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There is a water tank in the form of a rectangular parallelepiped of height 1.1 m and a square base of side 2 m. If a full tank of water is drained out completely in a long pipe of circular cross-sectional area of radius 1 cm, what should be the minimum length of the pipe, in km, to hold the entire water in it? (Take π = 22/7)
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Let's solve this step by step to determine the correct answer:
- Tank Volume Calculation:
- The tank is a rectangular parallelepiped with a height of 1.1 m and a square base of side 2 m.
- Volume of the tank = base area × height = (2 m × 2 m) × 1.1 m = 4.4 cubic meters.
- Pipe Volume Requirement:
- The pipe has a radius of 1 cm, which is 0.01 meters.
- Cross-sectional area of the pipe = (pi) r² = (22/7) × (0.01 m)² = 22/70,000 square meters.
- Pipe Length Calculation:
- Volume needed in the pipe = Volume of the tank = 4.4 cubic meters.
- Let L be the required length of the pipe.
- Volume of water = Cross-sectional area × Length of the pipe.
- 4.4 = (22/70,000) × L.
- Solving for L gives L = (4.4 × 70,000) / 22 = 14,000 meters = 14 km.
- Correct Answer: Option 3 - 14 km
By: Babita ProfileResourcesReport error
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