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 $10\ \Omega$ is introduced into the right gap, a null point is found at $60\ \text{cm}$ from left side. When $R_1$ and $R_2$ are connected in parallel and introduced into the left gap, a resistance of $3\ \Omega$ is introduced into the right gap to get null point at $40\ \text{cm}$ from left end. The product of $R_1R_2$ is ______ $\Omega^2$.
Numerical value type. Enter your answer.
Answer: 30
Series: $\dfrac{R_1+R_2}{10}=\dfrac{60}{40}\Rightarrow R_1+R_2=15\ \Omega$.
Parallel: $\dfrac{R_p}{3}=\dfrac{40}{60}\Rightarrow R_p=2\ \Omega$.
$R_p=\dfrac{R_1R_2}{R_1+R_2}\Rightarrow R_1R_2=2\times15=30\ \Omega^2$.
Solution by Sreeraj P, M.Sc Physics