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A container contains two immiscible liquids of density ρ1​ and ρ1(ρ2>ρ2). A capillary of radius r is inserted in the liquid so that its bottom reaches up to denser liquid. Denser liquid rises in capillary and attains height equal to h which is also equal to column length of lighter liquid. Assuming zero contact angle find the surface tension of the heavier liquid.


A

2πρ2gh

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B

rρ2gh2

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C

2πρ2-ρ1gh

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D

rρ2-ρ1gh2

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Solution

The correct option is D

rρ2-ρ1gh2


Step 1: Given data

  1. Container contains two immiscible liquids of density ρ1​ and

  1. ρ1(ρ2>ρ2)
  2. A capillary of radius r is inserted in the liquid
  3. Denser liquid rises in capillary and attains height equal to h which is also equal to column length of lighter liquid

Step 2: Formula used

  1. P=ρgh where P is the pressure ρ density of the liquid and h is the height

Step 3: Find the surface tension

The pressure difference at the points Aand B, just above and below the meniscus is

PA-PB=2Tr

Where T is the surface tension r is the radius of the capillary tube

However at the point A , the atmospheric pressure is acting.

PO=PA and PE=PA=PO

According to Pascal's Law PC=PD

PA-2Tr+ρ2gh+h'=PE+ρ1gh+ρ2gh'2Tr=ρ2-ρ1ghT=rρ2-ρ1gh2

Hence, option D is correct


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