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Current Question (ID: 10387)

Question:

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 $A$ is the area of the cross-section of the wire, the stress in the wire at a height $\frac{3L}{4}$ from its lower end is:

Options:

1. $\frac{W+W_1}{A}$
2. $\frac{4W+W_1}{3A}$
3. $\frac{3W+W_1}{4A}$
4. $\frac{\frac{3}{4}W+W_1}{A}$

Solution:

The length of the wire from the lower end is $L$. The weight of the wire is $W$. The weight suspended from the lower end is $W_1$. We need to find the stress in the wire at a height $\frac{3L}{4}$ from its lower end. The length of the wire segment below the point at height $\frac{3L}{4}$ from the lower end is $\frac{3L}{4}$. The weight of this segment of the wire will be $W' = \frac{W}{L} \times \frac{3L}{4} = \frac{3W}{4}$. The tension in the wire at this height is due to the weight $W_1$ suspended from the bottom and the weight of the wire segment below this point. Therefore, the net tension is: $\text{Net tension} = \text{Tension due to wire's own weight} + W_1 = \frac{3W}{4} + W_1$ The stress is defined as force per unit area. Here, the force is the net tension, and the area is $A$. $\text{Stress} = \frac{\text{Net tension}}{A} = \frac{\frac{3W}{4} + W_1}{A}$ This can also be written as: $\text{Stress} = \frac{3W + 4W_1}{4A}$ Comparing this with the given options, the closest match is option 4, which is $\frac{\frac{3}{4}W+W_1}{A}$.

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}