A bucket of water of mass `21 kg` is suspended by a rope wrapped around a solid cylinder `0.2 m` in diameter. The mass of the solid cylinder is `21 kg`. The bucket is released from rest. Which of the following statements are correct?

Options

A. The tension in the rope is `70 N`.
B. The acceleration of the bucket is `(20/3) m//s^(2)`
C. The acceleration of the bucket is independent of the mass of the bucket.
D. All of these

Explanation

To solve the problem step by step, we will analyze the forces acting on the bucket and the torque acting on the solid cylinder.

### Step 1: Identify the Forces Acting on the Bucket
The forces acting on the bucket are:
- The gravitational force (weight) acting downward: \( F_g = mg = 21 \, \text{kg} \times g \) (where \( g \approx 9.81 \, \text{m/s}^2 \))
- The tension \( T \) in the rope acting upward.

### Step 2: Write the Equation of Motion for the Bucket
Using Newton's second law, we can write the equation for the bucket:
\[ mg - T = ma \]
Substituting \( m = 21 \, \text{kg} \):
\[ 21g - T = 21a \tag{1} \]

### Step 3: Analyze the Solid Cylinder
The rope wrapped around the solid cylinder creates a torque \( \tau \) on the cylinder. The torque is given by:
\[ \tau = T \cdot R \]
where \( R \) is the radius of the cylinder. Since the diameter is \( 0.2 \, \text{m} \), the radius \( R = 0.1 \, \text{m} \).

### Step 4: Write the Torque Equation
The torque is also related to the angular acceleration \( \alpha \) of the cylinder:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia of the solid cylinder. For a solid cylinder, the moment of inertia is:
\[ I = \frac{1}{2} m R^2 = \frac{1}{2} \times 21 \, \text{kg} \times (0.1 \, \text{m})^2 = \frac{21 \times 0.01}{2} = 0.105 \, \text{kg m}^2 \]

### Step 5: Relate Linear and Angular Acceleration
The linear acceleration \( a \) of the bucket is related to the angular acceleration \( \alpha \) of the cylinder by:
\[ a = R \alpha \implies \alpha = \frac{a}{R} \]
Substituting this into the torque equation gives:
\[ T \cdot R = I \cdot \frac{a}{R} \]
Rearranging gives:
\[ T = \frac{I \cdot a}{R^2} \tag{2} \]

### Step 6: Substitute for \( I \)
Substituting \( I = 0.105 \, \text{kg m}^2 \) and \( R = 0.1 \, \text{m} \) into equation (2):
\[ T = \frac{0.105 \cdot a}{(0.1)^2} = \frac{0.105 \cdot a}{0.01} = 10.5a \]

### Step 7: Substitute \( T \) Back into Equation (1)
Substituting \( T = 10.5a \) into equation (1):
\[ 21g - 10.5a = 21a \]
Rearranging gives:
\[ 21g = 21a + 10.5a = 31.5a \implies a = \frac{21g}{31.5} \]

### Step 8: Calculate the Acceleration
Substituting \( g \approx 9.81 \, \text{m/s}^2 \):
\[ a = \frac{21 \times 9.81}{31.5} \approx \frac{205.01}{31.5} \approx 6.5 \, \text{m/s}^2 \]

### Step 9: Calculate the Tension
Using \( a \) to find \( T \):
\[ T = 10.5 \cdot a = 10.5 \cdot 6.5 \approx 68.25 \, \text{N} \]

### Conclusion
After analyzing the forces and calculating the acceleration and tension, we can conclude that:
- The acceleration of the bucket is approximately \( 6.5 \, \text{m/s}^2 \).
- The tension in the rope is approximately \( 68.25 \, \text{N} \).