Question 248: Properties of Matter

Question

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 ________ $$\mathrm{cm} / \mathrm{s}$$.

Accepted Answer

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.
The terminal velocity for a small spherical droplet falling through the air is given by the Stokes' law:
$$ v_t = \frac{2}{9} \frac{r^2 (ho - ho_{	ext{air}}) g}{\eta} $$
where:
$$v_t$$ is the terminal velocity.
$$r$$ is the radius of the droplet.
$$ho$$ is the density of the droplet.
$$ho_{	ext{air}}$$ is the density of the air.
$$g$$ is the acceleration due to gravity.
$$\eta$$ is the viscosity of the air.
Given that the radius of the small droplets is $$0.01 \ 	ext{mm}$$ and their terminal velocity is $$10 \ 	ext{cm/s}$$, we now need to determine the radius of the larger drop formed by the coalescence of 8 smaller droplets.
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:
$$ V = \frac{4}{3} \pi r^3 $$
Therefore, the volume of the large drop (V_large) can be calculated by:
$$ V_{	ext{large}} = 8 	imes V_{	ext{small}} = 8 	imes \left( \frac{4}{3} \pi r_{	ext{small}}^3 ight) $$
Let the radius of the larger drop be $$R$$. Then:
$$ \frac{4}{3} \pi R^3 = 8 	imes \left( \frac{4}{3} \pi r_{	ext{small}}^3 ight) $$
Simplifying, we get:
$$ R^3 = 8 r_{	ext{small}}^3 $$
Taking the cube root on both sides:
$$ R = 2 r_{	ext{small}} $$
Therefore, the radius of the larger drop is twice the radius of the smaller droplet:
$$ R = 2 	imes 0.01 \ 	ext{mm} = 0.02 \ 	ext{mm} $$
The terminal velocity of a droplet is proportional to the square of its radius. Therefore:
$$ v_{t_{	ext{large}}} \propto R^2 $$
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:
$$ v_{t_{	ext{large}}} = 10 \ 	ext{cm/s} 	imes \left( \frac{R}{r_{	ext{small}}} ight)^2 $$
$$ v_{t_{	ext{large}}} = 10 \ 	ext{cm/s} 	imes \left( \frac{0.02 \ 	ext{mm}}{0.01 \ 	ext{mm}} ight)^2 $$
$$ v_{t_{	ext{large}}} = 10 \ 	ext{cm/s} 	imes \left( 2 ight)^2 $$
$$ v_{t_{	ext{large}}} = 10 \ 	ext{cm/s} 	imes 4 $$
$$ v_{t_{	ext{large}}} = 40 \ 	ext{cm/s} $$
Therefore, the new terminal velocity of the larger drop will be 40 cm/s.

Properties of Matter

Integer / Numeric Answer Type