Question

The figure shows a meter bridge circuit with $\text{AB}=100\text{cm},X=12\Omega$ and $R=18\Omega$, and the jockey $J$ is at the balance point. If $R$ is now made $8\Omega$, through what distance will $J$ have to be moved to obtain a balanced bridge ?

Answer 1

In case $1:$

$X=12\Omega$ & $R=18\Omega$

Let the balanced point be at $l\text{cm}$ from point $\text{A}$.

$\text{AJ}=l;\text{BJ}=(100-l)\text{cm}$

Applying balance condition,

$\frac{X}{R}=\frac{l}{100-l}$

$\frac{12}{18}=\frac{l}{100-l}$

$3l=200-2l$

$\therefore l=\frac{200}{5}=40\text{cm}$

Now in case $2:$

$X=12\Omega$ and $R^{\prime }=8\Omega$

$\frac{X}{R^{\prime }}=\frac{l^{\prime }}{100-l^{\prime }}$

$\frac{12}{8}=\frac{l^{\prime }}{100-l^{\prime }}$

$2l^{\prime }=300-3l^{\prime }$

$\therefore l^{\prime }=\frac{300}{5}=60\text{cm}$

So, $\Delta l=l^{\prime }-l=60-40=20\text{cm}$

Hence, $J$ has to be moved by $20\text{cm}$ in order to achieve the balanced point.

Therefore, option $\left(b\right)$ is correct.