## When a big drop of water is formed from n small drops of water, the energy loss is 3 E , where E is the energy of the bigger drop…

Q: When a big drop of water is formed from n small drops of water, the energy loss is 3 E , where E is the energy of the bigger drop. If R is the radius of the bigger drop and r is the radius of the smaller drop, then number of smaller drops (n) is

Sol: The energy of n small drops – the energy of the bigger drop = Energy loss by bigger drop

n × 4 π r2 × T – 4 π R2 × T = 3 × 4 π R2 × T

n × 4 π r2 = 12 π R2 + 4 π R2

⇒ n = 4 R2/r2

## Number of droplets are combined isothermally to form a big drop, the ratio of initial and final surface energies of the system

Q: Number of droplets are combined isothermally to form a big drop, the ratio of initial and final surface energies of the system is

Sol: $\large \frac{U_i}{U_f} = \frac{n(4\pi r^2)T}{4\pi R^2 T}$

$\large \frac{U_i}{U_f} = \frac{n r^2}{R^2} = \frac{n r^2}{n^{(1/3 r)^2}}$

$\large \frac{U_i}{U_f} = \frac{n^{1/3}}{1}$

## A drop of radius R is split under isothermal conditions into ‘n’ droplets, each of radius ‘ r ’, the ratio of surface energies…

Q: A drop of radius R is split under isothermal conditions into ‘n’ droplets, each of radius ‘ r ’, the ratio of surface energies of big and each small drop is

Sol: $\large \frac{U_{big}}{U_{small}} = \frac{4\pi R^2 T}{4\pi r^2 T}$

$\large \frac{U_{big}}{U_{small}} = \frac{R^2}{r^2}$

$\large \frac{U_{big}}{U_{small}} = \frac{(n^{1/3}r)^2}{r^2}$

$\large \frac{U_{big}}{U_{small}} = \frac{n^{2/3}}{1}$

## A large number of liquid drops each of radius “ r ” merge to form a single spherical drop of radius “ R ”.

Q: A large number of liquid drops each of radius “ r ” merge to form a single spherical drop of radius “ R ”. If the energy released in the process is converted into the kinetic energy of the big drop formed. Find the speed of the big drop (d is density of the liquid)?

Sol: Energy released is $\large W = 3 V T [\frac{1}{r} – \frac{1}{R} ]$ …(i)

If V is volume of big drop, M the mass of the drop and ρ the density then

Kinetic energy $\large = \frac{1}{2}M v^2 = \frac{1}{2}(\rho V) v^2$ …(ii)

As (i) = (ii)

$\large 3 V T [\frac{1}{r} – \frac{1}{R} ] = \frac{1}{2}(\rho V) v^2$

$\large v = \sqrt{\frac{6T}{\rho}(\frac{1}{r} – \frac{1}{R})}$

## 1000 drops of a liquid each of diameter 4 mm coalesce to form a single large drop. If surface tension of liquid is 35 dyne/cm.

Q: 1000 drops of a liquid each of diameter 4 mm coalesce to form a single large drop. If surface tension of liquid is 35 dyne/cm. Calculate the energy evolved by the system in the process.

Sol: No. of drops n = 1000

⇒ n1/3 = 10 ; n2/3 = 100

Surface tension of liquid T = 35 dyne / cm

Radius of each small drop r = 2 mm = 0.2 cm

Energy evolved in merging W = 4 π r2 T [n – n2/3]

W = 4 × 22/7 × (2 × 10-1)2 35 [1000 – 100]

= 15840 ergs. ≈ 1.58 × 10-3 J

## A water drop of diameter 2 mm is split up into 10^9 identical water drops. Calculate the work done in this process.

Q: A water drop of diameter 2 mm is split up into 109 identical water drops. Calculate the work done in this process.
(The surface tension of water is 7.3 × 10-2 N/m)

Sol: Let a water drop of radius R be split up into 10^9 identical water drops each of radius r.

R = D/2 = 2/2 = 1 mm= 1 × 10-3 m

No. droplets n = 109 ; T = 7.3 × 10-2 N/m

W = 4 π R2 T [n1/3 -1]

= 4π (10-3 )2  × 7.3 × 10-2 [(109 )1/3 -1]

=9.17 × 10-4 J

## A soap bubble is blown to a radius of 3 cm. If it is to be further blown to a radius of 4 cm what is the work done ?

Q: A soap bubble is blown to a radius of 3 cm. If it is to be further blown to a radius of 4 cm what is the work done ?
(Surface tension of soap solution = 3.06 × 10-2 N/m

Sol: Initial radius of soap bubble R1 = 3cm = 3 × 10-2 m

Final radius of soap bubble R2 = 4cm = 4× 10-2 m

Work done in blowing soap bubble from radius R1 to R2 is

W = 8π (R22-R12) T

= 8 × 22/7 × 3.06 × 10-2 (16 – 9) × 10-4

= 176 × 3.06 × 10-6 J

= 539.6 × 10-6 J