1000 drops of water all of same size join together to form a single drop and the energy released raises the temperature of the drop. Given that T is the surface tension of water, r the radius of each small drop, ρ the density of liquid, J the mechanical equivalent of heat. What is the rise in the temperature

To find the rise in temperature when 1000 small water drops coalesce into a single large drop, we can analyze the conservation of volume and the change in surface energy.

Step 1: Relate the radius of the large drop to the radius of the small drops
Let $r$ be the radius of each small drop, and $R$ be the radius of the newly formed large drop.
Since the total volume remains conserved:
$V_{\text{final}}=N·V_{\text{initial}}$
where $N=1000$ is the number of drops.
Substitute the formula for the volume of a sphere:
$\frac{4}{3}\pi R^3=1000·\left(\frac{4}{3}\pi r^3\right)$
Simplifying this equation, we get:
$R^3=1000r^3\Rightarrow R=10r$

Step 2: Calculate the change in surface area
Initial surface area of 1000 small drops:
$A_i=1000·4\pi r^2$
Final surface area of the single large drop:
$A_f=4\pi R^2=4\pi {\left(10r\right)}^2=100·4\pi r^2$
The decrease in surface area ($\mathrm{\Delta A}$) is:
$\mathrm{\Delta A}=A_i-A_f=1000·4\pi r^2-100·4\pi r^2=900·4\pi r^2=3600\pi r^2$

Step 3: Calculate the energy released and heat produced
The energy released ($W$) due to the reduction in surface area is given by:
$W=T·\mathrm{\Delta A}=3600\pi r^{2}T$
This energy is converted into heat energy ($H$) in calories:
$H=\frac{W}{J}=\frac{3600\pi r^{2}T}{J}$

Step 4: Relate heat to the temperature rise
Let $\mathrm{\Delta \theta }$ be the rise in temperature. The heat absorbed by the water drop is:
$H=m·s·\mathrm{\Delta \theta }$
where:
- $m$ is the mass of the water drop, which is $\text{Volume}·\text{density}=\frac{4}{3}\pi R^3\rho =\frac{4}{3}\pi {\left(10r\right)}^3\rho =\frac{4000}{3}\pi r^3\rho$
- $s$ is the specific heat of water (equal to $1\text{cal/g}·\text{K}$ in CGS units).

Equating the heat expressions:
$\frac{3600\pi r^{2}T}{J}=\left(\frac{4000}{3}\pi r^3\rho \right)·\left(1\right)·\mathrm{\Delta \theta }$
Solving for $\mathrm{\Delta \theta }$:
$\mathrm{\Delta \theta }=\frac{3600\pi r^{2}T}{J}·\frac{3}{4000\pi r^3\rho }=\frac{2.7T}{Jr\rho }$

Since the resulting temperature rise of $\frac{2.7T}{Jr\rho }$ does not match any of the given options ("T/Jr", "10T/Jr", "100T/Jr"), the correct option is "None of these".