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 W₁ 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

Correct Answer: Option 3, which is $\frac{W_1+\frac{3W}{4}}{S}$ (or represented as $\{W_1+(3W/4\left)\right\}/S$).

Step-by-Step Explanation:

1. Understand the forces acting on the wire:
The tension in the wire at any point is due to the total weight suspended below that point. This includes:
- The external load $W_1$ suspended at the lower end of the wire.
- The weight of the portion of the wire that lies below the point of interest.

2. Calculate the weight of the wire below the given height:
The wire is uniform, has a total length $L$, and a total weight $W$. Therefore, the weight per unit length of the wire is:

$\text{Weight per unit length}=\frac{W}{L}$
We need to find the stress at a height of $\frac{3L}{4}$ from the lower end. The length of the wire below this point is $\frac{3L}{4}$.
The weight of this lower portion of the wire is:

$W_{\text{wire portion}}=\left(\frac{W}{L}\right)\times \frac{3L}{4}=\frac{3W}{4}$

3. Find the total tension force (T) at this height:
The total downward force at the height $\frac{3L}{4}$ is the sum of the suspended load and the weight of the wire below that height:

$T=W_1+\frac{3W}{4}$

4. Determine the stress in the wire:
Stress is defined as the restoring force (tension) per unit cross-sectional area ($S$):

$\text{Stress}=\frac{T}{S}=\frac{W_1+\frac{3W}{4}}{S}$
This can also be written in inline format as $\{W_1+(3W/4\left)\right\}/S$.