A small block of mass `4 kg` is attached to a cord passing through a hole in a horizontal frictionless surface. The block is originally revolving in a circle of radius of `5 m` about the hole, with a tangential velocity of `4 m//s`. The cord is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the cord is `200 N`.
Velocity of the block at the time of breaking of the string

Options

A. `10m//s`
B. `20 m//s`
C. `7.5 m//s`
D. `12.5m//s`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the conservation of angular momentum
The angular momentum of the block must remain constant since there are no external torques acting on it. The initial angular momentum \( L_1 \) is given by:
\[ L_1 = m \cdot v_1 \cdot r_1 \]
where:
- \( m = 4 \, \text{kg} \) (mass of the block)
- \( v_1 = 4 \, \text{m/s} \) (initial tangential velocity)
- \( r_1 = 5 \, \text{m} \) (initial radius)

### Step 2: Calculate the initial angular momentum
Substituting the values into the equation:
\[ L_1 = 4 \, \text{kg} \cdot 4 \, \text{m/s} \cdot 5 \, \text{m} = 80 \, \text{kg m}^2/\text{s} \]

### Step 3: Set up the equation for final angular momentum
When the radius is shortened to \( r_2 \) and the velocity becomes \( v_2 \), the final angular momentum \( L_2 \) is:
\[ L_2 = m \cdot v_2 \cdot r_2 \]
Since angular momentum is conserved, we have:
\[ L_1 = L_2 \implies 80 = 4 \cdot v_2 \cdot r_2 \]

### Step 4: Relate tension to centripetal force
The tension in the cord when the radius is \( r_2 \) is given as \( T_2 = 200 \, \text{N} \). This tension provides the necessary centripetal force:
\[ T_2 = \frac{m \cdot v_2^2}{r_2} \]
Substituting the values:
\[ 200 = \frac{4 \cdot v_2^2}{r_2} \]
From this, we can express \( r_2 \):
\[ r_2 = \frac{4 \cdot v_2^2}{200} = \frac{v_2^2}{50} \]

### Step 5: Substitute \( r_2 \) into the angular momentum equation
Now substitute \( r_2 \) back into the angular momentum equation:
\[ 80 = 4 \cdot v_2 \cdot \left(\frac{v_2^2}{50}\right) \]
This simplifies to:
\[ 80 = \frac{4 \cdot v_2^3}{50} \]
Multiplying both sides by 50:
\[ 4000 = 4 \cdot v_2^3 \]
Dividing both sides by 4:
\[ 1000 = v_2^3 \]

### Step 6: Solve for \( v_2 \)
Taking the cube root of both sides:
\[ v_2 = 10 \, \text{m/s} \]

### Final Answer
The velocity of the block at the time of breaking of the string is \( 10 \, \text{m/s} \).

---