MHT CET · Physics · Current Electricity
Two wires A and B of equal lengths are connected in left and right gap of a metre bridge, null point is obtained at \(40 \mathrm{~cm}\) from left end. Diameters of the wires \(\mathrm{A}\) and \(\mathrm{B}\) are in the ratio \(3: 1\). The ratio of specific resistance of \(A\) to that of \(B\) is
- A 8:1
- B 6:1
- C 4:1
- D 3:1
Answer & Solution
Correct Answer
(B) 6:1
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \ell_{\mathrm{A}} &=\ell_{\mathrm{B}} \quad \frac{r_{\mathrm{A}}}{r_{\mathrm{B}}}=\frac{3}{1} \\ \frac{R_{\mathrm{A}}}{R_{\mathrm{B}}} &=\frac{4}{6}=\frac{2}{3}=\frac{\rho_{\mathrm{A}} \ell_{\mathrm{B}}}{A_{\mathrm{A}}} \times \frac{A_{\mathrm{B}}}{\rho_{\mathrm{B}} \ell_{\mathrm{B}}} \\ &=\frac{\rho_{\mathrm{A}}}{\rho_{\mathrm{B}}} \times \frac{\ell_{\mathrm{A}}}{\ell_{\mathrm{B}}} \times \frac{r_{\mathrm{B}}^{2}}{r_{\mathrm{A}}^{2}} \\ \therefore \quad \frac{\rho_{\mathrm{A}}}{\rho_{\mathrm{B}}} &=\frac{2}{3} \times 9=\frac{6}{1} \end{aligned}\)
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