First slide

Surface tension and surface energy

Question

A spherical liquid drop of radius R is divided into eight equal droplets. If the surface tension is 7, then the work done in this process will be

Moderate

A

$2 π R^{2} T$
B

$3 π R^{2} T$
C

$2 π R T^{2}$
D

$4 π R^{2} T$

Solution

Radius of the larger drop = R
Suppose radius of the droplets = r
Since volume will remain constant,
$\begin{array}{l} \frac{4}{3} π R^{3} = 8 \times \frac{4}{3} π r^{2} (as no . of droplets = 8) \\ ∴ r = {(\frac{R^{3}}{8})}^{\frac{1}{3}} = \frac{R}{2} \end{array}$
Therefore, work done = increase in surface area x Surface tension
$\begin{matrix} = [8 \times 4 π {(\frac{R}{2})}^{2} - 4 π R^{2}] \times T \\ = (8 π R^{2} - 4 π R^{2}) \times T = 4 π R^{2} T \end{matrix}$

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