A disc – like reel with massless thread unwinds itself while falling vertically downwards. The acceleration of its fall is:

Options

A. Zero
B. g
C. `(2g)/3`
D. `g/2`

Explanation

To solve the problem of a disc-like reel unwinding itself while falling vertically downwards, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Disc:**
- The disc experiences two main forces:
- The gravitational force acting downwards, which is \( mg \) (where \( m \) is the mass of the disc and \( g \) is the acceleration due to gravity).
- The tension \( T \) in the thread acting upwards.

2. **Set Up the Equation of Motion:**
- According to Newton's second law, the net force acting on the disc can be expressed as:
\[ mg - T = ma \]
- Here, \( a \) is the acceleration of the disc downwards.

3. **Analyze the Torque on the Disc:**
- The tension in the thread creates a torque about the center of the disc. The torque \( \tau \) due to the tension is given by:
\[ \tau = T \cdot R \]
- Where \( R \) is the radius of the disc.

4. **Relate Torque to Angular Acceleration:**
- The torque is also related to the moment of inertia \( I \) and angular acceleration \( \alpha \) by:
\[ \tau = I \cdot \alpha \]
- For a disc, the moment of inertia \( I \) is:
\[ I = \frac{mR^2}{2} \]

5. **Express Angular Acceleration in Terms of Linear Acceleration:**
- The point of contact of the thread with the disc has zero acceleration, so we can relate linear acceleration \( a \) and angular acceleration \( \alpha \):
\[ a - \alpha R = 0 \implies \alpha = \frac{a}{R} \]

6. **Substitute Angular Acceleration into the Torque Equation:**
- Substitute \( \alpha \) into the torque equation:
\[ T \cdot R = \frac{mR^2}{2} \cdot \frac{a}{R} \]
- Simplifying this gives:
\[ T = \frac{ma}{2} \]

7. **Substitute Tension Back into the Equation of Motion:**
- Now substitute \( T \) back into the equation of motion:
\[ mg - \frac{ma}{2} = ma \]
- Rearranging gives:
\[ mg = ma + \frac{ma}{2} = \frac{3ma}{2} \]

8. **Solve for Acceleration \( a \):**
- From the equation \( mg = \frac{3ma}{2} \), we can solve for \( a \):
\[ a = \frac{2g}{3} \]

### Final Answer:
The acceleration of the fall of the disc-like reel is:
\[ \boxed{\frac{2g}{3}} \]