Mechanical Properties of Fluids NEET PYQ – Chapter-wise Previous Year Questions & Solutions

(3)

Let $n$ droplets each of radius $r$ coalesce to form a big drop of radius $R$ .

$∴$ $Volume of n droplets = Volume of big drop$

$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow n = \frac{R^{3}}{r^{3}}$ ...(i)

$Volume of big drop, V = \frac{4}{3} π R^{3}$ ...(ii)

Initial surface area of $n$ droplets,

$A_{i} = n \times 4 π r^{2} = \frac{R^{3}}{r^{3}} \times 4 π r^{2} (Using (i))$

$= 4 π \frac{R^{3}}{r} = (\frac{4}{3} π R^{3}) \frac{3}{r} = \frac{3 V}{r} (Using (ii))$

Final surface area of big drop,

$A_{f} = 4 π R^{2} = (\frac{4}{3} π R^{3}) \frac{3}{R} = \frac{3 V}{R} (Using (ii))$

Decrease in surface area

$Δ A = A_{i} - A_{f} = \frac{3 V}{r} - \frac{3 V}{R} = 3 V (\frac{1}{r} - \frac{1}{R})$

$∴$ $Energy released = Surface tension \times Decrease in surface area$

$= T \times Δ A = 3 V T (\frac{1}{r} - \frac{1}{R})$

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