When two resistances $R_{1} and R_{2}$ connected in series and introduced into the left gap of a meter bridge and a resistance of $10 Ω$ is introduced into the right gap, a null point is found at 60 cm from left side. When $R_{1} and R_{2}$ are connected in parallel and introduced into the left gap, a resistance of $3 Ω$ is introduced into the right gap to get null point at 40 cm from left end . The product of $R_{1} R_{2}$ is___ $Ω^{2}$

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Detailed Solution

Balancing condition in meter bridge is $\frac{R}{S} = \frac{l}{100 - l}$
R = Resistance in left gap
S = Resistance in right gap
l = balancing length
When $R, R_{2}$ are in series, $l = 60 cm, S = 10 Ω$
$\begin{aligned} \frac{R_{1} + R_{2}}{S} = \frac{60}{40} \\ \frac{R_{1} + R_{2}}{10} = \frac{3}{2} \end{aligned}$
$R_{1} + R_{2} = 15$ .....(1)
When $R, R_{2}$ are in parallel, $l = 40 cm, S = 3 Ω$
$\begin{array}{l} \frac{(R_{1} R_{2} / R_{1} + R_{2})}{3} = \frac{40}{60} \\ \Rightarrow R_{1} R_{2} = 2 (R_{1} + R_{2}) \end{array}$