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प्रश्न
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उत्तर
Plants die when salt is sprinkled over them due to plasmolysis of cells. The salt makes the solution hypertonic inside the plant cells which lead to shrinkage of cells or exosmosis.
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संबंधित प्रश्न
A plant cell placed in a certain solution got plasmolysed. What was the kind of solution?
Name the following:
The condition of a cell placed in a hypotonic solution.
Complete the following statements:
Hypotonic solution is one in which the solution kept outside the cell has lower solute concentration than ……………… the cell.
Name the following:
Two solutions having same concentrations.
Name the following:
Movement of molecules from a region of high concentration to a region of low concentration.
Choose the correct answer:
Cell turgidity is caused by _______________
Two potato cubes each 1 cm3 in size, were placed separately in two containers (A and B), the container A having water and the other (B) containing concentrated sugar solution. After 24 hours when the cubes were examined, those placed in water were found to be firm and had increased slightly in size and those placed in concentrated sugar solution were found to be soft and somewhat decreased in size. Use the above information to answer the questions that follow:
Account for the softness and decrease in size of the potato cubes which were placed in sugar solution.
Two potato cubes each 1 cm3 in size, were placed separately in two containers (A and B), the container A having water and the other (B) containing concentrated sugar solution. After 24 hours when the cubes were examined, those placed in water were found to be firm and had increased slightly in size and those placed in concentrated sugar solution were found to be soft and somewhat decreased in size. Use the above information to answer the questions that follow:
Name and define the physical process being investigated in this experiment.
Give Reasons for the following.
Freshwater fish cannot survive in seawater.
In the figure below ‘A’ shows a cell in the normal state and ‘B’ shows the same cell after leaving it in a certain solution for a few minutes.
(i) Describe the change which has occurred in the cell as seen in B.
(ii) Give the technical term for the condition of the cell as reached in B and as it was in A.
(iii) Define the process which led to this condition.
(iv) What was the solution-isotonic, hypotonic or hypertonic, in which the cell was kept?
(v) How can the cell in B, be brought back to its original condition?
(vi) Name the parts numbered 1 to 3.
