Question

A single drop of liquid is split into 8 identical drops. What will be the excess pressure in each drop?

Answer 1

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Answer 2

The excess pressure in each drop is $\dfrac{4T}{R}$

Explanation:

Given that,

Number or drop = 8

We need to calculate the radius of drop

Using formula of volume

Volume of big drop = Volume of small drops

$\dfrac{4}{3}\pi\times R^3=8\times\dfrac{4}{3}\pi\times r^3$

$R=2r$

$r=\dfrac{R}{2}$

We need to calculate the excess pressure in each drop

Using formula of excess pressure

$P=\dfrac{2T}{r}$

Put the value into the formula

$P=\dfrac{2T}{\dfrac{R}{2}}$

$P=\dfrac{4T}{R}$

Hence, The excess pressure in each drop is $\dfrac{4T}{R}$

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