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In a beaker with a solute having osmotic potential have, cell with O.P=15 and T.P= 4 , than which is value of D.P.D -
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Text solutionVerified
D.P.D = O.P - T.P
= 15 - 4
= 11
D.P.D = diffusion pressure deficit
O.P =osmotic pressure
T.P = turgor pressure
Water always flows from an area of lesser DPD towards an area having higher DPD.
= 15 - 4
= 11
D.P.D = diffusion pressure deficit
O.P =osmotic pressure
T.P = turgor pressure
Water always flows from an area of lesser DPD towards an area having higher DPD.
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Question Text | In a beaker with a solute having osmotic potential have, cell with O.P=15 and T.P= 4 , than which is value of D.P.D - |
Updated On | Apr 21, 2022 |
Topic | Transport in Plants |
Subject | Biology |
Class | Class 11 |
Answer Type | Text solution:1 Video solution: 2 |
Upvotes | 350 |
Avg. Video Duration | 4 min |