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Current Question (ID: 8814)

Question:
$\text{The coordinate of an object is given as a function of time by } x = 7t - 3t^2, \text{ where } x \text{ is in metres and } t \text{ is in seconds. Its average velocity over the interval } t = 0 \text{ to } t = 4 \text{ is will be:}$
Options:
  • 1. $5 \text{ m/s}$
  • 2. $-5 \text{ m/s}$
  • 3. $11 \text{ m/s}$
  • 4. $-11 \text{ m/s}$
Solution:
\text{Hint: Average velocity } = \frac{\Delta x}{\Delta t} \text{Step: Find the average velocity of the given object.} \text{At } t = 0: \quad x = 7(0) - 3(0)^2 = 0 \text{ m} \text{At } t = 4: \quad x = 7(4) - 3(4)^2 = 28 - 48 = -20 \text{ m} \text{Average velocity } = \frac{\Delta x}{\Delta t} = \frac{-20 - 0}{4 - 0} = \frac{-20}{4} = -5 \text{ m/s} \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
}