A lead sphere of mass ' m ' falls in viscous liquid with terminal velocity V 0 . Another lead sphere of mass ' 8 m ' but of same material will fall through the same liquid with t

Question

A lead sphere of mass ' m ' falls in viscous liquid with terminal velocity V 0 ​ . Another lead sphere of mass ' 8 m ' but of same material will fall through the same liquid with terminal velocity

TestHub · Accepted Answer

Terminal velocity of a spherical body of radius $r$ falling through a viscous liquid is given by

$V = \frac{2}{9} \frac{r ^{2} ( ρ - σ ) g}{η}$

Since the material and liquid are the same, $V \propto r^{2}$ .

Mass of the sphere is $m = \frac{4}{3} π r^{3} ρ$

This implies $m \propto r^{3}$ , or $r \propto m^{1/3}$ .

Therefore, $V \propto (m^{1/3})^{2} \Rightarrow V \propto m^{2/3}$ .

For the first sphere, $V_{1} = V_{0}$ and $m_{1} = m$ .

For the second sphere, $m_{2} = 8 m$ .

$\frac{V _{2}}{V _{1}} = (\frac{m _{2}}{m _{1}})^{2/3} = (\frac{8 m}{m})^{2/3} = (8)^{2/3} = 4$

$V_{2} = 4 V_{0}$

Answer: $4 V_{0}$

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