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

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 \( 22.5 \, \text{cm} \).