d
Assuming temperature remains constant at $300 K$
From $P _1 V _1= P _2 V _2$
$\frac{ P _1\left(\frac{ V _0}{2}\right)}{ T }=\frac{ P _1^{\prime}\left(\frac{ V _0}{2}- Ax \right)}{ T }$
$\begin{array}{l}\left( P _1^{\prime}- P _2^{\prime}\right) A = mg \\ {\left[\frac{ P _1\left(\frac{ V _0}{2}\right)}{\frac{ V _0}{2}- Ax }-\frac{ P _2\left(\frac{ V _0}{2}\right)}{\frac{ V _0}{2}+ Ax }\right] A = mg } \\ nRT \left[\frac{1}{4- x }-\frac{1}{4+ x }\right]= mg \\ (0.1)(8.3)\left[\frac{4+ x -4+ x }{16- x ^2}\right]= mg \\ 3\left(\frac{2 x }{16- x ^2}\right)=1 \\ 6 x =16- x ^2 \\ x ^2+6 x -16=0 \\ x =2 \\ \text { distance }=4+2=6 m \end{array}$
