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Two air bubble with radius $r_1$ and $r_2$ $(r_2>r_1)$ formed of the same liquid stick to each other to form a common interface. Therefore, the radius of curvature of the common surface is

(a) $\sqrt{r_1 r_2}$ (b) Infinity (c) $\frac{r_2}{r_1}\sqrt{{r_2}^2-{r_2}^2}$ (d) $\frac{r_1 r_2}{r_2 - r_1}$

Basic surface Tension

Relation between surface Tension and Radius of curvature

Suggested Reading

Source of the Problem

Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(d) $\frac{r_1 r_2}{r_2 - r_1}$

Hint 1

Hint 2

Hint 3

For a bubble of radius $r$ with double surface and whose inside pressure is $p_{in}$ and outside pressure is $p_{out}$ and also the surface tension is $T$, the relation between pressure and radius is,

$$p_{in} - p_{out} = \Delta p = \frac{4T}{r}$$

When 2 bubble merge on their common interface the pressure difference is Just the difference between the pressure(inside) of both two bubbles. Also, the surface tension remains same.

We know $p_1 - p_0 = \frac{4T}{r_1}$ and $p_2 - p_0 = \frac{4T}{r_2}$, Then,

$$ p_1 - p_2 = \frac{4T}{R} $$

here $R$ is the radius of curvature of the interface.

Hence,

$p_{1}-p_{2}$=$\left(p_{1}-p_{0}\right)-\left(p_{2}-p_{0}\right)$=$4 T\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)$

This gives us,

$$ \frac{1}{R} = \frac{1}{r_1}-\frac{1}{r_2} \to R= \frac{r_1 r_2}{r_2 - r_1} $$

Physics Olympiad Program at Cheenta

Try out this problem on the Surface Tension and Pressure from National Standard Examination in Physics 2015-2016.

Two air bubble with radius $r_1$ and $r_2$ $(r_2>r_1)$ formed of the same liquid stick to each other to form a common interface. Therefore, the radius of curvature of the common surface is

(a) $\sqrt{r_1 r_2}$ (b) Infinity (c) $\frac{r_2}{r_1}\sqrt{{r_2}^2-{r_2}^2}$ (d) $\frac{r_1 r_2}{r_2 - r_1}$

Basic surface Tension

Relation between surface Tension and Radius of curvature

Suggested Reading

Source of the Problem

Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(d) $\frac{r_1 r_2}{r_2 - r_1}$

Hint 1

Hint 2

Hint 3

For a bubble of radius $r$ with double surface and whose inside pressure is $p_{in}$ and outside pressure is $p_{out}$ and also the surface tension is $T$, the relation between pressure and radius is,

$$p_{in} - p_{out} = \Delta p = \frac{4T}{r}$$

When 2 bubble merge on their common interface the pressure difference is Just the difference between the pressure(inside) of both two bubbles. Also, the surface tension remains same.

We know $p_1 - p_0 = \frac{4T}{r_1}$ and $p_2 - p_0 = \frac{4T}{r_2}$, Then,

$$ p_1 - p_2 = \frac{4T}{R} $$

here $R$ is the radius of curvature of the interface.

Hence,

$p_{1}-p_{2}$=$\left(p_{1}-p_{0}\right)-\left(p_{2}-p_{0}\right)$=$4 T\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)$

This gives us,

$$ \frac{1}{R} = \frac{1}{r_1}-\frac{1}{r_2} \to R= \frac{r_1 r_2}{r_2 - r_1} $$

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