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Question Number 35124 by Tinkutara last updated on 15/May/18

Commented by Tinkutara last updated on 15/May/18

$I t c o n t a i n s n o α ?$
Commented by Tinkutara last updated on 15/May/18

Commented by ajfour last updated on 15/May/18

Commented by ajfour last updated on 16/May/18
$S(2πb)cos θ = (π/3){r^2 (h+y)−b^2 y}ρg ((h+y)/r) = (y/b) = (1/(tan α)) ⇒ r=b+htan α ⇒ S(2πb)cos (θ+α)=(π/(3tan α)){r^3 −b^3 }ρg r^3 =((6bStan αcos (θ+α))/(ρg))+b^3 h=(1/(tan α)){(((6bStan αcos (θ+α))/(ρg))+b^3 )^(1/3) −b} . if S is small , α is small too, then h=(b/(tan α)){1+(1/3)×((6Stan αcos (θ+α))/(b^2 ρg))−1} = ((2Scos (θ+α))/(ρgb)) . I have taken the apex angle to be 2α. If apex angle is α (as in your question) the answer indeed is then : h=((2Scos (𝛉+(𝛂/2)))/(𝛒bg)) .$
$S (2 π b) \cos θ = \frac{π}{3} {r^{2} (h + y) - b^{2} y} ρ g$ $\frac{h + y}{r} = \frac{y}{b} = \frac{1}{\tan α}$ $\Rightarrow r = b + h \tan α$ $\Rightarrow S (2 π b) \cos (θ + α) = \frac{π}{3 \tan α} {r^{3} - b^{3}} ρ g$ $r^{3} = \frac{6 b S \tan α \cos (θ + α)}{ρ g} + b^{3}$ $h = \frac{1}{\tan α} {{(\frac{6 b S \tan α \cos (θ + α)}{ρ g} + b^{3})}^{1 / 3} - b} .$ $i f S i s s m a l l, α i s s m a l l t o o, t h e n$ $h = \frac{b}{\tan α} {1 + \frac{1}{3} \times \frac{6 S \tan α \cos (θ + α)}{b^{2} ρ g} - 1}$ $= \frac{2 S \cos (θ + α)}{ρ g b} .$ $I h a v e t a k e n t h e a p e x a n g l e t o b e$ $2 α . I f a p e x a n g l e i s α (a s i n y o u r$ $q u e s t i o n) t h e a n s w e r i n d e e d i s$ $t h e n : h = \frac{2 S \cos (θ + \frac{α}{2})}{ρ b g} .$
Commented by Tinkutara last updated on 16/May/18
Answer given is dependent on alpha.
Commented by ajfour last updated on 16/May/18

$p o s t t h e a n s w e r p l e a s e .$
Commented by Tinkutara last updated on 16/May/18
Question from JEE Advanced 2014.
Commented by ajfour last updated on 16/May/18

$i c o m p a r e d t h e c a p i l l a r y i n t h e$ $d i a g r a m s n o w o n l y, s o r r y f o r$ $t h e r e v e r s e o r i e n t a t i o n, h o w e v e r$ $m y a n s w e r f o r h i s t h e s a m e!$
Commented by Tinkutara last updated on 16/May/18
I appreciate your solution Sir!
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