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 ___.

We need to find the value of $$x$$ (the distance from terminal $$A$$ to the balancing point $$P$$) where the galvanometer shows zero deflection in a meter bridge circuit.

1. Identify the Circuit Parameters

From the schematic on the page, the system forms a balanced Wheatstone bridge network:

• Resistance in the left gap ($$R_1$$) = $$12\ \Omega$$
• Resistance in the right gap ($$R_2$$) = $$6\ \Omega$$
• Total length of the uniform resistance wire $$AB$$ ($$L$$) = $$72\text{ cm}$$
• Distance from $$A$$ to the balancing point $$P$$ = $$x\text{ cm}$$
• Remaining distance from $$P$$ to $$B$$ = $$(72 - x)\text{ cm}$$

2. Apply the Condition for a Balanced Wheatstone Bridge

When the galvanometer shows zero deflection, no current flows through it. This implies that the potential at point $$C$$ is exactly equal to the potential at point $$P$$. For a uniform resistance wire, the resistance of each section is directly proportional to its length ($$R \propto L$$).

The balancing condition is given by the ratio:

$$\frac{R_1}{R_2} = \frac{\text{Resistance of segment } AP}{\text{Resistance of segment } PB}$$

Substituting the corresponding values and lengths into the formula:

$$\frac{12}{6} = \frac{x}{72 - x}$$

3. Solve for $$x$$

Simplify the fraction on the left side:

$$2 = \frac{x}{72 - x}$$

Cross-multiply to clear the denominator:

$$2 \cdot (72 - x) = x$$

$$144 - 2x = x$$

Isolate the variable $$x$$:

$$3x = 144$$

$$x = \frac{144}{3} = 48\text{ cm}$$

Conclusion

The value of $$x$$ to the nearest integer is 48.