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Current Question (ID: 19931)

Question:
$\text{In the simple harmonic motion of a pendulum, the square of the time period can be plotted against the length of the pendulum } (l) \text{ as:}$
Options:
  • 1. $T^2 \text{ vs } l \text{ (straight line passing through the origin)}$
  • 2. $T^2 \text{ vs } l \text{ (negative slope)}$
  • 3. $T^2 \text{ vs } l \text{ (parabolic curve)}$
  • 4. $T^2 \text{ vs } l \text{ (increasing curve)}$
Solution:
$\text{Hint: } T = 2\pi \sqrt{\frac{l}{g}}$ $\text{Step: Identify the graph between the square of the time period and the length of the pendulum.}$ $\text{The time period of the simple pendulum is given by:}$ $T = 2\pi \sqrt{\frac{l}{g}}$ $T^2 \propto l$ $\text{The graph represents a straight line passing through the origin, as shown below.}$ $T^2 \text{ vs } l \text{ (straight line passing through the origin)}$ $\text{Hence, option (1) is the correct answer.}$

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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
}