Respuesta :
Answer:
Option A)
[tex]\dfrac{dV}{dt} = k(4\pi r^2)[/tex]
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 =
[tex]\text{Volume of sphere} = V = \displaystyle\frac{4}{3}\pi r^3[/tex]
Surface area of rainfall drop =
[tex]\text{Surface area of sphere} = S =4\pi r^2[/tex]
[tex]\displaystyle\frac{dV}{dt} \propto 4\pi r^2\\\\\frac{dV}{dt} = k(4\pi r^2)\\\\\text{k is a constant of proportionality}[/tex]
Now,
Rate of evaporation =
[tex]\displaystyle\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3}\pi r^3) = 4\pi r^2\frac{dr}{dt}[/tex]
Equating the two values we get,
[tex]k(4\pi r^2) = 4\pi r^2\displaystyle\frac{dr}{dt}\\\\\frac{dr}{dt} = k[/tex]
Thus, the radius radius of the raindrop decreases at a constant rate. Because the evaporation is proportional to,
Option A)
[tex]\dfrac{dV}{dt} = k(4\pi r^2)[/tex]
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[tex]=V= \frac{4}{3}\pi r^3\\\\[/tex]
The surface area of a raindrop[tex]= S= 4\pi r^2[/tex]
[tex]\frac{dV}{dt} \propto 4\pi r^2\\\\\frac{dV}{dt}= K(4\pi r^2)\\\\[/tex]
k is a proportional factor.
Now,
Evaporation rate =
We get by equating the two values,
[tex]\to k(4\pi r^2)=4 \pi \ r^2 \frac{dr}{dt}\\\\[/tex]
As a result, the raindrop's radius diminishes at a consistent rate. Because evaporation is related to temperature, "[tex]\bold{\frac{dV}{dt}=K(4 \pi r^2)}[/tex]".
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