Fluid Mechanics

JEE Main level, single correct, medium

132 of 192

Question 132 of 192

A large number of water drops, each of radius rr, combine to have a drop of radius RR. If the surface tension is TT and mechanical equivalent of heat is JJ, the rise in heat energy per unit volume will be:

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Answer: (D) 3TJ(1r−1R)\dfrac{3T}{J}\left(\dfrac{1}{r}-\dfrac{1}{R}\right)

n×43πr3=43πR3n\times \dfrac{4}{3}\pi {r}^{3}=\dfrac{4}{3}\pi {R}^{3}
∴n1/3r=R∴{n}^{1/3}r=R
∴∴ Total change in surface energy
=(n(4πr2)−4πR2)T=\left(n\left(4\pi {r}^{2}\right)-4\pi {R}^{2}\right)T
⇒4πT(nr2−R2)\Rightarrow 4\pi T\left(n{r}^{2}-{R}^{2}\right)
∴∴ Heat energy
=4πT(nr2−R2)J×43πR3=3TJ(nr2R3−1R)=\dfrac{4\pi T\left(n{r}^{2}-{R}^{2}\right)}{J\times \dfrac{4}{3}\pi {R}^{3}}=\dfrac{3T}{J}\left(\dfrac{n{r}^{2}}{{R}^{3}}-\dfrac{1}{R}\right)
Put nr3=R3n{r}^{3}={R}^{3}
∴3TJ(1r−1R)∴\dfrac{3T}{J}\left(\dfrac{1}{r}-\dfrac{1}{R}\right)

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