In a meter bridge, null point is found at a distance of 20 cm from the end A, then the resistance of 10 Ω is replaced by another resistance of 20 Ω the null

33.3 cm

We know, \(\frac{X}{Y} = \frac{20 cm}{80 cm} \) and

$\Rightarrow \frac{X}{Y} = \frac{1}{4}$

$\Rightarrow Y = 4X = 40\Omega $

when 10$\Omega$ is replaced by 20$\Omega$

$\frac{20\Omega}{40\Omega} = \frac{l}{100-l}$

$ \frac{l}{100-l} = \frac{1}{2}$

$ 2l = 100-l$

$ l = \frac{100}{3} cm = 33.3cm$