Question

In a meter bridge experiment null point is obtained at $l$ cm from the left end. If the meter bridge wire is replaced by a wire of same material but twice the area of across-section, then the null point is obtained at a distance

• A. 21 cm from left end.
• B. $l$ cm from left end.
• C. $l/2$ cm from left end.
• D. $l/4$ cm from left end.

Hint:

Logic Tip: Meter bridge balance depends only on the ratio of lengths, provided the wire is uniform.

Answer 1

The Correct Option is B

Concept: In a metre bridge (Wheatstone bridge), at the null point: \[ \frac{R_1}{R_2} = \frac{R_{\text{left}{R_{\text{right} \] Since resistance of a wire is: \[ R = \rho \frac{l}{A} \] where $\rho$ is resistivity, $l$ is length, and $A$ is cross-sectional area.
Step 1: Express resistance ratio of wire segments \[ \frac{R_{\text{left}{R_{\text{right} = \frac{\rho \frac{l}{A{\rho \frac{100 - l}{A \] \[ = \frac{l}{100 - l} \]
Step 2: Observe area dependence The cross-sectional area $A$ cancels out: \[ \frac{R_{\text{left}{R_{\text{right} = \frac{l}{100 - l} \] Thus, the ratio depends only on lengths, not on thickness.
Step 3: Conclusion Since the balance condition depends only on the ratio of lengths, changing the thickness (area) of the wire does not affect the null point. Final Answer: \[ \boxed{\text{Null point remains unchanged \]