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:
- 22.5 cm
- 20 cm
- 62.5 cm
- 65 cm
The Correct Option is A
Approach Solution - 1
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:
- The balance condition:
- 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 \( L_1 = 40 \, \text{cm} \).
- We can express the balance condition as:
Solving for \( X \), we have:
- Rearranging yields:
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:
- Formula for parallel resistances:
Substituting \( X = 3 \, \Omega \):
- Calculate \( R_{\text{eff}} \):
With the new effective resistance, the balance length \( L_1' \) can be recalculated using the new balance condition:
- Recalculate the balance condition:
Substituting \( R_{\text{eff}} = 1.2 \, \Omega \), we get:
\[\frac{2}{1.2} = \frac{L_1'}{100 - L_1'}\]- Calculate \( L_1' \):
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.
Approach Solution -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 \( 22.5 \, \text{cm} \).
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