Physics - Fluid Question with Solution | TestHub

By the conservation of volume, the volume of the initial drop is equal to the total volume of the 216 droplets.

$\frac{4}{3}\pi R^3=216\times \frac{4}{3}\pi r^3$

$R^3=216r^3\Rightarrow R=6r\Rightarrow r=\frac{R}{6}$

The initial surface area of the drop is:

$A_1=4\pi R^2$

The final surface area of the 216 droplets is:

$A_2=216\times 4\pi r^2$

Substituting $r=\frac{R}{6}$:

$A_2=216\times 4\pi {\left(\frac{R}{6}\right)}^2=216\times 4\pi \left(\frac{R^2}{36}\right)=24\pi R^2$

The increase in surface area is:

$\Delta A=A_2-A_1=24\pi R^2-4\pi R^2=20\pi R^2$

The energy required to break the drop is given by the product of surface tension and the increase in surface area:

$E=T\Delta A=T(20\pi R^2)=20\pi T{R}^2$

Comparing this with the given expression $xT{R}^2$, we get:

$x=20\pi$

Answer: $20\pi$