Question 589: Current Electricity

Question

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 'A' to that of 'B' is

Accepted Answer

$$\begin{aligned}
& \frac{R_A}{R_B}=\frac{40}{60}=\frac{2}{3} \
& R=ho \cdot \frac{\ell}{A}=ho \cdot \frac{\ell}{\pi r^2}
\end{aligned}$$
Length of the wires are same.
$$\begin{aligned}
& 	herefore \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{ho_{\mathrm{A}}}{ho_{\mathrm{B}}}\left(\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}ight)^2 \
& 	herefore \frac{2}{3}=\frac{ho_{\mathrm{A}}}{ho_{\mathrm{B}}}\left(\frac{1}{3}ight)^2=\frac{ho_{\mathrm{A}}}{ho_{\mathrm{B}}} \cdot \frac{1}{9} \
& 	herefore \frac{ho_{\mathrm{A}}}{ho_{\mathrm{B}}}=6
\end{aligned}$$

Answer

$$6: 1$$

Answer

$$8: 1$$

Answer

$$16: 1$$

Answer

$$12: 1$$

Current Electricity

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