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3.Current Electricity
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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 $Y$ ........... $cm$
A
$50$
B
$80$
C
$40$
D
$70$
(AIEEE-2004)
Solution
(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 $ $l = 50\,cm$
Standard 12
Physics
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