Question

As a spherical raindrop falls, it reaches a layer of dry air and begins to evaporate at a rate that is proportional to its surface area Show that the radius of the raindrop decreases at a constant rate. Because the evaporation rate is proportional to the:_______.
a. dV/dt = k(4pir^2).
b. dS/dt = k(4pir^2).
c. dV/dt = k/4pir^2).
d. dV/dt = k(4/3pir^3).
e. dS/dt = k(4/3pir^3).

Answer 1

Answer:

Option A)

$\dfrac{dV}{dt} = k(4\pi r^2)$

Step-by-step explanation:

We are given the following in the question:

The rate of evaporation of rainfall drop is proportional to surface area.

Volume of rainfall drop =

$\text{Volume of sphere} = V = \displaystyle\frac{4}{3}\pi r^3$

Surface area of rainfall drop =

$\text{Surface area of sphere} = S =4\pi r^2$

$\displaystyle\frac{dV}{dt} \propto 4\pi r^2\\\\\frac{dV}{dt} = k(4\pi r^2)\\\\\text{k is a constant of proportionality}$

Now,

Rate of evaporation =

$\displaystyle\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3}\pi r^3) = 4\pi r^2\frac{dr}{dt}$

Equating the two values we get,

$k(4\pi r^2) = 4\pi r^2\displaystyle\frac{dr}{dt}\\\\\frac{dr}{dt} = k$

Thus, the radius radius of the raindrop decreases at a constant rate. Because the evaporation is proportional to,

Option A)

$\dfrac{dV}{dt} = k(4\pi r^2)$

Answer 2

Inside the question, you are given the following information:

An evaporation rate of a rainfall drop is proportional to its surface area.

The amount of rain that falls$=V= \frac{4}{3}\pi r^3$

The surface area of a raindrop$= S= 4\pi r^2$

$\frac{dV}{dt} \propto 4\pi r^2\\\\\frac{dV}{dt}= K(4\pi r^2)$

k is a proportional factor.

Now,

Evaporation rate =

We get by equating the two values,

$\to k(4\pi r^2)=4 \pi \ r^2 \frac{dr}{dt}$

As a result, the raindrop's radius diminishes at a consistent rate. Because evaporation is related to temperature, "$\bold{\frac{dV}{dt}=K(4 \pi r^2)}$".

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