Q106 / 150 · MHT CET 2020 (12 Oct Shift 1) · Physics

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
  • Correct answer
    C
    6: 1
  • D
    9: 1

Solution

$\begin{array}{l}\frac{R_{1}}{R_{2}}=\frac{40}{60}=\frac{2}{3}=\frac{\frac{\rho_{1} \ell_{1}}{A_{1}}}{\frac{\rho_{2} \ell_{2}}{A_{2}}}=\frac{\rho \pi r_{2}^{2}}{\rho_{2} \pi r_{1}^{2}}=\frac{\rho_{1} r_{2}^{2}}{\rho_{2} r_{1}^{2}} \quad \therefore \ell_{1}=\ell_{2} \\ \therefore \frac{\rho_{1}}{\rho_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}} \times \frac{2}{3}=\frac{6}{1}\end{array} \quad \because \frac{r_{1}}{r_{2}}=\frac{3}{1}$

Hence, the correct answer is option: 'C'.

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