Question

A water drop of radius of 1 cm is spread out into 1000 droplet calculate the work done in the process where t is equal to 7.2 into 10 to the power minus 2 newton per metre

Answer 1

Work done =force*diasplacement

This is my answer

Answer 2

Answer:

The work done in the process is 8.95×10⁻⁵J.

Explanation:

Let R be the radius of the big drop and r be the radius of each small drop. The volume of the drops will remain the same,

The volume of big drop= Volume of N small drops

$\frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{N} \times \frac{4}{3} \pi \mathrm{r}^{3}$

$\mathrm{R}=\mathrm{N}^{1 / 3} \mathrm{r} \quad \text$     (1)

And from the question, the value of N is 1000 and R is 1cm. So,

$1cm=\mathrm{1000}^{1 / 3} \mathrm{r} \quad \text$

$r=\frac{1}{10}$

$r=10^-^3m$     (2)

Increase in surface area(ΔA) = $4 \pi \mathrm{R}^{2}\left[\mathrm{~N}^{1 / 3}-1\right]$       (3)

Work done(W) is calculated as,

$W=T\Delta A$        (4)

Where,

T=surface tension

From the question, the value of T is 7.2×10⁻²N/m.

By placing all the required entities in equation (4) we get;

$W=7.2\times 10^-^2\times 4\pi\times R^2[N^1^/^3-1]$

$W=7.2\times 10^-^2\times 4\pi\times (10^-^2)^2[1000^1^/^3-1]$

$W=7.2\times 10^-^2\times 1243.44$

$W=8.95\times 10^-^5\ J$

So, the work done in the process is 8.95×10⁻⁵J.

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