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Current Question (ID: 19389)

Question:
$\text{A body of mass } 5 \text{ kg has the linear momentum of } 100 \text{ kg ms}^{-1} \text{ and acted upon by the force of } 2 \text{ N for } 2 \text{ seconds, then change in kinetic energy in Joule is:}$ $1. \ 100 \text{ J}$ $2. \ 90 \text{ J}$ $3. \ 204 \text{ J}$ $4. \ 81.6 \text{ J}$
Options:
  • 1. $100 \text{ J}$
  • 2. $90 \text{ J}$
  • 3. $204 \text{ J}$
  • 4. $81.6 \text{ J}$
Solution:
$\Delta p = F \times \Delta t$ $F \times t = \Delta P$ $\Rightarrow 2 \times 2 = P_t - 100$ $P_t = 104 \text{ kg ms}^{-1}$ $\Delta K = \frac{P_f^2}{2m} - \frac{P_i^2}{2m} = \frac{1}{2 \times 5} \times (104^2 - 100^2)$ $= \frac{1}{10} \times 4 \times 204 = 81.6 \text{ J}$

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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
}