Step 1: Understanding the Concept:
This problem involves the concept of rates of work, specifically applied to pipes and cisterns.
When a pump is filling a tank, it does "positive work."
When a leak is present, it does "negative work" by removing water while the pump adds it.
The total time taken to fill the tank increases because the net rate of filling is the difference between the pump's rate and the leak's rate.
To solve this, we must express all times in terms of hourly rates (part of the tank filled or emptied per hour).
Step 2: Key Formula or Approach:
1. Rate \( = \frac{1}{\text{Time}} \).
2. Net Rate \( = \text{Filling Rate} - \text{Leaking Rate} \).
3. Time taken by leak alone \( = \frac{1}{\text{Leaking Rate}} \).
Step 2: Detailed Explanation:
Let the rate at which the pump fills the tank be \( R_p \).
Since the pump can fill the tank in 2 hours, its rate is:
\[ R_p = \frac{1}{2} \text{ tank per hour.} \]
Now, let the rate of the leak be \( R_l \).
The combined time with the leak is 2 hours and 20 minutes. We must convert this into hours.
\[ 20 \text{ minutes} = \frac{20}{60} = \frac{1}{3} \text{ hour.} \]
Total combined time \( = 2 + \frac{1}{3} = \frac{7}{3} \text{ hours.} \]
The net filling rate \( R_{net} \) is the reciprocal of this combined time:
\[ R_{net} = \frac{1}{7/3} = \frac{3}{7} \text{ tank per hour.} \]
We know that the net rate is the pump's rate minus the leak's rate:
\[ R_{net} = R_p - R_l \]
\[ \frac{3}{7} = \frac{1}{2} - R_l \]
To find \( R_l \), rearrange the equation:
\[ R_l = \frac{1}{2} - \frac{3}{7} \]
Find a common denominator (14) to perform the subtraction:
\[ R_l = \frac{7}{14} - \frac{6}{14} = \frac{1}{14} \text{ tank per hour.} \]
This means the leak empties \( \frac{1}{14} \) of the tank every hour.
To empty the full tank, the time required \( T_l \) is:
\[ T_l = \frac{1}{R_l} = \frac{1}{1/14} = 14 \text{ hours.} \]
Thus, the leak would take exactly 14 hours to drain a completely full tank.
Step 3: Final Answer:
The leak can drain the tank in 14 hours.
This corresponds to Option (A).