Question

Two resistances are connected in the two gaps of a meter bridge. The balancing point is obtained at 20 cm. When a resistance of \( 15 \, \Omega \) is connected in series with the smaller resistance of the two, the balancing point shifts to 40 cm. The value of smaller resistance is

• A. \( 9 \, \Omega \)
• B. \( 12 \, \Omega \)
• C. \( 6 \, \Omega \)
• D. \( 4 \, \Omega \)

Hint:

In meter bridges, balancing points change when external resistances are connected in series or parallel, so make sure to apply the correct formula to find the unknown resistance.

Answer 1

The Correct Option is A

Let the smaller resistance be \( R \) and the larger resistance be \( R_2 \). For a meter bridge, the relation between the resistances and the balancing point is: \[ \frac{R}{R_2} = \frac{L}{100 - L} \] Where \( L \) is the balancing point. Initially, the balancing point is at \( L = 20 \, \text{cm} \), so: \[ \frac{R}{R_2} = \frac{20}{80} = \frac{1}{4} \] Thus, \( R = \frac{R_2}{4} \). When a \( 15 \, \Omega \) resistance is added in series with \( R \), the total resistance becomes \( R + 15 \), and the new balancing point is at \( L = 40 \, \text{cm} \). So: \[ \frac{R + 15}{R_2} = \frac{40}{60} = \frac{2}{3} \] From this, we solve for \( R \), and we get: \[ R = 9 \, \Omega \] Thus, the smaller resistance is \( \boxed{9 \, \Omega} \).