Question

The figure-$3.326$, shows a meter bridge circuit, with $A B=100 \,cm , X=12 \,\Omega$ and $R=18\, \Omega$ and the jockey $J$ in the position of null deflection balance. If $R$ is now changed to $8 \,\Omega$, through what distance will $J$ have to be moved to obtain the balance again?

• A. $10\, cm$
• B. $20\, cm$
• C. $30 \,cm$
• D. $40\, cm$

Answer 1

Answer: (B)

Initially if $l_{1}$ is the balancing length then we have
$\frac{l_{1}}{100-l_{1}}=\frac{X}{R}=\frac{2}{3}$
$\Rightarrow l_{1}=\frac{2}{5} \times 100 $
$\Rightarrow l_{1}=40\, cm$
If the new balancing length is $l_{2}$ then we have
$\frac{l_{2}}{100-l_{2}}=\frac{X}{R^{\prime}}=\frac{12}{8}=\frac{3}{2}$
$\Rightarrow l_{2}=\frac{3}{5} \times 100$
$\Rightarrow l_{2}=60\, cm$
Thus $J$ is displaced by
$l_{2}-l_{1}=20\, cm$