For the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

Options

A. `200 Omega`
B. `100 Omega`
C. `500 Omega`
D. `1000 Omega`

Explanation

To solve the problem, we need to analyze the circuit and apply the principles of current electricity. The key point here is that the galvanometer shows zero deflection, indicating that no current flows through it. This means that the potential difference across the galvanometer is zero, and we can use this information to find the value of the resistor \( R \).

### Step-by-Step Solution:

1. **Understanding Zero Deflection**:
- When the galvanometer shows zero deflection, it indicates that the current through it is zero. This means that the potential difference across the galvanometer is equal to zero.

2. **Analyzing the Circuit**:
- Let's denote the voltage of battery A as \( V_A = 12 \, \text{V} \) and the voltage of battery B as \( V_B = 2 \, \text{V} \).
- The galvanometer is connected in such a way that the two batteries are in two different branches of the circuit.

3. **Applying Kirchhoff's Voltage Law**:
- Since there is no current flowing through the galvanometer, we can analyze the two loops formed by the batteries and resistors.
- For the loop containing battery A and resistor \( R \):
\[ V_A = I \cdot (500 \, \Omega + R) \]
- Since \( I = 0 \) (no current through the galvanometer), we can set up the equation as:
\[ 12 = 0 \cdot (500 + R) \quad \text{(not useful since current is zero)} \]

4. **Voltage Across Resistor R**:
- The voltage across resistor \( R \) can be expressed as:
\[ V_R = I_R \cdot R \]
- Since the current through \( R \) is also zero, we need to consider the voltage across the other branch (with battery B):
\[ V_B = I_B \cdot 500 \, \Omega \]
- Since the galvanometer shows zero deflection, the voltage across the galvanometer must equal the voltage across \( R \).

5. **Setting Up the Equation**:
- The voltage across \( R \) must equal the voltage across the 500-ohm resistor:
\[ 2 = \frac{12}{500 + R} \cdot R \]

6. **Cross Multiplying**:
- Rearranging gives:
\[ 2(500 + R) = 12R \]
- Expanding this:
\[ 1000 + 2R = 12R \]

7. **Solving for R**:
- Rearranging the equation:
\[ 12R - 2R = 1000 \]
\[ 10R = 1000 \]
\[ R = \frac{1000}{10} = 100 \, \Omega \]

### Final Answer:
The value of the resistor \( R \) is \( 100 \, \Omega \).