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(A) For a meter bridge,the balancing condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$,where $l$ is the distance from end $A$.In the first cas

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$\frac{10}{3}\, \Omega, 5\, \Omega$

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(A) For a meter bridge,the balancing condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$,where $l$ is the distance from end $A$.In the first case,$l = 40\,cm$:$\frac{R_1}{R_2} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$$R_1 = \frac{2}{3} R_2$ ... $(1)$In the second case,$R_2$ is shunted by $10\, \Omega$,so the new resistance $R_2'$ is $\frac{10 R_2}{10 + R_2}$. The new balance point is $l' = 50\,cm$:$\frac{R_1}{R_2'} = \frac{50}{100-50} = \frac{50}{50} = 1$$R_1 = R_2' = \frac{10 R_2}{10 + R_2}$ ... $(2)$Equating $(1)$ and $(2)$:$\frac{2}{3} R_2 = \frac{10 R_2}{10 + R_2}$$2(10 + R_2) = 30$$20 + 2 R_2 = 30$$2 R_2 = 10$$R_2 = 5\, \Omega$Substituting $R_2 = 5\, \Omega$ into $(1)$:$R_1 = \frac{2}{3} 	imes 5 = \frac{10}{3}\, \Omega$Thus,$R_1 = \frac{10}{3}\, \Omega$ and $R_2 = 5\, \Omega$.

In the figure shown,for given values of $R_1$ and $R_2$,the balance point for the jockey is at $40\,cm$ from $A$. When $R_2$ is shunted by a resistance of $10\, \Omega$,the balance point shifts to $50\,cm$. Find $R_1$ and $R_2$. $(AB = 1\,m)$

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