The resistance in the left and right gaps of a meter bridge are $10 \Omega$ and $30 \Omega$ respectively. If…
The resistance in the left and right gaps of a meter bridge are $10 \Omega$ and $30 \Omega$ respectively. If the bridge is balanced, then the distance of the null point from the center of the wire is
$20 \mathrm{~cm}$
$30 \mathrm{~cm}$
$25 \mathrm{~cm}$
$40 \mathrm{~cm}$
Solution
The given ratio of resistors is:
$\frac{P}{Q}=\frac{10}{30}=\frac{1}{3}=\frac{l_1}{l_2}$
$\therefore l_2=3 l_1$
Since, $l_1+3 l_1=100 \mathrm{~cm}$
$\therefore l_1=25 \mathrm{~cm}$
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