A container has two immiscible liquids of densities ρ1 and ρ2(ρ2ρ1). A capillary 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 equal to h, which is equal to the column length of the lighter liquid. Assuming zero contact angle, find the surface tension of the heavier liquid?

Question

A container has two immiscible liquids of densities ρ1 and ρ2(ρ2>ρ1). A capillary 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 equal to h, which is equal to the column length of the lighter liquid. Assuming zero contact angle, find the surface tension of the heavier liquid? (A) (r ρ2 g h/2) (B) 2 π r ρ2 gh (C) (r/2)(ρ2-ρ1) g h (D) 2 π r (ρ2-ρ1) gh

Tardigrade Admin · Accepted Answer

Pressure of concave side is more than pressure at convex side

P_{0} - P_{A} = \frac{2 T}{r}

\Rightarrow P_{A} = P_{0} - \frac{2 T}{r}

\Rightarrow P_{C} = P_{0} - \frac{2 T}{r} + ρ_{2} g h = P_{B}

\Rightarrow P_{0} - \frac{2 T}{r} + ρ_{2} g h = P_{0} + ρ_{1} g h

\Rightarrow T = \frac{r}{2} (ρ_{2} - ρ_{1}) g h

$\frac{r ρ _{2} g h}{2}$

$2 π r ρ_{2} g h$

$\frac{r}{2} (ρ_{2} - ρ_{1}) g h$

$2 π r (ρ_{2} - ρ_{1}) g h$

2rρ2​gh​

2πrρ2​gh

2r​(ρ2​−ρ1​)gh

2πr(ρ2​−ρ1​)gh

Pressure of concave side is more than pressure at convex side P0​−PA​=r2T​ ⇒PA​=P0​−r2T​ ⇒PC​=P0​−r2T​+ρ2​gh=PB​ ⇒P0​−r2T​+ρ2​gh=P0​+ρ1​gh ⇒T=2r​(ρ2​−ρ1​)gh

$\frac{r ρ _{2} g h}{2}$

$2 π r ρ_{2} g h$

$\frac{r}{2} (ρ_{2} - ρ_{1}) g h$

$2 π r (ρ_{2} - ρ_{1}) g h$