Explanation

(a) Meter bridge is an arrangement which works on Wheatstone’s principle, so the balancing condition is
$\frac{R}{S} = \frac{l_{1}}{l_{2}}$
Where $l_{2} = 100 - l_{1}$
Ist case $R = X,\ S = Y,\ l_{1} = 20\ cm,\ l_{2} = 100 - 20 = 80cm$
$\therefore\ \ \ \ \ \frac{X}{Y} = \frac{20}{80}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
IInd Case Let the position null point is obtained at a distance $l$ from same end.
$\therefore R = 4X,\ S = Y,\ l_{1} = l,\ l_{2} = 100 - l$
So, from Eq. (i)
$\frac{4X}{Y} = \frac{l}{100 - l}$
$\Longrightarrow \frac{X}{Y} = \frac{l}{4(100 - l)}\ \ \ldots(ii)$
Therefore, form Eqs. (i) and (ii)
$\frac{l}{4(100 - l)} = \frac{20}{80}$
$\Longrightarrow \frac{l}{4(100 - l)} = \frac{1}{4}$
$\Longrightarrow \ \ l = 100 - l\ $
$\Longrightarrow \ \ 2l = 100$
Hence, $l$=50cm