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

Answer: (A)

22.5 cm

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 $22.5 \, \text{cm}$.