Fluid Mechanics

JEE Main level, single correct, medium

98 of 192

Question 98 of 192

Two water drops each of radius 'r' coalesce to from a bigger drop. If ' T ' is the surface tension, the surface energy released in this process is :

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Answer: (A) 4πr2 T[2−223]4 \pi \mathrm{r}^2 \mathrm{~T}\left[2-2^{\frac{2}{3}}\right]

2×43πR3=43πr3⇒r=21/3RUi=2×4πR2 TUf=4πr2 T=4πR2 T22/3∴ Heat lost =ui−uf=4πR2 T[2−22/3]\begin{aligned} & 2 \times \frac{4}{3} \pi \mathrm{R}^3=\frac{4}{3} \pi \mathrm{r}^3 \Rightarrow \mathrm{r}=2^{1 / 3} \mathrm{R} \\ & \mathrm{U}_{\mathrm{i}}=2 \times 4 \pi \mathrm{R}^2 \mathrm{~T} \\ & \mathrm{U}_{\mathrm{f}}=4 \pi \mathrm{r}^2 \mathrm{~T}=4 \pi \mathrm{R}^2 \mathrm{~T} 2^{2 / 3} \\ & \therefore \text { Heat lost }=\mathrm{u}_{\mathrm{i}}-\mathrm{u}_{\mathrm{f}}=4 \pi \mathrm{R}^2 \mathrm{~T}\left[2-2^{2 / 3}\right]\end{aligned}

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