JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2024
- Hard
- MCQ
The resistance per centimeter of a meter bridge wire is \(\mathrm{r}\), with \(\mathrm{X}\ \Omega\) resistance in left gap. Balancing length from left end is at \(40 \mathrm{~cm}\) with \(25\ \Omega\) resistance in right gap. Now the wire is replaced by another wire of \(2 \mathrm{r}\) resistance per centimeter. The new balancing length for same settings will be at _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(40 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
\(\frac{25}{\mathrm{r} \ell_1}=\frac{\mathrm{x}}{\mathrm{r} \ell_2}\) \(.....(i)\)
\(\frac{25}{2 \mathrm{r} \ell_1^{\prime}}=\frac{\mathrm{X}}{2 \mathrm{r} \ell_2^{\prime}}\) \(...(ii)\)
From \((i)\) and \((ii)\)
\(\ell_2^{\prime}=\ell_2=40 \mathrm{~cm}\)
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