JEE Main · 2014 · 06 Apr 2014 · Offline · Heat & Thermodynamics

While the tube is open the trapped column is 8 cm long at the atmospheric pressure of 76 cm of mercury. Once it is sealed and lifted a further 46 cm, the length available to the air is $8+46=54$ cm; call the new air column $\ell$. The mercury inside then stands $(54-\ell )$ cm high, so the gas pressure is $76-(54-\ell )=(22+\ell )$ cm of mercury. Boyle's law gives $76\times 8=(22+\ell )\ell$, that is ${\ell }^2+22\ell -608=0$, and solving, $\ell =\frac{-22+\sqrt{484+2432}}{2}=\frac{-22+54}{2}=16$ cm.