Two wires A and B of equal lengths are connec | vedclass

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(C) In a metre bridge, the condition for the null point is $\frac{R_A}{R_B} = \frac{l_1}{l_2}$, where $R_A$ and $R_B$ are the resistances of wires $A$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) In a metre bridge, the condition for the null point is $\frac{R_A}{R_B} = \frac{l_1}{l_2}$, where $R_A$ and $R_B$ are the resistances of wires $A$ and $B$, and $l_1, l_2$ are the lengths of the segments of the bridge wire.Given $l_1 = 40 	ext{ cm}$, then $l_2 = 100 - 40 = 60 	ext{ cm}$.So, $\frac{R_A}{R_B} = \frac{40}{60} = \frac{2}{3}$.The resistance of a wire is given by $R = ho \frac{L}{A} = ho \frac{L}{\pi (d/2)^2} = \frac{4 ho L}{\pi d^2}$.Since lengths $L$ are equal, $\frac{R_A}{R_B} = \frac{ho_A}{ho_B} 	imes \left( \frac{d_B}{d_A} ight)^2$.Given $\frac{d_A}{d_B} = \frac{3}{1}$, so $\frac{d_B}{d_A} = \frac{1}{3}$.Substituting the values: $\frac{2}{3} = \frac{ho_A}{ho_B} 	imes \left( \frac{1}{3} ight)^2$.$\frac{2}{3} = \frac{ho_A}{ho_B} 	imes \frac{1}{9}$.$\frac{ho_A}{ho_B} = \frac{2}{3} 	imes 9 = 6$.Thus, the ratio is $6:1$.

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