A uniform circular ring of radius $R$ is fixed in place. A particle is placed on the axis of the ring at a distance much greater than $R$ and allowed to fall towards the ring under the influence of the ring’s gravity. The particle achieves a maximum speed $v$. The ring is replaced with one of the same (linear) mass density but radius $2R$, and the experiment is repeated. What is the new maximum speed of the particle?

(a) $\frac{1}{2}v$

(b) $\frac{1}{\sqrt{2}}v$

(c) $v$

(d) $\sqrt{2}v$

(e) $2v$
1. **Understanding the problem**: We have a uniform circular ring of radius \( R \) and a particle placed on the axis of the ring at a distance much greater than \( R \). The particle falls towards the ring under the influence of the ring’s gravity and achieves a maximum speed \( v \). We need to determine the new maximum speed when the ring is replaced with one of the same linear mass density but radius \( 2R \).

2. **Linear mass density**: The linear mass density \( \lambda \) is given by:
   \[
   \lambda = \frac{M}{2\pi R}
   \]
   where \( M \) is the mass of the original ring.

3. **Mass of the new ring**: Since the linear mass density remains the same, the mass \( M' \) of the new ring with radius \( 2R \) is:
   \[
   \lambda = \frac{M'}{2\pi (2R)} = \frac{M'}{4\pi R}
   \]
   Equating the linear mass densities:
   \[
   \frac{M}{2\pi R} = \frac{M'}{4\pi R}
   \]
   Solving for \( M' \):
   \[
   M' = 2M
   \]

4. **Gravitational potential energy**: The gravitational potential energy \( U \) of a particle of mass \( m \) at a distance \( x \) from the center of a ring of mass \( M \) and radius \( R \) is given by:
   \[
   U = -\frac{GMm}{\sqrt{R^2 + x^2}}
   \]
   For \( x \gg R \), this simplifies to:
   \[
   U \approx -\frac{GMm}{x}
   \]

5. **Initial potential energy**: Initially, the particle is at a distance \( x \gg R \) from the ring, so the initial potential energy \( U_i \) is:
   \[
   U_i = -\frac{GMm}{x}
   \]

6. **Final potential energy**: When the particle reaches the ring, the final potential energy \( U_f \) is:
   \[
   U_f = -\frac{GMm}{R}
   \]

7. **Change in potential energy**: The change in potential energy \( \Delta U \) is:
   \[
   \Delta U = U_f - U_i = -\frac{GMm}{R} - \left(-\frac{GMm}{x}\right) \approx -\frac{GMm}{R}
   \]
   (since \( x \gg R \), the term \(\frac{GMm}{x}\) is negligible)

8. **Kinetic energy and maximum speed**: The change in potential energy is converted into kinetic energy. Thus, the kinetic energy \( K \) of the particle when it reaches the ring is:
   \[
   K = \frac{1}{2}mv^2 = \Delta U = \frac{GMm}{R}
   \]
   Solving for \( v \):
   \[
   v = \sqrt{\frac{2GM}{R}}
   \]

9. **New ring with radius \( 2R \)**: For the new ring with mass \( 2M \) and radius \( 2R \), the final potential energy \( U_f' \) is:
   \[
   U_f' = -\frac{G(2M)m}{2R} = -\frac{GMm}{R}
   \]
   The change in potential energy \( \Delta U' \) is:
   \[
   \Delta U' = U_f' - U_i' \approx -\frac{GMm}{R}
   \]
   The kinetic energy \( K' \) of the particle when it reaches the new ring is:
   \[
   K' = \frac{1}{2}m(v')^2 = \Delta U' = \frac{GMm}{R}
   \]
   Solving for \( v' \):
   \[
   v' = \sqrt{\frac{2GM}{R}}
   \]

10. **Conclusion**: The maximum speed \( v' \) of the particle when the ring is replaced with one of radius \( 2R \) is the same as the original maximum speed \( v \).

The final answer is \( \boxed{ v } \)