Fluid Mechanics

JEE Main level, single correct, medium

125 of 192

Question 125 of 192

A water drop of radius 1cm1cm is broken into 729729 equal droplets. If surface tension of water is 75dynecm−175dyne{cm}^{-1}, then the gain in surface energy upto first decimal place will be
[Given π=3.14\pi =3.14]

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Answer: (C) 7.5×10−4J7.5\times {10}^{-4}J

The volume of the bigger drop will be, V=43πR3V=\dfrac{4}{3}\pi {R}^{3}. If the radius of smaller drops is rr, using the conservation of volume,
n43πr3=43πR3⇒729×43πr3=43πR3⇒r=R9n\dfrac{4}{3}\pi {r}^{3}=\dfrac{4}{3}\pi {R}^{3} \Rightarrow 729\times \dfrac{4}{3}\pi {r}^{3}=\dfrac{4}{3}\pi {R}^{3} \Rightarrow r=\dfrac{R}{9}
The initial surface energy will be,
Ei=(4πR2)T{E}_{i}=\left(4\pi {R}^{2}\right)T
And final surface energy will be,
Ef=n(4πr2)T=729×4π(R9)2T{E}_{f}=n\left(4\pi {r}^{2}\right)T=729\times 4\pi {\left(\dfrac{R}{9}\right)}^{2}T=36πR2T=36\pi {R}^{2}T
Therefore, the change in the surface energy will be,
⇒ΔE=Ef−Ei=32πR2T=32×3.14×(1×10−2)2×75×10−3⇒ΔE=7.5×10−4J\Rightarrow \Delta E={E}_{f}-{E}_{i}=32\pi {R}^{2}T=32\times 3.14\times {\left(1\times {10}^{-2}\right)}^{2}\times 75\times {10}^{-3} \Rightarrow \Delta E=7.5\times {10}^{-4}J

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