Question

Eight spherical droplets, each of radius ${}^{\prime }{r}^{\prime }$ of a liquid of density ${}^{\prime }{\varphi }^{\prime }$ and surface tension ${}^{\prime }{T}^{\prime }$ coalesce to form one big drop. If ${}^{\prime }{s}^{\prime }$ in the specific heat of the liquid. Then the rise in the temperature of the liquid.

• A. $\frac{2T}{3r\rho s}$
• B. $\frac{3T}{r\rho s}$
• C. $\frac{3T}{2r\rho s}$
• D. $\frac{T}{r\rho s}$

Answer 1

Answer: (C)

$\frac{3T}{2r\rho s}$

Given

8 droplets of radius r join to make one big droplet

Density of liquid is $\rho$

Surface tension is T

Specific heat of liquid is 's'

Rise in temperature is ?

Solution

Let Radius of big drop is R

Volume of 8 small drops = Volume of Big drop

$8\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

$R=2r$

Increase in temperature will be because of Change in surface energy as There is change in surface area

Hence

$\Delta E=T\times \Delta A$

where T is surface tension

$\Delta E=T\times (4\pi R^2-8\times 4\pi r^2)$

putting R=2r

$\Delta E=-T\times 16\pi r^2$

Now

$\Delta E=m\times s\times \Delta t$

where m is mass of liquid

's' is specific heat of liquid

$\Delta t$ is change in temperature

$m=\rho \times volume$

$m=\rho \times \frac{4}{3}\pi R^2$

Putting values

$T\times 16\pi \times r^2=\rho \times \frac{4}{3}\pi 4r^2\times s\times \Delta t$

$\Delta t=\frac{3T}{2\rho \times s}$

The correct option is C