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Current Question (ID: 10401)

Question:

$\text{Copper of fixed volume } V \text{ is drawn into a wire of length } l. \text{ When this wire is subjected to a constant force } F, \text{ the extension produced in the wire is } \Delta l. \text{ Which of the following graphs is a straight line?}$

Options:

1. $\Delta l \text{ vs } \frac{1}{l}$
2. $\Delta l \text{ vs } l^2$
3. $\Delta l \text{ vs } \frac{1}{l^2}$
4. $\Delta l \text{ vs } l$

Solution:

\text{Young's modulus: } Y = \frac{\text{Tensile stress}}{\text{Tensile strain}} \text{The length of the wire increases from } l \text{ by } \Delta l Y = \frac{F/A}{\Delta l/l} \text{The volume: } V = A \times l \text{Rearranging: } \Delta l = \frac{Fl^2}{YV} \text{Therefore: } \Delta l \propto l^2 \text{A graph of } \Delta l \text{ vs } l^2 \text{ will be a straight line.}

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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
}