125 small water drops of same size fall through air with constant velocity $4 \mathrm{~cm} / \mathrm{s}$. They coalesce to form a big drop. The terminal velocity of the big drop is

Let a small water drop have radius $r$ and the large coalesced drop have radius $R$. Volume conservation requires that 125 small drops equal one large drop: $125 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$. Simplifying yields $125 r^3 = R^3$, so $R = 5r$.

For a spherical drop falling through a viscous medium, terminal velocity $v_t$ follows Stokes' Law, $v_t = \frac{2 r^2 g (\rho - \sigma)}{9 \eta}$, implying $v_t \propto r^2$. Let $v_{t,big}$ and $v_{t,small}$ denote the terminal velocities of the large and small drops, respectively. The ratio is $\frac{v_{t,big}}{v_{t,small}} = \frac{R^2}{r^2} = \frac{(5r)^2}{r^2} = 25$.

Given $v_{t,small} = 4\ \mathrm{cm/s}$, it follows that $v_{t,big} = 25 \times 4\ \mathrm{cm/s} = 100\ \mathrm{cm/s} = 1\ \mathrm{m/s}$.

Final answer: $\boxed{\text{B}}$