In a meter bridge experiment null point is obtained at $20\, cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$, then where will be the new position of the null point from the same end, if one decides to balance a resistance of $4X$ against $Y$ ........... $cm$
AIEEE 2004, Medium
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(a) In balancing condition, $\frac{{{R_1}}}{{{R_2}}} = \frac{{{l_1}}}{{{l_2}}} = \frac{{{l_1}}}{{100 - {l_1}}}$
$ \Rightarrow $ $\frac{X}{Y} = \frac{{20}}{{80}} = \frac{1}{4}$.....$(i)$
and $\frac{{4X}}{Y} = \frac{l}{{100 - l}}$.....$(ii)$
$ \Rightarrow $ $\frac{4}{4} = \frac{l}{{100 - l}}$$ \Rightarrow $ $l = 50\,cm$
$ \Rightarrow $ $\frac{X}{Y} = \frac{{20}}{{80}} = \frac{1}{4}$.....$(i)$
and $\frac{{4X}}{Y} = \frac{l}{{100 - l}}$.....$(ii)$
$ \Rightarrow $ $\frac{4}{4} = \frac{l}{{100 - l}}$$ \Rightarrow $ $l = 50\,cm$
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