where $\rho_{g}$ - density of silver

Answer: $0.15\frac{m}{s}$