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 900 angle at centre. Radius joining their interface makes an angle α with the vertical. Ratio d1d2 is

h1=R[1−sin⁡α];h2=R[1−cos⁡α]h21=Rsin⁡α+Rcos⁡αpressure at O level same from both sidesh1d1g=h2d1g+h21d2gd1R[1−sin⁡α]=d1R[1−cos⁡α]+d2R[sin⁡α+cos⁡α]d1[cos⁡α−sin⁡α]=d2[cos⁡α+sin⁡α]d1d2=cos⁡α+sin⁡αcos⁡α−sin⁡α=1+tan⁡α1−tan⁡α

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 900 angle at centre. Radius joining their interface makes an angle α with the vertical. Ratio d1d2 is

h1=R[1−sin⁡α];h2=R[1−cos⁡α]h21=Rsin⁡α+Rcos⁡αpressure at O level same from both sidesh1d1g=h2d1g+h21d2gd1R[1−sin⁡α]=d1R[1−cos⁡α]+d2R[sin⁡α+cos⁡α]d1[cos⁡α−sin⁡α]=d2[cos⁡α+sin⁡α]d1d2=cos⁡α+sin⁡αcos⁡α−sin⁡α=1+tan⁡α1−tan⁡α

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 90⁰ angle at centre. Radius joining their interface makes an angle $α$ with the vertical. Ratio $\frac{d_{1}}{d_{2}}$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1 + \sin α}{1 - \sin α}$

$\frac{1 + \sin α}{1 - \cos α}$

$\frac{1 + \cos α}{1 - \cos α}$

$\frac{1 + \tan α}{1 - \tan α}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} h_{1} = R [1 - \sin α]; h_{2} = R [1 - \cos α] \\ h_{2}^{1} = Rsin α + Rcos α \end{array}$
pressure at O level same from both sides
$\begin{array}{l} h_{1} d_{1} g = h_{2} d_{1} g + h_{2}^{1} d_{2} g \\ d_{1} R [1 - \sin α] = d_{1} R [1 - \cos α] + d_{2} R [\sin α + \cos α] \\ d_{1} [\cos α - \sin α] = d_{2} [\cos α + \sin α] \\ \frac{d_{1}}{d_{2}} = \frac{\cos α + \sin α}{\cos α - \sin α} = \frac{1 + \tan α}{1 - \tan α} \end{array}$