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Current Question (ID: 8838)

Question:
$\text{A body is moving along a straight line according to the equation of motion, } x = t^2 - 3t + 4, \text{ where } x \text{ is in metre and } t \text{ is in seconds.}$ $\text{What is the acceleration of the body when it comes to rest?}$
Options:
  • 1. $\text{zero}$
  • 2. $2 \text{ m/s}^2$
  • 3. $\frac{3}{2} \text{ m/s}^2$
  • 4. $1 \text{ m/s}^2$
Solution:
$\text{Hint: The acceleration of the body is given by } a(t) = \frac{d^2x(t)}{dt^2}$ $\text{Step 1: Find velocity of the body}$ $v = \frac{dx}{dt}$ $= 2t - 3$ $\text{Step 2: Find acceleration of the body}$ $a = \frac{dv}{dt}$ $= 2 \text{ m/s}^2$ $\text{Since acceleration is constant and independent of time, the acceleration remains } 2 \text{ m/s}^2 \text{ even when the body comes to rest.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}