Question

Two wires A and B of equal lengths are connected in left and right gap respectively 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 respectively, the ratio of specific resistance of A to that of B is

• A. \(2:1\)
• B. \(3:1\)
• C. \(6:1\)
• D. \(12:1\)

Hint:

Use the metre bridge balance condition, then write resistance in terms of resistivity, length and area.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a metre bridge, \(\dfrac{R_A}{R_B} = \dfrac{l}{100 - l}\), where \(l\) is the null point from the left end where \(A\) is connected.

Step 2: Use the null point.
\[ \frac{R_A}{R_B} = \frac{40}{60} = \frac{2}{3} \]

Step 3: Write resistances.
\(R = \rho\dfrac{L}{A}\) with area \(A = \dfrac{\pi d^2}{4}\). The wires have the same length, so
\[ \frac{R_A}{R_B} = \frac{\rho_A}{\rho_B}\times\frac{A_B}{A_A} = \frac{\rho_A}{\rho_B}\times\frac{d_B^2}{d_A^2} = \frac{\rho_A}{\rho_B}\times\frac{1}{9} \]

Step 4: Solve.
\[ \frac{\rho_A}{\rho_B} = 9\times\frac{2}{3} = 6 \]

Step 5: Check the options.
Using \(d_A : d_B\) instead of squares gives 2, which is option (A), a common error.

Final Answer:
The ratio of specific resistance A to B is \(6 : 1\), option (C). \[ \boxed{6 : 1} \]