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JEE Main 2014 · Archimedes' Principle & Floatation

Archimedes' Principle & Floatation question from JEE Main 2014

Q22JEE Main 2014Difficult

Archimedes' Principle & Floatation

An open glass tube is immersed in mercury in such a way that a length of 8 cm8\text{ cm} extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm46\text{ cm}. What will be length of the air column above mercury in the tube now? (Atmospheric pressure =76 cm= 76\text{ cm} of HgHg)

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Answer: (a)

  1. Initially air column 8 cm at 76 cm Hg. After raising, the tube top is 8+46=548 + 46 = 54 cm above the outside mercury level.
  2. Let air length be xx; mercury column inside above outside level =54−x= 54 - x, so air pressure =76−(54−x)=22+x= 76 - (54 - x) = 22 + x cm Hg.
  3. Boyle's law: 76×8=x(22+x)⇒x2+22x−608=076 \times 8 = x(22 + x) \Rightarrow x^2 + 22x - 608 = 0
  4. x=−22+484+24322=−22+542=16x = \dfrac{-22 + \sqrt{484 + 2432}}{2} = \dfrac{-22 + 54}{2} = 16 cm
  5. So the answer is (a) 16 cm.
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