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
Answer & explanation
Correct answer: option 2
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$