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_2` then the value of R in `Omega` is

Explanation

To solve the problem, we need to find the value of the resistor \( R \) when two identical batteries of internal resistance \( 1 \Omega \) each are connected in series and parallel across the resistor \( R \). We are given that the rate of heat produced in \( R \) when the batteries are in series is \( J_1 \) and when they are in parallel is \( J_2 \), with the relation \( J_1 = 2.25 J_2 \).

### Step-by-Step Solution:

1. **Determine the Current in Series Connection:**
- When the two batteries are connected in series, the total voltage \( V_T \) is \( 2V \) (where \( V \) is the voltage of each battery).
- The total internal resistance in series is \( 2 \Omega \) (since each battery has \( 1 \Omega \)).
- The total resistance in the circuit is \( R + 2 \).
- The current \( I_1 \) through the resistor \( R \) can be calculated using Ohm's law:
\[ I_1 = \frac{V_T}{R + 2} = \frac{2V}{R + 2} \]

2. **Calculate the Power (Heat) in Series:**
- The power \( J_1 \) (rate of heat produced in \( R \)) is given by:
\[ J_1 = I_1^2 R = \left(\frac{2V}{R + 2}\right)^2 R \]
- Simplifying this gives:
\[ J_1 = \frac{4V^2 R}{(R + 2)^2} \]

3. **Determine the Current in Parallel Connection:**
- When the two batteries are connected in parallel, the total voltage across \( R \) remains \( V \) (the same as one battery).
- The equivalent internal resistance \( R_{eq} \) of the two batteries in parallel is:
\[ R_{eq} = \frac{1}{\frac{1}{1} + \frac{1}{1}} = \frac{1}{2} \Omega \]
- The total resistance in this case is \( R + 0.5 \).
- The current \( I_2 \) through the resistor \( R \) can be calculated as:
\[ I_2 = \frac{V}{R + 0.5} \]

4. **Calculate the Power (Heat) in Parallel:**
- The power \( J_2 \) (rate of heat produced in \( R \)) is given by:
\[ J_2 = I_2^2 R = \left(\frac{V}{R + 0.5}\right)^2 R \]
- Simplifying this gives:
\[ J_2 = \frac{V^2 R}{(R + 0.5)^2} \]

5. **Relate \( J_1 \) and \( J_2 \):**
- We know from the problem statement that:
\[ J_1 = 2.25 J_2 \]
- Substituting the expressions for \( J_1 \) and \( J_2 \):
\[ \frac{4V^2 R}{(R + 2)^2} = 2.25 \cdot \frac{V^2 R}{(R + 0.5)^2} \]
- Canceling \( V^2 R \) from both sides (assuming \( V \neq 0 \) and \( R \neq 0 \)):
\[ \frac{4}{(R + 2)^2} = 2.25 \cdot \frac{1}{(R + 0.5)^2} \]

6. **Cross-Multiply and Simplify:**
- Cross-multiplying gives:
\[ 4(R + 0.5)^2 = 2.25(R + 2)^2 \]
- Expanding both sides:
\[ 4(R^2 + R + 0.25) = 2.25(R^2 + 4R + 4) \]
\[ 4R^2 + 4R + 1 = 2.25R^2 + 9R + 9 \]
- Rearranging gives:
\[ (4 - 2.25)R^2 + (4 - 9)R + (1 - 9) = 0 \]
\[ 1.75R^2 - 5R - 8 = 0 \]

7. **Solve the Quadratic Equation:**
- Using the quadratic formula \( R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ R = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1.75 \cdot (-8)}}{2 \cdot 1.75} \]
\[ R = \frac{5 \pm \sqrt{25 + 56}}{3.5} \]
\[ R = \frac{5 \pm \sqrt{81}}{3.5} \]
\[ R = \frac{5 \pm 9}{3.5} \]
- This gives two possible solutions:
\[ R = \frac{14}{3.5} = 4 \quad \text{(valid)} \]
\[ R = \frac{-4}{3.5} \quad \text{(not valid)} \]

### Final Answer:
Thus, the value of \( R \) is \( 4 \Omega \).