IIT-JEE Mains Question

Using Boyle's Law we can find the gas pressure in final condition.

${ p }_{ 1 }\times { V }_{ 1 }={ p }_{ 2 }\times { V }_{ 2 }$

The volume of the glass tube is equal to $A\times h$

${ p }_{ 1 }\times { A\times }{ h }_{ 1 }={ p }_{ 2 }\times { A\times h }_{ 2 }$

${ p }_{ 1 }\times { h }_{ 1 }={ p }_{ 2 }\times { h }_{ 2 }$

With ${ p }_{ 1 }$ is atmospheric pressure, ${ h }_{ 1 }=8cm$ , and ${ h }_{ 2 }=54cm-x$

$76(8)=p(54-x)$ .......... (i)

$p+x=76cmHg$ ............ (ii)

Substitute p from (ii) to (i), then do some algebra to get

${ x }_{ 1 }=92cm\\ { x }_{ 2 }=38cm$

Since $x<54cm$ , so $x={ x }_{ 2 }=38cm$

The length of air column is $\boxed{54-x=16cm}$