Understanding the Tank Filling and Emptying Problem

This problem involves a tank being filled by a pipe and simultaneously emptied by a leak located at a specific height. We need to determine the time it takes for the tank to go from being completely full to being one-fourth full under these conditions.

Analyzing the Rates

Let's first understand the rates at which the pipe fills and the leak empties the tank.

• The filling pipe can fill the entire tank in 4 hours. This means the filling rate is \(F = \frac{1 \text{ tank}}{4 \text{ hours}} = \frac{1}{4}\) tank per hour.
• The leak is located at one-fourth of the height from the bottom. The problem states that this leak "can empty upto that part in 2 hours". We interpret this to mean that the leak's capacity is such that it can empty the volume *above* the leak's height down to the leak's height, when starting from full, in 2 hours. The volume above the leak's height (which is at 1/4 height) when the tank is full is \(1 - \frac{1}{4} = \frac{3}{4}\) of the tank's total volume. So, the leak empties \(3/4\) of the tank volume in 2 hours.
• The leak rate is \(L = \frac{\frac{3}{4} \text{ tank}}{2 \text{ hours}} = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8}\) tank per hour. This leak rate is effective as long as the water level is above the leak's position (1/4 height).

Calculating the Net Rate of Change

The tank starts full (1 tank) and needs to reach one-fourth full (\(1/4\) tank). This means the tank volume needs to decrease by \(1 - \frac{1}{4} = \frac{3}{4}\) of its total capacity.

During this process (from full down to 1/4 full), the water level is always above the leak's height. Therefore, both the filling pipe and the leak are operating simultaneously.

• Filling rate \(F = \frac{1}{4}\) tank/hour.
• Leak rate \(L = \frac{3}{8}\) tank/hour.

Since the leak rate (\(3/8\)) is greater than the filling rate (\(1/4 = 2/8\)), the net effect is that the tank is emptying. The net emptying rate is:

\( \text{Net Rate} = L - F = \frac{3}{8} - \frac{1}{4} = \frac{3}{8} - \frac{2}{8} = \frac{1}{8} \text{ tank/hour} \)

So, the tank is emptying at a net rate of \(1/8\) tank per hour while the water level is above the leak.

Determining the Time Taken

We need the tank to go from being full (1 tank) to one-fourth full (\(1/4\) tank). The total volume that needs to be emptied is:

\( \text{Volume to Empty} = 1 - \frac{1}{4} = \frac{3}{4} \text{ tank} \)

Since the net emptying rate is \(1/8\) tank per hour, the time taken to empty \(3/4\) of the tank is:

\( \text{Time} = \frac{\text{Volume to Empty}}{\text{Net Emptying Rate}} = \frac{\frac{3}{4} \text{ tank}}{\frac{1}{8} \text{ tank/hour}} \)

Calculating the time:

\( \text{Time} = \frac{3}{4} \times \frac{8}{1} = \frac{24}{4} = 6 \text{ hours} \)

Therefore, it will take 6 hours for the tank to become one-fourth full from being initially full.

Revision Table: Key Concepts for Tank Problems

Additional Information on Tank Problems

Problems involving pipes and tanks often test your ability to work with rates and fractions. Key things to remember:

• Always calculate individual rates as "fraction of tank per unit time".
• Pay close attention to whether pipes are filling or emptying, and combine rates accordingly.
• If there's a leak at a certain height, consider how this affects the emptying process at different water levels. In this specific problem, the leak was effective throughout the process of the tank level dropping from full to 1/4 because the entire drop happens above the leak's height.
• Read the description of the leak's capacity carefully to correctly determine its rate. Different phrasings can imply different rate calculations.
• Ensure units are consistent (e.g., hours and tanks).

Solving these problems involves setting up the correct rates and calculating the net change over the required volume difference.