Two parallel plates at a distance d are submerged in water. (Surface tension = s and contact
angle $θ$ ) Water has density $ρ$ . The figure (a) shows the water rise and cross section view is in figure (b). Now the plates are pushed down so that the height of plates above water is $\frac{h}{2}$ as in figure (c).

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$h = \frac{2 s cos θ}{ρ dg}$

$cos θ^{'} = \frac{cos θ}{2}$

$R_{1} = \frac{d}{2 cos θ}$

$R_{2} = \frac{d}{4 cos θ}$

answer is A, B, C.

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Detailed Solution

Figure (a & b): Question Image

$\cos θ = \frac{r}{R_{1}} \Rightarrow R_{1} = \frac{d}{2 \cos θ} Δ P = S (\frac{1}{r / \cos θ} + \frac{1}{\infty}) = \frac{S \cos θ}{r} = \frac{2 S \cos θ}{d} \Rightarrow Δ P = h ρ g = \frac{2 S \cos θ}{d} \Rightarrow h = \frac{2 S \cos θ}{ρ g d}$

Figure (c): Question Image

$\cos θ' = \frac{r}{R_{2}} \Rightarrow R_{2} = \frac{r}{\cos θ'} = \frac{d}{2 \cos θ'} Δ P = S (\frac{1}{r / \cos θ'} + \frac{1}{\infty}) = \frac{S \cos θ'}{r} = \frac{2 S \cos θ'}{d} A s Δ P = \frac{ρ g h}{2} = \frac{2 S \cos θ'}{d} \Rightarrow \cos θ' = \frac{\cos θ}{2} A n s . A B C$