JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2020
- Hard
- Numerical
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

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Answer & Solution
Correct Answer
(D) \(40\)
Step-by-step Solution
Detailed explanation
In balancing
\(\frac{R}{S}=\frac{25}{75}\)
New resistance \(\mathrm{R}^{\prime}=\frac{\rho \ell}{\mathrm{A}}\)
\(=\frac{\rho \times \frac{\ell}{2}}{\frac{\mathrm{A}}{4}}=\frac{\rho \ell}{2} \times 4 \mathrm{A}\)
\(\mathrm{R}^{\prime}=2 \mathrm{R}\)
\(\frac{2 \mathrm{R}}{\mathrm{S}}=\frac{\ell}{100-\ell^{\prime}}\)
\(2 \times \frac{1}{3}=\frac{\ell^{\prime}}{100-\ell^{\prime}}=3 \ell^{\prime}=200-2 \ell^{\prime}\)
\(5 \ell '=200\)
\(\ell^{\prime}=40\)
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