Two wires ' $\mathrm{A}$ ' and ' $\mathrm{B}$ ' of equal length are connected in left and right gap…

Two wires ' $\mathrm{A}$ ' and ' $\mathrm{B}$ ' of equal length are connected in left and right gap respectively of meter bridge, null point is obtained at $40 \mathrm{~cm}$, from the left end. Diameters of the wires ' $A$ ' and 'B' are in the ratio $3: 1$ respectively, the ratio of specific resistance of ' $\mathrm{A}$ ' to that of ' $\mathrm{B}$ ' is
  1. $6: 1$
  2. $8: 1$
  3. $16: 1$
  4. $12: 1$

Solution

$\begin{aligned} & \frac{R_A}{R_B}=\frac{40}{60}=\frac{2}{3} \\ & R=\rho \cdot \frac{\ell}{A}=\rho \cdot \frac{\ell}{\pi r^2} \\ & \frac{R_A}{R_B}=\frac{\rho_A}{\rho_B}\left(\frac{r_B}{r_A}\right)^2 \\ & \frac{2}{3}=\frac{\rho_A}{\rho_B}\left(\frac{1}{3}\right)^2 \\ & \frac{\rho_A}{\rho_B}=6 \end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 1)

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