Question

A large number of water drops, each of radius $r$, combine to have a drop of radius $R$. If the surface tension is$T$ and mechanical equivalent of heat is $J$, the rise in heat energy per unit volume will be.

• A. $\raisebox{1ex}{$2T$}\!\left/ \!\raisebox{-1ex}{$rJ$}\right.$
• B. $\raisebox{1ex}{$3T$}\!\left/ \!\raisebox{-1ex}{$rJ$}\right.$
• C. $\raisebox{1ex}{$2T$}\!\left/ \!\raisebox{-1ex}{$J$}\right.\left(\frac{1}{r}-\frac{1}{R}\right)$
• D. $\raisebox{1ex}{$3T$}\!\left/ \!\raisebox{-1ex}{$J$}\right.\left(\frac{1}{r}-\frac{1}{R}\right)$

Answer 1

The correct option is D

$\raisebox{1ex}{$3T$}\!\left/ \!\raisebox{-1ex}{$J$}\right.\left(\frac{1}{r}-\frac{1}{R}\right)$

Step 1: Given data.

Radius of each water drop$=r$

Radius of combined number of water drop$=R$

Surface tension$=T$

Equivalent heat$=J$

Step 2: Finding the rise in heat energy per volume.

Formula used:

$H=\frac{w}{J}=m.s.∆T$ …..$\left(i\right)$

Where,

$H$ is amount of heat in unit joule.

$w$ is change in surface energy.

$m$ is mass in unit gram.

$s$ is specific heat in unite joule per gram.

$∆t$ is change in temperature in Celsius.

Now,

According to the question.

Volume of $n$number of each drops$=$Volume of large combined drop

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

$\Rightarrow$ $R^3=n{r}^3$

$\Rightarrow$ $n=\frac{R^3}{r^3}$ ……$\left(ii\right)$

Also,

$w=$surface tension $\times$ change in surface area

$w=T\left(n4\pi r^2-4\pi R^2\right)$

$\Rightarrow$$J.m.s.∆t=T\left(n4\pi r^2-4\pi R^2\right)$ $\left[fromequation\left(i\right)\right]$

$\Rightarrow$ $∆t=\frac{T\left(n4\pi r^2-4\pi R^2\right)}{J\times m\times s}$

$\Rightarrow$ $∆t=\frac{T\left({\displaystyle \frac{R^3}{r^3}}4\pi r^2-4\pi R^2\right)}{J\times m\times s}$ $\left[fromequation\left(ii\right)\right]$

For water $s=1$, then

$\Rightarrow$ $∆t=\frac{4\pi R^{2}T\left({\displaystyle \frac{R}{r}}-1\right)}{J\times d\times {\displaystyle \frac{4}{3}}\pi R^3\times 1}$ $\left[\because m=density\times volume\right]$

$\Rightarrow$ $∆t=\frac{3T\left({\displaystyle \frac{1}{r}}-{\displaystyle \frac{1}{R}}\right)}{Jd}$

For water density$=1$

Therefore,

$\Rightarrow$ $∆t=\frac{3T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$

Hence, option D is correct.