An open glass tube is immersed in mercury in such a way that a lenth of 8cm 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 46cm. What will be length of the air column above mercury in the tube now?
(Atmosphere pressure =76cm of Hg)

Answer

16

Explanation

In this question, the system is accelerating horizontally, i.e. no component of acceleration in vertical direction. Hence, the pressure in the vertical direction will remain unaffected.
i.e. `p = p_0 + rhogh`
Again, we have to use the concept that the pressure in the same level will be same.

For air trapped in tube, `p_1V_1 = rhog76`
`:. v_1 = A. 8` (A = area of cross section)
`implies P_2 = P_("atm") – rhog (54 – x)`
` = rhog (22+ x)`
`:. V_2 = A·x `
or `rhog76 xx 8A = pg (22 | x)Ax`
`x^2 + 22x – 78 xx 8 = 0 implies x = 16 cm `