Physics - Fluid Question with Solution | TestHub

Let the radius of the big drop be $R$. Since the volume remains conserved during the process of coalescence:

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

$R^3=2r^3\Rightarrow R=2^{1/3}r$

The initial surface energy of the two drops is:

$U_i=2\times (4\pi r^{2}T)=8\pi r^{2}T$

The final surface energy of the big drop is:

$U_f=4\pi R^{2}T=4\pi (2^{1/3}r{)}^{2}T=4\pi r^{2}T(2^{2/3})$

The surface energy released in this process is:

$\Delta U=U_i-U_f$

$\Delta U=8\pi r^{2}T-4\pi r^{2}T(2^{2/3})$

$\Delta U=4\pi r^{2}T(2-2^{2/3})$

Comparing this with the given expression $4\pi r^{2}Tx$, we get:

$x=2-2^{2/3}$

Answer: $\left(2-2^{2/3}\right)$