When a known standard resistor of 2.0 is connected to the 0.0cm end of a metre bridge, the balance point is found to be at 55.0cm

To solve this problem, we use the principle of the Metre Bridge, which is based on the Wheatstone Bridge circuit.

1. Analyze the question

• Standard resistor ($R$) = $2.0\ \Omega$
• Balance point ($l_1$) = $55.0\ \text{cm}$
• The total length of the bridge wire ($L$) = $100\ \text{cm}$
• We need to find the unknown resistor ($X$).

2. Explain the concept

In a metre bridge, the wire is divided into two segments by the balance point. The ratio of the resistances is equal to the ratio of the lengths of these segments:

Where:

• $R$ is the known resistor.
• $X$ is the unknown resistor.
• $l_1$ is the length from the 0.0cm end to the balance point.
• $l_2$ is the remaining length of the wire ($100\ \text{cm} - l_1$).

3. Show the solution

Step 1: Identify the lengths

• $l_1 = 55.0\ \text{cm}$
• $l_2 = 100\ \text{cm} - 55.0\ \text{cm} = 45.0\ \text{cm}$

Step 2: Apply the formula

Step 3: Solve for $X$ Cross-multiply to isolate $X$:

Rounding to two decimal places, we get $1.64\ \Omega$.

4. Identify the answer

Comparing our result to the given options, the correct choice is B.

5. Exam Tip

Always remember that the sum of the two lengths ($l_1 + l_2$) must equal $100\ \text{cm}$ because it is a metre bridge. If you ever forget the formula, just remember: "Resistor over Resistor equals Length over Length" (keeping the corresponding sides matched).