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JEE Main 2014 · 6 Apr · Shift 1, Physics question 13

An open glass tube is immersed in mercury in such a way that a length of 8 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 cm. What will be length of the air column above mercury in the tube now ?

(Atmospheric pressure =76=76 cm of Hg)

Choose one option
Answer and worked solution

Official answer: A

Initially the air column is 8 cm long at 76 cm of Hg. After raising, the closed top is 8+46=548+46=54 cm above the outside level. If the air column is LL, the mercury inside stands 54−L54-L above the outside level, so the air pressure is 76−(54−L)=22+L76-(54-L)=22+L.

Boyle's law:
76×8=L(22+L)  ⟹  L2+22L−608=0  ⟹  L=16 cm.76\times8=L(22+L)\implies L^2+22L-608=0\implies L=16\text{ cm}.

Topics:Ideal gas equationPressure and Pascal's law
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