There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle α with vertical. Ratio (d1/d2) is :

Question

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle &alpha; with vertical. Ratio (d1/d2) is : (A) (1+sinα/1-sinα) (B) (1+cosα/1-cosα) (C) (1+tanα/1-tanα) (D) (1+sinα/1-cosα)

Tardigrade Admin · Accepted Answer

Equating pressure at A <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2af24c2274ea356d63abee6e5728cd2e-.png" /> (R cos α+R sin α) d2 g=(R cos α-R sin α) d1 g ⇒ (d1/d2)=( cos α+ sin α/ cos α- sin α)=(1+ tan α/1- tan α)

Q. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $9 0^{\circ}$ angle at centre. Radius joining their interface makes an angle α with vertical. Ratio $\frac{d _{1}}{d _{2}}$ is :

$\frac{1 + s in α}{1 - s in α}$

$\frac{1 + cos α}{1 - cos α}$

$\frac{1 + t an α}{1 - t an α}$

$\frac{1 + s in α}{1 - cos α}$

Equating pressure at A

$(R cos α + R sin α) d_{2} g = (R cos α - R sin α) d_{1} g$
$\Rightarrow \frac{d _{1}}{d _{2}} = \frac{c o s α + s i n α}{c o s α - s i n α} = \frac{1 + t a n α}{1 - t a n α}$

Q. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1​ and d2​ are filled in the tube. Each liquid subtends 90∘ angle at centre. Radius joining their interface makes an angle α with vertical. Ratio d2​d1​​ is :

1−sinα1+sinα​

1−cosα1+cosα​

1−tanα1+tanα​

1−cosα1+sinα​