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

\(\cong 67\, \mathrm{cm}\)