Eight equal drops of water are falling through air with a steady...

The terminal velocity can be given by

${v}_{T} = \dfrac{2 ( ρ - \sigma ) {r}^{2}g}{9η}$.

Clearly, ${v}_{T} ∝ {r}^{2}$

Let the radius of the small drops be ${r}_{1}$ and the radius of the big drop be ${r}_{2}$.

So,

$8\left(\dfrac{4\pi}{3}\right){r}_{1}^{3} = \left(\dfrac{4\pi}{3}\right){r}_{2}^{3} ⇒ {r}_{1} = \dfrac{{r}_{2}}{2}$

Hence, taking the ratio
$\dfrac{{\left({v}_{T}\right)}_{1}}{{\left({v}_{T}\right)}_{2}} = {\left(\dfrac{{r}_{1}}{{r}_{2}}\right)}^{2}$
$⇒ \dfrac{10}{{\left({v}_{T}\right)}_{2}} = \dfrac{1}{4}$
$⇒ {\left({v}_{T}\right)}_{2} = 40cm{s}^{- 1}$