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Water and Life - Conservation of Water and Rainwater Harvesting

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Water is essential for all life forms; the chemical formula for water is H2OH_2O.

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Although about 71%71\% of the Earth's surface is covered with water, only a very small fraction (approximately 2.7%2.7\%) is freshwater, and less than 1%1\% of that is available for human use.

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Water scarcity is caused by the increasing population, industrialization, and unequal distribution of rainfall.

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Conservation of water involves the wise and economical use of water and avoiding its wastage.

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Rainwater Harvesting (RWH) is the technique of collecting and storing rainwater for future use rather than allowing it to run off.

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Rooftop Rainwater Harvesting: Rainwater is collected from the rooftop of the house through pipes into a storage tank or a pit to recharge the groundwater table.

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Another method of harvesting is to allow water to go into the ground directly from roadside drains that collect rainwater.

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Key slogan for Rainwater Harvesting: 'Catch water where it falls'.

📐Formulae

Percentage of Earth covered by water≈71%\text{Percentage of Earth covered by water} \approx 71\%

Available Freshwater<1% of total water\text{Available Freshwater} < 1\% \text{ of total water}

Total Water Wasted=Rate of Leakage×Time\text{Total Water Wasted} = \text{Rate of Leakage} \times \text{Time}

Water Saved=Total Water Used Previously−Total Water Used After Conservation\text{Water Saved} = \text{Total Water Used Previously} - \text{Total Water Used After Conservation}

💡Examples

Problem 1:

If a leaking tap drips at a rate of 20 mL20 \text{ mL} per minute, how much water is wasted in 11 hour?

Solution:

20 mL/min×60 min=1200 mL20 \text{ mL/min} \times 60 \text{ min} = 1200 \text{ mL} or 1.2 L1.2 \text{ L}.

Explanation:

To find the total volume wasted, multiply the rate of leakage (20 mL/min20 \text{ mL/min}) by the total time in minutes (1 hour=60 minutes1 \text{ hour} = 60 \text{ minutes}).

Problem 2:

Calculate the amount of rainwater collected from a roof of area 50 m250 \text{ m}^2 if the rainfall is 10 mm10 \text{ mm} (0.01 m0.01 \text{ m}).

Solution:

Volume=Area×Rainfall Height\text{Volume} = \text{Area} \times \text{Rainfall Height} Volume=50 m2×0.01 m=0.5 m3\text{Volume} = 50 \text{ m}^2 \times 0.01 \text{ m} = 0.5 \text{ m}^3 Since 1 m3=1000 L1 \text{ m}^3 = 1000 \text{ L}, 0.5 m3=500 L0.5 \text{ m}^3 = 500 \text{ L}.

Explanation:

The volume of water harvested is the product of the catchment area and the depth of the rainfall.