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

In this problem, we need to find the change in the balance length of a meter bridge when an unknown resistance is shunted with a known resistance. Let's solve this step-by-step:

A meter bridge works on the principle of a balanced Wheatstone bridge, where the balance condition is given by:

1. The balance condition: 

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

1. Where \( R_1 \) is the resistance in the left gap, \( R_2 \) is the resistance in the right gap, and \( L_1 \) and \( L_2 \) are the lengths of the bridge wire.

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

1. We can express the balance condition as:

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

Solving for \( X \), we have:

1. Rearranging yields:

\[X = 3 \, \Omega\]

Now, the unknown resistance \( X \) is shunted with a \( 2 \, \Omega \) resistor. When resistors are shunted (connected in parallel), the effective resistance \( R_{\text{eff}} \) is given by:

1. Formula for parallel resistances:

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

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

1. Calculate \( R_{\text{eff}} \):

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

With the new effective resistance, the balance length \( L_1' \) can be recalculated using the new balance condition:

1. Recalculate the balance condition:

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

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

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

1. Calculate \( L_1' \):

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

Solve for \( L_1' \):

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

Hence, the change in the balance length is:

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

Therefore, the balance length changes by 22.5 cm, which matches the given correct answer.

Answer 2

Given: - Resistance in the left gap (\( R \)) = \( 2 \, \Omega \) - Balance length (\( L_1 \)) = \( 40 \, \text{cm} \) - Shunting resistance (\( R_s \)) = \( 2 \, \Omega \)

Step 1: Calculating the Value of the Unknown Resistance (\( X \))

The balance condition of the Wheatstone bridge is given by:

\[ \frac{R}{X} = \frac{L_1}{100 - L_1} \]

Substituting the given values:

\[ \frac{2}{X} = \frac{40}{60} \] \[ X = \frac{2 \times 60}{40} \] \[ X = 3 \, \Omega \]

Step 2: Calculating the Equivalent Resistance when Shunted

When the unknown resistance \( X \) is shunted with \( R_s = 2 \, \Omega \), the equivalent resistance (\( X_{\text{sh}} \)) is given by:

\[ \frac{1}{X_{\text{sh}}} = \frac{1}{X} + \frac{1}{R_s} \]

Substituting the values:

\[ \frac{1}{X_{\text{sh}}} = \frac{1}{3} + \frac{1}{2} \] \[ \frac{1}{X_{\text{sh}}} = \frac{2 + 3}{6} = \frac{5}{6} \] \[ X_{\text{sh}} = \frac{6}{5} \, \Omega \]

Step 3: Calculating the New Balance Length (\( L_2 \))

Using the balance condition again:

\[ \frac{R}{X_{\text{sh}}} = \frac{L_2}{100 - L_2} \]

Substituting the values:

\[ \frac{2}{\frac{6}{5}} = \frac{L_2}{100 - L_2} \] \[ \frac{2 \times 5}{6} = \frac{L_2}{100 - L_2} \] \[ \frac{5}{3} = \frac{L_2}{100 - L_2} \]

Cross-multiplying:

\[ 5(100 - L_2) = 3L_2 \] \[ 500 - 5L_2 = 3L_2 \] \[ 500 = 8L_2 \] \[ L_2 = \frac{500}{8} = 62.5 \, \text{cm} \]

Step 4: Change in Balance Length

The change in balance length is given by:

\[ \Delta L = L_2 - L_1 \]

Substituting the values:

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

Conclusion:

The balance length changes by \( 22.5 \, \text{cm} \).