Question

Eight spherical drops of equal size are falling vertically through air with a terminal velocity $0.1{\text{ms}}^{-1}$. If the drops coalesce to form a large spherical drop, its terminal velocity would be

• A. $0.1{\text{ms}}^{-1}$
• B. $0.2{\text{ms}}^{-1}$
• C. $0.3{\text{ms}}^{-1}$
• D. $0.4{\text{ms}}^{-1}$

Answer 1

The correct option is D $0.4{\text{ms}}^{-1}$

We know that the terminal veocity $v_T$ of a body is given as-

$v_T=\frac{2r^2}{9\eta }(\rho -\sigma )g$

where $r$ is the radius, $\eta$ is the viscosity of the medium and $\rho$ is the density of the sphere and $\sigma$ is the density of the medium.

$\Rightarrow v_T\propto r^2......\left(1\right)$

Now, let the radius of the big drop be $R$ and the radius of the small drops be $r$. As $n$ small drops combine to form a big drop.

Volume of big drop $V_{\text{big}}=n{V}_{\text{small}}$

$\frac{4}{3}\pi R^3=n\frac{4}{3}\pi r^3$

$R^3=n{r}^3$

$\Rightarrow R=n^{1/3}r$

$\therefore \frac{R}{r}=n^{1/3}$

Now, considering the ratio of the two terminal velocities for the small drop and big drop we have from $\left(1\right)$

$\frac{v_T^{\prime }}{v_T}=\frac{R^2}{r^2}$;

$\frac{v_T^{\prime }}{v_T}=n^{2/3}$

$v_T^{\prime }=(8{)}^{2/3}{v}_T$ (Given $n=8$)

$v_T^{\prime }=2^2\times 0.1$

$v_T^{\prime }=0.4{\text{ms}}^{-1}$