The figure shows a meter-bridge circuit where X=12 Ω and R=18 Ω. The jockey J is at the null point. If R is made 8  Ω through what distance will the jockey J have to be moved to obtain null point again?

# The figure shows a meter-bridge circuit where $X=12\text{\hspace{0.17em}}\Omega$ and $R=18\text{\hspace{0.17em}}\Omega$. The jockey J is at the null point. If R is made $8\text{\hspace{0.17em}\hspace{0.17em}}\Omega$ through what distance will the jockey J have to be moved to obtain null point again?

1. A

10 cm

2. B

20 cm

3. C

30 cm

4. D

40 cm

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### Solution:

$\frac{\mathrm{AJ}}{100-\mathrm{AJ}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{X}}{\mathrm{R}}\text{\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}\frac{12}{18}\text{\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{AJ}=40\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}$

After changing R:

$\begin{array}{l}\frac{{\mathrm{AJ}}^{\text{'}}}{100-{\mathrm{AJ}}^{\text{'}}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}\frac{12}{8}\text{\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{AJ}}^{\text{'}}=60\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}\\ \mathrm{Shifting}\text{\hspace{0.17em}}:\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{AJ}}^{\text{'}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}20\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}\end{array}$

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