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Current Question (ID: 8882)

Question:
$\text{A body starts falling from height } h \text{ and if it travels a distance of } \frac{h}{2} \text{ during the last second of motion, then the time of flight is (in seconds):}$
Options:
  • 1. $\sqrt{2} - 1$
  • 2. $2 + \sqrt{2}$
  • 3. $\sqrt{2} + \sqrt{3}$
  • 4. $\sqrt{3} + 2$
Solution:
$\text{Hint: } H = \frac{1}{2}gt^2$ $\text{Step: Find the time of the flight.}$ $h = \frac{1}{2}gt^2 \quad \ldots (i)$ $\frac{h}{2} = \frac{1}{2}g(t-1)^2 \quad \ldots (ii)$ $\frac{1}{4}gt^2 = \frac{1}{2}g(t-1)^2$ $\frac{t}{\sqrt{2}} = t - 1$ $t\left(1 - \frac{1}{\sqrt{2}}\right) = 1$ $t = \frac{\sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1}$ $t = \sqrt{2}(\sqrt{2} + 1)$ $t = 2 + \sqrt{2}$ $\text{Therefore, the time of flight is } (2 + \sqrt{2}) \text{ s.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}