Question

Two wires 'A' and 'B' of equal lengths are connected in left and right gaps of a meter bridge, respectively. The null point is obtained at 40 cm from the 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. 3:1
• B. 1:1
• C. 6:1
• D. 9:1

Hint:

When comparing resistances in a meter bridge, remember that the resistance is proportional to the specific resistance and inversely proportional to the cross-sectional area.

Answer 1

The Correct Option is C

Step 1: Understanding the problem.
The meter bridge is used to compare the resistances of two wires. The null point position depends on the ratio of the resistances of the two wires, which in turn depends on the ratio of their specific resistances and their cross-sectional areas.

Step 2: Using the formula for resistance.
The resistance of a wire is given by \( R = \rho \frac{L}{A} \), where \( \rho \) is the specific resistance, \( L \) is the length of the wire, and \( A \) is the cross-sectional area. Since the wires are of equal length, the ratio of resistances is \( \frac{R_A}{R_B} = \frac{\rho_A A_B}{\rho_B A_A} \).

Step 3: Relating diameter to area.
The area of the cross-section of a wire is proportional to the square of its diameter. Given that the diameters of wires 'A' and 'B' are in the ratio 3:1, their areas will be in the ratio \( A_A : A_B = 9:1 \).

Step 4: Final Calculation.
Using the given information and solving for the ratio of specific resistances, we find that the ratio of the specific resistances \( \rho_A : \rho_B \) is 6:1.