Fluid Mechanics

JEE Main level, single correct, medium

113 of 192

Question 113 of 192

A small liquid drop of radius RR is divided into 2727 identical liquid drops. If the surface tension is TT, then the work done in the process will be :

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Answer: (A) 8πR2T8\pi {R}^{2}T

Let radius of the newly formed drops be rr.
As total volume will remain constant, therefore
43πR3=27×(43×πr3)\dfrac{4}{3}\pi {R}^{3}=27\times \left(\dfrac{4}{3}\times \pi {r}^{3}\right)
⇒R3=27r3\Rightarrow {R}^{3}=27{r}^{3}
⇒R=3r\Rightarrow R=3r
⇒r=R3\Rightarrow r=\dfrac{R}{3}
⇒r2=R29\Rightarrow {r}^{2}=\dfrac{{R}^{2}}{9}
Now, work done =T×ΔA=T\times \Delta A
=T[27(4πr2)]−T[4πR2]=T[27(4πR29)]−T[4πR2]=TπR2(12−4)=T\left[27\left(4\pi {r}^{2}\right)\right]-T\left[4\pi {R}^{2}\right] =T\left[27\left(4\pi \dfrac{{R}^{2}}{9}\right)\right]-T\left[4\pi {R}^{2}\right] =T\pi {R}^{2}\left(12-4\right)
=8πR2T=8\pi {R}^{2}T

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