Two cylindrical rods of same cross-section area and same length are connected in series to an ideal cell as shown.The resistivity of left rod is `rho` and that if right rod is `2 rho` Then the variation of potential and electric field at any point `P` distant `x` from left end of combined rod system are given by

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Answer

To solve the problem, we need to analyze the two cylindrical rods connected in series, taking into account their resistivities and how they affect the electric field and potential along the rods.

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let the length of each rod be \( L \).
- The resistivity of the left rod is \( \rho \).
- The resistivity of the right rod is \( 2\rho \).
- The cross-sectional area \( A \) of both rods is the same.

2. **Calculate the Resistance of Each Rod**:
- The resistance \( R \) of a rod is given by the formula:
\[ R = \frac{\rho L}{A} \]
- For the left rod:
\[ R_1 = \frac{\rho L}{A} \]
- For the right rod:
\[ R_2 = \frac{2\rho L}{A} \]

3. **Total Resistance in Series**:
- The total resistance \( R_{total} \) of the two rods in series is:
\[ R_{total} = R_1 + R_2 = \frac{\rho L}{A} + \frac{2\rho L}{A} = \frac{3\rho L}{A} \]

4. **Current Through the Rods**:
- If an ideal cell of voltage \( V \) is connected across the series combination, the current \( I \) through the rods can be calculated using Ohm's law:
\[ I = \frac{V}{R_{total}} = \frac{V A}{3\rho L} \]

5. **Electric Field in Each Rod**:
- The electric field \( E \) in a rod is given by:
\[ E = \frac{V}{L} \]
- For the left rod:
\[ E_1 = \frac{I R_1}{L} = \frac{I \cdot \frac{\rho L}{A}}{L} = \frac{I \rho}{A} \]
- For the right rod:
\[ E_2 = \frac{I R_2}{L} = \frac{I \cdot \frac{2\rho L}{A}}{L} = \frac{2I \rho}{A} \]

6. **Comparison of Electric Fields**:
- Since \( E_2 = 2E_1 \), the electric field is greater in the right rod than in the left rod.

7. **Potential Variation**:
- The potential drop across each rod can be calculated as:
- For the left rod:
\[ V_1 = E_1 \cdot L \]
- For the right rod:
\[ V_2 = E_2 \cdot L \]

8. **Total Potential**:
- The total potential across both rods is:
\[ V = V_1 + V_2 \]

9. **Conclusion**:
- The variation of potential and electric field at any point \( P \) (distance \( x \) from the left end) will show that the electric field is greater in the right rod, and the potential will be continuous across the junction of the two rods.