Answer
Rate of decrease in radius $=-\frac{kr}{2}$
Work Step by Step
Let $A$ denote the surface area of a meteor supposed to be in the form of a sphere
$A=4\pi r^2$
Where r is the radius of sphere
Let $\frac{dA}{dt}$ be the rate of decay of surface ares when meteor burns in air.
According to the condition
$\frac{dA}{dt} \propto -A$ Or
where negative sign indicates decay of area A
$\frac{dA}{dt}=-kA$
Putting $A=4\pi r^2$ in the above equation
$\frac{d(4\pi r^2)}{dt}=-4k \pi r^2$
$\frac{d(4 \pi r^2)}{dr}\frac{dr}{dt}=-4k\pi r^2$
$8 \pi r\frac{dr}{dt}=-4k \pi r^2$
$\frac{dr}{dt}=-\frac{4k \pi r^2}{8 \pi r}=-\frac{kr}{2}$
