A spherical raindrop evaporates at a rate proportional to its surface area. If originally its radius is $3…
A spherical raindrop evaporates at a rate proportional to its surface area. If originally its radius is $3 \mathrm{~mm}$ and 1 hour later it reduces to $2 \mathrm{~mm}$, then the expression for the radius $\mathrm{R}$ of the raindrop at any time $t$ is
$6R = t + 2$
$R(t+2)=6$
$R = 6(t + 2)$
$6R = t$
Solution
According to the given conditions, when $\mathrm{t}=0, \mathrm{R}=3$ and when $\mathrm{t}=1, \mathrm{R}=2$
This condition is satisfied by only option (B)