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Current Question (ID: 18910)

Question:
$\text{A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is } 90 \text{ s, } 91 \text{ s, } 95 \text{ s and } 92 \text{ s. If the minimum division in the measuring clock is } 1 \text{ s, then the reported mean time should be:}$
Options:
  • 1. $92 \pm 2 \text{ s}$
  • 2. $92 \pm 5 \text{ s}$
  • 3. $92 \pm 1.8 \text{ s}$
  • 4. $92 \pm 3 \text{ s}$
Solution:
$\text{Hint: Reported mean time = Mean value of time } \pm \text{ Mean absolute error.}$ $\text{Time period of 100 oscillations}$ $90 \text{ sec, } 91 \text{ sec, } 95 \text{ sec, } 92 \text{ sec}$ $\text{Mean value of time} = \frac{90 + 91 + 95 + 92}{4}$ $t_m = 92 \text{ sec}$ $\text{Absolute error}$ $|\Delta t_1| = |t_m - t_1| = 2 \text{ sec}$ $|\Delta t_2| = |t_m - t_2| = 1 \text{ sec}$ $|\Delta t_3| = |t_m - t_3| = 3 \text{ sec}$ $|\Delta t_4| = |t_m - t_4| = 0 \text{ sec}$ $\text{Mean absolute error} = \frac{2 + 1 + 3 + 0}{4} = 1.5 \text{ sec} \approx 2 \text{ s}$ $\text{Mean time } (92 \pm 2 \text{ sec})$

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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
}