First slide

Surface tension

Question

8 identical drops of water, each of radius 1 mm coalesce to form a single drop. Then heat generated in the process is [surface tension of water is 0.072 N/m]

Moderate

A

1.152 $π μ J$
B

6.528 $π μ J$
C

2.4 $π μ J$
D

32 $π μ J$

Solution

$\frac{4}{3} π R^{3} = 8 \times \frac{4}{3} π r^{3} \Rightarrow R = 2 r$
$U_{f} = Final surface energy = 4 π R^{2} . T = 4 π {(2 r)}^{2} T = 16 π r^{2} T$
$U_{i} = Initial surface energy = 8 \times 4 π r^{2} T = 32 π r^{2} T$
$∴$ Heat generated = Lost energy $= U_{i} - U_{f} = 32 π r^{2} T - 16 π r^{2} T$
$= 16 π \times {(10^{- 3})}^{2} \times 0.072 J = 1.152 π μ J$

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