Import Question JSON

Current Question (ID: 10341)

Question:

$\text{A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed } 10^8 \text{ N/m}^2\text{, what is the maximum load that the cable can support?}$

Options:

1. $7.06 \times 10^4 \text{ N}$
2. $5.03 \times 10^4 \text{ N}$
3. $1.09 \times 10^4 \text{ N}$
4. $17 \times 10^4 \text{ N}$

Solution:

\text{Hint: Stress is equal to force per unit area.} \text{Step 1: Use the equation of stress.} \text{Given:} \text{Radius of the steel cable, } r = 1.5 \text{ cm} = 0.015 \text{ m} \text{Maximum allowable stress} = 10^8 \text{ N/m}^2 \text{Step 2: Apply the stress formula.} \text{Maximum stress} = \frac{\text{Maximum force}}{\text{Area of cross section}} \text{Area of cross section} = \pi r^2 = \pi \times (0.015)^2 = \pi \times 0.000225 \text{ m}^2 \text{Step 3: Calculate the maximum force.} \text{Rearranging the stress formula:} \text{Maximum force} = \text{Maximum stress} \times \text{Area} \text{Maximum force} = 10^8 \times \pi \times (0.015)^2 \text{Maximum force} = 10^8 \times \pi \times 0.000225 \text{Maximum force} = 10^8 \times 3.14159 \times 0.000225 \text{Maximum force} = 7.065 \times 10^4 \text{ N} \approx 7.06 \times 10^4 \text{ N}

Import JSON File

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
}