Solution

This problem involves calculating the terminal velocity of a large raindrop formed after several smaller drops coalesce. We need to use principles of volume conservation and the relationship between terminal velocity and drop size.

Volume Conservation for Coalescing Drops

When $N$ small drops of radius $r$ coalesce to form a single large drop of radius $R$, the total volume remains constant.

Volume of one small drop: $ V_{small} = \frac{4}{3} \pi r^3 $

Volume of the large drop: $ V_{large} = \frac{4}{3} \pi R^3 $

Total volume conservation implies: $ N \times V_{small} = V_{large} $

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

$ N r^3 = R^3 $

Taking the cube root of both sides gives the relationship between the radii:

$ R = N^{1/3} r $

Given $N = 64$ and $r = 1$ mm:

$ R = (64)^{1/3} \times (1 \text{ mm}) = 4 \times 1 \text{ mm} = 4 \text{ mm} $

The radius of the larger drop is 4 mm.

Terminal Velocity Dependence on Radius

The terminal velocity ($v_t$) of a spherical object falling through a fluid is often approximated as being proportional to the square of its radius, especially under certain conditions (like those approximated by Stokes' law for small Reynolds numbers, or simplified models):

$ v_t \propto r^2 $

Therefore, the ratio of the terminal velocities of the large drop ($v_2$) and the small drops ($v_1$) is related to the ratio of their radii squared:

$ \frac{v_2}{v_1} = \left(\frac{R}{r}\right)^2 $

Calculating the New Terminal Velocity

We have the initial velocity $v_1 = 10$ cm/s, the initial radius $r = 1$ mm, and the final radius $R = 4$ mm.

Substitute these values into the velocity ratio equation:

$ \frac{v_2}{10 \text{ cm/s}} = \left(\frac{4 \text{ mm}}{1 \text{ mm}}\right)^2 $

$ \frac{v_2}{10 \text{ cm/s}} = (4)^2 = 16 $

Now, solve for $v_2$:

$ v_2 = 16 \times 10 \text{ cm/s} $

$ v_2 = 160 \text{ cm/s} $

The terminal velocity of the larger drop is 160 cm/s.