One end of a uniform wire of length L and of weight W is attached rigidly to a point in the roof and a weight `W_(1)` is suspended from its lower end. IF S is the area of cross-section of the wire, the stress in the wire at a height 3L/4 from its lower end is

Options

A. `W_(1)//S`
B. `(W_(1) + (W)/(4))S`
C. `(W_(1) + (3W)/(4))/S`
D. `(W_(1) + W)S`

Explanation

To solve the problem, we need to determine the stress in the wire at a height of \( \frac{3L}{4} \) from its lower end. Here's the step-by-step solution:

### Step 1: Understand the setup
- We have a uniform wire of length \( L \) and weight \( W \) attached to a roof at one end.
- A weight \( W_1 \) is suspended from the lower end of the wire.
- We need to find the stress in the wire at a height of \( \frac{3L}{4} \) from the lower end.

### Step 2: Identify the forces acting on the wire
- The total weight acting on the wire at the point \( \frac{3L}{4} \) from the lower end includes:
- The weight of the suspended mass \( W_1 \).
- The weight of the portion of the wire below the height \( \frac{3L}{4} \).

### Step 3: Calculate the weight of the wire below the height \( \frac{3L}{4} \)
- The length of the wire below the height \( \frac{3L}{4} \) is \( \frac{L}{4} \).
- The weight of this portion of the wire can be calculated as:
\[ \text{Weight of wire below } \frac{3L}{4} = \frac{W}{L} \cdot \frac{L}{4} = \frac{W}{4} \]
- This is because the weight of the wire is uniformly distributed along its length.

### Step 4: Calculate the total downward force at height \( \frac{3L}{4} \)
- The total force acting downwards at the height \( \frac{3L}{4} \) is the sum of the weight of the suspended mass and the weight of the wire below that point:
\[ F_{\text{total}} = W_1 + \frac{W}{4} \]

### Step 5: Calculate the stress in the wire
- Stress (\( \sigma \)) is defined as the force per unit area. Therefore, the stress at the height \( \frac{3L}{4} \) is given by:
\[ \sigma = \frac{F_{\text{total}}}{S} = \frac{W_1 + \frac{W}{4}}{S} \]

### Final Answer
Thus, the stress in the wire at a height of \( \frac{3L}{4} \) from its lower end is:
\[ \sigma = \frac{W_1 + \frac{W}{4}}{S} \]

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