Question

Two wires 'A' and 'B' of equal length are connected in left and right gap respectively of meter bridge, null point is obtained at 40 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

• A. 6 : 1
• B. 8 : 1
• C. 16 : 1
• D. 12 : 1

Hint:

In meter bridge, the ratio of resistances equals ratio of balancing lengths. Remember \(R \propto \rho / d^2\) for same length and material dependence.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Meter bridge null point at 40 cm from left. Wires have equal length, diameters ratio \(d_A : d_B = 3:1\). We need ratio of specific resistivities \(\rho_A : \rho_B\).

Step 2: Key Formula or Approach:
Resistance \(R = \rho \frac{L}{A}\). For equal length \(L\), \(R \propto \frac{\rho}{d^2}\) (since area \(\propto d^2\)).
Meter bridge condition: \(\frac{R_A}{R_B} = \frac{\ell}{100-\ell} = \frac{40}{60} = \frac{2}{3}\).

Step 3: Detailed Explanation:
\[ \frac{R_A}{R_B} = \frac{\rho_A}{\rho_B} \cdot \frac{d_B^2}{d_A^2} = \frac{\rho_A}{\rho_B} \cdot \left(\frac{1}{3}\right)^2 = \frac{\rho_A}{\rho_B} \cdot \frac{1}{9}. \] Set equal to \(\frac{2}{3}\): \(\frac{\rho_A}{\rho_B} \cdot \frac{1}{9} = \frac{2}{3}\) ⇒ \(\frac{\rho_A}{\rho_B} = \frac{2}{3} \times 9 = 6\).
Thus \(\rho_A : \rho_B = 6:1\).

Step 4: Final Answer:
Option (A) 6 : 1.