Import Question JSON

Current Question (ID: 10358)

Question:

$\text{The stress-strain curve for two materials } A \text{ and } B \text{ are as shown in the figure. Select the correct statement:}$

Options:

1. $\text{Material } A \text{ is less brittle and less elastic as compared to } B.$
2. $\text{Material } A \text{ is more ductile and less elastic as compared to } B.$
3. $\text{Material } A \text{ is less brittle and more elastic than } B.$
4. $\text{Material } B \text{ is more brittle and more elastic than } A.$

Solution:

$\text{Hint: Top most point in graph represents ultimate tensile strength}$ $\text{Step 1: Identify ultimate strength and fracture point from the graph}$ $\text{Material } A \text{ is less brittle than } B \text{ because it shows greater elongation before fracture (extends further along the strain axis).}$ $\text{Step 2: Identify Young's modulus from elastic region}$ $\text{The slope of the initial linear portion represents Young's modulus: Slope = Y}$ $\text{From the graph, the slope of material } A \text{ is steeper than material } B \text{ in the elastic region.}$ $\text{Therefore: } Y_A > Y_B$ $\text{This means material } A \text{ has higher Young's modulus, making it more elastic than material } B.$$ $\text{Material } A \text{ shows greater strain before failure (less brittle) and has a steeper slope in the elastic region (more elastic).}$

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
}