The figure shows a meter-bridge circuit, with $AB = 100\, cm$,$ X = 12\,\Omega$ and $R = 18\,\Omega$ , and the jockey $J$ in the position of balance. If $R$ is now made $8\,\Omega$ , through what distance will $J$ have to be moved to obtain balance? .............. $cm$
Diffcult
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The balancing condition for meter bridge is,
$\frac{X}{R}=\frac{l_{1}}{100-l_{1}}$
$X=R\left(\frac{l_{1}}{100-l_{1}}\right)$
$12=18\left(\frac{l_{1}}{100-l_{1}}\right)$
$1200=30 l_{1}$
$l_{1}=40 \mathrm{cm}$
When $R=8$ ohm,
$12=8\left(\frac{l_{2}}{100-l_{2}}\right)$
$1200=20 l_{2}$
$l_{2}=60 \mathrm{cm}$
Hence, distance through $J$ is shifted is
$l=l_{2}-l_{1}=20 \mathrm{cm}$
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