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Current Question (ID: 19278)

Question:
$\text{The potential energy function corresponding to a conservative force is given as } U(x, y, z) = \frac{3x^2}{2} + 5y + 6z, \text{ then the force at } x = 6 \text{ is } P \text{ N. The value of } P \text{ up to its nearest integral value is:}$ $1. \ 20$ $2. \ 40$ $3. \ 30$ $4. \ 60$
Options:
  • 1. $20$
  • 2. $40$
  • 3. $30$
  • 4. $60$
Solution:
$\text{Hint: } F = -\frac{dU}{dx}$ $F_x = -\frac{dV}{dx}$ $\vec{F} = -3x \hat{i} - 5 \hat{j} - 6 \hat{k}$ $|\vec{F}|_{x=6} = \sqrt{18^2 + 5^2 + 6^2}$ $= \sqrt{324 + 25 + 36}$ $= \sqrt{385}$ $\approx 19.62 \text{ N}$

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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
}