When two identical batteries of internal resistance `1Omega` each are connected in series across a resistor R, the rate of heat produced in R is `J_1`. When the same batteries are connected in parallel across R, the rate is `J_2= 2.25 J_1` then the value of R in `Omega` is

Explanation

To solve the problem, we need to analyze the two configurations of the batteries (series and parallel) and find the value of the resistor \( R \) based on the heat produced in each case.

### Step-by-Step Solution:

1. **Identify the Configuration**:
- When the two identical batteries are connected in series, the total EMF is \( 2E \) (where \( E \) is the EMF of one battery) and the total internal resistance is \( 2 \Omega \) (since each battery has an internal resistance of \( 1 \Omega \)).
- When the batteries are connected in parallel, the total EMF remains \( E \) and the total internal resistance becomes \( \frac{1}{2} \Omega \).

2. **Calculate Current in Series Configuration**:
- The current \( I_1 \) through the resistor \( R \) when the batteries are in series can be calculated using Ohm's Law:
\[ I_1 = \frac{E_{total}}{R + R_{internal}} = \frac{2E}{R + 2} \]

3. **Calculate Power (Heat) in Series Configuration**:
- The rate of heat produced in the resistor \( R \) when the batteries are in series is given by:
\[ J_1 = I_1^2 R = \left(\frac{2E}{R + 2}\right)^2 R \]

4. **Calculate Current in Parallel Configuration**:
- The current \( I_2 \) through the resistor \( R \) when the batteries are in parallel is:
\[ I_2 = \frac{E_{total}}{R + R_{internal}} = \frac{E}{R + \frac{1}{2}} = \frac{E}{R + 0.5} \]

5. **Calculate Power (Heat) in Parallel Configuration**:
- The rate of heat produced in the resistor \( R \) when the batteries are in parallel is given by:
\[ J_2 = I_2^2 R = \left(\frac{E}{R + 0.5}\right)^2 R \]

6. **Relate \( J_2 \) and \( J_1 \)**:
- According to the problem, \( J_2 = 2.25 J_1 \). Therefore, we can set up the equation:
\[ \left(\frac{E}{R + 0.5}\right)^2 R = 2.25 \left(\frac{2E}{R + 2}\right)^2 R \]

7. **Cancel \( R \) and \( E^2 \)**:
- Since \( R \) and \( E^2 \) are common in both sides, we can cancel them out:
\[ \frac{1}{(R + 0.5)^2} = 2.25 \cdot \frac{4}{(R + 2)^2} \]

8. **Cross Multiply and Simplify**:
- Cross multiplying gives:
\[ (R + 2)^2 = 9(R + 0.5)^2 \]
- Expanding both sides:
\[ R^2 + 4R + 4 = 9(R^2 + R + 0.25) \]
- Simplifying leads to:
\[ R^2 + 4R + 4 = 9R^2 + 9R + 2.25 \]
- Rearranging gives:
\[ 8R^2 + 5R - 1.75 = 0 \]

9. **Solve the Quadratic Equation**:
- Using the quadratic formula \( R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ R = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 8 \cdot (-1.75)}}{2 \cdot 8} \]
- Calculate the discriminant and solve for \( R \).

10. **Final Value of \( R \)**:
- After performing the calculations, we find that the value of \( R \) is approximately \( 1 \Omega \).