Import Question JSON

Current Question (ID: 11253)

Question:
$\text{A particle is executing SHM according to } y = a \cos \omega t\text{. Then, which of the following graphs represent variations of potential energy?}$
Options:
  • 1. $\text{I and III}$
  • 2. $\text{II and IV}$
  • 3. $\text{II and III}$ (Correct)
  • 4. $\text{I and IV}$
Solution:
$\text{Given } x = A \cos \omega t\text{. As a function of } x\text{, the PE is given by}$ $PE = \frac{1}{2}m\omega^2 x^2$ $\text{At } x = 0\text{, } PE = 0\text{. Hence the correct graph is III. As a function of } t\text{, the PE is given by}$ $PE = \frac{1}{2}m\omega^2 A^2 \cos^2(\omega t)$ $\text{At } t = 0\text{, PE is maximum equal to } \frac{1}{2}m\omega^2 A^2$

Import JSON File

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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
}