JEE Mains · Physics · STD 12 - 3. current electricity
In a meter bridge experiment \(\mathrm{S}\) is a standard resistance. \(\mathrm{R}\) is a resistance wire. It is found that balancing length is \(l=25 \;\mathrm{cm} .\) If \(\mathrm{R}\) is replaced by a wire of half length and half diameter that of \(\mathrm{R}\) of same material, then the balancing distance \(\left.l^{\prime} \text { (in } \mathrm{cm}\right)\) will now be

- A \(36\)
- B \(37\)
- C \(33\)
- D \(40\)
Answer & Solution
Correct Answer
(D) \(40\)
Step-by-step Solution
Detailed explanation
In balancing \(\frac{R}{S}=\frac{25}{75}\) New resistance \(\mathrm{R}^{\prime}=\frac{\rho \ell}{\mathrm{A}}\) \(=\frac{\rho \times \frac{\ell}{2}}{\frac{\mathrm{A}}{4}}=\frac{\rho \ell}{2} \times 4 \mathrm{A}\) \(\mathrm{R}^{\prime}=2 \mathrm{R}\)…
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