Physics - Fluid Question with Solution | TestHub

Let the radius of the original drop be $R$ and the radius of each smaller drop be $r$.

Equating the initial and final volumes:

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

$R^3=1000r^3\Rightarrow R=10r\Rightarrow r=\frac{R}{10}$

The initial surface energy is $u_i=T\times 4\pi R^2$.

The final total surface energy is $u_f=1000\times T\times 4\pi r^2$.

Substituting $r=\frac{R}{10}$:

$u_f=1000\times T\times 4\pi {\left(\frac{R}{10}\right)}^2=10\times T\times 4\pi R^2=10u_i$

The ratio is $\frac{u_f}{u_i}=10$.

Given $\frac{u_f}{u_i}=\frac{10}{x}$, we get $x=1$.

Answer: $1$