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 the length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

16 cm

22 cm

38 cm

6 cm

Solution

16 cm

Since the system is accelerating horizontally such that no component of acceleration in the vertical direction. Hence, the pressure in the vertical direction will remain unaffected.
$straight p subscript 1 space equals straight p subscript 0 space plus ρgh$
.
For air trapped in tube, p₁V₁ = p₂V₂
p₁ = p_atm = pg76
V₁ = A.8 [ A = area of cross section]
p₂ = patm - ρg(54-x) = ρg(22+x)
V₂ = A.x
ρg76 x 8A = ρg (22+x) Ax
x²+22x-78x 8
by solving, x = 16