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Question

A container has two immiscible liquids of densities ρ1 and ρ2(>ρ1). A capillary tube of radius r is inserted in the liquid so that its bottom reaches upto the denser liquid. The denser liquid rises in the capillary and attains a height h from the interface of the liquids, which is equal to the column length of the lighter liquid. Assuming angle of contact to be zero, the surface tension of heavier liquid is


A
2πrρ2gh
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B
ρ2rgh2
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C
r2(ρ2ρ1)gh
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D
2πr(ρ2ρ1)gh
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Solution

The correct option is C r2(ρ2ρ1)gh

The pressure difference at the points A and B, just above and below the meniscus is
PAPB=2Tr
However at the point A, the atmospheric pressure is acting.
P0=PA
and PE=PA=P0
According to Pascal's law, PC=PD
PA2Tr+ρ2g(h+h)=PE+ρ1gh+ρ2gh
2Tr=(ρ2ρ1)gh
T=r2(ρ2ρ1)gh

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