question_answer

3. 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 additional46 cm. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)   JEE Main  Solved  Paper-2014

A) 38 cm

B) 6 cm

C) 16 cm                   

D) 22 cm

Correct Answer: A

Solution :

\[P'(54-h)={{P}_{0}}8\]  (Boyel?s law) \[P'=\frac{{{P}_{0}}8}{(54-h)}\]                  \[P'+\rho gh={{P}_{0}}\] \[\frac{8{{P}_{0}}}{54-h}+\rho gh={{P}_{0}}\] \[\rho gh=\left( \frac{46-h}{54-h} \right){{P}_{0}}\]         \[({{P}_{0}}=\rho g(76))\] \[h=\left( \frac{46-h}{54-h} \right)(76)\] \[\Rightarrow \]\[54h-{{h}^{2}}=76\times 46-76h\] \[{{h}^{2}}-130h+76\times 46=0\] \[(h-38)(h-92)=0\] \[h=38,\underset{\text{Not}\,\text{possible}}{\mathop{\underset{\downarrow }{\mathop{92}}\,}}\,\]\[\Rightarrow \]\[h=38cm\]