To solve this problem, we need to determine the time it takes for all three pipes running together to fill the cistern. Here's a step-by-step explanation:
Step 1: Understanding the Given Information
- We have three pipes of diameters 2 cm, 3 cm, and 4 cm.
- The smallest pipe, with a diameter of 2 cm, can fill the cistern in 232 minutes.
- The rate of water flow is proportional to the square of the diameter of the pipe.
Step 2: Calculating the Rates of Flow
- Let's assume the rate of flow of the 2 cm diameter pipe is proportional to \(2^2 = 4\).
- The rates of the other pipes will be proportional to their respective diameters squared:
- The 3 cm pipe: \(3^2 = 9\)
- The 4 cm pipe: \(4^2 = 16\)
Step 3: Calculating the Time Taken by All Pipes Together
- The 2 cm pipe fills the cistern in 232 minutes, so its work done in one minute is: \(\frac{1}{232}\) of the cistern.
- Since the flow rates are proportional to the squares of the diameters, we calculate the combined rate as the sum of individual rates:
- Combined rate of flow per minute: \(\left(\frac{4}{4}\right) + \left(\frac{9}{4}\right) + \left(\frac{16}{4}\right) = 1 + \frac{9}{4} + \frac{16}{4} = 1 + 2.25 + 4\)
- This simplifies to: \(7.25\)
- Thus, the three pipes together fill \(7.25\) times their rate of the smallest pipe.
- Total time taken by all pipes together = \(\frac{232}{7.25} = 32 \, \text{minutes}\).
Conclusion
- The time taken by all pipes running together to fill the cistern is 32 minutes.
This matches the correct answer: 32.
Similar Questions
- A pipe can fill a sump with water in 2 hours. Because of a leak, it took $2 \frac{1}{3}$ hours to fill the sump. The leak can drain all the water of the sump in:
- Two pipes A and B can fill an empty cistern in 18 and 27 hours, respectively. Pipe C can drain the entire cistern in 45 hours when no other pipe is in operation. Initially, when the cistern was empty Pipe A and Pipe C were turned on. After a few hours Pipe A was turned off and Pipe B was turned on instantly. In all, it took 55 hours to fill the cistern. For how many hours was Pipe B turned on?
- Pipes A, B and C are attached to an empty cistern. While the first two can fill the cistern in 4 and 10 hours, respectively, the third can drain the cistern, when filled, in 6 hours. If all the three pipes are opened simultaneously when the cistern is half-full, how many hours will be needed to fill the cistern?
Important Questions from Pipe and Cistern
A pipe can fill a cistern in 20 minutes where as the cistern when full can be emptied by a leak in 28 minutes. When both are opened, The time taken to fill the cistern is:
‘A’ pipe can empty a tank in 20 minutes. The second pipe ‘B’ has a diameter twice as that of ‘A’. If both A & B pipe are attached to the tank how much time will be required to empty the tank?
Pipes A and B can empty a full tank in 16 hours and 24 hours, respectively. Pipe C alone can fill the empty tank in 4 hours. If A, B and C are opened together, the tank will be 35% full after :
Pipes A and B can fill a tank in 36 minutes and 45 minutes, respectively. Both these pipes were opened simultaneously. After 20 minutes, a leak at the bottom of the tank was spotted which was immediately sealed. The tank was full in another 15 minutes. The leak alone can empty the full tank in:
A cistern has a leak which would empty it in 6 hours. A tap is turned on which admits 10 litres of water per minute into the cistern. When it is full it is now emptied in 10 hours. What is the capacity (in litres) of the cistern?
