Question

A big drop of water whose diameter is 0.2 cm, is
broken into 27000 small drops of equal volume.
Work done in this process will be (surface tension
of water is 7 * 10-2 N/m).
(A) 5 x 105 joule (B)2.9 x 10-5 joule
(C) 2.55 x 10-5 joule (D) zero

Answer 1

The work done is $2.55\times10^{-5}\ Joule$

(C) is correct option.

Explanation:

Given that,

Diameter of big drop = 0.2 cm

Number of small drops = 27000

Surface tension $T =7\times10^{-2}\ N/m$

On breaking it into 27000 droplets, the volume would remain constant. Therefore, radius of each small drop,

We need to calculate the radius of small drop

Using formula of volume

$\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi\times 27000(r^3)$

$R^3=(30r)^3$

$R=30r$

$r=\dfrac{R}{30}$

$r=0.033R$

Now, change in surface area is

$\delta A=27000\times4\pi r^2-4\pi R^2$

$\delta A=4\pi(27000\times(0.033)^2R^2-R^2)$

$\delta A=4\pi(30R^2-R^2)$

$\delta A=4\pi\times29R^2$

$\delta A=116\pi R^2$

We need to calculate the work done in this process

Using formula of change in surface energy

$W=\delta U$

$W=T\delta A$

Put the value into the formula

$W=7\times10^{-2}\times116\pi R^2$

$W=7\times10^{-2}\times116\pi\times(0.1\times10^{-2})^2$

$W=0.0000255\ J$

$W=2.55\times10^{-5}\ Joule$

Hence, The work done is $2.55\times10^{-5}\ Joule$

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Topic : work done

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