Question

The time after which the cone is empty is:

• A. H/2k
• B. H/k
• C. H/3k
• D. 2H/k

Answer 1

Answer: (B)

H/k

Answer 2

Let r be liquid radius, h be liquid height.
$\frac{r}{h} = \frac{R}{H} \implies h = \frac{H}{R}r$.
Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (\frac{H}{R}r) = \frac{\pi H}{3R} r^3$.
Surface area $A = \pi r^2$.
Rate of evaporation $\frac{dV}{dt} = -kA = -k\pi r^2$.
Also, $\frac{dV}{dt} = \frac{\pi H}{3R} (3r^2 \frac{dr}{dt}) = \frac{\pi H}{R} r^2 \frac{dr}{dt}$.
Equating: $\frac{\pi H}{R} r^2 \frac{dr}{dt} = -k\pi r^2 \implies \frac{dr}{dt} = -\frac{kR}{H}$.
Integrate: $r(t) = -\frac{kR}{H}t + C$.
At $t=0$, $r=R \implies C=R$. So $r(t) = R - \frac{kR}{H}t = R(1 - \frac{k}{H}t)$.
Cone is empty when $r(t)=0$: $0 = R(1 - \frac{k}{H}T) \implies T = \frac{H}{k}$.