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Current Question (ID: 19925)

Question:
$\text{The relation between velocity } (v) \text{ and position } (x) \text{ of a particle moving along the x-axis is given by } 4v^2 = 50 - x^2. \text{ The time period of oscillatory motion of the particle is: (Use } \pi = \frac{22}{7})$
Options:
  • 1. $\frac{88}{9} \text{ s}$
  • 2. $\frac{88}{7} \text{ s}$
  • 3. $\frac{77}{9} \text{ s}$
  • 4. $\frac{77}{8} \text{ s}$
Solution:
$4v^2 = 50 - x^2$ $v^2 = \frac{1}{4}(50 - x^2)$ $v = \frac{1}{2} \sqrt{(50 - x^2)}$ $\text{comparing with equation of SHM}$ $v = \omega \sqrt{A^2 - x^2}$ $A^2 = 50 \Rightarrow A = 5\sqrt{2}$ $\omega = \frac{1}{2} = 0.5 \text{ rad/s}$ $T = \frac{2\pi}{\omega} = \frac{2\pi}{0.5} = 4\pi \text{ seconds}$ $T = 4 \left(\frac{22}{7}\right) = \frac{88}{7} \text{ seconds}$ $\text{So, } n = 7$

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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
}