Question

║⊕QUESTION⊕║
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PHYSICS
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2 liquid drops have the ratio of their radii 1:4. What is the ratio of excess pressure inside the drop?

Answer 1

Answer:

Ratio of excess pressure is 4 : 1

Explanation:

We Have :-

Ratio of Radius of two liquid drops = 1 : 4

To Find :-

Excess pressure inside the drop

Solution :-

Let the radii be x₁ and x₂

x₁ / x₂ = 1 / 4

Let α be the surface tension

Excess Pressure

Drop 1 = 4 α / x₁

Drop 2 = 4 α / x₂

Pressure

Drop 1 / Drop 2 = 4 / 1

Ratio = 4 : 1

Ratio of excess pressure is 4 : 1

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

$\huge {\tt {GIVEN :-}}$

• 2 liquid drops
• The ratio of their radii is 1 : 4

$\huge {\tt {TO\:FIND :-}}$

The ratio of the excess pressure inside it.

$\huge {\tt {SOLUTION :-}}$

Let a liquid drop, whose surface tension is $\tt {F_T}$ and the pressure inside the drop be P.

As we know,

$\leadsto\tt {\Delta P \times \pi {r}^{2} = F_T \times 2 \pi r}$

So,

$\leadsto\tt {\Delta P \times r= F_T \times 2 }$

$\leadsto \tt {\Delta P = \dfrac {F_T\times 2}{r}}$

Now,

The ratio of the excess pressure:-

$\leadsto \tt{\dfrac {\Delta P_1}{\Delta P_2}=\dfrac{r_2}{r_1}}$

(This is because the Surface tension is same)

$\leadsto \tt {\dfrac {\Delta P_1}{\Delta P_2}=\dfrac {4}{1}}$

So,

The ratio is :-

$\huge\leadsto{\tt {\boxed {4 : 1}}}$

$\rule {200}2$