A uniform chain of length / and mass mis hanging vertically from its ends A and B which are close together. At a given instant the end B is released. What is the tension at A when B has fallen a distance x `(xltl)`?

Options

A. `(mg)/(2)[1+(3x)/l]`
B. `mg[1+(2x)/(l)]`
C. `(mg)/(2)[1+(x)/(l)]`
D. `(mg)/(2)[1+(4x)/(l)]`

Answer

To find the tension at point A when end B of a uniform chain has fallen a distance \( x \) (where \( x < L \)), we can follow these steps:

### Step 1: Understand the System
We have a uniform chain of length \( L \) and mass \( m \) hanging vertically. When the end B is released, it begins to fall under the influence of gravity. The tension at point A will depend on the mass of the chain that is still hanging and the dynamics of the falling chain.

### Step 2: Determine the Mass of the Chain Still Hanging
When end B has fallen a distance \( x \), the length of the chain that is still hanging from point A is \( L - x \). The mass of this hanging portion can be calculated using the ratio of the lengths:
\[ m_{\text{hanging}} = \frac{m}{L} (L - x) = m \left(1 - \frac{x}{L}\right) \]

### Step 3: Calculate the Weight of the Hanging Portion
The weight of the hanging portion of the chain is given by:
\[ W = m_{\text{hanging}} \cdot g = m \left(1 - \frac{x}{L}\right) g \]

### Step 4: Consider the Dynamics of the Falling Chain
When end B falls a distance \( x \), it accelerates due to gravity. The tension at point A must support the weight of the hanging portion of the chain and provide the necessary force to accelerate the chain.

### Step 5: Apply Newton's Second Law
The net force acting on the hanging portion of the chain can be expressed as:
\[ T - W = m_{\text{hanging}} \cdot a \]
where \( T \) is the tension at point A, \( W \) is the weight of the hanging portion, and \( a \) is the acceleration of the chain.

### Step 6: Determine the Acceleration
Since the chain is falling freely, the acceleration \( a \) is equal to \( g \). Thus, we can rewrite the equation as:
\[ T - m \left(1 - \frac{x}{L}\right) g = m \left(1 - \frac{x}{L}\right) g \]

### Step 7: Solve for Tension \( T \)
Rearranging the equation gives us:
\[ T = 2m \left(1 - \frac{x}{L}\right) g \]

### Final Expression for Tension
The tension at point A when end B has fallen a distance \( x \) is:
\[ T = 2mg \left(1 - \frac{x}{L}\right) \]