Home
Class 12
BIOLOGY
Dry wooden stakes, if driven into a smal...

Dry wooden stakes, if driven into a small crack in a rock and then soaked, can develop enough pressure to split the rock . This pressure is developed due to

A

imbibition

B

exosomosis

C

plasmolysis

D

diffusion

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the pressure developed by dry wooden stakes when soaked in a crack in a rock, we can break down the explanation into clear steps: ### Step-by-Step Solution: 1. **Understanding the Setup**: - We have dry wooden stakes that are inserted into a small crack in a rock. When these stakes are soaked with water, they absorb the water. **Hint**: Consider what happens to materials when they come into contact with water. 2. **Nature of Wooden Stakes**: - The wooden stakes are made of dead plant cells. Even though they lack protoplasm, they can still absorb water due to their structure. **Hint**: Think about the properties of plant cells and their ability to interact with water. 3. **Process of Water Absorption**: - The process by which the dry wooden stakes take up water is called imbibition. This is a physical phenomenon where materials absorb liquid, leading to swelling. **Hint**: Recall the term that describes the absorption of water by dry materials. 4. **Development of Pressure**: - As the wooden stakes absorb water, they swell and create pressure within the crack. This pressure is known as imbibition pressure. **Hint**: Identify the specific type of pressure that results from the absorption of water by the wooden stakes. 5. **Outcome**: - The imbibition pressure generated is strong enough to exert force on the surrounding rock, potentially causing it to split. **Hint**: Consider the implications of this pressure on the physical structure of the rock. ### Final Answer: The pressure developed due to the soaking of dry wooden stakes in a crack in a rock is known as **imbibition pressure**. ---
Promotional Banner

Topper's Solved these Questions

  • THE LIVING WORLD

    TRUEMEN BIOLOGY ENGLISH|Exercise MULTIPLE CHOICE QUESTIONS|119 Videos
  • Viruses

    TRUEMEN BIOLOGY ENGLISH|Exercise ASSERTION AND REASON|21 Videos

Similar Questions

Explore conceptually related problems

Root pressure develops due to

Root pressure develops due to

Dry wooden stakes in cracks of a rock and soaked will develop pressure that wil split the rock. The phenomenon is

Dry wooden pieces fixed in the crevices of a rock and soaked with water split the rock. The pressure responsible for this is

Turgor pressure develops in epiblema cells of root due to

Assertion : Loss of water produces a negative hydrostatic pressure. Reason : Positive hydrostatic pressure is developed due to osmotic entry of water into it.

Each question contains STATEMENT-I(Assertion) and STATEMENT-2(Reason).the statement carefully and mark the correct answer accoring to the instrution given below: STATEMENT - 1 : Osmotic pressure is a colligative property. STATEMENT - 2 : Osmotic pressure is developed in a column due to osmosis.

The system shown in the figure is in equilibrium, where A and B are isomeric liquids and form an ideal solution at TK . Standard vapour pressures of A and B are P_(A)^(0) and P_(B)^(0) , respectively, at TK . We collect the vapour of A and B in two containers of volume V , first container is maintained at 2 T K and second container is maintained at 3T//2 . At the temperature greater than T K , both A and B exist in only gaseous form. We assume than collected gases behave ideally at 2 T K and there may take place an isomerisation reaction in which A gets converted into B by first-order kinetics reaction given as: Aoverset(k)rarrB , where k is a rate constant. In container ( II ) at the given temperature 3T//2 , A and B are ideal in nature and non reacting in nature. A small pin hole is made into container. We can determine the initial rate of effusion of both gases in vacuum by the expression r=K.(P)/(sqrt(M_(0))) where P= pressure differences between system and surrounding K= positive constant M_(0)= molecular weight of the gas Vapours of A and B are passed into a container of volume 8.21 L , maintained at 2T K , where T=50 K and after 5 min , moles of B=8//3 . The pressure developed into the cotainer after two half lives is

A skier starts from rest at point A and slides donw the hill without turning or breaking. The friction coefficient is mu When he stops at point B, his horizontal displacement is S. whalt is the height difference between points A and B? (The velocity of the skier is small so that the additional pressure on the snow due to the curvature can vbe neglected. Neglect also the friction of air and the dependence of mu on the velocity of the skier)

The main application of osmotic pressure measurement is in the determination of the molar mass of a substance which is either slightly soluble or has a very high molar mass such as proteins, polymers of various types and colloids.This is due to the fact that even a very small concentraion of the solution produces fairly large magnitude of osomotic pressure.In the laboratory the concentrations usually employed are of the order of 10^(-3) to 10^(-4) M.At concentration of 10^(-3) mol dm^(-3) , the magnitude of osmotic pressure of 300 K is : P=10^(-3)xx0.082xx300=0.0246 atm or 0.0246xx1.01325xx10^5=2492.595 Pa At this concentration, the values of other colligative properties such as boiling point elevation and depression in freezing point are too small to be determined experimentally. Further polymers have following two types of molar masses : (A) Number average molar mass (barM_n) , which is given by (undersetisumN_iM_i)/(undersetisumN_i) where N_i is the number of molecules having molar mass M_i . (B) Molar average molar mass (barM_m) , which is given by (undersetisumN_iM_i^2)/(undersetisumN_iM_i) Obviously the former is independent of the individual characteristics of the molecules and gives equal weightage to large and small molecules in the polymer sample.On the other hand later gives more weightage to the heavier molecules.Infact with the help of a colligative property only one type of molar mass of the polymer can be determined. One gram each of polymer A (molar mass=2000) and B(molar mass=6000) is dissolved in water to form one litre solution at 27^@C .The osmotic pressure of this solution will be :

TRUEMEN BIOLOGY ENGLISH-TRANSPORT IN PLANTS-Multiple Choice Questions
  1. Imbibition involves

    Text Solution

    |

  2. A bottle filled with previously moistened mustard seeds and water was ...

    Text Solution

    |

  3. Dry wooden stakes, if driven into a small crack in a rock and then soa...

    Text Solution

    |

  4. Maximum and minimum imbibing capacity is found in

    Text Solution

    |

  5. The psi(W) of imbibants is

    Text Solution

    |

  6. When a bottle of scent or ammonia is opened in one corner of a room, i...

    Text Solution

    |

  7. The relative rates of diffusion of substances are

    Text Solution

    |

  8. The rate of diffusion is dependent upon the permeability of the medium...

    Text Solution

    |

  9. The translocation of substances inside a cell and over short distances...

    Text Solution

    |

  10. The diffusion pressure of an individual substance during independent...

    Text Solution

    |

  11. Selectively (differentially ) permeable membrane is that which allows

    Text Solution

    |

  12. The membrane which allows only water molecules to pass through it and ...

    Text Solution

    |

  13. The value of osmotic potential of an electolyte is always

    Text Solution

    |

  14. The plasmalemma of the cell is a

    Text Solution

    |

  15. The main difference in DPD and Psi(W) is

    Text Solution

    |

  16. When a flaccid cell (e.g., grape raisin ) is placed in water

    Text Solution

    |

  17. Which has maximum diffusion pressure ?

    Text Solution

    |

  18. Which cell has maximum diffusion pressure deficit ?

    Text Solution

    |

  19. Cell sap is

    Text Solution

    |

  20. A cell placed in 0.5 M sugar solution was found to be with no change ...

    Text Solution

    |