For capillary tube
\(h=\frac{2 T}{r \rho g}\)
We can say
\(h \propto \frac{1}{r} \text { or } h \propto \frac{1}{d}\)
So, \(\frac{h_1}{h_2}=\frac{d_2}{d_1}\)
\(\Rightarrow \frac{4}{x}=\frac{d}{2 d}\)
\(\Rightarrow x=8 \,cm\)
$\left(g=10\, ms ^{-2}\right)$