Consider a 72 cm long wire AB as shown in the figure. The galvanometer jockey is placed at P on AB at a distance x cm from A. The galvanometer shows zero deflection.
The value of x, to the nearest integer, is ___________.

Solution

Physics sir · 4 steps

Zero deflection means a balanced Wheatstone bridge, so the two wire lengths split in the ratio 12 : 6.

1. Spot the bridge. The 12 Ω and 6 Ω resistors form one arm pair, and the wire pieces AP and PB form the other. With no current through G, C and P are at the same potential, so the bridge is balanced.
2. Wire resistance. A uniform wire has resistance proportional to its length. So R_(AP) ∝ x and R_(PB) ∝ (72 - x).

   12/6 = x/(72 - x)

3. Solve. x = 2(72 - x), so 3x = 144 and x = 48 cm.
4. So. The jockey balances at 48 cm from A.

The common slip: 24. Pairing the 12 Ω resistor with PB instead of AP gives x/(72 - x) = 6/12, which is 24, not 48.

The idea to keep. In a metre bridge, the resistor joined at the same end as a wire segment pairs with that segment: R_(left)/R_(right) = l/(L - l). Check which resistor sits on which side.

Answer from the NTA final answer key. Solution by PrepWiser.