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In a meter bridge experiment null point ...

In a meter bridge experiment null point is obtained at 20 cm. from one end of the wire when resistance X is balanced against another resistance Y. If XltY, then where will be the new position of the null point from the same end, if one deicdes to balance a resistance of 4 X against Y

A

40 cm

B

80 cm

C

50 cm

D

70 cm

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To solve the problem, we will use the principles of the meter bridge experiment. The meter bridge is based on the principle of Wheatstone bridge, where the ratio of the resistances is equal to the ratio of the lengths of the wire. ### Step-by-Step Solution: 1. **Understanding the Initial Setup:** - In the initial setup, we have a meter bridge with a null point obtained at 20 cm from one end when resistance \( X \) is balanced against resistance \( Y \). - This means that the length of wire corresponding to resistance \( X \) is 20 cm, and the length corresponding to resistance \( Y \) is \( 100 - 20 = 80 \) cm. 2. **Using the Balance Condition:** - According to the meter bridge principle, we have: \[ \frac{X}{Y} = \frac{L}{100 - L} \] - Substituting the known lengths: \[ \frac{X}{Y} = \frac{20}{80} = \frac{1}{4} \] - This implies that \( Y = 4X \). 3. **Setting Up for New Resistance:** - Now, we want to balance a resistance of \( 4X \) against \( Y \). - We need to find the new null point \( L \) when balancing \( 4X \) against \( Y \). 4. **Applying the Balance Condition Again:** - Using the same principle: \[ \frac{4X}{Y} = \frac{L}{100 - L} \] - Since we know from the previous calculation that \( Y = 4X \), we can substitute \( Y \) into the equation: \[ \frac{4X}{4X} = \frac{L}{100 - L} \] - This simplifies to: \[ 1 = \frac{L}{100 - L} \] 5. **Solving for \( L \):** - Cross-multiplying gives: \[ 100 - L = L \] - Rearranging this equation: \[ 100 = 2L \implies L = 50 \text{ cm} \] 6. **Conclusion:** - The new position of the null point when balancing a resistance of \( 4X \) against \( Y \) will be at 50 cm from the same end of the wire. ### Final Answer: The new position of the null point is at **50 cm** from the same end.

To solve the problem, we will use the principles of the meter bridge experiment. The meter bridge is based on the principle of Wheatstone bridge, where the ratio of the resistances is equal to the ratio of the lengths of the wire. ### Step-by-Step Solution: 1. **Understanding the Initial Setup:** - In the initial setup, we have a meter bridge with a null point obtained at 20 cm from one end when resistance \( X \) is balanced against resistance \( Y \). - This means that the length of wire corresponding to resistance \( X \) is 20 cm, and the length corresponding to resistance \( Y \) is \( 100 - 20 = 80 \) cm. ...
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