A hollow cylinder with a very thin wall (like a toilet paper tube) and a block are placed at rest at the top of a plane with inclination $\theta$ above the horizontal. The cylinder rolls down the plane without slipping and the block slides down the plane; it is found that both objects reach the bottom of the plane simultaneously. What is the coefficient of kinetic friction between the block and the plane?

(A) $0$

(B) $\frac{1}{3}\tan \theta$

(C) $\frac{1}{2}\tan \theta$

(D) $\frac{2}{3}\tan \theta$

(E) $\tan \theta$
1. **Determine the acceleration of the block:**

   For the block sliding down the plane, we use Newton's second law:
   \[
   F = ma
   \]
   The forces acting on the block are gravity, the normal force, and the kinetic friction force. The component of gravitational force parallel to the plane is \( mg \sin \theta \), and the kinetic friction force is \( \mu_k mg \cos \theta \). Therefore, the net force acting on the block is:
   \[
   ma_{\text{block}} = mg \sin \theta - \mu_k mg \cos \theta
   \]
   Solving for \( a_{\text{block}} \):
   \[
   a_{\text{block}} = g \sin \theta - \mu_k g \cos \theta
   \]

2. **Determine the acceleration of the hollow cylinder:**

   For the hollow cylinder rolling down the plane without slipping, we need to consider both translational and rotational motion. The forces acting on the cylinder are gravity, the normal force, and the friction force. The component of gravitational force parallel to the plane is \( mg \sin \theta \), and the friction force \( f \) acts up the plane.

   Using Newton's second law for translation:
   \[
   mg \sin \theta - f = ma_{\text{cylinder}}
   \]

   Using Newton's second law for rotation:
   \[
   \tau = I \alpha
   \]
   where \( \tau = fr \) (torque due to friction), \( I = mr^2 \) (moment of inertia of a hollow cylinder), and \( \alpha = \frac{a_{\text{cylinder}}}{r} \) (angular acceleration).

   Substituting these into the rotational equation:
   \[
   fr = mr^2 \left( \frac{a_{\text{cylinder}}}{r} \right)
   \]
   Simplifying:
   \[
   f = ma_{\text{cylinder}}
   \]

   Substituting \( f = ma_{\text{cylinder}} \) back into the translational equation:
   \[
   mg \sin \theta - ma_{\text{cylinder}} = ma_{\text{cylinder}}
   \]
   Solving for \( a_{\text{cylinder}} \):
   \[
   mg \sin \theta = 2ma_{\text{cylinder}}
   \]
   \[
   a_{\text{cylinder}} = \frac{1}{2} g \sin \theta
   \]

3. **Equate the accelerations:**

   Since both objects reach the bottom of the plane simultaneously, their accelerations must be equal:
   \[
   a_{\text{block}} = a_{\text{cylinder}}
   \]
   Substituting the expressions for \( a_{\text{block}} \) and \( a_{\text{cylinder}} \):
   \[
   g \sin \theta - \mu_k g \cos \theta = \frac{1}{2} g \sin \theta
   \]

4. **Solve for the coefficient of kinetic friction \( \mu_k \):**

   \[
   g \sin \theta - \mu_k g \cos \theta = \frac{1}{2} g \sin \theta
   \]
   \[
   g \sin \theta - \frac{1}{2} g \sin \theta = \mu_k g \cos \theta
   \]
   \[
   \frac{1}{2} g \sin \theta = \mu_k g \cos \theta
   \]
   \[
   \mu_k = \frac{\frac{1}{2} g \sin \theta}{g \cos \theta}
   \]
   \[
   \mu_k = \frac{1}{2} \frac{\sin \theta}{\cos \theta}
   \]
   \[
   \mu_k = \frac{1}{2} \tan \theta
   \]

Thus, the coefficient of kinetic friction is \( \boxed{\frac{1}{2} \tan \theta} \).