JEE-2015

Classical Mechanics Level 2

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d 1 d_1 and d 2 d_2 are filled in the tube. Each liquid subtends 9 0 ∘ 90^\circ angle at centre. Radius joining their interface makes an angle α \alpha with vertical. Ratio d 1 d 2 \frac {d_1}{d_2 } is:

1 + sin ⁡ α 1 − sin ⁡ α \frac {1 + \sin \alpha }{1 - \sin \alpha } 1 + tan ⁡ α 1 − tan ⁡ α \frac {1 + \tan \alpha }{1 - \tan \alpha } 1 + sin ⁡ α 1 − cos ⁡ α \frac {1 + \sin \alpha }{1 - \cos \alpha } 1 + cos ⁡ α 1 − cos ⁡ α \frac {1 + \cos \alpha }{1 - \cos \alpha }

Ramesh Goenka by Ramesh Goenka , 26, India

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2 solutions

Ramesh Goenka
Apr 6, 2015

7 Helpful 2 Interesting 0 Brilliant 0 Confused
Jake Lai
Apr 6, 2015

I've heard of the "JEE style" many a times here on Brilliant, and so I figure I might as well use it.

If d 1 = 0 d_{1} = 0 , then clearly α = − π 4 \alpha = -\frac{\pi}{4} . Since the only option that matches d 1 d 2 = 0 \frac{d_{1}}{d_{2}} = 0 is 1 + tan ⁡ α 1 − tan ⁡ α \frac{1+\tan \alpha}{1-\tan \alpha} , that must be the right answer.

6 Helpful 2 Interesting 0 Brilliant 0 Confused

Useful, thanks !

Sheetal Sahu - 3 years, 2 months ago

Hey there, sorry for a dumb doubt but how did you decide alpha's value when d_1 = 0?

DevUt ! - 2 years, 5 months ago

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