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A capillary tube of radius r is dipped vertically in a liquid of density d, surface tension T and angle of contact $θ$ then the pressure difference just below the two surface, one in the beaker and the other in the capillary tube, is

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$\frac{T}{r cosθ}$

$\frac{2 T}{r}$

$\frac{2 T cosθ}{r}$

$\frac{T cosθ}{r}$

answer is C.

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Detailed Solution

From the figure,

r is the radius of the tube,

R is the radius of the meniscus

then,

$R = \frac{r}{cosθ}$

the difference in pressure at two heights is given by,

$∆ P = \frac{2 T}{R} = \frac{2 Tcosθ}{r}$

Hence the correct answer is $\frac{2 Tcosθ}{r} .$

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