Question 602: Current Electricity

Question

Two wires $$A$$ and $$B$$ of equal lengths are connected in left and right gap of a meter bridge, null point is obtained at $$40 \mathrm{~cm}$$ from left end. Diameters of the wire $$A$$ and $$B$$ are in that ratio $$3: 1$$. The ratio of specific resistance of $$A$$ to the of $$B$$ is

Accepted Answer

In a meter bridge setup, the ratio of the resistances is directly proportional to the lengths of the wires if the wires are made up of the same material and have the same cross-sectional area. However, when the materials or the dimensions of the wires differ, we have to account for these differences to calculate the resistance.
The resistance ($R$) of a wire is given by the formula:
$$R = ho \frac{L}{A}$$
where $ho$ is the resistivity (or specific resistance) of the material, $L$ is the length of the wire, and $A$ is the cross-sectional area of the wire. For a wire of circular cross-section, $A = \pi \left(\frac{d}{2}ight)^2$.
In this case, both wires $A$ and $B$ have equal lengths but different diameters. Let's denote their diameters as $d_A$ and $d_B$ respectively, with the given ratio $d_A:d_B = 3:1$. Therefore, the cross-sectional areas of wires $A$ and $B$, denoted as $A_A$ and $A_B$ respectively, will be in the ratio:
$$\frac{A_A}{A_B} = \frac{\pi \left(\frac{d_A}{2}ight)^2}{\pi \left(\frac{d_B}{2}ight)^2} = \left(\frac{d_A}{d_B}ight)^2 = (3:1)^2 = 9:1$$
The resistance of wire $A$ to wire $B$ is given by:
$$\frac{R_A}{R_B} = \frac{ho_A \frac{L}{A_A}}{ho_B \frac{L}{A_B}}$$
Given that lengths of the two wires are the same and the resistances are inversely proportional to their respective cross-sectional areas, we have:
$$\frac{R_A}{R_B} = \frac{ho_A}{ho_B} \cdot \frac{A_B}{A_A}$$
At the balanced condition of the meter bridge:
$$\frac{R_A}{R_B} = \frac{l}{100-l}$$
where $l = 40 \, 	ext{cm}$ (the distance from the left end for the null point), giving $\frac{l}{100-l} = \frac{40}{60} = \frac{2}{3}$.
Thus,
$$\frac{ho_A}{ho_B} \cdot \frac{A_B}{A_A} = \frac{2}{3}$$
Substituting the ratio of $\frac{A_B}{A_A} = 1:9$ or $\frac{A_A}{A_B} = 9:1$ into the equation gives us:
$$\frac{ho_A}{ho_B} \cdot \frac{1}{9} = \frac{2}{3}$$
Thus, solving for $\frac{ho_A}{ho_B}$:
$$\frac{ho_A}{ho_B} = \frac{2}{3} 	imes 9 = 6$$
Therefore, the ratio of the specific resistance of $A$ to that of $B$ is $6:1$, which makes Option C the correct answer.

Answer

$$4: 1$$

Answer

$$3: 1$$

Answer

$$6: 1$$

Answer

$$8: 1$$

Current Electricity

Two wires $A and B of equal lengths are connected in left and right gap of a meter bridge, null point is obtained at 40 \mathrm{~cm} from left end. Diameters of the wire A and B are in that ratio 3: 1. The ratio of specific resistance of A to the of B$ is

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