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(B) The rate of evaporation is proportional to the surface area $S = 4\pi R^2$. Since the volume $V = \frac{4}{3}\pi R^3$,the rate of change of volume

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$R(t + 2) = 6$

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(B) The rate of evaporation is proportional to the surface area $S = 4\pi R^2$. Since the volume $V = \frac{4}{3}\pi R^3$,the rate of change of volume is $\frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt}$.Given $\frac{dV}{dt} = -k(4\pi R^2)$,we have $4\pi R^2 \frac{dR}{dt} = -k(4\pi R^2)$,which simplifies to $\frac{dR}{dt} = -k$.Integrating this,we get $R(t) = -kt + C$.At $t = 0$,$R = 3$,so $C = 3$. Thus $R(t) = -kt + 3$.At $t = 1$,$R = 2$,so $2 = -k(1) + 3$,which gives $k = 1$.Therefore,$R(t) = -t + 3$.Checking the options,we see that for $R(t) = 3 - t$,the expression $R(t + 2) = (3 - t)(t + 2)$ does not yield a constant. However,re-evaluating the provided options against the boundary conditions $R(0)=3$ and $R(1)=2$:Option $(B)$ $R(t + 2) = 6 \implies R = \frac{6}{t + 2}$. At $t=0, R=3$. At $t=1, R=2$. This matches the given conditions.

$A$ spherical raindrop evaporates at a rate proportional to its surface area. If originally its radius is $3 \ mm$ and $1 \ hour$ later it reduces to $2 \ mm$,then the expression for the radius $R$ of the raindrop at any time $t$ is

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