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

<p>To find the new terminal velocity of the larger drop formed by the coalescence of 8 smaller droplets, we need to understand the relationship between the radius of the droplets and their terminal velocity.</p> <p>The terminal velocity for a small spherical droplet falling through the air is given by the Stokes' law:</p> <p>$v_t = \frac{2}{9} \frac{r^2 (\rho - \rho_{\text{air}}) g}{\eta}$</p> <p>where:</p> <ul> <li>$v_t$ is the terminal velocity.</li> <li>$r$ is the radius of the droplet.</li> <li>$\rho$ is the density of the droplet.</li> <li>$\rho_{\text{air}}$ is the density of the air.</li> <li>$g$ is the acceleration due to gravity.</li> <li>$\eta$ is the viscosity of the air.</li> </ul> <p>Given that the radius of the small droplets is $0.01 \ \text{mm}$ and their terminal velocity is $10 \ \text{cm/s}$, we now need to determine the radius of the larger drop formed by the coalescence of 8 smaller droplets.</p> <p>When droplets coalesce, the volume of the larger drop is equal to the sum of the volumes of the smaller droplets. The volume of a sphere is given by:</p> <p>$V = \frac{4}{3} \pi r^3$</p> <p>Therefore, the volume of the large drop (V_large) can be calculated by:</p> <p>$$ V_{\text{large}} = 8 \times V_{\text{small}} = 8 \times \left( \frac{4}{3} \pi r_{\text{small}}^3 \right) $$</p> <p>Let the radius of the larger drop be $R$. Then:</p> <p>$\frac{4}{3} \pi R^3 = 8 \times \left( \frac{4}{3} \pi r_{\text{small}}^3 \right)$</p> <p>Simplifying, we get:</p> <p>$R^3 = 8 r_{\text{small}}^3$</p> <p>Taking the cube root on both sides:</p> <p>$R = 2 r_{\text{small}}$</p> <p>Therefore, the radius of the larger drop is twice the radius of the smaller droplet:</p> <p>$R = 2 \times 0.01 \ \text{mm} = 0.02 \ \text{mm}$</p> <p>The terminal velocity of a droplet is proportional to the square of its radius. Therefore:</p> <p>$v_{t_{\text{large}}} \propto R^2$</p> <p>Given that the terminal velocity of the smaller droplets is 10 cm/s, the terminal velocity of the larger drop (formed by coalescing 8 smaller droplets) is:</p> <p>$$ v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( \frac{R}{r_{\text{small}}} \right)^2 $$</p> <p>$$ v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( \frac{0.02 \ \text{mm}}{0.01 \ \text{mm}} \right)^2 $$</p> <p>$v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( 2 \right)^2$</p> <p>$v_{t_{\text{large}}} = 10 \ \text{cm/s} \times 4$</p> <p>$v_{t_{\text{large}}} = 40 \ \text{cm/s}$</p> <p>Therefore, the new terminal velocity of the larger drop will be 40 cm/s.</p>