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If 1000 droplets of water of surface tension 0.07 N m⁻¹, each having the same radius 1 mm, combine to form a single drop, then in the process the released surface energy is (take π=(22)/7)

Asked in JEE Main 31st Jan 1st Shift 2023 · Coalescence of drops

• (1) 8.8×10⁻⁵ J
• (2) 7.92×10⁻⁴ J
• (3) 7.92×10⁻⁶ J
• (4) 9.68×10⁻⁴ J

Step-by-step solution

Why the other options are wrong

• (1) comes from is the final big drop's own surface energy, 4π R²T=8.8×10⁻⁵ J, reported instead of the energy released.
• (3) comes from is the key with a factor of 100 lost, from slipping on r=1 mm=10⁻³ m when squaring it.
• (4) comes from adds the two surface energies, 4π T(10⁻³+10⁻⁴)=9.68×10⁻⁴ J, instead of subtracting them.

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