A: A free proton is stable but inside a nucleus , a proton gets converted into a neutron, position and neutrino `(p to n+e^(+)+v)`.
R: Inside a nucleus, neutron decay `(n to p + e^(-)+barv)` as well as proton decay are possible, since other nucleons can share energy nad momentum to conserve energy as well as momentum and both the decays are in dynamic equillibrium .

Answer

To solve the assertion and reason question, we need to analyze both statements carefully.

### Step 1: Understanding the Assertion
The assertion states that a free proton is stable, but when it is inside a nucleus, it can be converted into a neutron, a positron, and a neutrino. This process is known as beta-plus decay (β+ decay).

**Explanation:**
- A free proton (not bound in a nucleus) does not undergo decay because it is stable.
- However, when a proton is inside a nucleus, it can decay into a neutron due to the interactions with other nucleons (neutrons and protons) in the nucleus.

### Step 2: Understanding the Reason
The reason states that both neutron decay (beta-minus decay) and proton decay are possible inside a nucleus because other nucleons can share energy and momentum to conserve energy and momentum, and both decays are in dynamic equilibrium.

**Explanation:**
- Inside a nucleus, the balance of forces and the interactions between nucleons allow for both types of decay to occur.
- Neutron decay (n → p + e⁻ + ν̅) and proton decay (p → n + e⁺ + ν) can happen because the energy and momentum can be conserved through interactions with other nucleons.

### Step 3: Conclusion
Both the assertion and reason are correct, and the reason provides a correct explanation for the assertion.

### Final Answer:
- **Assertion (A)**: True
- **Reason (R)**: True
- **Explanation**: R is the correct explanation of A.