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A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2 . If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be?
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A water drop of 0.05cm3 is squeezed between two glass plates and sprea...
Surface Tension and Normal Force Calculation:
Given Data:
- Volume of water drop (V) = 0.05 cm3
- Area covered by water drop (A) = 40 cm2
- Surface tension of water (γ) = 70 dyne/cm
Calculating Height of Water Drop:
- Since V = A x h, where h is the height of the water drop
- Therefore, h = V / A = 0.05 cm3 / 40 cm2 = 0.00125 cm
Calculating Force Exerted by Surface Tension:
- The force exerted by surface tension (F) is given by F = γ x 2 x A
- F = 70 dyne/cm x 2 x 40 cm2 = 5600 dyne
Calculating Normal Force Required:
- The normal force required to separate the glass plates is equal and opposite to the force exerted by surface tension
- Therefore, the normal force required = 5600 dyne
Therefore, the normal force required to separate the glass plates will be 5600 dyne.
Given Data:
- Volume of water drop (V) = 0.05 cm3
- Area covered by water drop (A) = 40 cm2
- Surface tension of water (γ) = 70 dyne/cm
Calculating Height of Water Drop:
- Since V = A x h, where h is the height of the water drop
- Therefore, h = V / A = 0.05 cm3 / 40 cm2 = 0.00125 cm
Calculating Force Exerted by Surface Tension:
- The force exerted by surface tension (F) is given by F = γ x 2 x A
- F = 70 dyne/cm x 2 x 40 cm2 = 5600 dyne
Calculating Normal Force Required:
- The normal force required to separate the glass plates is equal and opposite to the force exerted by surface tension
- Therefore, the normal force required = 5600 dyne
Therefore, the normal force required to separate the glass plates will be 5600 dyne.
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A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2 . If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be?
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A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2 . If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2 . If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2 . If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be?.
A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2 . If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2 . If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2 . If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be?.
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