Question

A liquid drop of diameter 'D' is split into 27 droplets. It T is surface tension of the liquid the change in energy is ​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Change\:in\:energy=2\pi D^{2}T}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{given :}} \\ \tt: \implies Diameter \: of \: big \: drop = D \\ \\ \tt: \implies Small \: drops = 27 \\ \\ \tt: \implies Surface \: tension \: of \: big \: drop = T \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Change \: in \: energy =?$

• According to given question :

$\tt \circ \:Let \: radius \: of \: big \: drop \: be \: r_{1}\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Volume \: of \: big \: drop = 27 \times Volume \: of \: small \: drop \\ \\ \tt: \implies \frac{4}{3}\pi { r_{1}}^{3} = 27 \times \frac{4}{3} \pi {r}^{3} \\ \\ \tt: \implies r_{1}^{3} = {3}^{3} \times {r}^{3} \\ \\ \tt: \implies r_{1} = 3r - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Change \: in \: surface =27 \times small \: drop \: surface - big \: drop \: surface$

$\tt: \implies Change \: in \: surface = 27 \times 4\pi {r}^{2} - 4\pi { r_{1} }^{2} \\ \\ \tt: \implies Change \: in \: surface = 27 \times 4\pi { \bigg(\frac{ r_{1} }{3} \bigg) }^{2} - 4\pi {r_{1} }^{2} \\ \\ \tt: \implies Change \: in \: surface =8\pi { r_{1} }^{2} \\ \\ \tt \circ \: r_{1} = \frac{D}{2} \\ \\ \tt: \implies Change \: in \: surface = 2\pi D^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Change \: in \: energy =Change \: in \: surface \times Surface \: tension \\ \\ \tt: \implies Change \: in \: energy =2\pi D^{2} \times T\\ \\ \green{\tt: \implies Change \: in \: energy= 2\pi D^{2}T}$