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Current Question (ID: 8825)

Question:

$\text{The position } x \text{ of a particle moving along the x-axis varies with time } t \text{ as } x = 20t - 5t^2\text{, where } x \text{ is in meters and } t \text{ is in seconds. The particle reverses its direction of motion at:}$

Options:

1. $x = 40 \text{ m}$
2. $x = 10 \text{ m}$
3. $x = 20 \text{ m}$
4. $x = 30 \text{ m}$

Solution:

$\text{Hint: } v = \frac{dx}{dt}$ $\text{Step 1: Find the velocity of the particle.}$ $x = 20t - 5t^2$ $\Rightarrow v = \frac{dx}{dt} = 20 - 10t$ $\text{Step 2: Find the position where the particle reverses its direction.}$ $v = 0 = 20 - 10t$ $\Rightarrow t = 2 \text{ s}$ $\text{at, t=2 sec.}$ $x = 20t - 5t^2 = 20 \times 2 - 5 \times 4 = 20 \text{ m}$ $\text{Hence, option (3) is the correct answer.}$

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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
}