Discussion Forum

An open glass tube is immersed in mercury in such away 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 the mercury  in the tube now  ? (Atmospheric pressure =76 cm of Hg)

Answer:

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,    p₁  =p₀+ρgh

 Again, we have to use the concept that the pressure in the same level will be same.

For air trapped in tube, p₁V₁=p₂V₂

 $p_{1}=p_{atm}=\rho g76$

  V₁= A.8

  [A= area of cross-section]

$p_{2}=p_{atm}-\rho g(54-x)$

 $=\rho g(22+x)$

  V₂ =A.x

$\rho g 76\times 8A=\rho g(22+x)Ax$

    $x^{2}+22x-78\times 8=0$

  $\Rightarrow$                 $x=16 cm$