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The given circuit shows a uniform straight wire $A B$ of 40 cm length fixed at both ends. In order to get zero reading in the galvanometer $G$, the free end of $J$ is to be placed from $B$ at:
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24 cm
$\frac{8}{x}=\frac{12}{40-x}$
$\frac{2}{x}=\frac{3}{40-x}$
$80-2 x=3 x$
$16=x$
from B
$=40-16=24 \mathrm{~cm}$
$\frac{2}{x}=\frac{3}{40-x}$
$80-2 x=3 x$
$16=x$
from B
$=40-16=24 \mathrm{~cm}$
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