MHT CET · Physics · Current Electricity
Two wires 'A' and 'B' of equal lengths are connected in left and right gaps, of meter bridge, respectively. The null point is obtained at \(40 \mathrm{~cm}\) from left end. Diameters of the wires 'A' and 'B' are in the ratio \(3: 1\), the ratio of specific resistance of 'A' to that
of 'B' is
- A 3: 1
- B 1: 1
- C 6: 1
- D 9: 1
Answer & Solution
Correct Answer
(C) 6: 1
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}\frac{R_{1}}{R_{2}}=\frac{40}{60}=\frac{2}{3}=\frac{\frac{\rho_{1} \ell_{1}}{A_{1}}}{\frac{\rho_{2} \ell_{2}}{A_{2}}}=\frac{\rho \pi r_{2}^{2}}{\rho_{2} \pi r_{1}^{2}}=\frac{\rho_{1} r_{2}^{2}}{\rho_{2} r_{1}^{2}} \quad \therefore \ell_{1}=\ell_{2} \\ \therefore \frac{\rho_{1}}{\rho_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}} \times \frac{2}{3}=\frac{6}{1}\end{array}\)
\(\because \frac{r_{1}}{r_{2}}=\frac{3}{1}\)
\(\because \frac{r_{1}}{r_{2}}=\frac{3}{1}\)
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