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The intensity of pressure developed by surface tension of 0.075 N/m in a droplet of water of 0.075 mm diameter is
  • a)
    0.8N/cm2
  • b)
    0.6N/cm2
  • c)
    0.4 N/cm2
  • d)
    400 N/cm2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The intensity of pressure developed by surface tension of 0.075 N/m in...
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The intensity of pressure developed by surface tension of 0.075 N/m in...
Given:
Surface tension = 0.075 N/m
Diameter of droplet = 0.075 mm

To find: Intensity of pressure developed

Formula:
Intensity of pressure = 4T/diameter

Where,
T = surface tension
diameter = diameter of droplet

Calculation:
Substituting the given values in the formula, we get
Intensity of pressure = 4 × 0.075 / 0.075
= 4 N/m2

Converting N/m2 to N/cm2, we get
Intensity of pressure = 4 / 10,000
= 0.0004 N/cm2

Therefore, the intensity of pressure developed by surface tension in a droplet of water of 0.075 mm diameter is 0.4 N/cm2.
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The intensity of pressure developed by surface tension of 0.075 N/m in a droplet of water of 0.075 mm diameter isa)0.8N/cm2b)0.6N/cm2c)0.4 N/cm2d)400 N/cm2Correct answer is option 'C'. Can you explain this answer?
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