A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross-sections of different radii. When dipped in water vertically, the water rises in it to a height h , where the radius of its cross-section is b . If the surface tension of water is S , its density is ρ , and its contact angle with glass is θ , then the value of h will be ( g is the acceleration due to gravity)

Question

A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross-sections of different radii. When dipped in water vertically, the water rises in it to a height h , where the radius of its cross-section is b . If the surface tension of water is S , its density is ρ , and its contact angle with glass is θ , then the value of h will be ( g is the acceleration due to gravity) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-sf4vfnindjic9omh.jpg" /> (A) (2 S/b ρ g)cos (. θ -α .)  (B) (2 S/b ρ g)cos (θ + α ) (C) (2 S/b ρ g)cos (θ - (α /2)) (D) (2 S/b ρ g)cos (θ + (α /2))

Tardigrade Admin · Accepted Answer

If R be the meniscus radius, ⇒ Rcos (θ + (α /2))=b Excess pressure on concave side of meniscus =(2 S/R) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-sf4vfnindjicjcb4.jpg" /> hρ g=(2 S/R)=(2 S/b)cos (θ + (α /2)) ⇒ h=(2 S/b ρ g)cos (θ + (α /2))

$\frac{2 S}{b ρ g} cos (θ - α)$

$\frac{2 S}{b ρ g} cos (θ + α)$

$\frac{2 S}{b ρ g} cos (θ - \frac{α}{2})$

$\frac{2 S}{b ρ g} cos (θ + \frac{α}{2})$

If $R$ be the meniscus radius,
$\Rightarrow R cos (θ + \frac{α}{2}) = b$
Excess pressure on concave side of meniscus $= \frac{2 S}{R}$

$h ρ g = \frac{2 S}{R} = \frac{2 S}{b} cos (θ + \frac{α}{2})$
$\Rightarrow h = \frac{2 S}{b ρ g} cos (θ + \frac{α}{2})$

bρg2S​cos(θ−α)

bρg2S​cos(θ+α)

bρg2S​cos(θ−2α​)

bρg2S​cos(θ+2α​)

If R be the meniscus radius, ⇒Rcos(θ+2α​)=b Excess pressure on concave side of meniscus =R2S​ hρg=R2S​=b2S​cos(θ+2α​) ⇒h=bρg2S​cos(θ+2α​)

$\frac{2 S}{b ρ g} cos (θ - α)$

$\frac{2 S}{b ρ g} cos (θ + α)$

$\frac{2 S}{b ρ g} cos (θ - \frac{α}{2})$

$\frac{2 S}{b ρ g} cos (θ + \frac{α}{2})$

If $R$ be the meniscus radius,
$\Rightarrow R cos (θ + \frac{α}{2}) = b$
Excess pressure on concave side of meniscus $= \frac{2 S}{R}$

$h ρ g = \frac{2 S}{R} = \frac{2 S}{b} cos (θ + \frac{α}{2})$
$\Rightarrow h = \frac{2 S}{b ρ g} cos (θ + \frac{α}{2})$