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Two wires ‘ $\mathrm{A}^{\prime}$ and ‘ $\mathrm{B}$ ‘ 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 ${ }^{\prime} \mathrm{B}^{\prime} is$ A)3:1 B)1:1 C)6:1 D)9:1
Two wires ‘ $\mathrm{A}^{\prime}$ and ‘ $\mathrm{B}$ ‘ 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 ${ }^{\prime} \mathrm{B}^{\prime} is$
A)3:1
B)1:1
C)6:1
D)9:1
Physics
Two wires ‘ $\mathrm{A}^{\prime}$ and ‘ $\mathrm{B}$ ‘ 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 ${ }^{\prime} \mathrm{B}^{\prime} is$
A)3:1
B)1:1
C)6:1
D)9:1

Correct option is C)6:1

The specific resistance of a wire of diameter $d$ is
$
\rho=R \frac{A}{I}=\frac{R \pi d^{2}}{4 \mid}
$
$\rho \propto R \mathrm{~d}^{2} \ldots . .(\because I$ is constant)
At null point in a meter bridge, $\frac{R_{A}}{R_{B}}=\frac{40}{(100-40)}=\frac{2}{3}$
Given, $d_{A}: d_{B}=3:1$
$\therefore \frac{\rho_{\mathrm{A}}}{\rho_{\mathrm{B}}}=\frac{\mathrm{R}_{\mathrm{A}} \mathrm{d}_{\mathrm{A}}^{2}}{\mathrm{R}_{\mathrm{B}} \mathrm{d}_{\mathrm{B}}^{2}}=\frac{2}{3} \times\left(\frac{3}{1}\right)^{2}=\frac{6}{1}$ or $6:1$

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