    Question

# A container contains two immiscible liquids of density ${\rho }_{1}$​ and ${\rho }_{1}\left({\rho }_{2}>{\rho }_{2}\right)$. 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

$2\pi {\rho }_{2}gh$

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B

$\frac{r{\rho }_{2}gh}{2}$

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C

$2\pi \left({\rho }_{2}-{\rho }_{1}\right)gh$

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D

$\frac{r\left({\rho }_{2}-{\rho }_{1}\right)gh}{2}$

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Solution

## The correct option is D $\frac{r\left({\rho }_{2}-{\rho }_{1}\right)gh}{2}$Step 1: Given dataContainer contains two immiscible liquids of density ${\rho }_{1}$​ and ${\rho }_{1}\left({\rho }_{2}>{\rho }_{2}\right)$A capillary of radius $r$ is inserted in the liquidDenser liquid rises in capillary and attains height equal to $h$ which is also equal to column length of lighter liquid Step 2: Formula used$P=\rho gh$ where $P$ is the pressure $\rho$ density of the liquid and $h$ is the heightStep 3: Find the surface tension The pressure difference at the points $A$and $B$, just above and below the meniscus is${P}_{A}-{P}_{B}=\frac{2T}{r}$Where $T$ is the surface tension $r$ is the radius of the capillary tube However at the point $A$ , the atmospheric pressure is acting.$⇒{P}_{O}={P}_{A}$ and ${P}_{E}={P}_{A}={P}_{O}$According to Pascal's Law ${P}_{C}={P}_{D}$$⇒{P}_{A}-\frac{2T}{r}+{\rho }_{2}g\left(h+h\text{'}\right)={P}_{E}+{\rho }_{1}gh+{\rho }_{2}gh\text{'}\phantom{\rule{0ex}{0ex}}⇒\frac{2T}{r}=\left({\rho }_{2}-{\rho }_{1}\right)gh\phantom{\rule{0ex}{0ex}}⇒T=\frac{r\left({\rho }_{2}-{\rho }_{1}\right)gh}{2}$Hence, option D is correct  Suggest Corrections  0      Explore more