Question

Small water droplets of radius 0.01 mm are formed in the upper atmosphere and falling with a terminal velocity of 10 cm/s. Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new terminal velocity will be ………..cm/s.

Answer 1

Correct Answer: 40

To solve the problem of determining the new terminal velocity of the larger droplet formed by the coalescence of 8 smaller droplets, we can use the physics of terminal velocity and volume conservation. First, consider that the terminal velocity \( v \) of a droplet can be expressed as \( v \propto r^2 \), where \( r \) is the radius of the droplet. When droplets coalesce, their combined volume equals the volume of the larger drop. Given the radius \( r_0 = 0.01 \) mm for each of the smaller droplets and their terminal velocity \( v_0 = 10 \) cm/s, we can calculate the new radius of the larger droplet. The volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). Therefore, the total initial volume of 8 small droplets is \( 8 \times \frac{4}{3} \pi r_0^3 \) and the volume of the larger drop is \( \frac{4}{3} \pi R^3 \). Equating these, we get:

\( R^3 = 8r_0^3 \)

Solving for \( R \), we have:

\( R = 8^{1/3}r_0 = 2r_0 \)

Substituting \( r_0 = 0.01 \) mm, we find:

\( R = 2 \times 0.01 \text{ mm} = 0.02 \text{ mm} \)

Next, calculate the terminal velocity of the larger droplet, using \( v \propto r^2 \). The new terminal velocity \( V \) is given by:

\( \frac{V}{v_0} = \left(\frac{R}{r_0}\right)^2 \)

Substitute the values:

\( \frac{V}{10} = \left(\frac{0.02}{0.01}\right)^2 = 2^2 = 4 \)

Therefore, \( V = 4 \times 10 = 40 \text{ cm/s} \)

The calculated terminal velocity of the larger droplet is indeed 40 cm/s, which fits perfectly within the given range of 40 to 40 cm/s.

Answer 2

The terminal velocity (\(V_t\)) of a droplet is given by:
\[ V_t =\( \frac{2R^2 (\rho - \sigma) g}{9\eta}\), \]
where \(R\) is the radius of the droplet.
When 8 small droplets coalesce:
Mass of larger drop \(M \propto 8m\), \quad \(M = \(\frac{4}{3}\)\pi R^3 (\rho)\).
The radius of the larger drop is:
\[ R_{\text{large}}^3 = 8R_{\text{small}}^3 \implies R_{\text{large}} = 2R_{\text{small}}. \]
Since terminal velocity (\(V_t\)) is proportional to \(R^2\):
\[ V_t \propto R^2 \implies V_t^{\text{large}} = 4V_t^{\text{small}}. \]
For the smaller droplet:
\[ V_t^{\text{small}} = 10 \, \text{cm/s}. \]
Thus:
\[ V_t^{\text{large}} = 4 \cdot 10 = 40 \, \text{cm/s}. \]