Question

The unknown resistances are connected in two gaps of a metre bridge. The null point is at 20 cm from zero end. A resistance of 15 \( \Omega \) is connected in series with the smaller of the two. The null point shifts to 40 cm. The smaller resistance is

• A. 9 \( \Omega \)
• B. 7 \( \Omega \)
• C. 3 \( \Omega \)
• D. 5 \( \Omega \)

Hint:

In a meter bridge, the ratio of resistances is directly proportional to the ratio of the distances from the zero end to the null point.

Answer 1

The Correct Option is A

Step 1: Understanding the meter bridge concept.
The meter bridge uses the principle of the Wheatstone bridge. When the null point shifts, the ratio of resistances can be determined using the formula: \[ \frac{R_1}{R_2} = \frac{l_1}{l_2} \] where \( R_1 \) and \( R_2 \) are the resistances, and \( l_1 \) and \( l_2 \) are the lengths from the zero end to the null point and from the null point to the other end, respectively.
Step 2: Applying the given values.
Given, - Null point shifts from 20 cm to 40 cm, so \( l_1 = 20 \) cm, \( l_2 = 40 \) cm. - Resistance in series is \( 15 \, \Omega \). Using the formula, we get: \[ \frac{R_1}{R_2} = \frac{20}{40} = \frac{1}{2} \] Since \( R_2 = 15 \, \Omega \), we can calculate \( R_1 = 9 \, \Omega \).
Step 3: Conclusion.
The correct answer is (A), the smaller resistance is 9 \( \Omega \).