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Surface Tension Physics Mechanical Properties of Fluids Physics Class 11th

A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be

Option 1 -

$\frac{3 T}{J} (\frac{1}{r} - \frac{1}{R})$

Option 2 -

$\frac{2 T}{r J}$

Option 3 -

$\frac{3 T}{r J}$

Option 4 -

$\frac{2 T}{J} (\frac{1}{r} - \frac{1}{R})$

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Answered by

Payal Gupta | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

From volume conservation

$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R^{3} = n r^{3} . . . . . . . . . (1)$

Decrease in surface area = $n \times 4 π r^{2} - 4 π R^{2}$

$(Δ A) = 4 π [n r^{2} - R^{2}] = 4 π [\frac{n \times r^{3}}{r} - R^{2}] = 4 π [\frac{R^{3}}{r} - R^{2}] = 4 π R^{3} [\frac{1}{r} - \frac{1}{R}]$

Energy released (W) = $T \times Δ A = 4 π R^{3} T [\frac{1}{r} - \frac{1}{R}]$

Heat produced (Q) = $\frac{W}{J} = \frac{4 π T R^{3}}{J} [\frac{1}{r} - \frac{1}{R}]$

$N o w, Q = m s Δ θ$

$\Rightarrow \frac{4 π R^{3} T}{J} [\frac{1}{r} - \frac{1}{R}] = (\frac{4}{3} π R^{3}) \times | x | \times Δ θ = \frac{3 T}{J} [\frac{1}{r} - \frac{1}{R}]$

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