If a number of little droplets of water, all of the same radius ‘r’ coalesce to form a single drop of radius ‘R’, rise in temperature will be given by 18TηPS1r−1R where ‘T’ is the surface tension of water, ′ρ′ is density of water and 'η' is a dimensionless constant then find 'η'

E=msΔθ ⇒4πnr2−R2T=msΔθ ⇒Δθ=4πTmsnr2−R2 ⇒Δθ=4πT43πρsnr2nr−R2R3⇒Δθ=3Tρs1r−1R Rise in temperature: Δt=3Tρs1r−1R(J=1)Given that Δt=18Tρsη1r−1ROn comparing, η=6

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If a number of little droplets of water, all of the same radius ‘r’ coalesce to form a single drop of radius ‘R’, rise in temperature will be given by $\frac{18 T}{ηPS} (\frac{1}{r} - \frac{1}{R})$ where ‘T’ is the surface tension of water, $^{'} ρ^{'}$ is density of water and ' $η$ ' is a dimensionless constant then find ' $η$ '

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answer is 6.

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Detailed Solution

$\begin{aligned} E = msΔ θ \\ \Rightarrow 4 π ({nr}^{2} - R^{2}) T = msΔ θ \\ \Rightarrow Δ θ = \frac{4 π T}{ms} ({nr}^{2} - R^{2}) \\ \Rightarrow Δ θ = \frac{4 π T}{\frac{4}{3} π ρ s} (\frac{{nr}^{2}}{nr} - \frac{R^{2}}{R^{3}}) \\ \Rightarrow Δ θ = \frac{3 T}{ρs} (\frac{1}{r} - \frac{1}{R}) \end{aligned}$

Rise in temperature: $Δt = \frac{3 T}{ρs} (\frac{1}{r} - \frac{1}{R}) (J = 1)$
Given that $Δt = \frac{18 T}{ρsη} (\frac{1}{r} - \frac{1}{R})$
On comparing, $η = 6$