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Current Question (ID: 19531)

Question:

$\text{A steel wire having a radius of } 2.0 \text{ mm, carrying a load of } 4 \text{ kg, is hanging from a ceiling. Given that } g = 3.1\pi \text{ m/s}^2, \text{ what will be the tensile stress that would be developed in the wire?}$

Options:

1. $5.2 \times 10^6 \text{ N/m}^2$
2. $6.2 \times 10^6 \text{ N/m}^2$
3. $4.8 \times 10^6 \text{ N/m}^2$
4. $3.1 \times 10^6 \text{ N/m}^2$

Solution:

$\text{Hint: Tensile stress = Tensile force / Area}$ $\text{Step 1: Find the tensile force.}$ $T = mg$ $\text{Step 2: Find the tensile stress.}$ $\sigma = \frac{T}{A}$ $\Rightarrow \sigma = \frac{4 \times 3.1\pi}{\pi \times (2 \times 10^{-3})^2}$ $\Rightarrow \sigma = 3.1 \times 10^6 \text{ N/m}^{-2}$ $\text{Hence, option (4) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}