Fluid Mechanics

JEE Main level, single correct, medium

119 of 192

Question 119 of 192

If 10001000 droplets of water of surface tension 0.07Nm−10.07N{m}^{-1}. having same radius 1mm1mm each, combine to from a single drop. In the process the released surface energy is-
(Takeπ=227)\left(\text{Take}\pi =\dfrac{22}{7}\right)

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Answer: (B) 7.92×10−4J7.92\times {10}^{-4}J

Let the radius of bigger droplet is RR, then by volume conservation:
(43πR3)=1000(43π(1)3)\left(\dfrac{4}{3}\pi {R}^{3}\right)=1000\left(\dfrac{4}{3}\pi {\left(1\right)}^{3}\right)
⇒R=10mm\Rightarrow R=10mm
Final potential energy:
Uf=1000(4πR2T)=1000×4π(10×10−3)2T{U}_{f}=1000\left(4\pi {R}^{2}T\right)=1000\times 4\pi {\left(10\times {10}^{-3}\right)}^{2}T
Initial potential energy:
Ui=4π×(r)2T=4π×(10−3)2T{U}_{i}=4\pi \times {\left(r\right)}^{2}T=4\pi \times {\left({10}^{-3}\right)}^{2}T
Therefore, change in potential energy,
∆E=T×1000×4π(10−3)2−T×4π(10×10−3)2∆E=T\times 1000\times 4\pi {\left({10}^{-3}\right)}^{2}-T\times 4\pi {\left(10\times {10}^{-3}\right)}^{2}
∆E=4×π×7×10−2[1000−100]×10−6∆E=4\times \pi \times 7\times {10}^{-2}\left[1000-100\right]\times {10}^{-6}
∆E=7.92×10−4J∆E=7.92\times {10}^{-4}J

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