A rod of mass 5 kg is connected to the string at point B. The span of rod is along horizontal. The other end of the rod is hinged at point A. If the string is massless, then the reaction of hinge at the instant when string is cut, is (Take, `g=10 ms^(-2)`)

Options

A. 10.1 N
B. 12.5 N
C. 5 N
D. 15 N

Explanation

To solve the problem step by step, we will analyze the forces and torques acting on the rod when the string is cut.

### Step 1: Understand the System
We have a horizontal rod of mass \( m = 5 \, \text{kg} \) hinged at point A and connected to a massless string at point B. When the string is cut, the only forces acting on the rod are its weight and the reaction force at the hinge.

### Step 2: Identify Forces
The weight of the rod acts downward at its center of mass, which is located at a distance \( L/2 \) from point A. The weight \( W \) can be calculated as:
\[ W = mg = 5 \, \text{kg} \times 10 \, \text{m/s}^2 = 50 \, \text{N} \]

### Step 3: Calculate Torque
When the string is cut, the rod will start to rotate about point A due to the torque created by the weight of the rod. The torque \( \tau \) about point A due to the weight of the rod is given by:
\[ \tau = W \times \text{perpendicular distance} = W \times \frac{L}{2} \]
Substituting the value of \( W \):
\[ \tau = 50 \, \text{N} \times \frac{L}{2} \]

### Step 4: Moment of Inertia
The moment of inertia \( I \) of the rod about point A (the hinge) is given by:
\[ I = \frac{1}{3} m L^2 \]

### Step 5: Angular Acceleration
Using Newton's second law for rotation, we have:
\[ \tau = I \alpha \]
Substituting the expressions for torque and moment of inertia:
\[ 50 \times \frac{L}{2} = \frac{1}{3} m L^2 \alpha \]
This simplifies to:
\[ 25L = \frac{5}{3} L^2 \alpha \]
Cancelling \( L \) (assuming \( L \neq 0 \)):
\[ 25 = \frac{5}{3} L \alpha \]
Solving for \( \alpha \):
\[ \alpha = \frac{25 \times 3}{5L} = \frac{15}{L} \]

### Step 6: Linear Acceleration of the Center of Mass
The linear acceleration \( a \) of the center of mass (which is at \( L/2 \)) can be related to angular acceleration \( \alpha \) by:
\[ a = \alpha \cdot \frac{L}{2} = \frac{15}{L} \cdot \frac{L}{2} = \frac{15}{2} \, \text{m/s}^2 \]

### Step 7: Apply Newton's Second Law
Now, we can apply Newton's second law to the vertical motion of the center of mass:
\[ \text{Net Force} = m \cdot a \]
The forces acting on the rod are its weight \( mg \) downward and the reaction force \( R \) upward at the hinge:
\[ mg - R = ma \]
Substituting the values:
\[ 50 - R = 5 \cdot \frac{15}{2} \]
Calculating the right side:
\[ 50 - R = 5 \cdot 7.5 = 37.5 \]
Thus:
\[ R = 50 - 37.5 = 12.5 \, \text{N} \]

### Conclusion
The reaction force at the hinge when the string is cut is:
\[ R = 12.5 \, \text{N} \]