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Current Question (ID: 9580)

Question:
$\text{The } x \text{ and } y \text{ coordinates of the particle at any time are } x = 5t - 2t^2 \text{ and } y = 10t \text{ respectively, where } x \text{ and } y \text{ are in meters and } t \text{ in seconds. The acceleration of the particle at } t = 2 \text{ s is:}$
Options:
  • 1. $5\hat{i} \text{ m/s}^2$
  • 2. $-4\hat{i} \text{ m/s}^2$
  • 3. $-8\hat{j} \text{ m/s}^2$
  • 4. $0$
Solution:
$\text{Hint: } a = \frac{dv}{dt}$ $\text{Step: find the acceleration of the particle at.}$ $\text{Given that the } x \text{ and } y \text{ coordinates of a particle are } x = 5t - 2t^2 \text{ and } y = 10t$ $\Rightarrow v_x = \frac{dx}{dt} = 5 - 4t, \quad v_y = \frac{dy}{dt} = 10$ $\Rightarrow a_x = \frac{dv_x}{dt} = -4 \text{ m/s}^2, \quad a_y = \frac{dv_y}{dt} = 0$ $\vec{a} = a_x\hat{i} + a_y\hat{j}$ $= -4\hat{i} \text{ m/s}^2$ $\text{Hence, option (2) is the correct answer.}$

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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
}