Question

A meter bridge is used to determine the resistance of an unknown wire by measuring the balance point length \(l\). If the wire is replaced by another wire of same material but double the length and half the thickness, the balancing point is expected to be

• A. \(\frac{l}{8}\)
• B. \(\frac{l}{4}\)
• C. \(16l\)
• D. \(8l\)

Hint:

In meter bridge, balance length changes non-linearly with resistance. Always derive \(l'\) using ratio \(\frac{R}{S}=\frac{l}{100-l}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Problem
A meter bridge is used to measure the resistance of an unknown wire by balancing it against a known resistance. The balance point length, \( l \), is related to the resistance of the wire. If the wire is replaced by another wire of the same material but with changes in length and thickness, we need to analyze how the balance point will change.

Step 2: Key Concept
The resistance \( R \) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \] where: - \( R \) is the resistance, - \( \rho \) is the resistivity of the material (constant for the same material), - \( L \) is the length of the wire, and - \( A \) is the cross-sectional area of the wire. Since the material is the same, the resistivity \( \rho \) remains constant. The cross-sectional area \( A \) is proportional to the square of the radius \( r \) (since \( A = \pi r^2 \)). If the thickness of the wire is halved, the radius will also be halved, meaning \( A \) will be reduced by a factor of 4. Thus, the resistance will change as follows:

\[ R' = \rho \frac{2L}{A'} = \rho \frac{2L}{\frac{A}{4}} = 8 \times R \] The new resistance \( R' \) will be 8 times the original resistance \( R \).

Step 3: Relationship Between Resistance and Balance Point
The balance point in a meter bridge is inversely proportional to the resistance of the wire being measured. Since the new resistance is 8 times the original resistance, the balance point length will be reduced by a factor of 8.

Therefore, the new balance point will be: \[ \text{New Length} = \frac{l}{8} \] Step 4: Final Answer:
The balancing point is expected to be \( \frac{l}{8} \).

Answer: \( \frac{l}{8} \)