A big drop is formed by coalescing 1000 small identical drops of… — Solution | JEE Physics

MCQEasyJEE 2024Surface Tension & Capillarity

A big drop is formed by coalescing $1000$ small identical drops of water. If $E_1$ is the total surface energy of $1000$ small drops and $E_2$ is the surface energy of the single big drop, then the ratio $E_1:E_2$ is $x:1$ where $x$ equals:

A
$5$
B
$8$
C
$10$
D
$15$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A big drop is formed by coalescing $1000$ identical small water drops. Let the radius of each small drop be $r$ and the radius of the big drop be $R$ .

Find: The ratio $E_1:E_2 = x:1$ .

The surface energy of a droplet is given by

E = \sigma A

where $\sigma$ is the surface tension and $A$ is the surface area.

For one small drop,

A_1 = 4\pi r^2

So, for $1000$ small drops, the total surface area is

A_{\text{total}} = 1000 \times 4\pi r^2 = 4000\pi r^2

Hence, the total surface energy of the small drops is

E_1 = \sigma \times 4000\pi r^2

Using volume conservation

When $1000$ small drops coalesce, total volume remains constant.

Volume of one small drop:

V_{\text{small}} = \frac{4}{3}\pi r^3

Total volume of $1000$ small drops:

V_{\text{total}} = 1000 \times \frac{4}{3}\pi r^3

Volume of the large drop:

V_{\text{large}} = \frac{4}{3}\pi R^3

Direct scaling idea

Equating volumes,

\frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3

which gives

R^3 = 1000r^3 \Rightarrow R = 10r

Now surface area of the big drop is

A_2 = 4\pi R^2 = 4\pi (10r)^2 = 400\pi r^2

So its surface energy is

E_2 = \sigma \times 400\pi r^2

Therefore,

\frac{E_1}{E_2} = \frac{\sigma \times 4000\pi r^2}{\sigma \times 400\pi r^2} = 10

Thus,

E_1:E_2 = 10:1

Therefore, the correct option is C and $x = 10$ .

Why this shortcut works: on coalescing $n$ identical drops, radius scales as $n^{1/3}$ and surface area scales as $n^{2/3}$ . Hence the ratio of initial to final surface energy becomes $n/n^{2/3} = n^{1/3}$ . For $n = 1000$ , this is $10$ .

Common mistakes

Using surface area proportional to $r^3$ instead of $r^2$ is incorrect because $r^3$ belongs to volume, not area. Use surface area of a sphere as $4\pi r^2$ when calculating surface energy.
Not applying volume conservation during coalescence is wrong because the amount of water remains the same. First equate total volume of $1000$ small drops with the volume of the big drop to find $R$ .
Taking $R = 1000r$ instead of $R = 10r$ is incorrect because radii are related through the cube root from volume conservation. From $R^3 = 1000r^3$ , the correct result is $R = 10r$ .

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