Step-by-step Solution

1. Understand the relationship between Electric Field and Charge Density

   For a spherically symmetric charge distribution $\rho(r)$, the electric field $\vec{E}(r)$ is radial. We can use Gauss's Law to find the magnitude of the electric field $E(r)$ at a distance $r$ from the center. The differential form of Gauss's law in spherical coordinates for a radial field is: $\nabla \cdot \vec{E} = {\rho(r) \over \epsilon_0}$ ${1 \over r^2} {d \over dr} (r^2 E(r)) = {\rho(r) \over \epsilon_0} \quad (*)$

2. Analyze the given condition

   The problem states that the electric field $E$ within the nucleus ($0 \le r \le R$) is observed to be linearly dependent on $r$. This can be written as: $E(r) = k r$ where $k$ is a constant. (Note: The electric field must be zero at the center, $E(0)=0$, so there is no constant offset in the linear relationship).

3. Determine the required form of Charge Density

   We substitute the linear form of $E(r)$ into the differential form of Gauss's Law (equation $*$) to find the charge density $\rho(r)$ that produces such a field. $E(r) = kr \implies r^2 E(r) = kr^3$ Now, differentiate with respect to $r$: ${d \over dr} (r^2 E(r)) = {d \over dr} (kr^3) = 3kr^2$ Substitute this back into equation $(*)$: ${1 \over r^2} (3kr^2) = {\rho(r) \over \epsilon_0}$ $3k = {\rho(r) \over \epsilon_0} \implies \rho(r) = 3k\epsilon_0$ This result shows that for the electric field to be linearly dependent on $r$ throughout the nucleus, the charge density $\rho(r)$ must be constant for $0 \le r \le R$.

4. Compare the required $\rho(r)$ with the given model

   The problem provides a model for the charge density $\rho(r)$ as shown in the figure:

   • For $0 \le r \le a$, $\rho(r) = d$ (constant).
   • For $a \le r \le R$, $\rho(r)$ decreases linearly from $d$ to 0.

   For the condition from step 3 to be satisfied, the charge density must be constant over the entire volume of the nucleus, i.e., for all $r$ from 0 to $R$. The given model for $\rho(r)$ is only constant up to radius $a$.

5. Find the value of 'a' that satisfies the condition

   To make the charge density constant throughout the nucleus ($0 \le r \le R$), the region where the density is not constant, i.e., $a \le r \le R$, must be eliminated. This can only happen if the starting point of this region, $a$, coincides with the end point, $R$. Therefore, we must have: $a = R$ If $a=R$, the charge density becomes $\rho(r) = d$ for $0 \le r \le R$, and $\rho(r) = 0$ for $r > R$. This represents a uniformly charged sphere. For such a sphere, the electric field inside is indeed linear: $E(r) = \frac{\rho r}{3\epsilon_0} = \frac{d r}{3\epsilon_0}$, which matches the initial condition.

6. Conclusion

   The experimental observation that the electric field is linear with $r$ inside the nucleus implies a uniform charge density. For the given model of charge density to be uniform, the parameter 'a' must be equal to R.

Evaluation of Options

• A: $a=0$: $\rho(r)$ would decrease linearly from $d$ right from the center. This would not produce a linear E-field.
• B: $a=R/2$: For $r > R/2$, $\rho(r)$ is not constant, so $E(r)$ would not be linear in this region.
• C: $a=R$: $\rho(r)$ is constant ($d$) for $0 \le r \le R$. This produces a linear E-field $E(r) \propto r$ for the entire range.
• D: $a=2R/3$: Similar to B, this leads to a non-linear E-field for $r > 2R/3$.

Thus, the only value of 'a' consistent with the observation is $a=R$.