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 solid cylinder, and derive the necessary equations to find the acceleration of the bucket and the tension in the rope.

### Step 1: Identify the Forces Acting on the Bucket
The forces acting on the bucket of water are:
- The gravitational force (weight) acting downward, \( W = mg = 21 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 205.71 \, \text{N} \).
- 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 of motion for the bucket:
\[ mg - T = ma \]
Substituting the values we have:
\[ 205.71 - T = 21a \quad \text{(1)} \]

### Step 3: Analyze the Solid Cylinder
The solid cylinder is rotating due to the tension in the rope. The torque \( \tau \) caused by the tension is given by:
\[ \tau = T \cdot r \]
where \( r = 0.1 \, \text{m} \) (the radius of the cylinder). The moment of inertia \( I \) of a solid cylinder is:
\[ I = \frac{1}{2} m r^2 = \frac{1}{2} \times 21 \, \text{kg} \times (0.1 \, \text{m})^2 = 0.105 \, \text{kg m}^2 \]
Using the relation \( \tau = I \alpha \) and knowing that \( \alpha = \frac{a}{r} \), we can write:
\[ T \cdot 0.1 = I \cdot \frac{a}{0.1} \]
Substituting \( I \):
\[ T \cdot 0.1 = 0.105 \cdot \frac{a}{0.1} \]
Simplifying gives:
\[ T = \frac{0.105 \cdot a}{0.01} = 10.5a \quad \text{(2)} \]

### Step 4: Substitute Equation (2) into Equation (1)
Now we can substitute \( T \) from equation (2) into equation (1):
\[ 205.71 - 10.5a = 21a \]
Combining terms gives:
\[ 205.71 = 31.5a \]
Solving for \( a \):
\[ a = \frac{205.71}{31.5} \approx 6.53 \, \text{m/s}^2 \]

### Step 5: Calculate the Tension \( T \)
Now we can find the tension \( T \) using equation (2):
\[ T = 10.5a = 10.5 \times 6.53 \approx 68.5 \, \text{N} \]

### Conclusion
The acceleration of the bucket is approximately \( 6.53 \, \text{m/s}^2 \) and the tension in the rope is approximately \( 68.5 \, \text{N} \).