MHT CET · Physics · Current Electricity
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
- A \(6: 1\)
- B \(8: 1\)
- C \(16: 1\)
- D \(12: 1\)
Answer & Solution
Correct Answer
(A) \(6: 1\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \frac{R_A}{R_B}=\frac{40}{60}=\frac{2}{3} \\
& R=\rho \cdot \frac{\ell}{A}=\rho \cdot \frac{\ell}{\pi r^2} \\
& \frac{R_A}{R_B}=\frac{\rho_A}{\rho_B}\left(\frac{r_B}{r_A}\right)^2 \\
& \frac{2}{3}=\frac{\rho_A}{\rho_B}\left(\frac{1}{3}\right)^2 \\
& \frac{\rho_A}{\rho_B}=6
\end{aligned}
\)
\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}
\)
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