Physics - Fluid Question with Solution | TestHub

Let the radius of the small drop be $r$ and the radius of the large drop be $R$.

Since $1000$ small drops combine to form $1$ large drop, the total volume remains conserved.

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

$R^3=1000r^3\Rightarrow R=10r$

The surface energy of a drop is given by $E=T\times A$, where $T$ is the surface tension and $A$ is the surface area.

Surface energy of $1$ large drop, $E_L=T\times 4\pi R^2$

Surface energy of $1$ small drop, $E_S=T\times 4\pi r^2$

The ratio of their surface energies is:

$\frac{E_L}{E_S}=\frac{T\times 4\pi R^2}{T\times 4\pi r^2}={\left(\frac{R}{r}\right)}^2$

Substituting $R=10r$:

$\frac{E_L}{E_S}={\left(\frac{10r}{r}\right)}^2=100$

The required ratio is $100:1$.

Answer: $100:1$