Question

A large drop is formed by combining 27 small drops of water what will be the change in surface energy what will be the ratio of surface energy of big drop to the surface energy of the small drop

Answer 1

Answer:

Explanation:

Let radius of smaller drops be r

Volume of bigger drop = 27× volume of smaller drops

4/3piR³=27×4/3pi ×r³

R= 3r

Surface erg of bigger/surface erg of smaller=

Surface tension × surface area(bigger)/surface tension × surface area(smaller)

=(T×4piR²)/(T×4pir²)

=(R/r)²

=9

Answer 2

The ratio of surface energy of big drop to the surface energy of the small drop 9:1.

Explanation:

The surface energy of the drop is given as

$E=(4\pi R^2)T$

Here T is the surface tension and R is the radius of the drop.

Let r is the radius of each single drop.

Since volume of liquid remains same, we have

$27 \frac{4}{3}\pi r^3= \frac{4}{3}\pi R^3$

$r=\frac{R}{3}$.

The surface energy of the small drop,

$E_r=(4\pi r^2)T$

or

$E_r=(4\pi (\frac{R}{3})^2 )T$

Therefore ratio,

$\frac{E}{E_r} =\frac{(4\pi R^2)T}{(4\pi (\frac{R}{3})^2 )T}$

$\frac{E}{E_r}=\frac{9}{1}$.

Thus, the ratio of surface energy of big drop to the surface energy of the small drop 9:1.

#Learn More: surface energy.

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