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JEE Main 2023 (31 Jan, Shift 1) · Splitting & Merging of Drops

Splitting & Merging of Drops question from JEE Main 2023 (31 Jan, Shift 1)

Q40JEE Main 2023 · 31 JAN · SHIFT 1Medium

Splitting & Merging of Drops

If 1000 droplets of water of surface tension 0.07 N/m0.07\text{ N/m}, having same radius 1 mm1\text{ mm} each, combine to form a single drop. In the process the released surface energy is :-

(Take π=227)\left(\text{Take } \pi = \frac{22}{7}\right)

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Answer: (a)

  1. 1000 r3=R3⇒R=10r=10−2 m1000\, r^3 = R^3 \Rightarrow R = 10r = 10^{-2}\text{ m}.
  2. Energy released =4πT(1000r2−R2)=4π(0.07)(10−3−10−4)= 4\pi T(1000 r^2 - R^2) = 4\pi(0.07)(10^{-3} - 10^{-4})
  3. =4⋅227⋅0.07⋅9×10−4=7.92×10−4 J= 4 \cdot \frac{22}{7} \cdot 0.07 \cdot 9 \times 10^{-4} = 7.92 \times 10^{-4}\text{ J}
  4. So the answer is (a) 7.92×10−4 J7.92 \times 10^{-4}\text{ J}.
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