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In the figure shown for gives values of $R_1$ and $R_2$ the balance point for Jockey is at $40\,cm$ from $A$. When $R_2$ is shunted by a resistance of $10\, \Omega$ , balance shifts to $50\,cm.$ $R_1$ and $R_2$ are $(AB = 1 \,m)$
$\frac{{10}}{3} \, \Omega , 5\, \Omega$
$20\, \Omega , 30\, \Omega$
$10\, \Omega , 15\, \Omega$
$5\, \Omega , \frac{{15}}{2}\, \Omega$
Solution
For balancing condition we can write,
$R_{1}=R_{2} \frac{l}{100-l}$
in first case,
$R_{1}=R_{2} \frac{40}{100-40}=R_{2} \frac{4}{6}=R_{2} \frac{2}{3}$$…(1)$
in second case,
$R_{1}=\left(\frac{10 R_{2}}{10+R_{2}}\right) \frac{l}{100-l}=\left(\frac{10 R_{2}}{10+R_{2}}\right)\left(\frac{50}{100-50}\right)$
$R_{1}=\frac{10 R_{2}}{10+R_{2}}$
from$(1)$
$R_{2} \frac{2}{3}=\frac{10 R_{2}}{10+R_{2}}$
$-10 R_{2}+2 R_{2}^{2}=0$
$2 R_{2}=10$
$R_{2}=5 \Omega$
hence, $R_{1}=R_{2} \frac{2}{3}=(5) \frac{2}{3}=\frac{10}{3} \Omega$
