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The resistance per centimeter of a meter bridge wire is r , with XΩ resistance in the left gap. Balancing length from the left end is at 40 cm with 25cm resistance in the right gap. Now the wire is replaced by another wire of 2r resistance per centimeter. The new balancing length for the same settings will be at
  • a)
    20 cm
  • b)
    10 cm
  • c)
    80 cm
  • d)
    40 cm
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The resistance per centimeter of a meter bridge wire is r, with X&Omeg...
Analysis:
Given:
Resistance per centimeter of the meter bridge wire = r
Resistance in the left gap (X) = X ohms
Balancing length from the left end = 40 cm
Resistance in the right gap = 25 ohms

Solution:

Original Situation:
Let the total resistance of the wire be R ohms.
Since the balancing length is at 40 cm, the ratio of resistances in the two gaps is inversely proportional to the lengths from the left end. Therefore, we have:
X / 25 = 40 / (100 - 40)
X = 25 * 40 / 60
X = 16.67 ohms
Now, we know that the total resistance R of the wire is given by:
R = r * 100
R = 100r ohms
Since the balancing length is at 40 cm, the total resistance R is divided into two parts at the balancing point in the ratio of 40:60. Therefore, we have:
X = R * (40 / 100)
R = X * (100 / 40)
R = 16.67 * (100 / 40)
R = 41.67 ohms

New Situation:
Resistance per centimeter of the new wire = 2r
Let the new total resistance of the wire be R' ohms.
Since the balancing length will change, the new balancing length for the same settings will be:
R' = 16.67 * (100 / 40)
R' = 41.67 ohms
Since the resistance per centimeter of the new wire is 2r, the new total resistance R' is given by:
R' = 2r * L
41.67 = 2r * L
Since the balancing length is 40 cm:
L = 40 / 100 = 0.4 m
Substitute L = 0.4 in the equation:
41.67 = 2r * 0.4
41.67 = 0.8r
r = 41.67 / 0.8
r = 52.09 ohms
Therefore, the new balancing length for the same settings will be:
R' = 52.09 * 0.4
R' = 20.83 ohms
Hence, the correct answer is option D) 40 cm.
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The resistance per centimeter of a meter bridge wire is r, with XΩresistance in the left gap. Balancing length from the left end is at 40 cmwith 25cmresistance in the right gap. Now the wire is replaced by another wire of 2rresistance per centimeter. The new balancing length for the same settings will be ata)20 cmb)10 cmc)80 cmd)40 cmCorrect answer is option 'D'. Can you explain this answer?
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