Question

The diameter of one drop of water is 0.2
cm. The work done in breaking one
drop into 1000 equal droplets will be:-
(surface tension of water = 7 * 10?
N/m)​

Answer 1

Given :

Diameter of one drop of water = 0.2 cm

Surface tension = 7 × 10^2 N/m

To find :

Work done in breaking one

drop into 1000 equal droplets

Solution :

Let R and r be tha radius of big and small droplet respectively

• Volume in whole process would remain same

Therefore,

(4/3)×pi×R^3 = 1000×(4/3)×pi×r^3

r = R/10 = 0.01 cm

• Change in surface area is given as

∆A = 1000×4×pi×r^2 - 4×pi×R^2

= 4×pi×(10R^2 - R^2)

= 36×pi×R^2

• Therefore, work done would be

W = ∆U = T∆A

W = 7×10^-2×36×pi×10^-6

W = 7.9 × 10^-6 J

Answer 2

Given :

The diameter of one drop of water is 0.2

cm.

We are breaking one

drop into 1000 equal droplets.

To find :

Work done in this process.

Solution :

It is given that

Old diameter of a drop of water = 0.2 cm

so,

Old radius of a drop of water = half of its diameter

R = 0.1 cm

When we break the drop of water into 1000 droplets,

its volume will not change but radius of small droplets will change and they will be same with each other because all droplets are same.

Let the radius of droplets = r

so, as the volumes are same so,

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

so,

${R}^{3} = 1000 \times {r}^{3}$

Taking cube root on both sides,

R = 10r

and

r = 0.1R

Putting radius of drop R = 0.1 , we get

$r = 0.1 \times 0.1$

so,

Radius of a droplet = 0.01 cm.

Surface area of old drop = 4πR²

Surface area of one new droplet = 4πr²

Surface area of all new droplets :

$= 1000 \times4 \pi {r}^{2}$

= 4000πr²

So,

Change in surface areas ∆A = 4000πr² - 4πR²

we know that,

r = 0.01, and R = 0.1

so

$\bf \Delta A = 4000 \pi ({0.01}^{2}) - 4\pi ({0.1}^{2} ) = 0.36\pi$

So,

∆A = 0.36π cm²

changing units in meters,

$\bf\Delta A = 36\pi \times {10}^{ - 6} \: {m}^{2}$

We know that,

Work done is equal to change in surface energy.

W = ∆U

where

U = Surface energy

W = Work done.

and

Change in surface energy

∆U = W = T∆A.

where T = Surface tension of liquid,

We know that value of T = 0.07 Newton per meter

(Typing error in question),

So

Work done :

$\bf = 0.07 \times 36\pi \times {10}^{ - 6} = 7.92 \times {10}^{ - 6} \\ \bf \: 7.92 \: \: \mu \: joules \:$

So,

Work done = 7.92 μJ