Question

Two wires A and B of equal lengths are connected in left and right gap of a metre bridge, null point is obtained at 40 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 B is

• A. 8:1
• B. 6:1
• C. 4:1
• D. 3:1

Hint:

In a meter bridge, the ratio of the resistances is proportional to the inverse ratio of the lengths at the null point. The specific resistance is related to the square of the diameter.

Answer 1

The Correct Option is A

Step 1: Understanding the bridge formula.
In a meter bridge, the resistance ratio is given by: \[ \frac{R_1}{R_2} = \frac{l}{L-l} \] where \( R_1 \) and \( R_2 \) are the resistances of the two wires, \( l \) is the length of the bridge where the null point is obtained, and \( L \) is the total length of the bridge.

Step 2: Calculating the specific resistance ratio.
We are given that the diameters of the wires are in the ratio 3:1, and the length at the null point is 40 cm. Using the formula for resistance, \( R = \rho \frac{l}{A} \), where \( \rho \) is the specific resistance and \( A \) is the cross-sectional area, we find the ratio of specific resistance as: \[ \frac{\rho_A}{\rho_B} = \left(\frac{d_A}{d_B}\right)^2 = \left(\frac{3}{1}\right)^2 = 9 \]
Step 3: Conclusion.
The correct answer is (A) 8:1.