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Current Question (ID: 19312)

Question:
$\text{A particle of mass } m \text{ moves on a straight line with its velocity}$ $\text{increasing with distance according to the equation } v = \alpha \sqrt{x}, \text{ where}$ $\alpha \text{ is a constant. The total work done by all the forces applied on the}$ $\text{particle during its displacement from } x = 0 \text{ to } x = d, \text{ will be:}$
Options:
  • 1. $\frac{m}{2\alpha^2 d}$
  • 2. $\frac{md}{2\alpha^2}$
  • 3. $\frac{m\alpha^2 d}{2}$
  • 4. $2m\alpha^2 d$
Solution:
$\text{Given } v = \alpha \sqrt{x}$ $\text{Kinetic energy } K = \frac{1}{2}mv^2 = \frac{1}{2}m(\alpha \sqrt{x})^2 = \frac{1}{2}m\alpha^2 x$ $\text{Work done } W = \Delta K = K_f - K_i = \frac{1}{2}m\alpha^2 d - 0 = \frac{m\alpha^2 d}{2}$

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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
}