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Life Processes in Plants - Transport of water and minerals

Grade 7CBSE

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

🔑Concepts

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Xylem: This is the vascular tissue responsible for the transport of water and dissolved minerals from the roots to all parts of the plant. It forms a continuous network of channels.

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Root Hairs: These are extensions of the root epidermal cells that significantly increase the surface area (AA) of the root for the absorption of water and minerals.

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Osmosis: Water moves from the soil (higher water concentration) into the root hairs (lower water concentration) through a semi-permeable membrane by the process of osmosis.

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Transpiration: The loss of water in the form of vapor through the stomata present on the surface of leaves. It creates a cooling effect for the plant.

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Suction Pull (Transpiration Pull): As water evaporates from the leaves during transpiration, it creates a suction force that pulls water upwards from the roots through the xylem, similar to drinking through a straw.

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Cohesion and Adhesion: These forces help maintain a continuous column of water molecules within the xylem vessels, preventing the water column from breaking.

📐Formulae

Rate of Absorption=Total Water Absorbed (ml)Time Taken (hr)\text{Rate of Absorption} = \frac{\text{Total Water Absorbed (ml)}}{\text{Time Taken (hr)}}

Transpiration Pull∝Rate of Transpiration\text{Transpiration Pull} \propto \text{Rate of Transpiration}

Water Transport Pathway: Soil→Root Hair→Root Xylem→Stem Xylem→Leaf Xylem→Stomata (as vapor)\text{Water Transport Pathway: Soil} \rightarrow \text{Root Hair} \rightarrow \text{Root Xylem} \rightarrow \text{Stem Xylem} \rightarrow \text{Leaf Xylem} \rightarrow \text{Stomata (as vapor)}

💡Examples

Problem 1:

In an experiment, a potted plant absorbed 250 ml250 \text{ ml} of water over a period of 10 hours10 \text{ hours}. Calculate the rate of water absorption in ml/hr\text{ml/hr}.

Solution:

Rate=250 ml10 hr=25 ml/hr\text{Rate} = \frac{250 \text{ ml}}{10 \text{ hr}} = 25 \text{ ml/hr}

Explanation:

To find the rate, we divide the total quantity of water absorbed by the total time taken. This gives the average amount of water the plant takes up per hour.

Problem 2:

During a hot afternoon, a tall tree loses water through transpiration at a faster rate. If the total water lifted is 1200 units1200 \text{ units} and it is distributed equally among 44 main branches, how much water reaches each branch?

Solution:

1200÷4300\begin{array}{r} 1200 \\ \div 4 \\ \hline 300 \end{array} Each branch receives 300 units300 \text{ units} of water.

Explanation:

The transpiration pull acts on the entire xylem column. If the flow is uniform, the total volume lifted is divided by the number of primary distribution points (branches).