Question

The figure shows a metre-bridge circuit, with $AB=100cm$, $X=12\Omega$ 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?

• A. 10 cm
• B. 20 cm
• C. 30 cm
• D. 40 cm

Answer 1

The correct option is B 20 cm

The balancing condition for meter bridge is,
$\frac{X}{R}=\frac{l_1}{100-l_1}$
$X=R\left(\frac{l_1}{100-l_1}\right)$
$12=18\left(\frac{l_1}{100-l_1}\right)$

$1200=30l_1$
$l_1=40$ cm.
When $R=8$ ohm,

$12=8\left(\frac{l_2}{100-l_2}\right)$

$1200=20l_2$
$l_2=60$ cm.
Hence, distance through $J$ is shifted is
$l=l_2-l_1=20$ cm.