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Current Question (ID: 10405)

Question:
$\text{A vessel of } 1 \times 10^{-3} \text{ m}^3 \text{ volume contains oil. When a pressure of } 1.2 \times 10^5 \text{ N/m}^2 \text{ is applied on it, then volume decreases by } 0.3 \times 10^{-6} \text{ m}^3\text{. The bulk modulus of oil is:}$
Options:
  • 1. $1 \times 10^6 \text{ N/m}^2$
  • 2. $2 \times 10^7 \text{ N/m}^2$
  • 3. $4 \times 10^8 \text{ N/m}^2$
  • 4. $6 \times 10^{10} \text{ N/m}^2$
Solution:
$\text{Given:}$ $V = 1 \times 10^{-3} \text{ m}^3$ $\Delta P = 1.2 \times 10^5 \text{ N/m}^2$ $\Delta V = 0.3 \times 10^{-6} \text{ m}^3$ $\text{Using the bulk modulus formula:}$ $B = \frac{\Delta P}{\Delta V/V}$ $\text{Substituting the values:}$ $B = \frac{1.2 \times 10^5 \times 1 \times 10^{-3}}{0.3 \times 10^{-6}}$ $B = \frac{1.2 \times 10^5 \times 1 \times 10^{-3}}{0.3 \times 10^{-6}} = 4 \times 10^8 \text{ N/m}^2$

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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
}