Question

A steel ball of mass m falls in a viscous medium with a terminal velocity V. Another steel ball of mass 64m will fall through the same liquid with terminal velocity

• A. V
• B. 4V
• C. 8V
• D. 16V

Answer 1

Answer: (D)

16V

Answer 2

Terminal velocity , $v_{1} = \frac{2}{9} r^{2} \frac{\left(\rho-\sigma\right)}{\eta} g$
$i.e., v_{1} \propto r^{2}$
$r_1$ = Radius of ball of mass $m$
$V$ = Terminal velocity of this ball
$r_2$ = Radius of ball of mass 64 m
$V'$ = Terminal velocity of larger ball
Now, mass= density $\times$ volume.
Volume ohhe larger ball is 64 times that of the smaller ball.
So, $r_2 = (64)^{1/3} r_1$
or , $r_2 = 4r_1$
$\therefore\:\: \frac{V}{V'} = \frac{r^{2}_{1}}{\left(4r_{1}\right)^{2}} = \frac{1}{16}$
or, $V' = 16 V$