A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T , then the work done in the process will be :

Question

TestHub · Accepted Answer

Let radius of the newly formed drops be $r$ .

As total volume will remain constant, therefore

$\frac{4}{3} π R^{3} = 27 \times (\frac{4}{3} \times π r^{3})$

$\Rightarrow R^{3} = 27 r^{3}$

$\Rightarrow R = 3 r$

$\Rightarrow r = \frac{R}{3}$

$\Rightarrow r^{2} = \frac{R^{2}}{9}$

Now, work done $= T \times Δ A$

$= T [27 (4 π r^{2})] - T [4 π R^{2}] = T [27 (4 π \frac{R^{2}}{9})] - T [4 π R^{2}] = T π R^{2} (12 - 4)$

$= 8 π R^{2} T$

Physics - Fluid Question with Solution | TestHub