Question

After a tap above an empty rectangular tank has been opened, the tank is filled with water at a constant rate in time $T_1$. After the tap has been closed, poking a small hole at the bottom of the tank empties it in another duration of time $T_2$. If the tap above the tank is now opened again, for what ratio(s) of $\frac{T_1}{T_2}$ will the tank overflow?

• A. 0.25
• B. 0.1
• C. 0.6
• D. 1

Answer 1

Answer: (A), (B)

(A), (B)

Answer 2

Let $V$ be the volume of the tank. The inflow rate is constant: $R_{in} = \frac{V}{T_1}$. By Torricelli's law, the outflow rate is proportional to $\sqrt{h}$, where $h$ is the water height. The time to empty, $T_2$, is related to the average outflow rate. The maximum outflow rate, when the tank is full, is $R_{out,max} = \frac{2V}{T_2}$. Overflow occurs if the inflow rate is greater than the maximum outflow rate: $R_{in} > R_{out,max}$. Substituting the expressions: $\frac{V}{T_1} > \frac{2V}{T_2}$. This simplifies to $\frac{1}{T_1} > \frac{2}{T_2}$, or $\frac{T_1}{T_2} < \frac{1}{2} = 0.5$. Therefore, the tank overflows if $\frac{T_1}{T_2}$ is less than 0.5. Options (A) 0.25 and (B) 0.1 satisfy this condition.