c
From question, $\frac{x}{y}=\frac{40}{100-40}=\frac{2}{3}$
$\Rightarrow x=\frac{2}{3} y$
Again, $\frac{3 x}{y}=\frac{z}{100-z}$
or $\frac{3 \times \frac{2 y}{3}}{y}=\frac{z}{100-z}$
Solving we get $Z=67\, \mathrm{cm}$
Therefore new position of null point
$\cong 67\, \mathrm{cm}$