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)

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
$V^{\prime }$ = 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 , $r_2=4r_1$
$\therefore \frac{V}{V^{\prime }}=\frac{r_1^2}{{\left(4r_1\right)}^2}=\frac{1}{16}$
or, $V^{\prime }=16V$