Question

In a metre-bridge, when a resistance of $2 \, \Omega$ is in the left gap and the unknown resistance in the right gap, the balance length is found to be 40 cm. On shunting the unknown resistance with $2 \, \Omega$, the balance length changes by:

• A. 22.5 cm
• B. 20 cm
• C. 62.5 cm
• D. 65 cm

Answer 1

The Correct Option is A

This problem requires calculating the alteration in the balance length of a meter bridge when an unknown resistance is connected in parallel with a known resistance. The solution proceeds as follows:

A meter bridge operates on the principle of a balanced Wheatstone bridge, with the balance condition defined by:

1. Balance Condition:

\[\frac{R_1}{R_2} = \frac{L_1}{L_2}\]

1. Variables: \( R_1 \) represents the resistance in the left gap, \( R_2 \) represents the resistance in the right gap, and \( L_1 \) and \( L_2 \) denote the lengths of the bridge wire segments.

Initially, the left gap contains a resistance of \( 2 \, \Omega \), the right gap contains the unknown resistance \( X \), and the balance length is \( L_1 = 40 \, \text{cm} \).

1. The balance condition is expressed as:

\[\frac{2}{X} = \frac{40}{60}\]

Solving for \( X \):

1. Rearranging the equation gives:

\[X = 3 \, \Omega\]

Subsequently, the unknown resistance \( X \) is shunted (connected in parallel) with a \( 2 \, \Omega \) resistor. The effective resistance \( R_{\text{eff}} \) for parallel resistors is calculated using:

1. Parallel Resistance Formula:

\[\frac{1}{R_{\text{eff}}} = \frac{1}{X} + \frac{1}{2}\]

Substituting \( X = 3 \, \Omega \):

1. Effective Resistance Calculation:

\[\frac{1}{R_{\text{eff}}} = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}\]\[R_{\text{eff}} = \frac{6}{5} \, \Omega = 1.2 \, \Omega\]

With the updated effective resistance, the new balance length \( L_1' \) is determined by the modified balance condition:

1. Balance Condition Recalculation:

\[\frac{2}{R_{\text{eff}}} = \frac{L_1'}{100 - L_1'}\]

Substituting \( R_{\text{eff}} = 1.2 \, \Omega \):

\[\frac{2}{1.2} = \frac{L_1'}{100 - L_1'}\]

1. New Balance Length Calculation:

\[\frac{5}{3} = \frac{L_1'}{100 - L_1'}\]\[5(100 - L_1') = 3L_1'\]

Solving for \( L_1' \):

\[500 - 5L_1' = 3L_1'\]\[500 = 8L_1'\]\[L_1' = \frac{500}{8} = 62.5 \, \text{cm}\]

The change in the balance length is calculated as:

\[\Delta L = L_1' - L_1 = 62.5 \, \text{cm} - 40 \, \text{cm} = 22.5 \, \text{cm}\]

Consequently, the balance length exhibits a change of 22.5 cm.