JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2023
- Hard
- Numerical
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 \(.............\Omega\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(30\)
Step-by-step Solution
Detailed explanation
\(\frac{ R _1+ R _2}{10}=\frac{60}{40}=\frac{3}{2} \Rightarrow R _1+ R _2=15\)
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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