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Current Question (ID: 8886)

Question:

$\text{The motion of a particle along a straight line is described by the equation: } x = 8 + 12t - t^3, \text{ where } x \text{ is in metre and } t \text{ is in second. The retardation of the particle when its velocity becomes zero is:}$

Options:

1. $24 \text{ ms}^{-2}$
2. $\text{zero}$
3. $6 \text{ ms}^{-2}$
4. $12 \text{ ms}^{-2}$

Solution:

$\text{Hint: } v = \frac{dx}{dt}$ $\text{Step: Find the retardation of the particle when its velocity becomes zero.}$ $\text{Given, } x = 8 + 12t - t^3$ $\text{We know } v = \frac{dx}{dt}$ $\text{So, } v = 12 - 3t^2$ $\text{When } v = 0, t = 2 \text{ sec}$ $\text{and } a = \frac{dv}{dt}$ $\text{and } a = -6t$ $\text{At } t = 2\text{s,}$ $a = -12 \text{ m/s}^2$ $\text{Therefore, the retardation of the particle is } 12 \text{ m/s}^2.$ $\text{Hence, option (4) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}