When two resistance $R_1$ 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 $60 cm$ 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 cm from left end. The product of $R_1 R_2$ is $.............\Omega$
JEE MAIN 2023, Diffcult
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$\frac{ R _1+ R _2}{10}=\frac{60}{40}=\frac{3}{2} \Rightarrow R _1+ R _2=15$
Now $\frac{ R _1 R _2}{\left( R _1+ R _2\right) \times 3}=\frac{40}{60}=\frac{2}{3} \Rightarrow R _1 R _2=30$
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